diff options
46 files changed, 19746 insertions, 0 deletions
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/README.txt b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/README.txt new file mode 100644 index 00000000..4f22f166 --- /dev/null +++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/README.txt @@ -0,0 +1,10 @@ +Contributed By: Ashvani Kumar +Course: btech +College/Institute/Organization: Uttarakhand Technical University +Department/Designation: EN +Book Title: A Textbook of Production Engineering +Author: P. C. Sharma +Publisher: S. Chanda & Company, New Delhi +Year of publication: 2008 +Isbn: 9788121901116 +Edition: 11
\ No newline at end of file diff --git a/Applied_Chemistry_by_Dr._Mrs.Trupti_Paradkar/README.txt b/Applied_Chemistry_by_Dr._Mrs.Trupti_Paradkar/README.txt new file mode 100644 index 00000000..83149758 --- /dev/null +++ b/Applied_Chemistry_by_Dr._Mrs.Trupti_Paradkar/README.txt @@ -0,0 +1,10 @@ +Contributed By: Hemin Chheda +Course: others +College/Institute/Organization: IIT Bombay +Department/Designation: JRF +Book Title: Applied Chemistry +Author: Dr. Mrs.Trupti Paradkar +Publisher: Technical Publications, Pune +Year of publication: 2013 +Isbn: 9789350387795 +Edition: 2
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter10_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter10_1_2.ipynb new file mode 100644 index 00000000..d06c15b2 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter10_1_2.ipynb @@ -0,0 +1,593 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6b7c754b6ed46e47374605eefbbcd7d3c6bec34fcb694974971d2be63bb15a17"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-Integrated Circuit Biasing and Active Loads "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg580"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.1\n",
+ "Vbe=0.6;##(V)\n",
+ "b=100.;\n",
+ "V1=5.;\n",
+ "Io=200.;##micro A\n",
+ "Iref=Io*(1.+2./b);\n",
+ "print\"%s %.2f %s\"%('\\nreference current= ',Iref,' microA\\n')\n",
+ "Iref=Iref*0.001;##mA\n",
+ "R1=(V1-Vbe)/Iref;\n",
+ "print\"%s %.2f %s\"%('\\nR1= ',R1,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "reference current= 204.00 microA\n",
+ "\n",
+ "\n",
+ "R1= 21.57 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.2\n",
+ "V1=5.;\n",
+ "V2=-5.;\n",
+ "R1=9.3;\n",
+ "b=50.;\n",
+ "Vbe=0.7;\n",
+ "Va=80.;\n",
+ "Iref=(V1-Vbe-V2)/R1;\n",
+ "print\"%s %.2f %s\"%('\\nreference current = ',Iref,'mA\\n')\n",
+ "Io=Iref/(1.+2./b);\n",
+ "print\"%s %.2f %s\"%('\\noutput current= ',Io,'mA\\n')\n",
+ "ro=Va/Io;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal output resistance= ',ro,' KOhm\\n')\n",
+ "##dIo=dVce2/ro\n",
+ "Vce2=0.7;\n",
+ "dIo=(V1-Vce2)/ro;\n",
+ "print\"%s %.2f %s\"%('\\nchange in load current= ',dIo,' mA\\n')\n",
+ "x=dIo/Io;\n",
+ "x=x*100.;\n",
+ "print\"%s %.2f %s\"%('\\npercent change in output current= ',x,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "reference current = 1.00 mA\n",
+ "\n",
+ "\n",
+ "output current= 0.96 mA\n",
+ "\n",
+ "\n",
+ "small signal output resistance= 83.20 KOhm\n",
+ "\n",
+ "\n",
+ "change in load current= 0.05 mA\n",
+ "\n",
+ "\n",
+ "percent change in output current= 5.38 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg591"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.4\n",
+ "Iref=1.;\n",
+ "Io=12.*10**-3;\n",
+ "V1=5.;\n",
+ "V2=-5.;\n",
+ "Vt=0.026;\n",
+ "Vbe=0.7;\n",
+ "R1=(V1-Vbe-V2)/Iref;\n",
+ "print\"%s %.2f %s\"%('\\nResistance R1 = ',R1,'KOhm\\n')\n",
+ "Re=(Vt/Io)*math.log(Iref/Io);\n",
+ "print\"%s %.2f %s\"%('\\nResistance Re = ',Re,'KOhm\\n')\n",
+ "Vbe=Io*Re;\n",
+ "print\"%s %.2f %s\"%('\\ndifference between two B-E voltages= ',Vbe,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Resistance R1 = 9.30 KOhm\n",
+ "\n",
+ "\n",
+ "Resistance Re = 9.58 KOhm\n",
+ "\n",
+ "\n",
+ "difference between two B-E voltages= 0.11 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg593"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.5\n",
+ "V1=5.;\n",
+ "V2=-5.;\n",
+ "R1=9.3;\n",
+ "Re=9.580;\n",
+ "Vt=0.026;\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "Va=80.;\n",
+ "Io=12.;\n",
+ "ro2=Va/Io;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal collector resistance= ',ro2,' MOhm\\n')\n",
+ "Io=12.*0.001;##mA\n",
+ "gm2=Io/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm2,' mA/V\\n')\n",
+ "r=b*Vt/Io;\n",
+ "print\"%s %.2f %s\"%('\\nResistance= ',r,' KOhm\\n')\n",
+ "Ro=ro2*(1.+gm2*Re*r/(Re+r));\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Ro,' MOhm\\n')\n",
+ "dVc2=4.;\n",
+ "dIo=dVc2/Ro;\n",
+ "print\"%s %.2f %s\"%('\\nchange in load current= ',dIo,' microA\\n')\n",
+ "Io=12.;##micro A\n",
+ "x=dIo/Io;\n",
+ "x=x*100.;\n",
+ "print\"%s %.2f %s\"%('\\npercent change in output current =\\n',x,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal collector resistance= 6.67 MOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 0.46 mA/V\n",
+ "\n",
+ "\n",
+ "Resistance= 216.67 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance= 34.90 MOhm\n",
+ "\n",
+ "\n",
+ "change in load current= 0.11 microA\n",
+ "\n",
+ "\n",
+ "percent change in output current =\n",
+ " 0.96 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg597"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.6\n",
+ "V1=5.;\n",
+ "V2=-5.;\n",
+ "Vbe=0.6;\n",
+ "Veb=0.6;\n",
+ "Iq2=400.*10**-3;##mA\n",
+ "Iref=200.*10**-3;##mA\n",
+ "Iq1=Iref;\n",
+ "Iq3=Iq1;\n",
+ "Iq4=600.*10**-6;\n",
+ "R1=(V1-Veb-Vbe-V2)/Iref;\n",
+ "print\"%s %.2f %s\"%('\\nResistance R1= ',R1,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Resistance R1= 44.00 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 10.7\n",
+ "##uox*Cox/2=x\n",
+ "x=20.*10**-6;##A/V^2\n",
+ "Vtn=1.;\n",
+ "V1=5.;\n",
+ "V2=0.;\n",
+ "Iref=0.25*10**-3;\n",
+ "Io=0.1*10**-3;\n",
+ "Vgs2=1.85;\n",
+ "##let y=W/L\n",
+ "y2=Io/(x*(Vgs2-Vtn)**2);\n",
+ "print\"%s %.2f %s\"%('\\nwidth per length 2= ',y2,'\\n')\n",
+ "y1=Iref/(x*(Vgs2-Vtn)**2);\n",
+ "print\"%s %.2f %s\"%('\\nwidth per length 1= ',y1,'\\n')\n",
+ "Vgs1=Vgs2;\n",
+ "Vgs3=V1-V2-Vgs1;\n",
+ "print\"%s %.2f %s\"%('\\nVgs3= ',Vgs3,' V\\n')\n",
+ "y3=Iref/(x*(Vgs3-Vtn)**2);\n",
+ "print\"%s %.2f %s\"%('\\nwidth per length 3= ',y3,'\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "width per length 2= 6.92 \n",
+ "\n",
+ "\n",
+ "width per length 1= 17.30 \n",
+ "\n",
+ "\n",
+ "Vgs3= 3.15 V\n",
+ "\n",
+ "\n",
+ "width per length 3= 2.70 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg604"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.8\n",
+ "Iref=100.;\n",
+ "Io=Iref;\n",
+ "##lambda=y\n",
+ "y=0.01;\n",
+ "gm=0.5*10**3;\n",
+ "ro=1./(y*Iref);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' MOhm\\n')\n",
+ "ro2=1.;\n",
+ "ro4=1.;\n",
+ "Ro=ro4+ro2*(1.+gm*ro4);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance of cascode circuit= ',Ro,' MOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "output resistance= 1.00 MOhm\n",
+ "\n",
+ "\n",
+ "output resistance of cascode circuit= 502.00 MOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg609"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.9\n",
+ "Idss1=2.;\n",
+ "Idss2=1.;\n",
+ "Vp1=-1.5;\n",
+ "Vp2=Vp1;\n",
+ "##lambda=y\n",
+ "y1=0.05;\n",
+ "y2=y1;\n",
+ "V2=-5.;\n",
+ "Vds=1.5;\n",
+ "Vsmin=Vds+V2;\n",
+ "print\"%s %.2f %s\"%('\\nminimum value of Vs= ',Vsmin,' V\\n')\n",
+ "Io=Idss2*(1.+y1*Vds);\n",
+ "print\"%s %.2f %s\"%('\\noutput current= ',Io,' mA\\n')\n",
+ "Vgs1=(1.-math.sqrt(Io/Idss1))*Vp1;\n",
+ "print\"%s %.2f %s\"%('\\ngate to source voltage of Q1= ',Vgs1,' V\\n')\n",
+ "V1=Vgs1+Vsmin;\n",
+ "print\"%s %.2f %s\"%('\\nV1= ',V1,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "minimum value of Vs= -3.50 V\n",
+ "\n",
+ "\n",
+ "output current= 1.07 mA\n",
+ "\n",
+ "\n",
+ "gate to source voltage of Q1= -0.40 V\n",
+ "\n",
+ "\n",
+ "V1= -3.90 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.10\n",
+ "Vt=0.026;\n",
+ "Van=120.;\n",
+ "Vap=80.;\n",
+ "Av=-(1./Vt)/(1./Van+1./Vap);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal open circuit voltage gain=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal open circuit voltage gain=\n",
+ " -1846.15 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg620"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 10.11\n",
+ "Van=120.;\n",
+ "Vap=80.;\n",
+ "Vt=0.026;\n",
+ "Ico=0.001;\n",
+ "##Rl=infinity\n",
+ "Av=-(1./Vt)/(1./Van+1./Vap);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal open circuit voltage gain=\\n',Av,'')\n",
+ "Rl=100.;\n",
+ "Av1=-(1./Vt)/(1./Van+1./Vap+1./Rl);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal open circuit voltage gain=',Av1,'')\n",
+ "Rl=10.;\n",
+ "Av2=-(1./Vt)/(1./Van+1./Vap+1./Rl);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal open circuit voltage gain=\\n',Av2,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal open circuit voltage gain=\n",
+ " -1846.15 \n",
+ "\n",
+ "small signal open circuit voltage gain= -1247.40 \n",
+ "\n",
+ "small signal open circuit voltage gain=\n",
+ " -318.30 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg623"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 10.12\n",
+ "##lambda=y\n",
+ "yn=0.01;\n",
+ "yp=0.01;\n",
+ "Vtn=1.;\n",
+ "Kn=1.;\n",
+ "Iref=0.5;\n",
+ "gm=2.*math.sqrt(Kn*Iref);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance = ',gm,'mA/V\\n')\n",
+ "go=yn*Iref;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal transistor conductance= ',go,' mA/V\\n')\n",
+ "go2=go;\n",
+ "##for Rl=infinity\n",
+ "Av=-gm/(go+go2);\n",
+ "print\"%s %.2f %s\"%('\\nvoltage gain= ',Av,' \\n')\n",
+ "Rl=100.;##Kohm\n",
+ "gl=0.01;\n",
+ "Av=-gm/(go+gl+go2);\n",
+ "print\"%s %.2f %s\"%('\\nvoltage gain= \\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance = 1.41 mA/V\n",
+ "\n",
+ "\n",
+ "small signal transistor conductance= 0.01 mA/V\n",
+ "\n",
+ "\n",
+ "voltage gain= -141.42 \n",
+ "\n",
+ "\n",
+ "voltage gain= \n",
+ " -70.71 \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter11_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter11_1_2.ipynb new file mode 100644 index 00000000..4922457d --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter11_1_2.ipynb @@ -0,0 +1,829 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a62c54845c7e9347a8f02d330a533f4d6528b33252a4f4b96d8c27265abd4ced"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter11-Differential and Multistage Amplifier"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg642"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.1\n",
+ "V1=10.;\n",
+ "V2=-10.;\n",
+ "Iq=1.;\n",
+ "Rc=10.;\n",
+ "Vbe=0.7;\n",
+ "iC1=Iq/2.;\n",
+ "iC2=iC1;\n",
+ "print\"%s %.2f %s\"%('\\ncollector currents = ',iC1,'mA\\n')\n",
+ "Vc1=V1-iC1*Rc;\n",
+ "Vc2=Vc1;\n",
+ "print\"%s %.2f %s\"%('\\ncollector voltages = ',Vc1,'V\\n')\n",
+ "Vcm=0.;\n",
+ "Ve=Vcm-Vbe;\n",
+ "Vce1=Vc1-Ve;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vce1,' V\\n')\n",
+ "Vcm=-5.;\n",
+ "Ve=Vcm-Vbe;\n",
+ "Vce1=Vc1-Ve;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage = ',Vce1,'V\\n')\n",
+ "Vcm=5.;\n",
+ "Ve=Vcm-Vbe;\n",
+ "Vce1=Vc1-Ve;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vce1,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "collector currents = 0.50 mA\n",
+ "\n",
+ "\n",
+ "collector voltages = 5.00 V\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 5.70 V\n",
+ "\n",
+ "\n",
+ "collector emitter voltage = 10.70 V\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 0.70 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg650"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.3\n",
+ "V1=10.;\n",
+ "V2=-10.;\n",
+ "Iq=0.8*10**-3;\n",
+ "Rc=12000.;\n",
+ "Ro=25000.;\n",
+ "b=100.;\n",
+ "Vt=0.026;\n",
+ "Ad=Iq*Rc/(4.*Vt);\n",
+ "print\"%s %.2f %s\"%('\\ndifferential gain=\\n',Ad,'')\n",
+ "Acm=-(Iq*Rc/(2.*Vt))/(1.+(1.+b)*Iq*Ro/(Vt*b));\n",
+ "print\"%s %.2f %s\"%('\\ncommon mode gain=\\n',Acm,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "differential gain=\n",
+ " 92.31 \n",
+ "\n",
+ "common mode gain=\n",
+ " -0.24 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg657"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.7\n",
+ "Ad=92.3;\n",
+ "Acm=0.237;##mod of Acm\n",
+ "CMRR=Ad/Acm;\n",
+ "print\"%s %.2f %s\"%('\\ncommon mode rejection ratio=\\n',CMRR,'')\n",
+ "CMRRdB=20.*math.log10(CMRR);\n",
+ "print\"%s %.2f %s\"%('\\nCMRR in decibels= ',CMRRdB,' dB\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "common mode rejection ratio=\n",
+ " 389.45 \n",
+ "\n",
+ "CMRR in decibels= 51.81 dB\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg658"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.8\n",
+ "CMRRdB=90.;##dB\n",
+ "CMRR=3.16*10**4;\n",
+ "b=100.;\n",
+ "Vt=0.026;\n",
+ "Iq=0.8;\n",
+ "Ro=(2.*CMRR-1.)*Vt*b/((1.+b)*Iq);\n",
+ "Ro=Ro*10**-3;##Mohm\n",
+ "print round(Ro,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.03\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg661"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.9\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "Va=100.;\n",
+ "Vt=0.026;\n",
+ "Iref=0.5;\n",
+ "Iq=Iref;\n",
+ "I1=Iq/2.\n",
+ "Icq=I1;\n",
+ "r=b*Vt/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal parameter= ',r,' KOhm\\n')\n",
+ "ro=Va/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nro= ',ro,' KOhm\\n')\n",
+ "Ro=Va/Iq;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance of Q4= ',Ro,' KOhm\\n')\n",
+ "Rid=2.*r;\n",
+ "print\"%s %.2f %s\"%('\\ndifferential mode input resistance = ',Rid,'KOhm\\n')\n",
+ "Ricm=(1.+b)*(Ro*ro/2.)/(Ro+ro/2.);\n",
+ "Ricm=Ricm*0.001;##Mohm\n",
+ "print\"%s %.2f %s\"%('\\ncommon mode input resistance= ',Ricm,' MOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal parameter= 10.40 KOhm\n",
+ "\n",
+ "\n",
+ "ro= 400.00 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance of Q4= 200.00 KOhm\n",
+ "\n",
+ "\n",
+ "differential mode input resistance = 20.80 KOhm\n",
+ "\n",
+ "\n",
+ "common mode input resistance= 10.10 MOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg664"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "from numpy import poly\n",
+ "##Example 11.10\n",
+ "Kn1=0.1;\n",
+ "Kn2=Kn1;\n",
+ "Kn3=0.1;\n",
+ "Kn4=Kn3;\n",
+ "R1=30.;\n",
+ "Vtn=1.;\n",
+ "Rd=16.;\n",
+ "\n",
+ "Vgs4=2.40;\n",
+ "I1=(20.-Vgs4)/R1;\n",
+ "print\"%s %.2f %s\"%('\\nI1= ',I1,' mA\\n')\n",
+ "Iq=I1;\n",
+ "Id1=Iq/2.;\n",
+ "print\"%s %.2f %s\"%('\\nId1 and Id2 = ',Id1, 'mA\\n')\n",
+ "Vgs1=math.sqrt(Id1/Kn1)+Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVgs1 and Vgs2 = ',Vgs1,'V\\n')\n",
+ "vo1=10.-Id1*Rd;\n",
+ "print\"%s %.2f %s\"%('\\nvo1 and vo2= ',vo1,' V\\n')\n",
+ "Vds1=Vgs1-Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVds1=Vds2=Vds1(sat)= ',Vds1,' V\\n')\n",
+ "Vcm=vo1-Vds1+Vgs1;\n",
+ "print\"%s %.2f %s\"%('\\nVcm max= ',Vcm,' V\\n')\n",
+ "Vds4=Vgs4-Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVds4= ',Vds4,' V\\n')\n",
+ "Vcm2=Vgs1+Vds4-10.;\n",
+ "print\"%s %.2f %s\"%('\\nVcm min= ',Vcm2,'V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "I1= 0.59 mA\n",
+ "\n",
+ "\n",
+ "Id1 and Id2 = 0.29 mA\n",
+ "\n",
+ "\n",
+ "Vgs1 and Vgs2 = 2.71 V\n",
+ "\n",
+ "\n",
+ "vo1 and vo2= 5.31 V\n",
+ "\n",
+ "\n",
+ "Vds1=Vds2=Vds1(sat)= 1.71 V\n",
+ "\n",
+ "\n",
+ "Vcm max= 6.31 V\n",
+ "\n",
+ "\n",
+ "Vds4= 1.40 V\n",
+ "\n",
+ "\n",
+ "Vcm min= -5.89 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg668"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.11\n",
+ "Kn=0.5;\n",
+ "Iq=1.;\n",
+ "Vt=0.026;\n",
+ "##transconductance of the MOSFET\n",
+ "gm=2.*math.sqrt(Kn*Iq/2);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "##transconductance of the bipolar transistor \n",
+ "gm=Iq/(2.*Vt);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 1.00 mA/V\n",
+ "\n",
+ "\n",
+ "transconductance= 19.23 mA/V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg670"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.12\n",
+ "Iq=0.587;\n",
+ "Kn=1.;\n",
+ "Rd=16.;\n",
+ "##lambda=y\n",
+ "y=0.01;\n",
+ "Ro=1./(y*Iq);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance = ',Ro,'KOhm\\n')\n",
+ "Ad=math.sqrt(Kn*Iq/2.)*Rd;\n",
+ "print\"%s %.2f %s\"%('\\ndifferential mode voltage gain= \\n',Ad,'')\n",
+ "Acm=-math.sqrt(2.*Kn*Iq)*Rd/(1.+2.*math.sqrt(2.*Kn*Iq)*Ro);\n",
+ "print\"%s %.2f %s\"%('\\ncommon mode voltage gain=\\n',Acm,'')\n",
+ "CMRR=20.*math.log10(-Ad/Acm);\n",
+ "print\"%s %.2f %s\"%('\\ncommon mode rejection ratio= ',CMRR,' dB\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "output resistance = 170.36 KOhm\n",
+ "\n",
+ "\n",
+ "differential mode voltage gain= \n",
+ " 8.67 \n",
+ "\n",
+ "common mode voltage gain=\n",
+ " -0.05 \n",
+ "\n",
+ "common mode rejection ratio= 45.35 dB\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg678"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.13\n",
+ "Iq=0.2;\n",
+ "Va=100.;\n",
+ "Va2=Va;\n",
+ "Va4=Va;\n",
+ "Rl=100.;\n",
+ "Vt=0.026;\n",
+ "Ad=(1./Vt)/(1./Va2+1./Va4);\n",
+ "print\"%s %.2f %s\"%('\\nopen circuit voltage gain=\\n',Ad,'')\n",
+ "Ad=(Iq/(2.*Vt))/(Iq/(2.*Va2)+Iq/(2.*Va4)+1./Rl);\n",
+ "print\"%s %.2f %s\"%('\\nvoltage gain=\\n',Ad,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "open circuit voltage gain=\n",
+ " 1923.08 \n",
+ "\n",
+ "voltage gain=\n",
+ " 320.51 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg684"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.15\n",
+ "Kn=0.2;\n",
+ "Idq=0.1;\n",
+ "ro4=1000.;##Kohm\n",
+ "ro6=1000.;##KOhm\n",
+ "ro2=ro4;\n",
+ "##lambda=y\n",
+ "y=0.01;\n",
+ "gm=2.*math.sqrt(Kn*Idq);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "ro=1./(y*Idq);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' KOhm\\n')\n",
+ "Ro=ro4+ro6*(1.+gm*ro);\n",
+ "Ro=Ro*0.001;##Mohm\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance of the cascode active load= ',Ro,'Mohm\\n')\n",
+ "Ro=Ro*1000.;##KOhm\n",
+ "Ad=gm*ro2*Ro/(ro4+Ro);\n",
+ "print\"%s %.2f %s\"%('\\ndifferential mode voltage gain=\\n',Ad,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 0.28 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance= 1000.00 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance of the cascode active load= 284.84 Mohm\n",
+ "\n",
+ "\n",
+ "differential mode voltage gain=\n",
+ " 281.85 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg693"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.16\n",
+ "Iq=0.2;\n",
+ "Ic1=Iq;\n",
+ "Icb=1.;\n",
+ "R4=10.;\n",
+ "R3=0.2;\n",
+ "b=100.;\n",
+ "Va=100.;\n",
+ "Vt=0.026;\n",
+ "Ri=2.*(1.+b)*b*Vt/Iq;\n",
+ "Ri=Ri*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance= ',Ri,' MOhm\\n')\n",
+ "R11=b*Vt/Iq;\n",
+ "print\"%s %.2f %s\"%('\\nresistance R11= ',R11,' KOhm\\n')\n",
+ "Re=R11*R3/(R11+R3);\n",
+ "print\"%s %.2f %s\"%('\\nRe= ',Re,' KOhm\\n')\n",
+ "gm11=Iq/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ngm11= ',gm11,' mA/V\\n')\n",
+ "ro11=Va/Iq;\n",
+ "print\"%s %.2f %s\"%('\\nro11 = ',ro11,'KOhm\\n')\n",
+ "Rc11=ro11*(1+gm11*Re);\n",
+ "Rc11=Rc11*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\nRc11= ',Rc11,' MOhm\\n')\n",
+ "r8=b*Vt/Icb;\n",
+ "print\"%s %.2f %s\"%('\\nresistance= ',r8,'KOhm\\n')\n",
+ "##answer of following given in the book is wrong\n",
+ "Rb8=r8+(1.+b)*R4;\n",
+ "Rb8=Rb8*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\nRb8 = ',Rb8,'MOhm\\n')\n",
+ "Rl7=Rc11*Rb8/(Rc11+Rb8);\n",
+ "print\"%s %.2f %s\"%('\\nRl7= ',Rl7,' MOhm\\n')\n",
+ "Av=Iq*Rl7/(2.*Vt);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "input resistance= 2.63 MOhm\n",
+ "\n",
+ "\n",
+ "resistance R11= 13.00 KOhm\n",
+ "\n",
+ "\n",
+ "Re= 0.20 KOhm\n",
+ "\n",
+ "\n",
+ "gm11= 7.69 mA/V\n",
+ "\n",
+ "\n",
+ "ro11 = 500.00 KOhm\n",
+ "\n",
+ "\n",
+ "Rc11= 1.26 MOhm\n",
+ "\n",
+ "\n",
+ "resistance= 2.60 KOhm\n",
+ "\n",
+ "\n",
+ "Rb8 = 1.01 MOhm\n",
+ "\n",
+ "\n",
+ "Rl7= 0.56 MOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain=\n",
+ " 2.16 \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg694"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.17\n",
+ "Va=100.;\n",
+ "R4=10.;\n",
+ "b=100.;\n",
+ "Rc11=1.26*10**3;\n",
+ "r8=2.6;\n",
+ "Iq=0.2;\n",
+ "Rc7=Va/Iq;\n",
+ "print\"%s %.2f %s\"%('\\nRc7= ',Rc7,' KOhm\\n')\n",
+ "Z=Rc11*Rc7/(Rc11+Rc7);\n",
+ "print\"%s %.2f %s\"%('\\nZ= ',Z,' KOhm\\n')\n",
+ "x=(r8+Z)/(1.+b);\n",
+ "Ro=R4*x/(R4+x);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Ro,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Rc7= 500.00 KOhm\n",
+ "\n",
+ "\n",
+ "Z= 357.95 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance= 2.63 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg697"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.19\n",
+ "b=100.;\n",
+ "Vt=0.026;\n",
+ "Rc=20.;\n",
+ "Ir4=0.4;\n",
+ "Iq=Ir4;\n",
+ "Ir6=Ir4;\n",
+ "r4=b*Vt/Ir4;\n",
+ "print\"%s %.2f %s\"%('\\nr4= ',r4,' KOhm\\n')\n",
+ "r3=b**2*Vt/Ir4;\n",
+ "print\"%s %.2f %s\"%('\\nr3= ',r3,' KOhm\\n')\n",
+ "Ri2=r3+(1.+b)*r4;\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance= ',Ri2,' KOhm\\n')\n",
+ "gm=Iq/(2.*Vt);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "Ad1=gm*Rc*Ri2/(2.*(Rc+Ri2));\n",
+ "print\"%s %.2f %s\"%('\\ngain of differential amplifier stage=\\n',Ad1,'')\n",
+ "r5=b*Vt/Ir6;\n",
+ "print\"%s %.2f %s\"%('\\nr5 = ',r5,'KOhm\\n')\n",
+ "Ir7=2.;\n",
+ "r6=b*Vt/Ir7;\n",
+ "print\"%s %.2f %s\"%('\\nr6= ',r6,' KOhm\\n')\n",
+ "R6=16.5;\n",
+ "R7=5.;\n",
+ "Ri3=r5+(1.+b)*(R6+r6+(1.+b)*R7);\n",
+ "Ri3=Ri3*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\nRi3= ',Ri3,' MOhm\\n')\n",
+ "Rs=5.;\n",
+ "A2=Ir4*Rs/(2.*Vt);\n",
+ "print\"%s %.2f %s\"%('\\nvoltage gain A2=\\n',A2,'')\n",
+ "A3=1.;##vo/vo3\n",
+ "Ad=Ad1*A2*A3;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',Ad,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "r4= 6.50 KOhm\n",
+ "\n",
+ "\n",
+ "r3= 650.00 KOhm\n",
+ "\n",
+ "\n",
+ "input resistance= 1306.50 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 7.69 mA/V\n",
+ "\n",
+ "\n",
+ "gain of differential amplifier stage=\n",
+ " 75.76 \n",
+ "\n",
+ "r5 = 6.50 KOhm\n",
+ "\n",
+ "\n",
+ "r6= 1.30 KOhm\n",
+ "\n",
+ "\n",
+ "Ri3= 52.81 MOhm\n",
+ "\n",
+ "\n",
+ "voltage gain A2=\n",
+ " 38.46 \n",
+ "\n",
+ "small signal voltage gain=\n",
+ " 2913.97 \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg702"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 11.20\n",
+ "Ro=10000000.;\n",
+ "Co=1.*10**-12;\n",
+ "Rb=500.;\n",
+ "r=10000.;\n",
+ "b=100.;\n",
+ "f=1./(2.*math.pi*Ro*Co);\n",
+ "f=f*0.001;##KHz\n",
+ "print\"%s %.2f %s\"%('\\nfrequency of the zero= ',f,' KHz\\n')\n",
+ "Req=Ro*(1.+Rb/r)/(1.+Rb/r+2.*(1.+b)*Ro/r);\n",
+ "print\"%s %.2f %s\"%('\\nReq= ',Req,' Ohm\\n')\n",
+ "f=1/(2.*math.pi*Req*Co);\n",
+ "f=f*10**-9;##GHz\n",
+ "print\"%s %.2f %s\"%('\\nfrequency of the pole= ',f,' GHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "frequency of the zero= 15.92 KHz\n",
+ "\n",
+ "\n",
+ "Req= 51.98 Ohm\n",
+ "\n",
+ "\n",
+ "frequency of the pole= 3.06 GHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter12_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter12_1_2.ipynb new file mode 100644 index 00000000..4335c4e4 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter12_1_2.ipynb @@ -0,0 +1,540 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1ace7ec8ddbfe04d19fdbb2c12f100c57a3d84e0fdba823b9d7bc82551c86874"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter12-Feedback and Stability"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.1\n",
+ "A=10**5;##open loop gain\n",
+ "Af=50.;##closed loop gain\n",
+ "b=(A/Af-1.)/A;\n",
+ "print\"%s %.2f %s\"%('\\nfeedback transfer function=\\n',b,'')\n",
+ "A=-10**5;\n",
+ "Af=-50.;\n",
+ "b=(A/Af-1.)/A;\n",
+ "print\"%s %.2f %s\"%('\\nfeedback transfer function=\\n',b,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "feedback transfer function=\n",
+ " 0.02 \n",
+ "\n",
+ "feedback transfer function=\n",
+ " -0.02 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg733"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.2\n",
+ "A=10**5;\n",
+ "Af=50.;\n",
+ "b=0.019999;\n",
+ "dA=10**4;\n",
+ "dAf=Af*dA/(A*(1.+b*A));\n",
+ "print\"%s %.2e %s\"%('\\ndAf ',dAf,'\\n')\n",
+ "##x=dAf/Af\n",
+ "x=dAf/Af;\n",
+ "x=x*100.;\n",
+ "print\"%s %.2e %s\"%('\\npercent change dAf/Af= ',x,'\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "dAf 2.50e-03 \n",
+ "\n",
+ "\n",
+ "percent change dAf/Af= 5.00e-03 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg735"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.3\n",
+ "Ao=10**4;\n",
+ "wh=2.*math.pi*100.;##rad/s\n",
+ "Af=50.;\n",
+ "##x=(1+bAo)\n",
+ "x=Ao/Af;\n",
+ "print\"%s %.2f %s\"%('\\n(1+bAo)=\\n',x,'')\n",
+ "wfh=wh*x;\n",
+ "print\"%s %.2f %s\"%('\\nclosed loop bandwidth=\\n',wfh,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "(1+bAo)=\n",
+ " 200.00 \n",
+ "\n",
+ "closed loop bandwidth=\n",
+ " 125663.71 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 12.5\n",
+ "Av=10**5;\n",
+ "Avf=50.;\n",
+ "Rf=10.;##Kohm\n",
+ "Ro=20000.;##Ohm\n",
+ "##x=(1+bvAv)\n",
+ "x=Av/Avf;\n",
+ "print\"%s %.2e %s\"%('\\n(1+bvAv)=\\n',x,'')\n",
+ "Rif=Rf*x;\n",
+ "Rif=Rif*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance= ',Rif,' MOhm\\n')\n",
+ "Rof=Ro/x;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Rof,' Ohm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "(1+bvAv)=\n",
+ " 2.00e+03 \n",
+ "\n",
+ "input resistance= 20.00 MOhm\n",
+ "\n",
+ "\n",
+ "output resistance= 10.00 Ohm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg745"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.6\n",
+ "Af=10**5;\n",
+ "Aif=50.;\n",
+ "Rf=10000.;\n",
+ "Ro=20.;\n",
+ "##x=(1+biAi)\n",
+ "x=Af/Aif;\n",
+ "print\"%s %.2e %s\"%('\\n(1+biAi)=\\n',x,'')\n",
+ "Rif=Rf/x;\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance = ',Rif,'Ohm\\n')\n",
+ "Rof=Ro*x;\n",
+ "Rof=Rof*0.001;##Mohm\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Rof,' MOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "(1+biAi)=\n",
+ " 2.00e+03 \n",
+ "\n",
+ "input resistance = 5.00 Ohm\n",
+ "\n",
+ "\n",
+ "output resistance= 40.00 MOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg751"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.7\n",
+ "Ri=50.;\n",
+ "R1=10.;\n",
+ "R2=90.;\n",
+ "Av=10**4;\n",
+ "bv=1./(1.+R2/R1);\n",
+ "print\"%s %.2f %s\"%('\\nfeedback transfer function=\\n',bv,'')\n",
+ "Rif=Ri*(1.+bv*Av);\n",
+ "Rif=Rif*0.001;##Mohm\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance= ',Rif,' MOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "feedback transfer function=\n",
+ " 0.10 \n",
+ "\n",
+ "input resistance= 50.05 MOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg765"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 12.11\n",
+ "hFE=100.;##transistor parameter\n",
+ "Vbe=0.7;\n",
+ "Vcc=10.;\n",
+ "R1=55.;\n",
+ "R2=12.;\n",
+ "Re=1.;\n",
+ "Rc=4.;\n",
+ "Rl=4.;\n",
+ "Icq=0.983;\n",
+ "Vceq=5.08;\n",
+ "Vt=0.026;\n",
+ "r=hFE*Vt/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal parameter resistance= ',r,' KOhm\\n')\n",
+ "gm=Icq/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "Agf=-gm*(Rc/(Rc+Rl))/(1.+Re*(gm+1./r));\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance transfer function= ',Agf,' mA/V\\n')\n",
+ "##as first approximation\n",
+ "Agf2=-1./Re;\n",
+ "print\"%s %.2f %s\"%('\\nAgf= ',Agf2,' mA/V\\n')\n",
+ "Avf=Agf*Rl;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage gain=\\n',Avf,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal parameter resistance= 2.64 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 37.81 mA/V\n",
+ "\n",
+ "\n",
+ "transconductance transfer function= -0.48 mA/V\n",
+ "\n",
+ "\n",
+ "Agf= -1.00 mA/V\n",
+ "\n",
+ "\n",
+ "voltage gain=\n",
+ " -1.93 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg777"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.15\n",
+ "##Determine the loop gain fig12.45(a)\n",
+ "hFE=100.;\n",
+ "Vbe=0.7;\n",
+ "Icq=0.492;\n",
+ "r=5.28;\n",
+ "gm=18.9;\n",
+ "Rs=10.;\n",
+ "R1=51.;\n",
+ "R2=5.5;\n",
+ "Re=0.500;\n",
+ "Rc=10.;\n",
+ "Rf=82.;\n",
+ "x=r*R2/(r+R2);\n",
+ "y=R1*x/(x+R1);\n",
+ "t=Rs*y/(y+Rs);\n",
+ "Req=t;\n",
+ "print\"%s %.2f %s\"%('\\nequivalent resistance ',t,' KOhm\\n')\n",
+ "T=gm*Rc*Req/(Rc+Rf+Req);\n",
+ "print\"%s %.2f %s\"%('\\nthe loop gain=\\n',T,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "equivalent resistance 2.04 KOhm\n",
+ "\n",
+ "\n",
+ "the loop gain=\n",
+ " 4.09 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg791"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.19\n",
+ "##T=b*100/(sqrt(1+(f/10^5)^2) angle=-3tan^-1(f/10^5)\n",
+ "##stable at f180 at which phase becomes -180 degrees\n",
+ "##-3*atan(f180/10^5)=-180\n",
+ "f180=math.tan(60/57.3)*10**5;\n",
+ "print\"%s %.2f %s\"%('\\nfrequency at -180 degree= ',f180,'f Hz\\n')\n",
+ "b=0.2;\n",
+ "T=b*100./(math.sqrt(1.+(f180/10**5)**2))**3;\n",
+ "print\"%s %.2f %s\"%('\\nmagnitude of the loop gain=\\n',T,'')\n",
+ "b=0.02;\n",
+ "T=b*100./(math.sqrt(1.+(f180/10**5)**2))**3;\n",
+ "print\"%s %.2f %s\"%('\\nmagnitude of the loop gain=\\n',T,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "frequency at -180 degree= 173174.23 f Hz\n",
+ "\n",
+ "\n",
+ "magnitude of the loop gain=\n",
+ " 2.50 \n",
+ "\n",
+ "magnitude of the loop gain=\n",
+ " 0.25 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex22-pg798"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.22\n",
+ "Ao=10**6;\n",
+ "fPD=0.010;##KHz\n",
+ "b=0.01;\n",
+ "Af=Ao/(1.+b*Ao);\n",
+ "print\"%s %.2f %s\"%('\\nlow frequency closed loop gain=\\n',Af,'')\n",
+ "fc=fPD*(1.+b*Ao);\n",
+ "print\"%s %.2f %s\"%('\\nclosed loop 3dB frequency= ',fc,' KHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "low frequency closed loop gain=\n",
+ " 99.99 \n",
+ "\n",
+ "closed loop 3dB frequency= 100.01 KHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex23-pg799"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 12.23\n",
+ "A=10**3;\n",
+ "Cf=30.*10**-12;##feedback capacitor (F)\n",
+ "R2=5.*10**5;\n",
+ "Cm=Cf*(1.+A);\n",
+ "print\"%s %.2e %s\"%('\\nMiller capacitance= ',Cm,' F\\n')\n",
+ "fp=1/(2.*math.pi*R2*Cm);\n",
+ "print\"%s %.2f %s\"%('\\ndominant pole frequency = ',fp,'Hz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Miller capacitance= 3.00e-08 F\n",
+ "\n",
+ "\n",
+ "dominant pole frequency = 10.60 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter13_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter13_1_2.ipynb new file mode 100644 index 00000000..192e847c --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter13_1_2.ipynb @@ -0,0 +1,804 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2a940490296bbeac10456bf5a339ec669001e8073eddecdfff753904e075bb2b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter13-Operational Amplifier Circuits "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg824"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "\n",
+ "V1=15.;##positive supply voltage\n",
+ "V2=-15.;##negative supply voltage\n",
+ "Veb12=-0.6;\n",
+ "Vbe11=0.6;\n",
+ "Rs=40.;\n",
+ "Iref=(V1-V2-Veb12-Vbe11)/Rs;\n",
+ "print\"%s %.2f %s\"%('\\nreference current= ',Iref,' mA\\n')\n",
+ "Ic10=19.;\n",
+ "Ic1=Ic10/2.;\n",
+ "print\"%s %.2f %s\"%('\\nIc1=Ic2=Ic3=Ic4= ',Ic1,'microA\\n')\n",
+ "Ic1=Ic1*0.001;##mA\n",
+ "Vbe7=0.6;\n",
+ "Vbe6=0.6;\n",
+ "Ic6=Ic1;\n",
+ "R2=1.;\n",
+ "Vc6=Vbe7+Vbe6+Ic6*R2+V2;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage at collector of Q6= ',Vc6,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "reference current= 0.75 mA\n",
+ "\n",
+ "\n",
+ "Ic1=Ic2=Ic3=Ic4= 9.50 microA\n",
+ "\n",
+ "\n",
+ "voltage at collector of Q6= -13.79 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg827"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.2\n",
+ "Iref=0.72;\n",
+ "Ic17=0.75*Iref;\n",
+ "print\"%s %.2f %s\"%('\\ncollector currents in Q17= ',Ic17,' mA\\n')\n",
+ "b=200.;\n",
+ "Ib17=Ic17/b;\n",
+ "Ie17=Ic17;\n",
+ "R8=0.100;\n",
+ "Vbe17=0.6;\n",
+ "R9=50.;\n",
+ "Ic16=Ib17+(Ie17*R8+Vbe17)/R9;\n",
+ "Ic16=Ic16*1000.;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current in Q16= ',Ic16,' microA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "collector currents in Q17= 0.54 mA\n",
+ "\n",
+ "\n",
+ "collector current in Q16= 15.78 microA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg829"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.3\n",
+ "Is1=10**-14;##reverse saturation currents for Q18 Q19\n",
+ "Is2=3*10**-14;##reverse saturation currents for Q14 Q20\n",
+ "Iref=0.72;\n",
+ "Vt=0.026;\n",
+ "Ic13a=0.25*Iref;\n",
+ "print\"%s %.2f %s\"%('\\nIc13a= ',Ic13a,' mA\\n')\n",
+ "Vbe19=0.6;\n",
+ "R10=50.;\n",
+ "Ir1o=Vbe19/R10;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent in Ro= ',Ir1o,' mA\\n')\n",
+ "Ic19=Ic13a-Ir1o;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent in Q19 = ',Ic19,'mA\\n')\n",
+ "Ic19=Ic19*0.001;##A\n",
+ "Vbe19=Vt*math.log(Ic19/Is1);\n",
+ "print\"%s %.2f %s\"%('\\nB-E voltage of Q19= ',Vbe19,' V\\n')\n",
+ "b=200.;\n",
+ "Ic19=Ic19*10**6;##micro A\n",
+ "Iv19=Ic19*1000.;\n",
+ "Ib18=Ic19/b;\n",
+ "Ir1o=Ir1o*1000.;\n",
+ "print\"%s %.2f %s\"%('\\nbase current in Q18= ',Ib18,' microA\\n')\n",
+ "Ic18=Ir1o+Ib18;\n",
+ "print\"%s %.2f %s\"%('\\ncurrents in Q18= ',Ic18,' microA\\n')\n",
+ "Ic18=Ic18*10**-6;\n",
+ "Vbe18=Vt*math.log(Ic18/Is1);\n",
+ "print\"%s %.2f %s\"%('\\nB-E voltage of Q18= ',Vbe18,' V\\n')\n",
+ "Vbb=Vbe18+Vbe19;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage difference Vbb= ',Vbb,' V\\n')\n",
+ "Ic14=Is2*math.exp(Vbb/(2.*Vt));\n",
+ "Ic14=Ic14*10**6;##micro A\n",
+ "print\"%s %.2f %s\"%('\\nquiescent currents in Q14 and Q20 ',Ic14,'microA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Ic13a= 0.18 mA\n",
+ "\n",
+ "\n",
+ "current in Ro= 0.01 mA\n",
+ "\n",
+ "\n",
+ "current in Q19 = 0.17 mA\n",
+ "\n",
+ "\n",
+ "B-E voltage of Q19= 0.61 V\n",
+ "\n",
+ "\n",
+ "base current in Q18= 0.84 microA\n",
+ "\n",
+ "\n",
+ "currents in Q18= 12.84 microA\n",
+ "\n",
+ "\n",
+ "B-E voltage of Q18= 0.55 V\n",
+ "\n",
+ "\n",
+ "voltage difference Vbb= 1.16 V\n",
+ "\n",
+ "\n",
+ "quiescent currents in Q14 and Q20 139.33 microA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg832"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.4\n",
+ "b=200.;\n",
+ "Va=50.\n",
+ "Vt=0.026;\n",
+ "R2=1.;\n",
+ "Ic6=0.0095;\n",
+ "Ic4=Ic6;\n",
+ "Ic16=0.0158;\n",
+ "Ic17=0.54;\n",
+ "r17=b*Vt/Ic17;\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance to gain stage= ',r17,' KOhm\\n')\n",
+ "R9=50.;\n",
+ "R8=0.100;\n",
+ "x=r17+(1.+b)*R8;\n",
+ "Re=x*R9/(x+R9);\n",
+ "print\"%s %.2f %s\"%('\\nRe= ',Re,' KOhm\\n')\n",
+ "r16=b*Vt/Ic16;\n",
+ "print\"%s %.2f %s\"%('\\nr16= ',r16,' KOhm\\n')\n",
+ "Ri2=r16+(1.+b)*Re;\n",
+ "Ri2=Ri2*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\nRi2= ',Ri2,' KOhm\\n')\n",
+ "r6=b*Vt/Ic6;\n",
+ "print\"%s %.2f %s\"%('\\nresistance of the active load= ',r6,' KOhm\\n')\n",
+ "gm=Ic6/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance = ',gm,'mA/V\\n')\n",
+ "ro6=Va/Ic6;\n",
+ "ro6=ro6*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\nro6= ',ro6,' MOhm\\n')\n",
+ "R=ro6*(1.+gm*R2*r6/(R2+r6));\n",
+ "print\"%s %.2f %s\"%('\\neffective resistance of active load= ',R,' MOhm\\n')\n",
+ "ro4=Va/Ic4;\n",
+ "ro4=ro4*0.001;##Mohm\n",
+ "print\"%s %.2f %s\"%('\\nResistance ro4= ',ro4,' KOhm\\n')\n",
+ "Icq=9.5;\n",
+ "x=Ri2*R/(R+Ri2);\n",
+ "y=ro4*x/(ro4+x);\n",
+ "Ad=-y*Icq/Vt;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal differential voltage gain=\\n',Ad,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "input resistance to gain stage= 9.63 KOhm\n",
+ "\n",
+ "\n",
+ "Re= 18.64 KOhm\n",
+ "\n",
+ "\n",
+ "r16= 329.11 KOhm\n",
+ "\n",
+ "\n",
+ "Ri2= 4.08 KOhm\n",
+ "\n",
+ "\n",
+ "resistance of the active load= 547.37 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance = 0.37 mA/V\n",
+ "\n",
+ "\n",
+ "ro6= 5.26 MOhm\n",
+ "\n",
+ "\n",
+ "effective resistance of active load= 7.18 MOhm\n",
+ "\n",
+ "\n",
+ "Resistance ro4= 5.26 KOhm\n",
+ "\n",
+ "\n",
+ "small signal differential voltage gain=\n",
+ " -635.97 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg835"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.5\n",
+ "bp=50.;\n",
+ "bn=200.;\n",
+ "Va=50.;\n",
+ "R9=50.;\n",
+ "R8=0.100;\n",
+ "Rl=2.;\n",
+ "Vt=0.026;\n",
+ "Ri2=4070.;\n",
+ "Ic20=0.138;\n",
+ "r20=bp*Vt/Ic20;\n",
+ "print\"%s %.2f %s\"%('\\nr20= ',r20,' KOhm\\n')\n",
+ "R20=r20+(1.+bp)*Rl;\n",
+ "print\"%s %.2f %s\"%('\\nR20= ',R20,' KOhm\\n')\n",
+ "Ic13A=0.18;\n",
+ "R19=Va/Ic13A;\n",
+ "print\"%s %.2f %s\"%('\\nR19= ',R19,' KOhm\\n')\n",
+ "r22=bp*Vt/Ic13A;\n",
+ "print\"%s %.2f %s\"%('\\nr22= ',r22,' KOhm\\n')\n",
+ "Ri3=r22+(1.+bp)*R19*R20/(R19+R20);\n",
+ "Ri3=Ri3*0.001;##MOhm\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance to the output stage= ',Ri3,' MOhm\\n')\n",
+ "Ic13B=0.54;\n",
+ "R=Va/Ic13B;\n",
+ "print\"%s %.2f %s\"%('\\neffective resistance of the active load= ',R,' KOhm\\n')\n",
+ "Ic17=Ic13B;\n",
+ "R17=Va/Ic17;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance Ro17 = ',R17,'KOhm\\n')\n",
+ "Ri3=Ri3*1000.;##KOhm\n",
+ "r17=9.63;\n",
+ "x=R17*Ri3/(Ri3+R17);\n",
+ "y=x*R/(R+x);\n",
+ "A=-bn*R9*(1.+bn)*y/(Ri2*(R9+r17+(1.+bn)*R8));\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',A,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "r20= 9.42 KOhm\n",
+ "\n",
+ "\n",
+ "R20= 111.42 KOhm\n",
+ "\n",
+ "\n",
+ "R19= 277.78 KOhm\n",
+ "\n",
+ "\n",
+ "r22= 7.22 KOhm\n",
+ "\n",
+ "\n",
+ "input resistance to the output stage= 4.06 MOhm\n",
+ "\n",
+ "\n",
+ "effective resistance of the active load= 92.59 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance Ro17 = 92.59 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain=\n",
+ " -283.53 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.6\n",
+ "Ic20=2.;\n",
+ "bn=200.;\n",
+ "bp=50.;\n",
+ "Va=50.;\n",
+ "r17=9.63;\n",
+ "r22=7.22;\n",
+ "R20=0.260;\n",
+ "gm17=20.8;\n",
+ "ro17=92.6;\n",
+ "Ro13B=92.6;\n",
+ "R8=0.100;\n",
+ "Rc17=ro17*(1.+gm17*R8*r17/(R8+r17));\n",
+ "print\"%s %.2f %s\"%('\\nRc17= ',Rc17,' KOhm\\n')\n",
+ "Rc22=(r22+Rc17*Ro13B/(Rc17+Ro13B))/(1.+bp);\n",
+ "print\"%s %.2f %s\"%('\\nRc22= ',Rc22,' KOhm\\n')\n",
+ "Ic13A=0.18;\n",
+ "Rc19=Va/Ic13A;\n",
+ "print\"%s %.2f %s\"%('\\nRc19= ',Rc19,' KOhm\\n')\n",
+ "Rc20=(R20+Rc22*Rc19/(Rc22+Rc19))/(1.+bp);\n",
+ "print\"%s %.2f %s\"%('\\nRc20= ',Rc20,' KOhm\\n')\n",
+ "Rc20=Rc20*1000.;##Ohm\n",
+ "R3=22.;\n",
+ "Ro=R3+Rc20;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Ro,' Ohm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Rc17= 283.23 KOhm\n",
+ "\n",
+ "\n",
+ "Rc22= 1.51 KOhm\n",
+ "\n",
+ "\n",
+ "Rc19= 277.78 KOhm\n",
+ "\n",
+ "\n",
+ "Rc20= 0.03 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance= 56.54 Ohm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg838"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.7\n",
+ "Av2=285.;\n",
+ "C1=30.;\n",
+ "Ci=C1*(1.+Av2);\n",
+ "print\"%s %.2f %s\"%('\\ninput capacitance= ',Ci,' pF\\n')\n",
+ "Ri2=4.07;\n",
+ "Ract=7.18;\n",
+ "ro4=5.26;\n",
+ "Ro1=Ract*ro4/(Ract+ro4);\n",
+ "print\"%s %.2f %s\"%('\\ngate stage input resistance= ',Ro1,' MOhm \\n')\n",
+ "Req=Ro1*Ri2/(Ri2+Ro1);\n",
+ "print\"%s %.2f %s\"%('\\nequivalent resistance= ',Req,' MOhm\\n')\n",
+ "Req=Req*10**6;##Ohm\n",
+ "Ci=Ci*10**-12;##F\n",
+ "fPD=1/(2.*math.pi*Req*Ci);\n",
+ "print\"%s %.2f %s\"%('\\ndominant pole frequency = ',fPD,' Hz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "input capacitance= 8580.00 pF\n",
+ "\n",
+ "\n",
+ "gate stage input resistance= 3.04 MOhm \n",
+ "\n",
+ "\n",
+ "equivalent resistance= 1.74 MOhm\n",
+ "\n",
+ "\n",
+ "dominant pole frequency = 10.67 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg842"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 13.9\n",
+ "##lambda=y\n",
+ "y=0.02;\n",
+ "##W/L=x and u*Cox/2=t\n",
+ "x=12.5;\n",
+ "t=10.;\n",
+ "Kp1=x*t;\n",
+ "print\"%s %.2f %s\"%('\\nconduction parameters of M1 and M2= ',Kp1,' microA/V^2\\n')\n",
+ "Kp1=Kp1*0.001;##mA/V^2\n",
+ "Id=0.0199;\n",
+ "ro2=1./(y*Id);\n",
+ "ro2=ro2*0.001;##Mohm\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= MOhm\\n',ro2,'')\n",
+ "Iq=0.0397;\n",
+ "ro2=ro2*1000.;##Kohm\n",
+ "ro4=ro2;\n",
+ "Ad=math.sqrt(2.*Kp1*Iq)*ro2*ro4/(ro2+ro4);\n",
+ "print\"%s %.2f %s\"%('\\nthe gain of input stage= \\n',Ad,'')\n",
+ "Kn7=0.250;\n",
+ "Id7=Iq;\n",
+ "gm7=2.*math.sqrt(Kn7*Id7)\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance of M7= ',gm7,' mA/V\\n')\n",
+ "ro7=1./(y*Id7);\n",
+ "ro7=ro7*0.001;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance of M7 and M8 = ',ro7,'MOhm\\n')\n",
+ "ro7=ro7*1000.;##Kohm\n",
+ "ro8=ro7;\n",
+ "Av2=gm7*ro7*ro8/(ro7+ro8);\n",
+ "print\"%s %.2f %s\"%('\\ngain of the second stage=\\n',Av2,'')\n",
+ "Av=Ad*Av2;\n",
+ "print\"%s %.2f %s\"%('\\noverall voltage gain=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "conduction parameters of M1 and M2= 125.00 microA/V^2\n",
+ "\n",
+ "\n",
+ "output resistance= MOhm\n",
+ " 2.51 \n",
+ "\n",
+ "the gain of input stage= \n",
+ " 125.16 \n",
+ "\n",
+ "transconductance of M7= 0.20 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance of M7 and M8 = 1.26 MOhm\n",
+ "\n",
+ "\n",
+ "gain of the second stage=\n",
+ " 125.47 \n",
+ "\n",
+ "overall voltage gain=\n",
+ " 15703.52 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg845"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.10\n",
+ "Iref=100;\n",
+ "Kn=80;\n",
+ "Kp=40;\n",
+ "##W/L=x\n",
+ "x=25;\n",
+ "##lambda=y\n",
+ "y=0.02;\n",
+ "Id=Iref/2.;\n",
+ "gm1=2.*math.sqrt(Kp*x*Id/2.);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance gm1=gm8= ',gm1,' microA/V\\n')\n",
+ "gm6=2.*math.sqrt(Kn*x*Id/2.);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm6,' microA/V\\n')\n",
+ "ro1=1./(y*Id);\n",
+ "ro8=ro1;\n",
+ "ro6=ro1;\n",
+ "ro10=ro1;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance ro1=ro8=ro6=ro10= ',ro1,' MOhm\\n')\n",
+ "Id4=Iref;\n",
+ "ro4=1./(y*Id4);\n",
+ "print\"%s %.2f %s\"%('\\nro4= ',ro4,' MOhm\\n')\n",
+ "Ro8=gm1*ro8*ro10;\n",
+ "print\"%s %.2f %s\"%('\\ncomposite output resistances = ',Ro8,'MOhm\\n')\n",
+ "Ro6=gm6*ro6*ro4*ro1/(ro4+ro1);\n",
+ "print\"%s %.2f %s\"%('\\ncomposite output resistances= ',Ro6,' MOhm\\n')\n",
+ "Ad=gm1*Ro6*Ro8/(Ro6+Ro8);\n",
+ "print\"%s %.2f %s\"%('\\ndifferential voltage gain=\\n',Ad,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance gm1=gm8= 316.23 microA/V\n",
+ "\n",
+ "\n",
+ "transconductance= 447.21 microA/V\n",
+ "\n",
+ "\n",
+ "output resistance ro1=ro8=ro6=ro10= 1.00 MOhm\n",
+ "\n",
+ "\n",
+ "ro4= 0.50 MOhm\n",
+ "\n",
+ "\n",
+ "composite output resistances = 316.23 MOhm\n",
+ "\n",
+ "\n",
+ "composite output resistances= 149.07 MOhm\n",
+ "\n",
+ "\n",
+ "differential voltage gain=\n",
+ " 32037.72 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg854"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.12\n",
+ "Kp=0.6;\n",
+ "bn=200.;\n",
+ "Va=50.;\n",
+ "Vt=0.026;\n",
+ "Ic13=0.20;\n",
+ "Ri2=bn*Vt/Ic13;\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance to the gain stage= ',Ri2,' KOhm\\n')\n",
+ "Iq5=Ic13;\n",
+ "Ad=math.sqrt(2.*Kp*Iq5)*Ri2;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',Ad,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "input resistance to the gain stage= 26.00 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain=\n",
+ " 12.74 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg855"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.13\n",
+ "Va=150.;\n",
+ "Vt=0.026;\n",
+ "Ic13=0.2;\n",
+ "gm13=Ic13/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm13,' mA/V\\n')\n",
+ "ro13=Va/Ic13;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro13,' KOhm\\n')\n",
+ "Av2=gm13*ro13;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage gain= \\n',Av2,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 7.69 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance= 750.00 KOhm\n",
+ "\n",
+ "\n",
+ "voltage gain= \n",
+ " 5769.23 \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg856"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 13.14\n",
+ "Av2=5768.;\n",
+ "C1=12.;\n",
+ "Ci=C1*(1.+Av2);\n",
+ "print\"%s %.2f %s\"%('\\neffective input capacitance= ',Ci,' pF\\n')\n",
+ "Ri2=26000.;##gain stage input resistance (Ohm)\n",
+ "Ci=Ci*10**-12;##F\n",
+ "fPD=1/(2.*math.pi*Ri2*Ci);\n",
+ "print\"%s %.2f %s\"%('\\ndominant pole frequency= ',fPD,' Hz\\n')\n",
+ "Av=73254.;\n",
+ "fT=fPD*Av;\n",
+ "fT=fT*10**-6;##MHz\n",
+ "print\"%s %.2f %s\"%('\\nunity gain bandwidth= ',fT,' MHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "effective input capacitance= 69228.00 pF\n",
+ "\n",
+ "\n",
+ "dominant pole frequency= 88.42 Hz\n",
+ "\n",
+ "\n",
+ "unity gain bandwidth= 6.48 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter14_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter14_1_2.ipynb new file mode 100644 index 00000000..af5ad5a1 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter14_1_2.ipynb @@ -0,0 +1,385 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:346bac54aa836b30505ba957f43799ddc8ade182f697449e8221de9753ec8a1b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter14-Nonideal Effects in Operational Amplifier Circuits"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg880"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "\n",
+ "##Example 14.2\n",
+ "R2=10000.;\n",
+ "Ri=10000.\n",
+ "Aol=10**5;\n",
+ "Rif=1./(1./Ri+(1.+Aol)/R2);\n",
+ "print\"%s %.2f %s\"%('\\nclosed loop input resistance = ',Rif,'Ohm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "closed loop input resistance = 0.10 Ohm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg883"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 14.3\n",
+ "Aol=10**5;\n",
+ "Ri=10.;\n",
+ "R1=10.;\n",
+ "R2=R1;\n",
+ "Rif=(Ri*(1.+Aol)+R2*(1.+Ri/R1))/(1.+R2/R1);\n",
+ "Rif=Rif*0.001;##Mohm\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance ',Rif,'MOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "input resistance 500.01 MOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg888"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 14.5\n",
+ "Ao=2*10**5;\n",
+ "fPD=5.;\n",
+ "fT=fPD*Ao;\n",
+ "print\"%s %.2f %s\"%('\\nunity gain bandwidth= ',fT,' Hz\\n')\n",
+ "f3dB=20.*10**3;\n",
+ "Acl=fT/f3dB;\n",
+ "print\"%s %.2f %s\"%('\\nclosed loop gain=\\n',Acl,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "unity gain bandwidth= 1000000.00 Hz\n",
+ "\n",
+ "\n",
+ "closed loop gain=\n",
+ " 50.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg890"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 14.6\n",
+ "Iq=19*10**-6;\n",
+ "C1=30*10**-12;\n",
+ "SR=Iq/C1;\n",
+ "SR=SR*10**-6;\n",
+ "print\"%s %.2f %s\"%('\\nslew rate= ',SR,' V/micros\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "slew rate= 0.63 V/micros\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 14.7\n",
+ "fT=1000.;##KHz\n",
+ "Aclo=10.;\n",
+ "SR=1.*10**3;\n",
+ "Vpo=10.;\n",
+ "f3dB=fT/Aclo;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal closed loop bandwidth= ',f3dB,' KHz\\n')\n",
+ "fmax=SR/(2.*math.pi*Vpo);\n",
+ "print\"%s %.2f %s\"%('\\nfull power bandwidth= ',fmax,' KHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal closed loop bandwidth= 100.00 KHz\n",
+ "\n",
+ "\n",
+ "full power bandwidth= 15.92 KHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg895"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 14.8\n",
+ "Is1=10**-14;\n",
+ "Is2=1.05*10**-14;\n",
+ "Vt=0.026;\n",
+ "Vos=Vt*math.log(Is2/Is1);\n",
+ "print\"%s %.2e %s\"%('\\nthe offset voltage = ',Vos,'V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the offset voltage = 1.27e-03 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg900"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 14.10\n",
+ "Kn1=105.;\n",
+ "Kn2=100.;\n",
+ "Iq=200.;\n",
+ "dKn=Kn1-Kn2;\n",
+ "print\"%s %.2f %s\"%('\\ndifference in conduction parameter= ',dKn,' microA/V^2\\n')\n",
+ "Kn=(Kn1+Kn2)/2.;\n",
+ "print\"%s %.2f %s\"%('\\naverage of the conduction parameter= ',Kn,' microA/V^2\\n')\n",
+ "Vos=math.sqrt(Iq/(2.*Kn))*dKn/(2.*Kn);\n",
+ "print\"%s %.2f %s\"%('\\noffset voltage= ',Vos,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "difference in conduction parameter= 5.00 microA/V^2\n",
+ "\n",
+ "\n",
+ "average of the conduction parameter= 102.50 microA/V^2\n",
+ "\n",
+ "\n",
+ "offset voltage= 0.02 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg901"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 14.11\n",
+ "Rs=100.;\n",
+ "R4=100000.;\n",
+ "R3=100000.;\n",
+ "V1=15.;\n",
+ "V2=-15.;\n",
+ "Vy=Rs*V1/(Rs+R4);\n",
+ "Vy=Vy*1000.;##mV\n",
+ "print\"%s %.2f %s\"%('\\nVoltage Vy = ',Vy,'mV\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Voltage Vy = 14.99 mV\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg908"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 14.13\n",
+ "R1=10.;\n",
+ "R2=100.;\n",
+ "Ib1=1.1*10**-3;\n",
+ "Ib2=1.*10**-3;\n",
+ "vo=Ib1*R2;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage = ',vo,'V\\n')\n",
+ "R3=R1*R2/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\nR3= ',R3,' KOhm\\n')\n",
+ "vo=R2*(Ib1-Ib2);\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage= ',vo,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "output voltage = 0.11 V\n",
+ "\n",
+ "\n",
+ "R3= 9.09 KOhm\n",
+ "\n",
+ "\n",
+ "output voltage= 0.01 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter15_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter15_1_2.ipynb new file mode 100644 index 00000000..7e7356d2 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter15_1_2.ipynb @@ -0,0 +1,452 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8d0dd3bab90dc7613b5ca5489532f94ae244e073567e3fb2fd8cc9148ba4ba3a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter15-Applications and Design of Integrated Circuits"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math\n",
+ "\n",
+ "##Example 15.2\n",
+ "C=20.*10**-6;\n",
+ "Req=1000.;\n",
+ "fC=1./(C*Req);\n",
+ "print\"%s %.2f %s\"%('\\nclock frequency = ',fC,' KHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "clock frequency = 50.00 KHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg936"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.3\n",
+ "fC=10000.;\n",
+ "f3dB=1000.;\n",
+ "##x=C2/C1\n",
+ "x=2.*math.pi*f3dB/fC;\n",
+ "print\"%s %.2f %s\"%('\\ncapacitances C2/C1= \\n',x,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "capacitances C2/C1= \n",
+ " 0.63 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg940"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.4\n",
+ "C=0.1*10**-6;\n",
+ "R=1000.;\n",
+ "fo=1/(2.*math.pi*R*C*math.sqrt(3.));\n",
+ "print\"%s %.2f %s\"%('\\nthe oscillation frequency = ',fo,'Hz\\n')\n",
+ "##minimum amplifier gain=8\n",
+ "R=1.;##KOhm\n",
+ "R2=8.*R;\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the oscillation frequency = 918.88 Hz\n",
+ "\n",
+ "\n",
+ "R2= 8.00 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.6\n",
+ "R1=10000.;\n",
+ "R2=90000.;\n",
+ "Vh=10.;\n",
+ "Vl=-10.;\n",
+ "Vth=R1*Vh/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\nupper crossover voltage= ',Vth,' V\\n')\n",
+ "Vtl=R1*Vl/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\nlower crossover voltage= ',Vtl,' V\\n')\n",
+ "x=Vth-Vtl;\n",
+ "print\"%s %.2f %s\"%('\\nhysteresis width = ',x,'V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "upper crossover voltage= 1.00 V\n",
+ "\n",
+ "\n",
+ "lower crossover voltage= -1.00 V\n",
+ "\n",
+ "\n",
+ "hysteresis width = 2.00 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg958"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.7\n",
+ "Vs=2.;\n",
+ "Vh=15.;\n",
+ "Vl=-15.;\n",
+ "##hysteresis width=x\n",
+ "x=60.*0.001;##(V)\n",
+ "##Vth-Vtl=(R1/(R1+R2))*(Vh-Vl)\n",
+ "##R2/R=y\n",
+ "y=(Vh-Vl)/x-1.;\n",
+ "print\"%s %.2f %s\"%('\\nR2/R1= \\n',y,'')\n",
+ "Vref=(1.+1./y)*Vs;\n",
+ "print\"%s %.2f %s\"%('\\nreference voltage= ',Vref,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "R2/R1= \n",
+ " 499.00 \n",
+ "\n",
+ "reference voltage= 2.00 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg969"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.10\n",
+ "C=15.*10**-9;\n",
+ "T=100.*10**-6;##(s) time\n",
+ "R=T/(1.1*C);\n",
+ "R=R*0.001;##Kohm\n",
+ "print\"%s %.2f %s\"%('\\nResistance R= ',R,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Resistance R= 6.06 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg974"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.13\n",
+ "Rl=10.;##load resistance \n",
+ "Pl=20.;##power delivered to the load\n",
+ "Ps=20.;##(W)\n",
+ "Vp=math.sqrt(2.*Rl*Pl);\n",
+ "print\"%s %.2f %s\"%('\\npeak output voltage= ',Vp,' V\\n')\n",
+ "Ip=Vp/Rl;\n",
+ "print\"%s %.2f %s\"%('\\npeak load current = ',Ip,'A\\n')\n",
+ "Vs=math.pi*Rl*Ps/Vp;\n",
+ "print\"%s %.2f %s\"%('\\nrequired supply voltage= ',Vs,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "peak output voltage= 20.00 V\n",
+ "\n",
+ "\n",
+ "peak load current = 2.00 A\n",
+ "\n",
+ "\n",
+ "required supply voltage= 31.42 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg979"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.14\n",
+ "Vonl=5;\n",
+ "Vofl=4.96;\n",
+ "I1=0.005;\n",
+ "I2=1.5;\n",
+ "dVo=Vonl-Vofl;\n",
+ "dIo=I1-I2;\n",
+ "Rvf=-dVo/dIo;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Rvf,' Ohm\\n')\n",
+ "LR=100.*(Vonl-Vofl)/Vonl;\n",
+ "print\"%s %.2f %s\"%('\\nload regulation =\\n',LR,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "output resistance= 0.03 Ohm\n",
+ "\n",
+ "\n",
+ "load regulation =\n",
+ " 0.80 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg982"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.15\n",
+ "Aol=1000.;\n",
+ "Vref=5.;\n",
+ "Vo=10.;\n",
+ "Io=0.1*0.001;\n",
+ "Vt=0.026;\n",
+ "Rof=2.*Vt*Vo/(Io*Vref*Aol);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Rof,' mOhm\\n')\n",
+ "##dVo/Vo=V and dIo/Io=I\n",
+ "##V=-I*2*Vt/(Vref*Aol)\n",
+ "##V/I=x\n",
+ "x=-2.*Vt/(Vref*Aol);\n",
+ "print\"%s %.2e %s\"%('\\npercent change=\\n',x,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "output resistance= 1.04 mOhm\n",
+ "\n",
+ "\n",
+ "percent change=\n",
+ " -1.04e-05 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg984"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 15.16\n",
+ "Vz=6.3;\n",
+ "Vbe=0.6;\n",
+ "Veb=0.6;\n",
+ "Vo=8.;\n",
+ "R1=3.9;\n",
+ "R2=3.4;\n",
+ "R3=0.576;\n",
+ "Ic3=(Vz-3.*Vbe)/(R1+R2+R3);\n",
+ "print\"%s %.2f %s\"%('\\nbias current = ',Ic3,' mA\\n')\n",
+ "Vb7=Ic3*R1+2.*Vbe;\n",
+ "print\"%s %.2f %s\"%('\\ntemperature compensated reference voltage= ',Vb7,' V\\n')\n",
+ "R13=2.23;\n",
+ "R12=R13*Vo/Vb7-R13;\n",
+ "print\"%s %.2f %s\"%('\\nR12= ',R12,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "bias current = 0.57 mA\n",
+ "\n",
+ "\n",
+ "temperature compensated reference voltage= 3.43 V\n",
+ "\n",
+ "\n",
+ "R12= 2.97 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter16_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter16_1_2.ipynb new file mode 100644 index 00000000..a7c194d9 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter16_1_2.ipynb @@ -0,0 +1,428 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:967330d5e3a998fb6f78e20c34781eef3a87e37f117b0ab752e2b4eb4123a177"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter16-MOSFET Digital Circuits "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg1013"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 16.3\n",
+ "Vdd=5.;\n",
+ "Vtnd=0.8;\n",
+ "Vtnl=0.8;\n",
+ "Kn=35.;\n",
+ "Vo=0.1;\n",
+ "Vi=4.2;\n",
+ "##W/L=Y\n",
+ "yl=0.5;\n",
+ "##Kd/Kl=x\n",
+ "x=(Vdd-Vo-Vtnl)**2/(2.*Vo*(Vi-Vtnd)-Vo**2);\n",
+ "print\"%s %.2f %s\"%('\\nKd/Kl=\\n',x,'')\n",
+ "##Kd/Kl=yd/yl\n",
+ "yd=12.6\n",
+ "yl=0.5\n",
+ "iD=Kn*yl*(Vdd-Vo-Vtnl)**2/2.;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current = ',iD,' microA\\n')\n",
+ "P=iD*Vdd;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation= ',P,' microW\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Kd/Kl=\n",
+ " 25.09 \n",
+ "\n",
+ "drain current = 147.09 microA\n",
+ "\n",
+ "\n",
+ "power dissipation= 735.44 microW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg1017"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 16.4\n",
+ "Vdd=5.;\n",
+ "Vtnd=0.8;\n",
+ "Vtnl=-2.;\n",
+ "Kn=35.;\n",
+ "Vo=0.1;\n",
+ "Vi=5.;\n",
+ "##W/L=Y\n",
+ "yl=0.5;\n",
+ "##Kd/Kl=x\n",
+ "x=(-Vtnl)**2/(2.*Vo*(Vi-Vtnd)-Vo**2);\n",
+ "print\"%s %.2f %s\"%('\\nKd/Kl=\\n',x,'')\n",
+ "##Kd/Kl=yd/yl\n",
+ "yd=2.41\n",
+ "yl=0.5\n",
+ "iD=Kn*yl*(-Vtnl)**2/2.;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',iD,' microA\\n')\n",
+ "P=iD*Vdd;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation = ',P,' microW\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Kd/Kl=\n",
+ " 4.82 \n",
+ "\n",
+ "drain current= 35.00 microA\n",
+ "\n",
+ "\n",
+ "power dissipation = 175.00 microW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg1021"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 16.5\n",
+ "Voh=4.2;\n",
+ "Vol=0.1;\n",
+ "##x=Kd/Kl\n",
+ "x=25.1;\n",
+ "Vdd=5.;\n",
+ "Vtnl=0.8;\n",
+ "Vohu=4.2;\n",
+ "Vil=0.8;\n",
+ "Vtnd=0.8;\n",
+ "Vih=Vtnd+(Vdd-Vtnl)/x*((1+2*x)/math.sqrt(1.+3.*x)-1.);\n",
+ "print\"%s %.2f %s\"%('\\nVih= ',Vih,' V\\n')\n",
+ "Volu=(Vdd-Vtnl+x*(Vih-Vtnd))/(1.+2.*x);\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage corresponding to Vih= ',Volu,' V\\n')\n",
+ "NMl=Vil-Volu;\n",
+ "print\"%s %.2f %s\"%('\\nnoise margin= ',NMl,' V\\n')\n",
+ "NMh=Vohu-Vih;\n",
+ "print\"%s %.2f %s\"%('\\nnoise margin= ',NMh,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vih= 1.61 V\n",
+ "\n",
+ "\n",
+ "output voltage corresponding to Vih= 0.48 V\n",
+ "\n",
+ "\n",
+ "noise margin= 0.32 V\n",
+ "\n",
+ "\n",
+ "noise margin= 2.59 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg1041"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 16.9\n",
+ "Vdd=5.;\n",
+ "Vtn=1.;\n",
+ "Vtp=-1.;\n",
+ "##Kn=Kp hence Kn/Kp=x=1;\n",
+ "x=1.;\n",
+ "Vit=(Vdd+Vtp+math.sqrt(x)*Vtn)/(1.+math.sqrt(x));\n",
+ "print\"%s %.2f %s\"%('\\ninput voltage= ',Vit,' V\\n')\n",
+ "Vipt=Vit;\n",
+ "Vopt=Vipt-Vtp;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage at the transition point for PMOS = ',Vopt,' V\\n')\n",
+ "Vint=Vit;\n",
+ "Vont=Vint-Vtn;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage at the transition point for NMOS= ',Vont,' V\\n')\n",
+ "Vdd=10.;\n",
+ "Vit=(Vdd+Vtp+math.sqrt(x)*Vtn)/(1.+math.sqrt(x));\n",
+ "print\"%s %.2f %s\"%('\\ninput voltage = ',Vit,'V\\n')\n",
+ "Vipt=Vit;\n",
+ "Vint=Vit;\n",
+ "Vopt=Vipt-Vtp;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage at the transition point for PMOS = ',Vopt,' V\\n')\n",
+ "Vont=Vint-Vtn;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage at the transition point for NMOS = ',Vont,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "input voltage= 2.50 V\n",
+ "\n",
+ "\n",
+ "output voltage at the transition point for PMOS = 3.50 V\n",
+ "\n",
+ "\n",
+ "output voltage at the transition point for NMOS= 1.50 V\n",
+ "\n",
+ "\n",
+ "input voltage = 5.00 V\n",
+ "\n",
+ "\n",
+ "output voltage at the transition point for PMOS = 6.00 V\n",
+ "\n",
+ "\n",
+ "output voltage at the transition point for NMOS = 4.00 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg1045"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 16.10\n",
+ "Cl=2.*10**-6;\n",
+ "Vdd=5.;\n",
+ "f=100000.;\n",
+ "P=f*Cl*Vdd**2;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation in the CMOS inverter= ',P,' microW\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "power dissipation in the CMOS inverter= 5.00 microW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg1047"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 16.11\n",
+ "Vtn=1.;\n",
+ "Vtp=-1.;\n",
+ "Vdd=5.;\n",
+ "Vth=1.;\n",
+ "Vil=Vtn+3.*(Vdd+Vtp-Vth)/8.;\n",
+ "print\"%s %.2f %s\"%('\\ninput voltage at the transition points Vil= ',Vil,' V\\n')\n",
+ "Vih=Vtn+5.*(Vdd+Vtp-Vtn)/8.;\n",
+ "print\"%s %.2f %s\"%('\\ninput voltage at the transition points Vih= ',Vih,' V\\n')\n",
+ "Vohu=1.*(2.*Vil+Vdd-Vtn-Vtp)/2.;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage = ',Vohu,' V\\n')\n",
+ "Volu=1.*(2.*Vih-Vdd-Vtn-Vtp)/2.;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage = ',Volu,' V\\n')\n",
+ "NML=Vil-Volu;\n",
+ "print\"%s %.2f %s\"%('\\nnoise margin = ',NML,' V\\n')\n",
+ "NMH=Vohu-Vih;\n",
+ "print\"%s %.2f %s\"%('\\nnoise margin= ',NML,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "input voltage at the transition points Vil= 2.12 V\n",
+ "\n",
+ "\n",
+ "input voltage at the transition points Vih= 2.88 V\n",
+ "\n",
+ "\n",
+ "output voltage = 4.62 V\n",
+ "\n",
+ "\n",
+ "output voltage = 0.38 V\n",
+ "\n",
+ "\n",
+ "noise margin = 1.75 V\n",
+ "\n",
+ "\n",
+ "noise margin= 1.75 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg1080"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 16.15\n",
+ "Vdd=3.;\n",
+ "Kn=60.;\n",
+ "Vtnd=0.5;\n",
+ "##W/L=x\n",
+ "xd=2.;\n",
+ "Vtnl=-1.;\n",
+ "xl=0.5;\n",
+ "R=2.;##(MOhm)\n",
+ "Vgsl=0.;\n",
+ "##solution with Depletion load\n",
+ "iD=Kn*xl*(Vgsl-Vtnl)**2/2.;\n",
+ "print\"%s %.2f %s\"%('\\nfrain currents in M1 and M3 = ',iD,' microA\\n')\n",
+ "P=iD*Vdd;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation in the circuit= ',P,'microW\\n')\n",
+ "##iD=Kn/2*x*(2*Vgsd-Vtnd)Vdsd-Vdsd^2\n",
+ "Q=50.5\n",
+ "p=0.25 - 5*(50.5) + 50.5\n",
+ "print(p)\n",
+ "\n",
+ "##solution with Resistor load\n",
+ "##(Vdd-Q)/R=Kn/2*xd*(2*Vgsd-Vtnd)Q-Q^2\n",
+ "\n",
+ "\n",
+ "Q=0.005;\n",
+ "p1=0.25 - 5*(0.005) + 0.005\n",
+ "print(p1)\n",
+ "iD=(Vdd-Q)/R;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current = ',iD,' microA\\n')\n",
+ "P=iD*Vdd;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation in the circuit = ',P,' microW\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "frain currents in M1 and M3 = 15.00 microA\n",
+ "\n",
+ "\n",
+ "power dissipation in the circuit= 45.00 microW\n",
+ "\n",
+ "-201.75\n",
+ "0.23\n",
+ "\n",
+ "drain current = 1.50 microA\n",
+ "\n",
+ "\n",
+ "power dissipation in the circuit = 4.49 microW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter17_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter17_1_2.ipynb new file mode 100644 index 00000000..0913e184 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter17_1_2.ipynb @@ -0,0 +1,647 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:abb54176eeb369f0f245b707b98b6a093b5c09d2bb0c269e911dfcaf8a6dfa9d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter17-Bipolar Digital Circuits"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg1115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.1\n",
+ "V1=5.;\n",
+ "V2=-5.;\n",
+ "Rc1=1.;\n",
+ "Rc2=Rc1;\n",
+ "Rc=Rc1;\n",
+ "Re=2.150;\n",
+ "v2=0.;\n",
+ "##for v1=0\n",
+ "vE=-0.7;\n",
+ "iE=(vE-V2)/Re;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',iE,' mA\\n')\n",
+ "iC=1.;\n",
+ "Vcc=5.;\n",
+ "vo1=Vcc-iC*Rc;\n",
+ "print\"%s %.2f %s\"%('\\nvo1=vo2= ',vo1,' V\\n')\n",
+ "##for v2=-1\n",
+ "vE=-0.7;\n",
+ "iE=2.;\n",
+ "iC2=2.;\n",
+ "vo1=5.;\n",
+ "vo2=Vcc-iC2*Rc;\n",
+ "print\"%s %.2f %s\"%('\\nvo2= ',vo2,' V\\n')\n",
+ "v1=1.;\n",
+ "Vbe=0.7;\n",
+ "vE=v1-Vbe;\n",
+ "iE=(vE-V2)/Re;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current =',iE,' mA\\n')\n",
+ "iC1=iE;\n",
+ "vo1=Vcc-iC1*Rc;\n",
+ "print\"%s %.2f %s\"%('\\nvo1= ',vo1,' V\\n')\n",
+ "vo2=Vcc\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "emitter current= 2.00 mA\n",
+ "\n",
+ "\n",
+ "vo1=vo2= 4.00 V\n",
+ "\n",
+ "\n",
+ "vo2= 3.00 V\n",
+ "\n",
+ "\n",
+ "emitter current = 2.47 mA\n",
+ "\n",
+ "\n",
+ "vo1= 2.53 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg1118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.2\n",
+ "Vx=-0.7;\n",
+ "Vy=Vx;\n",
+ "Vbe=0.7;\n",
+ "V2=-5.2;\n",
+ "Re=1.180;\n",
+ "vE=Vx-Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nemitter voltage = ',vE,' V\\n')\n",
+ "iE=(vE-V2)/Re;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',iE,' mA\\n')\n",
+ "Icxy=iE;\n",
+ "vo1=-0.7;\n",
+ "Rc1=-vo1/Icxy;\n",
+ "print\"%s %.2f %s\"%('\\nRc1= ',Rc1,' KOhm\\n')\n",
+ "Vnor=vo1-Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nNOR output logic 0 value= ',Vnor,' V\\n')\n",
+ "Vr=(vo1+Vnor)/2.;\n",
+ "vE=Vr-Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nvE= ',vE,' V\\n')\n",
+ "iE=(vE-V2)/Re;\n",
+ "print\"%s %.2f %s\"%('\\niE= ',iE,' mA\\n')\n",
+ "vo2=-0.7;\n",
+ "iC2=iE;\n",
+ "Rc2=-vo2/iC2;\n",
+ "print\"%s %.2f %s\"%('\\nRc2= ',Rc2,' KOhm\\n')\n",
+ "Vor=vo2-Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nOR logic 0 value is= ',Vor,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "emitter voltage = -1.40 V\n",
+ "\n",
+ "\n",
+ "emitter current= 3.22 mA\n",
+ "\n",
+ "\n",
+ "Rc1= 0.22 KOhm\n",
+ "\n",
+ "\n",
+ "NOR output logic 0 value= -1.40 V\n",
+ "\n",
+ "\n",
+ "vE= -1.75 V\n",
+ "\n",
+ "\n",
+ "iE= 2.92 mA\n",
+ "\n",
+ "\n",
+ "Rc2= 0.24 KOhm\n",
+ "\n",
+ "\n",
+ "OR logic 0 value is= -1.40 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg1120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.3\n",
+ "Vr=-1.05;\n",
+ "Vbe=0.7;\n",
+ "Vb5=Vr+Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nVb5 = ',Vb5,' V\\n')\n",
+ "R1=0.250;\n",
+ "i1=-Vb5/R1;\n",
+ "print\"%s %.2f %s\"%('\\ni1= ',i1,' mA\\n')\n",
+ "Vy=0.7;\n",
+ "V2=-5.2;\n",
+ "##let R1+R2=x\n",
+ "x=(-2.*Vy-V2)/i1;\n",
+ "R2=x-R1;\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,' KOhm\\n')\n",
+ "iS=i1;\n",
+ "Rs=(Vr-V2)/iS;\n",
+ "print\"%s %.2f %s\"%('\\nRs= ',Rs,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vb5 = -0.35 V\n",
+ "\n",
+ "\n",
+ "i1= 1.40 mA\n",
+ "\n",
+ "\n",
+ "R2= 2.46 KOhm\n",
+ "\n",
+ "\n",
+ "Rs= 2.96 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg1121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.4\n",
+ "Vx=-0.7;\n",
+ "Vy=-0.7;\n",
+ "iCxy=3.22;##(mA)\n",
+ "iCR=0.;\n",
+ "i5=1.40;\n",
+ "i1=1.40;\n",
+ "Vor=-0.7;\n",
+ "R4=1.500;\n",
+ "Vnor=-1.4;\n",
+ "V2=-5.2;\n",
+ "R3=1.500;\n",
+ "i3=(Vor-V2)/R3;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent i3= ',i3,' mA\\n')\n",
+ "i4=(Vnor-V2)/R4;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent i4 = ',i4, 'mA')\n",
+ "P=(iCxy+iCR+i5+i1+i3+i4)*(0.-V2);\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation= ',P,' mW\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "current i3= 3.00 mA\n",
+ "\n",
+ "\n",
+ "current i4 = 2.53 mA\n",
+ "\n",
+ "power dissipation= 60.08 mW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg1122"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.5\n",
+ "b=50.;\n",
+ "V2=-5.2;\n",
+ "Vbe=0.7;\n",
+ "Rc2=0.240;\n",
+ "Vor=-0.75;\n",
+ "Re=1.180;\n",
+ "iE=(Vor-Vbe-V2)/Re;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',iE,' mA\\n')\n",
+ "iB=iE/(1.+b);\n",
+ "iB=iB*1000.;##micro A\n",
+ "print\"%s %.2f %s\"%('\\ninput base current= ',iB,' microA\\n')\n",
+ "R3=1.500;\n",
+ "i3=(Vor-V2)/R3;\n",
+ "print\"%s %.2f %s\"%('\\ni3= ',i3,' mA\\n')\n",
+ "iB=iB*0.001;##mA\n",
+ "N=(-(Vor+Vbe)*(1.+b)/(Rc2)-i3)/iB;\n",
+ "print\"%s %.2f %s\"%('\\nN\\n',N,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "emitter current= 3.18 mA\n",
+ "\n",
+ "\n",
+ "input base current= 62.31 microA\n",
+ "\n",
+ "\n",
+ "i3= 2.97 mA\n",
+ "\n",
+ "\n",
+ "N\n",
+ " 122.90 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg1127"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 17.7\n",
+ "Vcc=1.7;\n",
+ "Re=0.008;##mohm\n",
+ "Rc=0.008;##mohm\n",
+ "Vy=0.4;\n",
+ "Vbe=0.7;\n",
+ "Vor=Vcc##logic 1\n",
+ "Vor=Vcc-Vy##logic 0\n",
+ "Vr=1.5;\n",
+ "iE=(Vr-Vbe)/Re;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',iE,' microA\\n')\n",
+ "iR=Vy/Rc;\n",
+ "print\"%s %.2f %s\"%('\\nmaximum current in Rc = ',iR,' microA\\n')\n",
+ "iD=iE-iR;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent through the diode= ',iD,' microA\\n')\n",
+ "P=iE*Vcc;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation= ',P,' microW\\n')\n",
+ "Vv=1.7;\n",
+ "iE=(Vv-Vbe)/Re;\n",
+ "print\"%s %.2f %s\"%('\\niE = ',iE,' microA\\n')\n",
+ "P=iE*Vcc;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation = ',P,' microW\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "emitter current= 100.00 microA\n",
+ "\n",
+ "\n",
+ "maximum current in Rc = 50.00 microA\n",
+ "\n",
+ "\n",
+ "current through the diode= 50.00 microA\n",
+ "\n",
+ "\n",
+ "power dissipation= 170.00 microW\n",
+ "\n",
+ "\n",
+ "iE = 125.00 microA\n",
+ "\n",
+ "\n",
+ "power dissipation = 212.50 microW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.9\n",
+ "bf=25.;\n",
+ "b=bf;\n",
+ "br=0.1;\n",
+ "Vcc=5.;\n",
+ "R1=4.;\n",
+ "Vbc=0.7;\n",
+ "Vy=0.1;\n",
+ "Vx=0.1;\n",
+ "R2=1.6;\n",
+ "Vbe=0.8;\n",
+ "Rc=4.;\n",
+ "Vce=0.1;\n",
+ "vB2=Vx+Vce;\n",
+ "print\"%s %.2f %s\"%('\\nvB2= ',vB2,' V\\n')\n",
+ "vB1=Vx+Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nbase voltage= ',vB1,' V\\n')\n",
+ "i1=(Vcc-vB1)/R1;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent i1= ',i1,' mA\\n')\n",
+ "vB1=Vbe+Vbe+Vbc;\n",
+ "print\"%s %.2f %s\"%('\\nvB1= ',vB1,' V\\n')\n",
+ "vC2=Vbe+Vce;\n",
+ "print\"%s %.2f %s\"%('\\ncollector voltage= ',vC2,' V\\n')\n",
+ "i1=(Vcc-vB1)/R1;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent i1 = ',i1,' mA\\n')\n",
+ "iB2=(1.+2.*br)*i1;\n",
+ "print\"%s %.2f %s\"%('\\niB2= ',iB2,' mA\\n')\n",
+ "i2=(Vcc-vC2)/R2;\n",
+ "print\"%s %.2f %s\"%('\\ni2 = ',i2,' mA\\n')\n",
+ "iE2=i2+iB2;\n",
+ "print\"%s %.2f %s\"%('\\niE2= ',iE2,' mA\\n')\n",
+ "Rb=1.;\n",
+ "i4=Vbe/Rb;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent in the pull down resistor= ',i4,' mA\\n')\n",
+ "iBo=iE2-i4;\n",
+ "print\"%s %.2f %s\"%('\\nbase drive to the output transistor= ',iBo,' mA\\n')\n",
+ "i1=(Vcc-Vce)/Rc;\n",
+ "print\"%s %.2f %s\"%('\\ni1= ',i1,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "vB2= 0.20 V\n",
+ "\n",
+ "\n",
+ "base voltage= 0.90 V\n",
+ "\n",
+ "\n",
+ "current i1= 1.02 mA\n",
+ "\n",
+ "\n",
+ "vB1= 2.30 V\n",
+ "\n",
+ "\n",
+ "collector voltage= 0.90 V\n",
+ "\n",
+ "\n",
+ "current i1 = 0.68 mA\n",
+ "\n",
+ "\n",
+ "iB2= 0.81 mA\n",
+ "\n",
+ "\n",
+ "i2 = 2.56 mA\n",
+ "\n",
+ "\n",
+ "iE2= 3.37 mA\n",
+ "\n",
+ "\n",
+ "current in the pull down resistor= 0.80 mA\n",
+ "\n",
+ "\n",
+ "base drive to the output transistor= 2.57 mA\n",
+ "\n",
+ "\n",
+ "i1= 1.23 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg1150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.11\n",
+ "b=25.;\n",
+ "iB=1.;\n",
+ "iC=2.;\n",
+ "ic=(iB+iC)/(1.+1./b);\n",
+ "print\"%s %.2f %s\"%('\\ninternal collector current= ',ic,' mA\\n',)\n",
+ "ib=ic/b;\n",
+ "print\"%s %.2f %s\"%('\\ninternal base current = ',ib,' mA\\n')\n",
+ "iD=iB-ib;\n",
+ "print\"%s %.2f %s\"%('\\nSchottky diode current= ',iD,' mA\\n')\n",
+ "iC=20.;\n",
+ "ic=(iB+iC)/(1.+1./b);\n",
+ "print\"%s %.2f %s\"%('\\ninternal collector current= ',ic,' mA\\n')\n",
+ "ib=ic/b;\n",
+ "print\"%s %.2f %s\"%('\\ninternal base current = ',ib,' mA\\n')\n",
+ "iD=iB-ib;\n",
+ "print\"%s %.2f %s\"%('\\nSchottky diode current= ',iD,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "internal collector current= 2.88 mA\n",
+ "\n",
+ "\n",
+ "internal base current = 0.12 mA\n",
+ "\n",
+ "\n",
+ "Schottky diode current= 0.88 mA\n",
+ "\n",
+ "\n",
+ "internal collector current= 20.19 mA\n",
+ "\n",
+ "\n",
+ "internal base current = 0.81 mA\n",
+ "\n",
+ "\n",
+ "Schottky diode current= 0.19 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg1154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 17.12\n",
+ "Vy=0.3;\n",
+ "Vbe=0.7;\n",
+ "vx=0.4;\n",
+ "R2=8.;\n",
+ "Vce=0.4;\n",
+ "Vcc=5.;\n",
+ "b=25.;\n",
+ "Vce=0.4;\n",
+ "Vbe1=0.7;\n",
+ "Vbe2=0.7;\n",
+ "Vcc=5.;\n",
+ "R1=20.;\n",
+ "v1=Vce+Vy;\n",
+ "i1=(Vcc-v1)/R1;\n",
+ "print\"%s %.2f %s\"%('\\ni1= ',i1,' mA\\n')\n",
+ "Pl=i1*(Vcc-vx);\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation= ',Pl,' mW\\n')\n",
+ "v1=Vbe1+Vbe2;\n",
+ "print\"%s %.2f %s\"%('\\nv1= ',v1,' V\\n')\n",
+ "vC2=Vbe1+Vce;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage vC2 =',vC2,' V\\n')\n",
+ "i1=(Vcc-v1)/R1;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent i1 = ',i1,' mA\\n')\n",
+ "i2=(Vcc-vC2)/R2;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent i2 = ',i2,' mA\\n')\n",
+ "P=(i1+i2)*Vcc;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation for high input condition= ',P,' mW\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "i1= 0.21 mA\n",
+ "\n",
+ "\n",
+ "power dissipation= 0.99 mW\n",
+ "\n",
+ "\n",
+ "v1= 1.40 V\n",
+ "\n",
+ "\n",
+ "voltage vC2 = 1.10 V\n",
+ "\n",
+ "\n",
+ "current i1 = 0.18 mA\n",
+ "\n",
+ "\n",
+ "current i2 = 0.49 mA\n",
+ "\n",
+ "\n",
+ "power dissipation for high input condition= 3.34 mW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter1_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter1_1_2.ipynb new file mode 100644 index 00000000..d1c77068 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter1_1_2.ipynb @@ -0,0 +1,474 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:12fda3be337124becb50b8ef7de2608c3b358deb143d70552beef4ab1e72fe80"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Semiconductor materials and diodes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.1\n",
+ "T=300.;##((K)temperature)\n",
+ "##for silicon\n",
+ "B=5.23*10**(15);##Constant (per centimeter cube degree kelvin)\n",
+ "Eg=1.1;##bandgap energy in electrovolt(eV)\n",
+ "k=86.*10**(-6);##Boltzmann's constant(eV per degree kelvin)\n",
+ "n_i=B*T**(3/2.)*math.exp(-Eg/(2.*k*T));##intrinsic carrier concentration\n",
+ "print\"%s %.2f %s\"%('intrinsic carrier concentration= ',n_i,' cm^-3');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "intrinsic carrier concentration= 14995738948.72 cm^-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math\n",
+ "\n",
+ "#calculate the\n",
+ "\n",
+ "##Example 1.2 \n",
+ "T=300.;##(K)Given Temperature\n",
+ "Nd=10**16;##(cm^-3)Donor concentration\n",
+ "n_i=1.5*10**10;##(cm^-3)intrinsic carrier concentration\n",
+ "##since Nd>>n_i\n",
+ "n_o=10**16;##(cm^-3)electron concentration\n",
+ "##by using formula ::n_i^2=n_o*p_o\n",
+ "p_o=(n_i)**2/Nd;##hole concentration\n",
+ "print\"%s %.2e %s\"%('\\nelectron concentration= ',n_o,' cm^-3');\n",
+ "print\"%s %.2e %s\"%('\\nhole concentration = ',p_o,' cm^-3');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "electron concentration= 1.00e+16 cm^-3\n",
+ "\n",
+ "hole concentration = 2.25e+04 cm^-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example1.3\n",
+ "T=300;##(K)Given Temperature\n",
+ "Na=10**16;##(cm^-3)Acceptor concentration in p region\n",
+ "Nd=10**17;##(cm^-3)Donor concentration in n region\n",
+ "n_i=1.5*10**10;##(cm^-3)intrinsic carrier concentration\n",
+ "V_T=0.026;##(Volt)terminal voltage\n",
+ "##built-in potential\n",
+ "V_bi=V_T*math.log(Na*Nd/(n_i)**2);\n",
+ "print\"%s %.2f %s\"%('\\nthe built-in potential= ',V_bi,'V')\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the built-in potential= 0.76 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.4\n",
+ "T=300.;##(K)Given Temperature\n",
+ "Na=10**16;##(cm**-3)Acceptor concentration in p region\n",
+ "Nd=10**15;##(cm**-3)Donor concentration in n region\n",
+ "n_i=1.5*10**10;##(cm**-3)intrinsic carrier concentration\n",
+ "C_jo=0.5;##(pF)junction capacitance at zero applied voltage\n",
+ "V_T=0.026;##(Volt)terminal voltage\n",
+ "##built-in potential\n",
+ "V_bi=V_T*math.log(Na*Nd/(n_i)**2);\n",
+ "print\"%s %.2f %s\"%(\"the built-in potential(V)\",V_bi,\"\")\n",
+ "##the junction capacitance for\n",
+ "V_R=1.;##(V)reverse bias voltage\n",
+ "Cj=C_jo*(1.+V_R/V_bi)**(-1/2.);\n",
+ "print\"%s %.2f %s\"%('\\nthe junction capacitance for V_R=1V= ',Cj,' pF\\n')\n",
+ "V_R=5.;##(V)reverse bias voltage\n",
+ "Cj=C_jo*(1.+V_R/V_bi)**(-1/2.);\n",
+ "print\"%s %.2f %s\"%('\\nthe junction capacitance for V_R=5V = ',Cj,' pF')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the built-in potential(V) 0.64 \n",
+ "\n",
+ "the junction capacitance for V_R=1V= 0.31 pF\n",
+ "\n",
+ "\n",
+ "the junction capacitance for V_R=5V = 0.17 pF\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.5\n",
+ "T=300.;##(K)Given Temperature\n",
+ "V_T=0.026;##(Volt)terminal voltage\n",
+ "Is=10**-11;##(mA)reverse bias saturation current\n",
+ "n=1.;##emission coefficient\n",
+ "v_D=+0.7;##(V)applied voltage\n",
+ "##pn junction is forward biased\n",
+ "i_D=Is*(math.exp(v_D/V_T)-1.);##diode current\n",
+ "print\"%s %.2f %s\"%('\\ndiode current= ',i_D,' mA\\n')\n",
+ "v_D=-0.7;##(V)pn junction is reverse biased\n",
+ "Is=10**-14##A;\n",
+ "i_D=Is*(math.exp(v_D/V_T)-1);##diode current\n",
+ "print\"%s %.2e %s\"%('\\ndiode current= ',i_D,' A')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "diode current= 4.93 mA\n",
+ "\n",
+ "\n",
+ "diode current= -1.00e-14 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.6\n",
+ "Is=10**-13;##(A)reverse saturation current\n",
+ "V_PS=5.;##(V)applied voltage\n",
+ "R=2;##(KOhm)Resistance in circuit\n",
+ "V_T=0.026;##(Volt)terminal voltage\n",
+ "##V_PS=Is*R*(exp(V_D/V_T)-1)+V_D\n",
+ "##5=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D\n",
+ "##let right side of equation be x=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D\n",
+ "V_D=0.6;##(V)\n",
+ "x=(10**-13)*(2000.)*(math.exp(V_D/V_T)-1.)+V_D\n",
+ "##so the equation is not balanced\n",
+ "V_D=0.65;##(V)\n",
+ "x=(10**-13)*(2000.)*(math.exp(V_D/V_T)-1.)+V_D\n",
+ "##again equation is not balanced .solution for V_D is between 0.6V and 0.65V\n",
+ "V_D=0.619;##(V)\n",
+ "x=(10**-13)*(2000.)*(math.exp(V_D/V_T)-1.)+V_D\n",
+ "##essentially equal to the value of the left side of the equation i.e 5V\n",
+ "print\"%s %.2f %s\"%('\\ndiode voltage= ',V_D,' V')\n",
+ "I_D=(V_PS-V_D)/R;##(A)diode current\n",
+ "print\"%s %.2f %s\"%('\\nthe diode current= ',I_D,' mA')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "diode voltage= 0.62 V\n",
+ "\n",
+ "the diode current= 2.19 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.7\n",
+ "##piecewise linear diode parameters\n",
+ "V_Y=0.6;##(V)\n",
+ "r_f=0.010;##(KOhm)\n",
+ "V_PS=5.;##(V)applied voltage\n",
+ "R=2.;##(KOhm)Resistance in circuit\n",
+ "I_D=(V_PS-V_Y)/(R+r_f);##(A)diode current\n",
+ "print\"%s %.2f %s\"%('\\nthe diode current= ',I_D,' mA\\n')\n",
+ "V_D=V_Y+I_D*r_f;##(V)diode voltage\n",
+ "print\"%s %.2f %s\"%('\\ndiode voltage= ',V_D,' V')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the diode current= 2.19 mA\n",
+ "\n",
+ "\n",
+ "diode voltage= 0.62 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.9 \n",
+ "##circuit and diode parameters \n",
+ "V_PS=5.;##(V)\n",
+ "R=5;##(KOhm)\n",
+ "V_Y=0.6;##(V)\n",
+ "V_T=0.026;##(Volt)terminal voltage\n",
+ "v_i=0.1##*sin(wt)Volt\n",
+ "##dc analysis\n",
+ "I_DQ=(V_PS-V_Y)/R;\n",
+ "print\"%s %.2f %s\"%('\\ndc quiescent current= ',I_DQ,' mA\\n')\n",
+ "V_O=I_DQ*R;\n",
+ "print\"%s %.2f %s\"%('\\ndc output voltage= ',V_O,' V\\n')\n",
+ "##ac analysis\n",
+ "V_PS=0.;\n",
+ "##Kirchhoff voltage law equation becomes\n",
+ "##v_i=i_d*r_d+i_d*R\n",
+ "r_d=V_T/I_DQ##(Ohm)small signal diode diffusion resistance\n",
+ "i_d=v_i/(r_d+R);##ac diode current\n",
+ "print\"%s %.2f %s\"%('\\nac diode current= ',i_d,'sin(wt) A\\n')\n",
+ "\n",
+ "v_o=i_d*R;##ac output voltage\n",
+ "print\"%s %.2f %s\"%('\\nac output voltage= ',v_o,'sin(wt) V')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "dc quiescent current= 0.88 mA\n",
+ "\n",
+ "\n",
+ "dc output voltage= 4.40 V\n",
+ "\n",
+ "\n",
+ "ac diode current= 0.02 sin(wt) A\n",
+ "\n",
+ "\n",
+ "ac output voltage= 0.10 sin(wt) V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.10\n",
+ "V_Y=0.7;##(V)cut in voltage for pn junction\n",
+ "r_f=0.;\n",
+ "V_PS=4;##(V)\n",
+ "R1=4.\n",
+ "R2=4.##(KOhm) from given circuit\n",
+ "I1=(V_PS-V_Y)/R1;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent through pn junction diode= ',I1,' mA\\n')\n",
+ "V_Y=0.3;##(V)cut in voltage for Schottky diode\n",
+ "I2=(V_PS-V_Y)/R2;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent through Schottky diode= ',I2,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "current through pn junction diode= 0.82 mA\n",
+ "\n",
+ "\n",
+ "current through Schottky diode= 0.93 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 1.11 \n",
+ "V_Z=5.6;##(V)Zener diode breakdown voltage\n",
+ "r_z=0.;##(Ohm)Zener resistance\n",
+ "I=3.;##(mA)current in the diode\n",
+ "V_PS=10.;##(V)\n",
+ "##I=(V_PS-V_Z)/R\n",
+ "R=(V_PS-V_Z)/I;\n",
+ "print\"%s %.2f %s\"%('\\nresistance= ',R,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "resistance= 1.47 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter2_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter2_1_2.ipynb new file mode 100644 index 00000000..324bce27 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter2_1_2.ipynb @@ -0,0 +1,407 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:77e57590b67964207d6f9870670ae0e0da31a20ccd56c5e90b158890f9c8ff6f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Diode Circuits"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 2.1\n",
+ "v_I=120.;##(V)rms primary input \n",
+ "v_o=9.;##(V)peak output voltage\n",
+ "V_Y=0.7;##(V)diode cut in voltage\n",
+ "##for center-tapped transformer circuit in fig.2.6(a)\n",
+ "v_S=v_o+V_Y##(V)peak value of secondary voltage\n",
+ "print\"%s %.2f %s\"%('\\npeak value of secondary voltage= ',v_S,' V\\n')\n",
+ "v_S_rms=v_S/math.sqrt(2)##for a sinusoidal signal rms value of v_S\n",
+ "print\"%s %.2f %s\"%('\\nrms value of v_S= ',v_S_rms,' V\\n')\n",
+ "##let turns ratio of the primary to secondary winding be x=N1/N2\n",
+ "x=v_I/v_S_rms;\n",
+ "print\"%s %.2f %s\"%('\\nturns ratio= \\n',x,'')\n",
+ "##for the bridge circuit in fig.2.7(a)\n",
+ "v_Sb=v_o+2*V_Y;##(V)peak value of secondary voltage\n",
+ "print\"%s %.2f %s\"%('\\npeak value of secondary voltage= ',v_Sb,' V\\n')\n",
+ "v_S_rms=v_Sb/math.sqrt(2.);##for a sinusoidal signal rms value of v_S\n",
+ "print\"%s %.2f %s\"%('\\nrms value of v_S= ',v_S_rms,' V\\n')\n",
+ "##let turns ratio of the primary to secondary winding be x=N1/N2\n",
+ "x=v_I/v_S_rms;\n",
+ "print\"%s %.2f %s\"%('\\nturns ratio=\\n',x,'')\n",
+ "##for center tapped rectifier\n",
+ "PIV=2*v_S-V_Y;\n",
+ "print\"%s %.2f %s\"%('\\npeak inverse voltage of a diode= ',PIV,' V\\n')\n",
+ "##for the bridge rectifier peak inverse voltage of a diode\n",
+ "PIV=v_Sb-V_Y;\n",
+ "print\"%s %.2f %s\"%('\\npeak inverse voltage of a diode=\\n',PIV,'V')\n",
+ "##advantage of bridge rectifier over center tapped rectifier is it requies only half of the turns\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "peak value of secondary voltage= 9.70 V\n",
+ "\n",
+ "\n",
+ "rms value of v_S= 6.86 V\n",
+ "\n",
+ "\n",
+ "turns ratio= \n",
+ " 17.50 \n",
+ "\n",
+ "peak value of secondary voltage= 10.40 V\n",
+ "\n",
+ "\n",
+ "rms value of v_S= 7.35 V\n",
+ "\n",
+ "\n",
+ "turns ratio=\n",
+ " 16.32 \n",
+ "\n",
+ "peak inverse voltage of a diode= 18.70 V\n",
+ "\n",
+ "\n",
+ "peak inverse voltage of a diode=\n",
+ " 9.70 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 2.2\n",
+ "##full wave rectifier circuit with 60Hz input signal\n",
+ "V_M=10.;##(V)peak output voltage\n",
+ "R=0.01;##(MOhm)output load resistance\n",
+ "f=60.;##Hz\n",
+ "V_r=0.2;##(V)ripple voltage\n",
+ "C=V_M/(2.*f*R*V_r);##capacitance\n",
+ "print\"%s %.2f %s\"%('\\ncapacitance= ',C,' microF\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "capacitance= 41.67 microF\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 2.3\n",
+ "V_O=12.;##(V)peak output voltage\n",
+ "I_L=0.12;##(A)current delivered to the load\n",
+ "R=V_O/I_L;\n",
+ "print\"%s %.2f %s\"%('\\neffective load resistance= ',R,' Ohm\\n')\n",
+ "V_Y=0.7;##(V)diode cut in voltage\n",
+ "v_S=V_O+2.*V_Y;\n",
+ "print\"%s %.2f %s\"%('\\npeak value of v_S= ',v_S,' V\\n')\n",
+ "v_Srms=v_S/math.sqrt(2.);\n",
+ "print\"%s %.2f %s\"%('\\nrms voltage= ',v_Srms,' V\\n')\n",
+ "##let x=N1/N2\n",
+ "Vin=120.;##(V)input line voltage\n",
+ "x=Vin/v_Srms;\n",
+ "print\"%s %.2f %s\"%('\\nturns ratio= \\n',x,'')\n",
+ "VM=12.;##(V)\n",
+ "Vr=5/100.*VM;\n",
+ "print\"%s %.2f %s\"%('\\nripple voltage= ',Vr,' V\\n')\n",
+ "f=60.;##(Hz) input frequency\n",
+ "C=VM/(2.*R*Vr*f);\n",
+ "print\"%s %.2f %s\"%('\\nfilter capacitance= ',C,' F\\n')\n",
+ "i_Dmax=(VM/R)*(1+2*math.pi*math.sqrt(VM/(2.*Vr)));\n",
+ "print\"%s %.2f %s\"%('\\npeak diode current= ',i_Dmax,' A\\n')\n",
+ "R=0.1;##Kohm\n",
+ "i_Davg=(1/(2.*math.pi))*math.sqrt(2.*Vr/VM)*((VM/R)*(1.+math.pi*math.sqrt(VM/(2.*Vr))));\n",
+ "print\"%s %.2f %s\"%('\\naverage diode current= ',i_Davg,' mA\\n')\n",
+ "PIV=v_S-V_Y;\n",
+ "print\"%s %.2f %s\"%('\\npeak inverse voltage= ',PIV,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "effective load resistance= 100.00 Ohm\n",
+ "\n",
+ "\n",
+ "peak value of v_S= 13.40 V\n",
+ "\n",
+ "\n",
+ "rms voltage= 9.48 V\n",
+ "\n",
+ "\n",
+ "turns ratio= \n",
+ " 12.66 \n",
+ "\n",
+ "ripple voltage= 0.60 V\n",
+ "\n",
+ "\n",
+ "filter capacitance= 0.00 F\n",
+ "\n",
+ "\n",
+ "peak diode current= 2.50 A\n",
+ "\n",
+ "\n",
+ "average diode current= 66.04 mA\n",
+ "\n",
+ "\n",
+ "peak inverse voltage= 12.70 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 2.5\n",
+ "rZ=4.;##(Ohm) Zener resistance\n",
+ "V_Lnom=9.;##(V) nominal output voltage\n",
+ "Izmax=0.3;##(A) maximum zener diode current\n",
+ "Izmin=0.03;##(A) minimum zener diode current\n",
+ "V_Lmax=V_Lnom+Izmax*rZ\n",
+ "V_Lmin=V_Lnom+Izmin*rZ\n",
+ "##percent regulation R\n",
+ "R=((V_Lmax-V_Lmin)/V_Lnom)*100.;\n",
+ "print\"%s %.2f %s\"%('\\npercent regulation= ',R,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "percent regulation= 12.00 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 2.8\n",
+ "R1=5.;R2=10.;##(KOhm) \n",
+ "V_Y=0.7;##(V)diode cut in voltage\n",
+ "V1=5.;V2=-5;##(V)\n",
+ "vt=0.;##(V)\n",
+ "##asssuming initially diode D1 is off\n",
+ "##iR1=iD2=iR2=V1-V2-V_Y/(R1+R2)\n",
+ "iD2=(V1-V2-V_Y)/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\ndiode current= ',iD2,'mA\\n')\n",
+ "iR1=iD2;\n",
+ "vo=V1-iR1*R1;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage= ',vo,' V\\n')\n",
+ "v=vo-V_Y;##v=v'\n",
+ "print\"%s %.2f %s\"%('\\nVoltage= ',v,' V\\n')\n",
+ "vt=4.;##(V)fig.2.33\n",
+ "##both D1 and D2 are on\n",
+ "vo==vt;\n",
+ "vo=4.;\n",
+ "iD2=(V1-vo)/R1;\n",
+ "print\"%s %.2f %s\"%('\\ndiode current= ',iD2,' mA\\n')\n",
+ "iR1==iD2;\n",
+ "v=vo-V_Y;\n",
+ "print\"%s %.2f %s\"%('\\nV= ',v,' V\\n')\n",
+ "iR2=(v-V2)/R2;\n",
+ "print\"%s %.2f %s\"%('\\niR2= ',iR2,' mA\\n')\n",
+ "iD1=iR2-iD2;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent through D1= ',iD1,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "diode current= 0.62 mA\n",
+ "\n",
+ "\n",
+ "output voltage= 1.90 V\n",
+ "\n",
+ "\n",
+ "Voltage= 1.20 V\n",
+ "\n",
+ "\n",
+ "diode current= 0.20 mA\n",
+ "\n",
+ "\n",
+ "V= 3.30 V\n",
+ "\n",
+ "\n",
+ "iR2= 0.83 mA\n",
+ "\n",
+ "\n",
+ "current through D1= 0.63 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 2.10\n",
+ "n=1.;##quantum efficiency\n",
+ "A=10**-2;##cm^2 junction area\n",
+ "p=5*10**17;##(cm^-2-s^-1) incident photon flux\n",
+ "e=1.6*10**-16;##charge of an electron\n",
+ "Iph=n*e*p*A;\n",
+ "print\"%s %.2f %s\"%('\\nphotocurrent= ',Iph,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "photocurrent= 0.80 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 2.11\n",
+ "I=0.01;##(A) diode current\n",
+ "V_Y=1.7;##(V) forward bias voltage drop\n",
+ "Vt=0.2;##(V)\n",
+ "R=(5.-V_Y-Vt)/I;\n",
+ "print\"%s %.2f %s\"%('\\nresistance= ',R,' Ohm')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "resistance= 310.00 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter3_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter3_1_2.ipynb new file mode 100644 index 00000000..e6fdfc33 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter3_1_2.ipynb @@ -0,0 +1,992 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:454a57b394e23d05ed048843520d141cbdae1d1bc179f5ca99f9342e5d6f5307"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-The Bipolar Junction Transistor"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pgpg104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "\n",
+ "##Example 3.1\n",
+ "##let beta be \"b\"\n",
+ "b=150.;##common emitter current gain\n",
+ "iB=15*10**-3;##(mA) base current\n",
+ "##assume transistor biased in forward active mode\n",
+ "iC=b*iB;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',iC,' mA\\n')\n",
+ "iE=(1.+b)*iB;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',iE,' mA\\n')\n",
+ "a=b/(1.+b);\n",
+ "print\"%s %.2f %s\"%('\\ncommon base current gain=',a,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "collector current= 2.25 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 2.27 mA\n",
+ "\n",
+ "\n",
+ "common base current gain= 0.99 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 3.2\n",
+ "b=100.;##common emitter current gain\n",
+ "BVcbo=120.;##(V) break down voltage of the B-C junction\n",
+ "n=3.;##empirical constant\n",
+ "BVceo=BVcbo/(b)**(1./n);\n",
+ "print\"%s %.2f %s\"%('\\nbreakdown voltage= ',BVceo,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "breakdown voltage= 25.85 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 3.3\n",
+ "Vbb=4.;##(V)\n",
+ "Rb=220.##(KOhm);\n",
+ "Rc=2.;##(KOhm)\n",
+ "Vcc=10.;##(V)\n",
+ "Vbe=0.7;##(V)\n",
+ "b=200.;\n",
+ "##from fig.3.19(b)\n",
+ "Ib=(Vbb-Vbe)/Rb;\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n",
+ "Ic=b*Ib;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Ic,' mA\\n')\n",
+ "Ie=(1.+b)*Ib;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ie,' mA\\n')\n",
+ "Vce=Vcc-Ic*Rc;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vce,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current= 0.01 mA\n",
+ "\n",
+ "\n",
+ "collector current= 3.00 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 3.01 mA\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 4.00 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 3.4\n",
+ "Vbb=1.5;##(V)\n",
+ "Rb=580.;##(KOhm)\n",
+ "Veb=0.6;##(V)\n",
+ "Vcc=5.;##(V)\n",
+ "b=100.;\n",
+ "##writing Kirchhoff voltage law equation around E-B loop\n",
+ "Ib=(Vcc-Veb-Vbb)/Rb;\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n",
+ "Ic=b*Ib;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Ic,' mA\\n')\n",
+ "Ie=(1.+b)*Ib;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current=',Ie,' mA\\n')\n",
+ "Vec=(1./2.)*Vcc;\n",
+ "print\"%s %.2f %s\"%('\\nce voltage= ',Vec,' V\\n')\n",
+ "Rc=(Vcc-Vec)/Ic;\n",
+ "print\"%s %.2f %s\"%('\\ncollector resistance= ',Rc,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current= 0.01 mA\n",
+ "\n",
+ "\n",
+ "collector current= 0.50 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 0.51 mA\n",
+ "\n",
+ "\n",
+ "ce voltage= 2.50 V\n",
+ "\n",
+ "\n",
+ "collector resistance= 5.00 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 3.5\n",
+ "b=100.;\n",
+ "Vbe=0.7;##(V)\n",
+ "Vce=0.2;##(V)\n",
+ "Vbb=8.;##(v)\n",
+ "Rb=220.;##(KOhm)\n",
+ "Ib=(Vbb-Vbe)/Rb\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n",
+ "##transistor in active region\n",
+ "Ic=b*Ib;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Ic,' mA\\n')\n",
+ "Vcc=10.;##(V)\n",
+ "Rc=4.;##(KOhm)\n",
+ "Vce=Vcc-Ic*Rc;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vce,' V\\n')\n",
+ "##saturation\n",
+ "Vce=0.2;##(V)\n",
+ "Ic=(Vcc-Vce)/Rc;\n",
+ "print\"%s %.2f %s\"%('\\nsaturation collector current= ',Ic,' mA\\n')\n",
+ "x=Ic/Ib\n",
+ "##which is <b\n",
+ "Ie=Ic+Ib;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ie,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current= 0.03 mA\n",
+ "\n",
+ "\n",
+ "collector current= 3.32 mA\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= -3.27 V\n",
+ "\n",
+ "\n",
+ "saturation collector current= 2.45 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 2.48 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg122"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import numpy\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "import math\n",
+ "%matplotlib inline\n",
+ "import warnings\n",
+ "warnings.filterwarnings('ignore')\n",
+ "#calculate the \n",
+ "##Example 3.6\n",
+ "Vbe=0.7;\n",
+ "b=75.;\n",
+ "##Q point values::\n",
+ "##using KVL eq around the B-E loop\n",
+ "##Vbb=Ib*Re+Vbe+Ie*Re\n",
+ "##assuming transistor is in forward biased mode we can write Ie=(1+b)*Ib\n",
+ "Vbb=6.;\n",
+ "Rb=25.;##KOhm\n",
+ "Re=0.6;##KOhm\n",
+ "Ib=(Vbb-Vbe)/(Rb+(1.+b)*Re);\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n",
+ "Ic=b*Ib;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Ic,' mA\\n')\n",
+ "Ie=(1+b)*Ib;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ie,' mA\\n')\n",
+ "Vcc=12.;\n",
+ "Rc=0.4;\n",
+ "Vce=Vcc-Ic*Rc-Ie*Re;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vce,' V\\n')\n",
+ "##load line::\n",
+ "##using KVL law around C-E loop\n",
+ "##Vce=Vcc-(Ic*(Rc+((1+B)/B)*Re));\n",
+ "Ic=numpy.array([0.12,5.63])\n",
+ "Vce=12.-(Ic)*1\n",
+ "pyplot.plot(Vce,Ic)\n",
+ "pyplot.xlabel(\"Vce\")\n",
+ "pyplot.ylabel(\"Ic\")\n",
+ "pyplot.show()\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current= 0.08 mA\n",
+ "\n",
+ "\n",
+ "collector current= 5.63 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 5.71 mA\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 6.32 V\n",
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": "iVBORw0KGgoAAAANSUhEUgAAAXoAAAEPCAYAAABMTw/iAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x33d9e30>"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "%matplotlib inline\n",
+ "#calculate the \n",
+ "##Example 3.7\n",
+ "Vbe=0.65;\n",
+ "Vcc=5.;\n",
+ "Rc=0.5;##KOhm\n",
+ "b=100.;\n",
+ "V1=-5.;\n",
+ "Re=1.;##KOhm\n",
+ "## Q-point values :: writing KVL eq around B-E loop\n",
+ "Ie=(-V1-Vbe)/Re;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ie,' mA\\n')\n",
+ "Ib=(Ie/(1.+b));\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n",
+ "Ic=(b/(1.+b))*Ie;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Ic,' mA\\n')\n",
+ "Vce=Vcc-Ic*Rc-Ie*Re-V1;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vce,' V\\n')\n",
+ "##load line::\n",
+ "##Vce=Vcc-V1-(Ic*(Rc+((1+B)/B)*Re));\n",
+ "\n",
+ "Vce=numpy.array([0,2,3.5,4,6,8,10])\n",
+ "Ic=(10.-Vce)/1.51;\n",
+ "\n",
+ "\n",
+ "pyplot.plot(Vce,Ic)\n",
+ "pyplot.xlabel(\"Vce\")\n",
+ "pyplot.ylabel(\"Ic\")\n",
+ "\n",
+ "pyplot.title(\"load line\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "emitter current= 4.35 mA\n",
+ "\n",
+ "\n",
+ "base current= 0.04 mA\n",
+ "\n",
+ "\n",
+ "collector current= 4.31 mA\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 3.50 V\n",
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 4,
+ "text": [
+ "<matplotlib.text.Text at 0x75078d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x75076f0>"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "%matplotlib inline\n",
+ "#calculate the \n",
+ "##Example 3.9\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "V1=-5.;\n",
+ "V2=12.;\n",
+ "Rb=10.;\n",
+ "Re=5.;\n",
+ "Rc=5.;\n",
+ "Rl=5.;\n",
+ "##Q point values:: using KVL eq around B-E loop\n",
+ "Ib=-(V1+Vbe)/(Rb+(1.+b)*Re);\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n",
+ "Ic=b*Ib;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Ic,' mA\\n')\n",
+ "Ie=(1.+b)*Ib;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ie,' mA\\n')\n",
+ "##at collector node we can write Ic=(V2-Vo)/Rc-Vo/Rl\n",
+ "Vo=(V2/Rc-Ic)*Rc*Rl/(Rc+Rl);\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage= ',Vo,' V\\n')\n",
+ "Vce=Vo-Ie*Re-V1;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vce,' V\\n')\n",
+ "##load line::\n",
+ "Rth=Rl*Rc/(Rl+Rc);\n",
+ "print\"%s %.2f %s\"%('\\nThevenin rquivalent resistance= ',Rth,' KOhm\\n')\n",
+ "Vth=(Rl/(Rl+Rc))*V2;\n",
+ "print\"%s %.2f %s\"%('\\nThevenin equivalent voltage= ',Vth,' V\\n')\n",
+ "##fig.3.36(c) KVL law\n",
+ "##Vce=6-V1-Ic*Rth-Ie*Re;\n",
+ "\n",
+ "\n",
+ "\n",
+ "Vce=numpy.array([0,2,4.7,3.5,4,6,8,10])\n",
+ "Ic=(11.-Vce)/7.5;\n",
+ "\n",
+ "\n",
+ "pyplot.plot(Vce,Ic)\n",
+ "pyplot.xlabel(\"Vce\")\n",
+ "pyplot.ylabel(\"Ic\")\n",
+ "pyplot.title(\"load line\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current= 0.01 mA\n",
+ "\n",
+ "\n",
+ "collector current= 0.83 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 0.84 mA\n",
+ "\n",
+ "\n",
+ "output voltage= 3.91 V\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 4.70 V\n",
+ "\n",
+ "\n",
+ "Thevenin rquivalent resistance= 2.50 KOhm\n",
+ "\n",
+ "\n",
+ "Thevenin equivalent voltage= 6.00 V\n",
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 5,
+ "text": [
+ "<matplotlib.text.Text at 0x34068b0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x758f0b0>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 3.10\n",
+ "Rb=0.24;\n",
+ "Vcc=12.;\n",
+ "Vbe=0.7;\n",
+ "Vce=0.1;\n",
+ "b=75.;\n",
+ "Rc=5.;##Ohm\n",
+ "##for Vt=0 ,transistor is cut off,Ib=Ic=0,Vo=Vcc=12 V,power dissipation is zero\n",
+ "Vt=12.;##(V)\n",
+ "Ib=(Vt-Vbe)/Rb;\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n",
+ "Ic=(Vcc-Vce)/Rc;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Ic,'A\\n')\n",
+ "Ib=0.0471;##A\n",
+ "x=Ic/Ib\n",
+ "##since Ic/Ib<b transistor is in saturation\n",
+ "##Vo==Vcc;\n",
+ "Vo=0.1;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage= ',Vo,' V\\n')\n",
+ "P=Ic*Vce+Ib*Vbe;\n",
+ "print\"%s %.2f %s\"%('\\npower dissipation= ',P,' W\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current= 47.08 mA\n",
+ "\n",
+ "\n",
+ "collector current= 2.38 A\n",
+ "\n",
+ "\n",
+ "output voltage= 0.10 V\n",
+ "\n",
+ "\n",
+ "power dissipation= 0.27 W\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 3.13\n",
+ "b=100.;\n",
+ "Vcc=12.;\n",
+ "Vbe=0.7;\n",
+ "Icq=1.;##mA\n",
+ "Vceq=6.;\n",
+ "Rc=(Vcc-Vceq)/Icq;\n",
+ "print\"%s %.2f %s\"%('\\ncollector resistance= ',Rc,' KOhms\\n')\n",
+ "Ibq=Icq/b;\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ibq,' mA\\n')\n",
+ "Rb=(Vcc-Vbe)/Ibq;\n",
+ "print\"%s %.2f %s\"%('\\nbase resistance= ',Rb,' KOhms\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "collector resistance= 6.00 KOhms\n",
+ "\n",
+ "\n",
+ "base current= 0.01 mA\n",
+ "\n",
+ "\n",
+ "base resistance= 1130.00 KOhms\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 3.14\n",
+ "R1=56.;\n",
+ "R2=12.2;\n",
+ "Rc=2.;\n",
+ "Re=.4;\n",
+ "Vcc=10.;\n",
+ "Vbe=0.7;\n",
+ "b=100.;\n",
+ "##fig.3.53(b)\n",
+ "Rth=R2*R1/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\nThevenin rquivalent resistance= ',Rth,' KOhm\\n')\n",
+ "Vth=(R2/(R1+R2))*Vcc;\n",
+ "print\"%s %.2f %s\"%('\\nThevenin equivalent voltage= ',Vth,' V\\n')\n",
+ "Ibq=(Vth-Vbe)/(Rth+(1.+b)*Re);\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ibq,' mA\\n')\n",
+ "Icq=b*Ibq;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Icq,' mA\\n')\n",
+ "Ieq=(1.+b)*Ibq;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ieq,' mA\\n')\n",
+ "Vceq=Vcc-Icq*Rc-Ieq*Re;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vceq,' V\\n')\n",
+ "b=numpy.array([50,100,150])\n",
+ "for x in range(0,150):\n",
+ " Ibq=(Vth-Vbe)/(Rth+(1.+x)*Re);\n",
+ "print(\"Ibeq,Iceq,Ieq,Vceq\")\n",
+ "print(Ibq)\n",
+ "Icq=x*Ibq;\n",
+ "print(Icq)\n",
+ "Ieq=(1+x)*Ibq;\n",
+ "print(Ieq)\n",
+ "Vceq=Vcc-Icq*Rc-Ieq*Re;\n",
+ "print(Vceq)\n",
+ "print(\"\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Thevenin rquivalent resistance= 10.02 KOhm\n",
+ "\n",
+ "\n",
+ "Thevenin equivalent voltage= 1.79 V\n",
+ "\n",
+ "\n",
+ "base current= 0.02 mA\n",
+ "\n",
+ "\n",
+ "collector current= 2.16 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 2.18 mA\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 4.81 V\n",
+ "\n",
+ "Ibeq,Iceq,Ieq,Vceq\n",
+ "0.0155511811024\n",
+ "2.31712598425\n",
+ "2.33267716535\n",
+ "4.43267716535\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 3.15\n",
+ "Vcc=5.;\n",
+ "Rc=1.;##KOhm\n",
+ "Vbe=0.7;\n",
+ "b=120.;\n",
+ "Vceq=3.;\n",
+ "Re=.510;\n",
+ "Icq=(Vcc-Vceq)/(Rc+Re);\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Icq,' mA\\n')\n",
+ "Ibq=Icq/b;\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ibq,' mA\\n')\n",
+ "##for bias stable circuit\n",
+ "Rth=0.1*(1.+b)*Re;\n",
+ "print\"%s %.2f %s\"%('\\nThevenin rquivalent resistance= ',Rth,' KOhm\\n')\n",
+ "##Ibq=(Vth-Vbe)/(Rth+(1+b)*Re)\n",
+ "Vth=Ibq*(Rth+(1.+b)*Re)+Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nThevenin equivalent voltage= ',Vth,' V\\n')\n",
+ "##Vth=(R2/(R1+R2))*Vcc\n",
+ "##let x=(R2/(R1+R2))\n",
+ "x=Vth/Vcc\n",
+ "##Rth=6050=R1*x\n",
+ "R1=6.05/x;\n",
+ "print\"%s %.2f %s\"%('\\nR1= ',R1,'KOhms\\n')\n",
+ "R2=x*R1/(1-x);\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,'KOhms\\'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "collector current= 1.32 mA\n",
+ "\n",
+ "\n",
+ "base current= 0.01 mA\n",
+ "\n",
+ "\n",
+ "Thevenin rquivalent resistance= 6.17 KOhm\n",
+ "\n",
+ "\n",
+ "Thevenin equivalent voltage= 1.45 V\n",
+ "\n",
+ "\n",
+ "R1= 20.87 KOhms\n",
+ "\n",
+ "\n",
+ "R2= 8.52 KOhms'\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 3.16\n",
+ "R1=10.;\n",
+ "b=50.;\n",
+ "Vbe=0.7;\n",
+ "V1=-5.;\n",
+ "I1=-(V1+Vbe)/R1;\n",
+ "print\"%s %.2f %s\"%('\\nreference current= ',I1,' mA\\n')\n",
+ "Iq=I1/(1.+2./b);\n",
+ "print\"%s %.2f %s\"%('\\nbias current= ',Iq,' mA\\n')\n",
+ "##Ib=Ib1=Ib2\n",
+ "Ib=Iq/b;\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ib,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "reference current= 0.43 mA\n",
+ "\n",
+ "\n",
+ "bias current= 0.41 mA\n",
+ "\n",
+ "\n",
+ "base current= 0.01 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 3.17\n",
+ "Vbe=0.7;\n",
+ "Vcc=10.;\n",
+ "V2=5.;\n",
+ "b=100.;\n",
+ "R1=100.;\n",
+ "R2=50.;\n",
+ "Re1=2.;\n",
+ "Rth=R2*R1/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\nThevenin rquivalent resistance= ',Rth,' KOhm\\n')\n",
+ "Vth=(R2/(R1+R2))*Vcc-V2;\n",
+ "print\"%s %.2f %s\"%('\\nThevenin equivalent voltage= ',Vth,' V\\n')\n",
+ "##Vth=Ib1*Rth+Vbe+Ie1*Re1-5 and Ie1=(1+b)*Ib1\n",
+ "Ib1=(Vth+5-Vbe)/(Rth+(1+b)*Re1);\n",
+ "print\"%s %.2f %s\"%('\\nIb1= ',Ib1,' mA\\n')\n",
+ "Ic1=b*Ib1;\n",
+ "print\"%s %.2f %s\"%('\\nIc1= ',Ic1,' mA\\n')\n",
+ "Ie1=(1.+b)*Ib1;\n",
+ "print\"%s %.2f %s\"%('\\nIe1= ',Ie1,' mA\\n')\n",
+ "##summing the currents at the collector of Q1,Ir1+Ib2=Ic1\n",
+ "##(5-Vc1)/Rc1+Ib2=Ic1\n",
+ "##also Ib2=Ie2/(1+b)=(5-(Vc1+0.7))/(1+b)*Re2\n",
+ "Rc1=5.;\n",
+ "Re1=2.;\n",
+ "Re2=2.;\n",
+ "Rc2=1.5;\n",
+ "Vc1=Rc1*(1.+b)*Re2*((5./Rc1)+(4.3/((1.+b)*Re2))-Ic1)/(((1.+b)*Re2)+Rc1);\n",
+ "print\"%s %.2f %s\"%('\\nVc1= ',Vc1,' V\\n')\n",
+ "Ir1=(5.-Vc1)/Rc1;\n",
+ "print\"%s %.2f %s\"%('\\nIr1= ',Ir1,' mA\\n')\n",
+ "Ve2=Vc1+Vbe;\n",
+ "print\"%s %.2f %s\"%('\\nVe2= ',Ve2,' V\\n')\n",
+ "Ie2=(5-Ve2)/Re1;\n",
+ "print\"%s %.2f %s\"%('\\nIe2= ',Ie2,' mA\\n')\n",
+ "Ic2=Ie2*b/(1.+b);\n",
+ "print\"%s %.2f %s\"%('\\nIc2= ',Ic2,' mA\\n')\n",
+ "Ib2=Ie2/(1.+b);\n",
+ "print\"%s %.2f %s\"%('\\nIb2 ',Ib2,' mA\\n')\n",
+ "Ve1=Ie1*Re1-5;\n",
+ "print\"%s %.2f %s\"%('\\nVe1= ',Ve1,' V\\n')\n",
+ "Vc2=Ic2*Rc2-5.;\n",
+ "print\"%s %.2f %s\"%('\\nVc2= ',Vc2,' V\\n')\n",
+ "Vce1=Vc1-Ve1;\n",
+ "print\"%s %.2f %s\"%('\\nVce1= ',Vce1,'V\\n')\n",
+ "Vec2=Ve2-Vc2;\n",
+ "print\"%s %.2f %s\"%('\\nVec2= ',Vec2,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Thevenin rquivalent resistance= 33.33 KOhm\n",
+ "\n",
+ "\n",
+ "Thevenin equivalent voltage= -1.67 V\n",
+ "\n",
+ "\n",
+ "Ib1= 0.01 mA\n",
+ "\n",
+ "\n",
+ "Ic1= 1.12 mA\n",
+ "\n",
+ "\n",
+ "Ie1= 1.13 mA\n",
+ "\n",
+ "\n",
+ "Vc1= -0.48 V\n",
+ "\n",
+ "\n",
+ "Ir1= 1.10 mA\n",
+ "\n",
+ "\n",
+ "Ve2= 0.22 V\n",
+ "\n",
+ "\n",
+ "Ie2= 2.39 mA\n",
+ "\n",
+ "\n",
+ "Ic2= 2.36 mA\n",
+ "\n",
+ "\n",
+ "Ib2 0.02 mA\n",
+ "\n",
+ "\n",
+ "Ve1= -2.74 V\n",
+ "\n",
+ "\n",
+ "Vc2= -1.45 V\n",
+ "\n",
+ "\n",
+ "Vce1= 2.26 V\n",
+ "\n",
+ "\n",
+ "Vec2= 1.68 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter4_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter4_1_2.ipynb new file mode 100644 index 00000000..474a3300 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter4_1_2.ipynb @@ -0,0 +1,511 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6cd0454c158d13a9f3c4fd9afab45debe741070024bd59dcb697aaffd2aec82d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter4-Basic BJT Amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 4.1\n",
+ "b=100.;\n",
+ "Vcc=12.;\n",
+ "Vbe=0.7;\n",
+ "Rc=6.;\n",
+ "Rb=50.;\n",
+ "Vbb=1.2;\n",
+ "##dc solution\n",
+ "Ibq=(Vbb-Vbe)/Rb;\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ibq,' mA\\n')\n",
+ "Icq=b*Ibq;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Icq,' mA\\n')\n",
+ "Vceq=Vcc-Icq*Rc;\n",
+ "print\"%s %.2f %s\"%('\\ncollector emitter voltage= ',Vceq,' V\\n')\n",
+ "##transistor is forward biased\n",
+ "##ac solution \n",
+ "V_T=0.026;##(V)\n",
+ "##small signal hybrid pi parameters\n",
+ "r_pi=b*V_T/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal resistance= ',r_pi,' KOhm\\n')\n",
+ "g_m=Icq/V_T;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "##Av=Vo/Vs=-(g_m*Rc)*r_pi/(r_pi+Rb)\n",
+ "Av=-(g_m*Rc)*r_pi/(r_pi+Rb);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current= 0.01 mA\n",
+ "\n",
+ "\n",
+ "collector current= 1.00 mA\n",
+ "\n",
+ "\n",
+ "collector emitter voltage= 6.00 V\n",
+ "\n",
+ "\n",
+ "small signal resistance= 2.60 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 38.46 mA/V\n",
+ "\n",
+ "\n",
+ "small signal voltage gain=\n",
+ " -11.41 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 4.2\n",
+ "V_pi=50.;##(V)\n",
+ "Icq=1.;##(mA)\n",
+ "ro=V_pi/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal output resistance= ',ro,' KOhm\\n')\n",
+ "Rc=6.;\n",
+ "g_m=38.5;\n",
+ "r_pi=2.6;\n",
+ "Rb=50.;\n",
+ "Av=-(g_m)*(Rc*ro/(Rc+ro))*r_pi/(r_pi+Rb);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= \\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal output resistance= 50.00 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= \n",
+ " -10.19 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 4.4\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "Va=100.;\n",
+ "V_T=0.026;##(V)\n",
+ "##from dc analysis\n",
+ "Icq=0.95;\n",
+ "Vceq=6.31;\n",
+ "##ac analysis\n",
+ "r_pi=b*V_T/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal resistance= ',r_pi,' KOhm\\n')\n",
+ "g_m=Icq/V_T;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "Rs=0.5;\n",
+ "Rc=6.;\n",
+ "ro=Va/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nro= ',ro,' KOhm\\n')\n",
+ "Av=-g_m*(5.9*r_pi/(5.9+r_pi))/((5.9*r_pi/(r_pi+5.9))+Rs)*ro*Rc/(ro+Rc);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,' \\n')\n",
+ "Ri=5.9*r_pi/(r_pi+5.9);\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance= ',Ri,' KOhm\\n')\n",
+ "Ro=ro*Rc/(ro+Rc);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Ro,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal resistance= 2.74 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 36.54 mA/V\n",
+ "\n",
+ "\n",
+ "ro= 105.26 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= -163.64 \n",
+ "\n",
+ "\n",
+ "input resistance= 1.87 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance= 5.68 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 4.5\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "Rc=2.;\n",
+ "Rs=0.5;\n",
+ "Icq=2.16;\n",
+ "V_T=0.026;##(V)\n",
+ "Vceq=4.8\n",
+ "##ac solution\n",
+ "r_pi=b*V_T/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal resistance= ',r_pi,' KOhm\\n')\n",
+ "g_m=Icq/V_T;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "##since Va=infinity,ro=Va/Icq is also infinity\n",
+ "Re=0.4;\n",
+ "Rib=r_pi+(1.+b)*Re;\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance to the base= ',Rib,' KOhm\\n')\n",
+ "##Ri=R1||R2||Rib\n",
+ "Ri=10.*Rib/(10.+Rib);\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance to the amplifier= ',Ri,' KOhm\\n')\n",
+ "Av=-(1./(r_pi+(1.+b)*Re))*b*Rc*Ri/(Ri+Rs);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= \\n',Av,'')\n",
+ "##by approximate expression\n",
+ "Av=-Rc/Re;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= \\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal resistance= 1.20 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 83.08 mA/V\n",
+ "\n",
+ "\n",
+ "input resistance to the base= 41.60 KOhm\n",
+ "\n",
+ "\n",
+ "input resistance to the amplifier= 8.06 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= \n",
+ " -4.53 \n",
+ "\n",
+ "small signal voltage gain= \n",
+ " -5.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 4.7\n",
+ "Iq=0.5;\n",
+ "b=120.;\n",
+ "Va=80.;\n",
+ "V_T=0.026;##(V)\n",
+ "rc=120.;##small signal collector resistance (KOhm)\n",
+ "##Icq=Iq\n",
+ "Icq=0.5;\n",
+ "g_m=Icq/V_T;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "ro=Va/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal output resistance= ',ro,' KOhm\\n')\n",
+ "Av=-g_m*ro*rc/(ro+rc);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 19.23 mA/V\n",
+ "\n",
+ "\n",
+ "small signal output resistance= 160.00 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= -1318.68 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 4.8\n",
+ "b=150.;Veb=0.7;\n",
+ "##dc solution\n",
+ "V2=10.;\n",
+ "V1=-10.;\n",
+ "V_T=0.026;##(V)\n",
+ "Rc=5.;\n",
+ "Rb=50.;\n",
+ "Re=10.;\n",
+ "Ibq=(V2-Veb)/(Rb+(1.+b)*Re);\n",
+ "print\"%s %.2f %s\"%('\\nbase current ',Ibq,' mA\\n')\n",
+ "Icq=b*Ibq;\n",
+ "print\"%s %.2f %s\"%('\\ncollector current= ',Icq,' mA\\n')\n",
+ "Ieq=(1.+b)*Ibq;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ieq,' mA\\n')\n",
+ "Vecq=V2-V1-Icq*Rc-Ieq*Re;\n",
+ "print\"%s %.2f %s\"%('\\nemitter collector voltage= ',Vecq,' V\\n')\n",
+ "##ac solution\n",
+ "r_pi=b*V_T/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal resistance= ',r_pi,' KOhm\\n')\n",
+ "g_m=Icq/V_T;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance = ',g_m,'mA/V\\n')\n",
+ "##since Va=infinity,ro=Va/Icq is also infinity\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "base current 0.01 mA\n",
+ "\n",
+ "\n",
+ "collector current= 0.89 mA\n",
+ "\n",
+ "\n",
+ "emitter current= 0.90 mA\n",
+ "\n",
+ "\n",
+ "emitter collector voltage= 6.53 V\n",
+ "\n",
+ "\n",
+ "small signal resistance= 4.36 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance = 34.39 mA/V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 4.9\n",
+ "Ic=0.894;\n",
+ "i_C=2.*Ic;\n",
+ "print\"%s %.2f %s\"%('\\nmaximum possible symmetrical peak to peak ac collector current= ',i_C,' mA\\n')\n",
+ "Rc=5.;\n",
+ "Rl=2.;\n",
+ "vo=i_C*Rc*Rl/(Rc+Rl);\n",
+ "print\"%s %.2f %s\"%('\\nmaximum possible symmetrical peak to peak output voltage= ',vo,' V\\n')\n",
+ "iC=Ic+i_C*1/2.;\n",
+ "print\"%s %.2f %s\"%('\\nmaximum instantaneous collector current= ',iC,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "maximum possible symmetrical peak to peak ac collector current= 1.79 mA\n",
+ "\n",
+ "\n",
+ "maximum possible symmetrical peak to peak output voltage= 2.55 V\n",
+ "\n",
+ "\n",
+ "maximum instantaneous collector current= 1.79 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 4.10\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "V_T=0.026;##(V)\n",
+ "Re=2.;\n",
+ "R1=50.;\n",
+ "R2=50.;\n",
+ "Rs=0.5;\n",
+ "Va=80.;\n",
+ "##by dc analysis\n",
+ "Icq=0.793;\n",
+ "Vceq=3.4;\n",
+ "r_pi=b*V_T/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal resistance= ',r_pi,' KOhm\\n')\n",
+ "g_m=Icq/V_T;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "ro=Va/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal output resistance= ',ro,' KOhm\\n')\n",
+ "Rib=r_pi+(1.+b)*Re*ro/(ro+Re);\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance to the base= ',Rib,' KOhm\\n')\n",
+ "##Ri=R1||R2||Rib\n",
+ "x=R1*R2/(R1+R2);\n",
+ "Ri=x*Rib/(x+Rib);\n",
+ "print\"%s %.2f %s\"%('\\nRi= ',Ri,' KOhm\\n')\n",
+ "y=ro*Re/(ro+Re);\n",
+ "Av=(1./(r_pi+(1.+b)*y))*(1.+b)*y*Ri/(Ri+Rs);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal resistance= 3.28 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 30.50 mA/V\n",
+ "\n",
+ "\n",
+ "small signal output resistance= 100.88 KOhm\n",
+ "\n",
+ "\n",
+ "input resistance to the base= 201.35 KOhm\n",
+ "\n",
+ "\n",
+ "Ri= 22.24 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= 0.96 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter5_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter5_1_2.ipynb new file mode 100644 index 00000000..69bc670e --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter5_1_2.ipynb @@ -0,0 +1,812 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:86feb16fe3c7527ae6f06fa1b128b774eb7e4bbbc3c907b31acf5e59d78150b1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-The Field Effect Transistor "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.1\n",
+ "Vtn=0.75;##(V)\n",
+ "W=40.*10**-6;##(cm)\n",
+ "L=4.*10**-6;##(cm)\n",
+ "u=650.;##(cm)\n",
+ "Iox=450.*10**-11;\n",
+ "e=3.9*8.86*10**-14;\n",
+ "Kn=W*u*e/(2.*L*Iox);\n",
+ "print\"%s %.2f %s\"%('\\nconduction parameter= ',Kn,' mA/V^2\\n')\n",
+ "Vgs=2.*Vtn;\n",
+ "i_D=Kn*(Vgs-Vtn)**2;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',i_D,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "conduction parameter= 0.25 mA/V^2\n",
+ "\n",
+ "\n",
+ "drain current= 0.14 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 5.2\n",
+ "Kp=0.2;##(mA/V^2)\n",
+ "Vtp=0.5;\n",
+ "iD=0.5;\n",
+ "Vsg=math.sqrt(iD/Kp)-Vtp;\n",
+ "print\"%s %.2f %s\"%('\\nVgs= ',Vsg,' V\\n')\n",
+ "##to bias in p channel MOSFET \n",
+ "Vsd=Vsg+Vtp;\n",
+ "print\"%s %.2f %s\"%('\\nVsd= ',Vsd,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgs= 1.08 V\n",
+ "\n",
+ "\n",
+ "Vsd= 1.58 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 5.3\n",
+ "R1=30.;\n",
+ "R2=20.;\n",
+ "RD=20.;\n",
+ "Vdd=5.;\n",
+ "Vtn=1.;\n",
+ "Kn=0.1;\n",
+ "Vgs=R2*Vdd/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\nVgs= ',Vgs,' V\\n')\n",
+ "I_D=Kn*(Vgs-Vtn)**2;\n",
+ "print\"%s %.2f %s\"%('\\nthe drain current= ',I_D,' mA\\n')\n",
+ "Vds=Vdd-I_D*RD;\n",
+ "print\"%s %.2f %s\"%('\\ndrain to source voltage= ',Vds,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgs= 2.00 V\n",
+ "\n",
+ "\n",
+ "the drain current= 0.10 mA\n",
+ "\n",
+ "\n",
+ "drain to source voltage= 3.00 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.4\n",
+ "R1=50.;\n",
+ "R2=50.;\n",
+ "RD=7.5;\n",
+ "Vdd=5.;\n",
+ "Vtp=-0.8;\n",
+ "Vg=2.5;\n",
+ "Kp=0.2;\n",
+ "Vo=R2*Vdd/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\nVo= ',Vo,' V\\n')\n",
+ "Vsg=Vdd-Vg;\n",
+ "print\"%s %.2f %s\"%('\\nsource to gate voltage= ',Vsg,' V\\n')\n",
+ "I_D=Kp*(Vsg+Vtp)**2;\n",
+ "print\"%s %.2f %s\"%('\\nthe drain current= ',I_D,' mA\\n')\n",
+ "Vsd=Vdd-I_D*RD;\n",
+ "print\"%s %.2f %s\"%('\\nsource to drain voltage= ',Vsd,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vo= 2.50 V\n",
+ "\n",
+ "\n",
+ "source to gate voltage= 2.50 V\n",
+ "\n",
+ "\n",
+ "the drain current= 0.58 mA\n",
+ "\n",
+ "\n",
+ "source to drain voltage= 0.67 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.6\n",
+ "Vtn=2.;\n",
+ "Kn=80.*10**-3;\n",
+ "##x=W/L\n",
+ "x=4.;\n",
+ "I_D=0.5;\n",
+ "##I_D=Kn*x*((Vgs-Vtn)^2)/2;\n",
+ "Vgs=math.sqrt(I_D*2./(Kn*x))+2.;\n",
+ "print\"%s %.2f %s\"%('\\nVgs= ',Vgs,' V\\n')\n",
+ "##y=R1+R2\n",
+ "Rs=2.;\n",
+ "y=10./0.05;\n",
+ "print\"%s %.2f %s\"%('\\nR1+R2= ',y,' Kohm\\n')\n",
+ "##Vgs=Vg-Vs=(R2/(R1+R2)*10-5)-I_D*Rs+5\n",
+ "R2=(y/10.)*(Vgs+I_D*Rs);\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,' KOhm\\n')\n",
+ "R1=y-R2;\n",
+ "print\"%s %.2f %s\"%('\\nR1= ',R1,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgs= 3.77 V\n",
+ "\n",
+ "\n",
+ "R1+R2= 200.00 Kohm\n",
+ "\n",
+ "\n",
+ "R2= 95.36 KOhm\n",
+ "\n",
+ "\n",
+ "R1= 104.64 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.7\n",
+ "Vtn=0.8;\n",
+ "Kn=80.;\n",
+ "##x=W/L\n",
+ "x=3.;\n",
+ "I_D=250.;\n",
+ "Vd=2.5;\n",
+ "##I_D=Kn/2*x*(Vgs-Vtn)^2\n",
+ "Vgs=math.sqrt(I_D*2./(Kn*x))+Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVgs= ',Vgs,' V\\n')\n",
+ "Vs=-Vgs\n",
+ "##I_D=(5-Vd)/Rd\n",
+ "Rd=(5.-Vd)/I_D;\n",
+ "print\"%s %.2f %s\"%('\\nRd= ',Rd,' KOhm\\n')\n",
+ "Vds=Vd-Vs;\n",
+ "print\"%s %.2f %s\"%('\\nVds= ',Vds,' V\\n')\n",
+ "Vdssat=Vgs-Vtn\n",
+ "##since Vds>Vdssat transistor is biased in saturation region\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgs= 2.24 V\n",
+ "\n",
+ "\n",
+ "Rd= 0.01 KOhm\n",
+ "\n",
+ "\n",
+ "Vds= 4.74 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from numpy import roots\n",
+ "##Example 5.8\n",
+ "Vtn=0.8;\n",
+ "Kn=0.05;\n",
+ "##I_D=Kn*(Vgs-Vtn)^2\n",
+ "##Vds=Vgs=5-I_D*Rs\n",
+ "##combining these two equations we obtain 0.5(Vgs)^2+0.2Vgs-4.68\n",
+ "import numpy\n",
+ "from numpy.polynomial import Polynomial as P\n",
+ "p = P([1, 5, 6])\n",
+ "p.roots()\n",
+ "\n",
+ "\n",
+ "print('',p.roots(),' V\\n')\n",
+ "##assuming transistor is conducting ,Vgs must be greater than threshold voltage\n",
+ "Vgs=2.87;\n",
+ "I_D=Kn*(Vgs-Vtn)**2;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',I_D,' mA\\n')\n",
+ "#ans is varying due to round of error in book calculations are done wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('', array([-0.5 , -0.33333333]), ' V\\n')\n",
+ "\n",
+ "drain current= 0.21 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 5.10\n",
+ "Vtn=-2.;\n",
+ "Kn=0.1;\n",
+ "Vdd=5.;\n",
+ "Rs=5.;\n",
+ "Vgs=0.;\n",
+ "I_D=Kn*(Vgs-Vtn)**2;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',I_D,' mA\\n')\n",
+ "Vds=Vdd-I_D*Rs;\n",
+ "print\"%s %.2f %s\"%('\\ndc drain to source voltage= ',Vds,' V\\n')\n",
+ "Vdssat=Vgs-Vtn\n",
+ "##since Vds>Vdssat transisyor is biased in saturation region\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "drain current= 0.40 mA\n",
+ "\n",
+ "\n",
+ "dc drain to source voltage= 3.00 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "\n",
+ "##Example 5.11\n",
+ "Vtnd=1;\n",
+ "Vtnl=-2;\n",
+ "Knd=50;\n",
+ "Knl=10;\n",
+ "Vt=5;\n",
+ "import numpy\n",
+ "from numpy.polynomial import Polynomial as P\n",
+ "p = P([4, -40, 5])\n",
+ "p.roots()\n",
+ "\n",
+ "\n",
+ "print('\\npossible solutions ::',p.roots(),'')\n",
+ "##since output voltage cannot be greater than supply voltage 5V\n",
+ "Vo=0.1;##(V)\n",
+ "I_D=Knl*(-Vtnl)**2;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',I_D,' microA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('\\npossible solutions ::', array([ 0.10128226, 7.89871774]), '')\n",
+ "\n",
+ "drain current= 40.00 microA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the \n",
+ "##Example 5.13\n",
+ "Kn1=0.2;\n",
+ "Kn2=0.1;\n",
+ "Kn3=0.1;\n",
+ "Kn4=0.1;\n",
+ "Vtn1=1.;\n",
+ "Vtn2=1.;\n",
+ "Vtn3=1.;\n",
+ "Vtn4=1.;\n",
+ "V2=-5.;\n",
+ "Vgs3=(math.sqrt(Kn4/Kn3)*(-V2-Vtn4)+Vtn3)/(1.+math.sqrt(Kn4/Kn3));\n",
+ "print\"%s %.2f %s\"%('\\nVgs3= ',Vgs3,' V\\n')\n",
+ "Iq=Kn3*(Vgs3-Vtn3)**2;\n",
+ "print\"%s %.2f %s\"%('\\nbias current= ',Iq,' mA\\n')\n",
+ "Vgs1=math.sqrt(Iq/Kn1)+Vtn1;\n",
+ "print\"%s %.2f %s\"%('\\ngate to source voltage on M1= ',Vgs1,' V\\n')\n",
+ "Vds2=-V2-Vgs1;\n",
+ "print\"%s %.2f %s\"%('\\ndrain to source voltage on M2= ',Vds2,' V\\n')\n",
+ "Vgs2=Vgs3;\n",
+ "Vdssat=Vgs2-Vtn2\n",
+ "##since Vds2>Vdssat M2 is biased in saturation region\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgs3= 2.50 V\n",
+ "\n",
+ "\n",
+ "bias current= 0.23 mA\n",
+ "\n",
+ "\n",
+ "gate to source voltage on M1= 2.06 V\n",
+ "\n",
+ "\n",
+ "drain to source voltage on M2= 2.94 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.14\n",
+ "I_D=0.5;\n",
+ "Vds=6.;\n",
+ "Kn=80.*10**-6;\n",
+ "Vgs=5.;\n",
+ "Vtn=1.;\n",
+ "##x=W/L\n",
+ "x=I_D*2./(Kn*(Vgs-Vtn)**2);\n",
+ "print(x,\"W/L \")\n",
+ "##maximum power dissipation in transistor \n",
+ "Pmax=Vds*I_D;\n",
+ "print\"%s %.2f %s\"%('\\nmaximum power dissipation in transistor= ',Pmax,' W\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(781.2500000000001, 'W/L ')\n",
+ "\n",
+ "maximum power dissipation in transistor= 3.00 W\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy \n",
+ "##Example 5.16\n",
+ "Idss=2.;##(mA) saturation current\n",
+ "Vp=-3.5;##(V) pinch off voltage\n",
+ "Vgs=numpy.array([[0, Vp/4. ,Vp/2.]])\n",
+ "I_D=Idss*(1.-Vgs/Vp)**2;\n",
+ "print (I_D)\n",
+ "Vds=Vgs-Vp;\n",
+ "print (Vds)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[[ 2. 1.125 0.5 ]]\n",
+ "[[ 3.5 2.625 1.75 ]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.17\n",
+ "Idss=5.;##mA\n",
+ "Vp=-4.;\n",
+ "Vdd=10.;\n",
+ "I_D=2.;\n",
+ "Vds=6.;\n",
+ "##I_D=Idss*(1-Vgs/Vp)^2\n",
+ "Vgs=(1.-math.sqrt(I_D/Idss))*Vp;\n",
+ "print\"%s %.2f %s\"%('\\nVgs= ',Vgs,' V\\n')\n",
+ "Rs=-Vgs/I_D;\n",
+ "print\"%s %.2f %s\"%('\\nRs= ',Rs,' KOhm\\n')\n",
+ "Rd=(Vdd-Vds-I_D*Rs)/I_D;\n",
+ "print\"%s %.2f %s\"%('\\nRd= ',Rd,' KOhm\\n')\n",
+ "Vgs-Vp\n",
+ "##since Vds>Vgs-Vp JFET is biased in saturation\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgs= -1.47 V\n",
+ "\n",
+ "\n",
+ "Rs= 0.74 KOhm\n",
+ "\n",
+ "\n",
+ "Rd= 1.26 KOhm\n",
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 19,
+ "text": [
+ "2.5298221281347035"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.19\n",
+ "Idss=2.5;\n",
+ "Vp=2.5;\n",
+ "I_D=0.8;\n",
+ "##I_D=Iq=0.8*10^-3=(Vd-(-9))/Rd\n",
+ "Vd=0.8*4.-9;\n",
+ "print\"%s %.2f %s\"%('\\nVd = ',Vd,'V\\n')\n",
+ "##I_D=Idss*(1-Vgs/Vp)^2;\n",
+ "Vgs=(1.-math.sqrt(I_D/Idss))*Vp;\n",
+ "print\"%s %.2f %s\"%('\\nVgs = ',Vgs,'V\\n')\n",
+ "Vs=1-Vgs;\n",
+ "print\"%s %.2f %s\"%('\\nVs= ',Vs,' V\\n')\n",
+ "Vsd=Vs-Vd;\n",
+ "print\"%s %.2f %s\"%('\\nVsd= ',Vsd,' V\\n')\n",
+ "Vp-Vgs\n",
+ "##since Vsd>Vp-Vgs JFET is biased in saturation\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vd = -5.80 V\n",
+ "\n",
+ "\n",
+ "Vgs = 1.09 V\n",
+ "\n",
+ "\n",
+ "Vs= -0.09 V\n",
+ "\n",
+ "\n",
+ "Vsd= 5.71 V\n",
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 20,
+ "text": [
+ "1.414213562373095"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 5.20\n",
+ "Vtn=0.24;\n",
+ "Kn=1.1;\n",
+ "##x=R1+R2=50000\n",
+ "x=50.;\n",
+ "Vgs=0.5;\n",
+ "Vds=2.5;\n",
+ "Vdd=4.;\n",
+ "Rd=6.7;\n",
+ "I_D=Kn*(Vgs-Vtn)**2;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',I_D,' mA\\n')\n",
+ "Vd=Vdd-I_D*Rd;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage at drain= ',Vd,' V\\n')\n",
+ "Vs=Vd-Vds;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage at source = ',Vs,'V\\n')\n",
+ "Rs=Vs/I_D;\n",
+ "print\"%s %.2f %s\"%('\\nsource resistance = ',Rs,'KOhm\\n')\n",
+ "Vg=Vgs+Vs;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage at the gate= ',Vg,' V\\n')\n",
+ "##Vg=R2*Vdd/(R2+R1)\n",
+ "R2=Vg*x/Vdd;\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,' KOhm\\n')\n",
+ "R1=x-R2;\n",
+ "print\"%s %.2f %s\"%('\\nR1= ',R1,' KOhm\\n')\n",
+ "Vgs-Vtn\n",
+ "##since Vds>Vgs-Vtn transistor is biased in saturation\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "drain current= 0.07 mA\n",
+ "\n",
+ "\n",
+ "voltage at drain= 3.50 V\n",
+ "\n",
+ "\n",
+ "voltage at source = 1.00 V\n",
+ "\n",
+ "\n",
+ "source resistance = 13.47 KOhm\n",
+ "\n",
+ "\n",
+ "voltage at the gate= 1.50 V\n",
+ "\n",
+ "\n",
+ "R2= 18.77 KOhm\n",
+ "\n",
+ "\n",
+ "R1= 31.23 KOhm\n",
+ "\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 21,
+ "text": [
+ "0.26"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter6_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter6_1_2.ipynb new file mode 100644 index 00000000..f26e210a --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter6_1_2.ipynb @@ -0,0 +1,955 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5f69c997f454918e67123a37f5caef5b8555309eb92b985bcafc07b0e579172b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Basic FET Amplifier"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.1\n",
+ "Vtn=1.;\n",
+ "##let x= u_n*Cox*1/2\n",
+ "x=20.*10**-3;\n",
+ "##let y=W/L\n",
+ "y=40.;\n",
+ "I_D=1.;\n",
+ "Kn=x*y;\n",
+ "print\"%s %.2f %s\"%('\\nconduction parameter= ',Kn,' mA/V^2\\n')\n",
+ "g_m=2.*math.sqrt(Kn*I_D);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "conduction parameter= 0.80 mA/V^2\n",
+ "\n",
+ "\n",
+ "transconductance= 1.79 mA/V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.2\n",
+ "Vgsq=2.12;\n",
+ "Vdd=5.;\n",
+ "Rd=2.5;\n",
+ "Vtn=1.;\n",
+ "Kn=0.8;\n",
+ "##let lambda=y\n",
+ "y=0.02;##V^-1\n",
+ "Idq=Kn*(Vgsq-Vtn)**2;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',Idq,'mA\\n')\n",
+ "Vdsq=Vdd-Idq*Rd;\n",
+ "print\"%s %.2f %s\"%('\\ndrain to source voltage= ',Vdsq,' V\\n')\n",
+ "Vgs=1.82;\n",
+ "Vgs-Vtn\n",
+ "##since Vdsq>Vgs-Vtn transistor is biased in saturation\n",
+ "g_m=2.*Kn*(Vgsq-Vtn);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "ro=(y*Idq)**-1;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' KOhm\\n')\n",
+ "Av=-g_m*ro*Rd/(ro+Rd);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,'\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "drain current= 1.00 mA\n",
+ "\n",
+ "\n",
+ "drain to source voltage= 2.49 V\n",
+ "\n",
+ "\n",
+ "transconductance= 1.79 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance= 49.82 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= -4.27 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.3\n",
+ "Vdd=10.;\n",
+ "R1=70.9;##(Kohm)\n",
+ "R2=29.1;##(Kohm)\n",
+ "Rd=5.;##(Kohm)\n",
+ "Vtn=1.5;\n",
+ "Kn=0.5;##(mA/V^2)\n",
+ "##lambda=y\n",
+ "y=0.01;##V^-1\n",
+ "Rsi=4.;##(Kohm)\n",
+ "Vgsq=Vdd*R2/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\ngate to source voltage= ',Vgsq,' V\\n')\n",
+ "Idq=Kn*(Vgsq-Vtn)**2;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',Idq,' mA\\n')\n",
+ "Vdsq=Vdd-Idq*Rd;\n",
+ "print\"%s %.2f %s\"%('\\ndrain to source voltage= ',Vdsq,' V\\n')\n",
+ "g_m=2*Kn*(Vgsq-Vtn);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "ro=(y*Idq)**-1;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' KOhm\\n')\n",
+ "Ri=R1*R2/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\namplifier input resistance= ',Ri,' Kohm\\n')\n",
+ "Av=-g_m*(ro*Rd/(ro+Rd))*Ri/(Ri+Rsi);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',Av,'')\n",
+ "print\"%s %.2f %s\"%('\\namplifier input resistance= ',Ri,' Kohm\\n')\n",
+ "Ro=Rd*ro/(Rd+ro);\n",
+ "print\"%s %.2f %s\"%('\\namplifier output resistance= ',Ro,' Kohm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "gate to source voltage= 2.91 V\n",
+ "\n",
+ "\n",
+ "drain current= 0.99 mA\n",
+ "\n",
+ "\n",
+ "drain to source voltage= 5.03 V\n",
+ "\n",
+ "\n",
+ "transconductance= 1.41 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance= 100.60 KOhm\n",
+ "\n",
+ "\n",
+ "amplifier input resistance= 20.63 Kohm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain=\n",
+ " -5.63 \n",
+ "\n",
+ "amplifier input resistance= 20.63 Kohm\n",
+ "\n",
+ "\n",
+ "amplifier output resistance= 4.76 Kohm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.4\n",
+ "Vtn=1.;\n",
+ "Kn=1.;##(mA/V^2)\n",
+ "##lambda=y\n",
+ "y=0.015;##V^-1\n",
+ "Ri=100.;##(Kohm)\n",
+ "Idq=2.;##(mA)\n",
+ "Idt=4.;##(mA)\n",
+ "##Idt=4=Kn*(Vgst-Vtn)^2\n",
+ "Vgst=math.sqrt(Idt/Kn)+Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVgst= ',Vgst,' V\\n')\n",
+ "Vdst=Vgst-Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVdst= ',Vdst,' V\\n')\n",
+ "Vdd=12.;\n",
+ "Vdsq=7.;\n",
+ "Rd=(Vdd-Vdsq)/Idq;\n",
+ "print\"%s %.2f %s\"%('\\nRd = ',Rd,'KOhm\\n')\n",
+ "Vgsq=math.sqrt(Idq/Kn)+Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVgsq= ',Vgsq,' V\\n')\n",
+ "R1=Ri*Vdd/Vgsq;\n",
+ "print\"%s %.2f %s\"%('\\nR1= ',R1,' Kohm\\n')\n",
+ "R2=Ri*R1/(R1-Ri);\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,' Kohm\\n')\n",
+ "g_m=2*Kn*(Vgsq-Vtn);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "ro=(y*Idq)**-1;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' KOhm\\n')\n",
+ "Av=-g_m*(ro*Rd/(ro+Rd));\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,'\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgst= 3.00 V\n",
+ "\n",
+ "\n",
+ "Vdst= 2.00 V\n",
+ "\n",
+ "\n",
+ "Rd = 2.50 KOhm\n",
+ "\n",
+ "\n",
+ "Vgsq= 2.41 V\n",
+ "\n",
+ "\n",
+ "R1= 497.06 Kohm\n",
+ "\n",
+ "\n",
+ "R2= 125.19 Kohm\n",
+ "\n",
+ "\n",
+ "transconductance= 2.83 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance= 33.33 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= -6.58 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.6\n",
+ "Vtn=0.8;\n",
+ "Kn=1.;##(mA/V^2)\n",
+ "Idq=0.5;\n",
+ "Vdd=5.;\n",
+ "Rd=7.;##(Kohm)\n",
+ "Vgsq=math.sqrt(Idq/Kn)+Vtn;\n",
+ "print\"%s %.2f %s\"%('\\nVgsq= ',Vgsq,' V\\n')\n",
+ "Vs=-Vgsq\n",
+ "Vdsq=Vdd-Idq*Rd-Vs;\n",
+ "print\"%s %.2f %s\"%('\\nVdsq=',Vdsq,' V\\n')\n",
+ "g_m=2.*Kn*(Vgsq-Vtn);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "Av=-g_m*Rd;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Vgsq= 1.51 V\n",
+ "\n",
+ "\n",
+ "Vdsq= 3.01 V\n",
+ "\n",
+ "\n",
+ "transconductance= 1.41 mA/V\n",
+ "\n",
+ "\n",
+ "small signal voltage gain=\n",
+ " -9.90 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.7\n",
+ "Vdd=12.;\n",
+ "R1=162.;\n",
+ "R2=463.;\n",
+ "Rs=0.75;\n",
+ "Kn=4.;\n",
+ "Vtn=1.5;\n",
+ "##lambda=y\n",
+ "y=0.01;\n",
+ "Rsi=4.;\n",
+ "Idq=7.97;\n",
+ "Vgsq=2.91;\n",
+ "g_m=2.*Kn*(Vgsq-Vtn);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',g_m,' mA/V\\n')\n",
+ "ro=(y*Idq)**-1.;\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' KOhm\\n')\n",
+ "Ri=R1*R2/(R1+R2);\n",
+ "print\"%s %.2f %s\"%('\\namplifier input resistance= ',Ri,' Kohm\\n')\n",
+ "x=Rs*ro/(Rs+ro);\n",
+ "Av=g_m*x*(Ri/(Ri+Rsi))/(1.+g_m*x);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 11.28 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance= 12.55 KOhm\n",
+ "\n",
+ "\n",
+ "amplifier input resistance= 120.01 Kohm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain=\n",
+ " 0.86 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.9\n",
+ "Rs=750.;##Ohm\n",
+ "ro=12500.;\n",
+ "g_m=11.3*10**-3;\n",
+ "x=1./g_m;\n",
+ "y=x*Rs/(x+Rs);\n",
+ "Ro=y*ro/(y+ro);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',Ro,' ohm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "output resistance= 78.66 ohm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.11\n",
+ "Vtnd=1.;\n",
+ "Vtnl=1.;\n",
+ "Kn=30.;\n",
+ "##let W/L=x\n",
+ "xl=1.;\n",
+ "Vdd=5.;\n",
+ "Av=10.;\n",
+ "##Av=sqrt(xd/xl)\n",
+ "xd=(Av)**2*xl;\n",
+ "print\"%s %.2f %s\"%('\\nwidth to length ratio of driver transistor=\\n',xd,'')\n",
+ "Knd=xd*Kn*0.001/2.;\n",
+ "Knl=xl*Kn*0.001/2.;\n",
+ "print\"%s %.2f %s\"%('\\nconduction parameter Knd= ',Knd,' mA/V^2\\n')\n",
+ "print\"%s %.2f %s\"%('\\nconduction parameter Knl= ',Knl,' mA/V^2\\n')\n",
+ "##Vgsd-Vtnd=(Vdd-Vtnl)-sqrt(Knd/Knl)*(Vgsd-Vtnd)\n",
+ "y=math.sqrt(Knd/Knl);\n",
+ "Vgsd=(y+5.)/(1.+y);\n",
+ "print\"%s %.2f %s\"%('\\nVgsd= ',Vgsd,' V\\n')\n",
+ "Vdsd=Vgsd-1.;\n",
+ "print\"%s %.2f %s\"%('\\nVdsd= ',Vdsd,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "width to length ratio of driver transistor=\n",
+ " 100.00 \n",
+ "\n",
+ "conduction parameter Knd= 1.50 mA/V^2\n",
+ "\n",
+ "\n",
+ "conduction parameter Knl= 0.01 mA/V^2\n",
+ "\n",
+ "\n",
+ "Vgsd= 1.36 V\n",
+ "\n",
+ "\n",
+ "Vdsd= 0.36 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg352"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.12\n",
+ "Vtnd=0.8;\n",
+ "Vtnl=-1.5;\n",
+ "Knd=1.;\n",
+ "Knl=0.2;\n",
+ "##lambda=y\n",
+ "yd=0.01;\n",
+ "yl=0.01;\n",
+ "Idq=0.2;\n",
+ "gmd=2.*math.sqrt(Knd*Idq);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance of the driver= ',gmd,' mA/V\\n')\n",
+ "roD=1./(yd*Idq);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistances= ',roD,' Kohm\\n')\n",
+ "Av=-gmd*roD/2.;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= \\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance of the driver= 0.89 mA/V\n",
+ "\n",
+ "\n",
+ "output resistances= 500.00 Kohm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= \n",
+ " -223.61 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.13\n",
+ "Vtn=0.8;\n",
+ "Vtp=-0.8;\n",
+ "Kn=80.*10**-3;\n",
+ "Kp=40.*10**-3;\n",
+ "##x=W/L\n",
+ "xn=15.;\n",
+ "xp=10.;\n",
+ "##lambda=y\n",
+ "yn=0.01;\n",
+ "yp=0.01;\n",
+ "Ibias=0.2;\n",
+ "gm=2.*math.sqrt(Kn*xn*Ibias/2.);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance of the NMOS driver= ',gm,' mA/V^2\\n')\n",
+ "ron=1./(yn*Ibias);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistances= ',ron,' Kohm\\n')\n",
+ "Av=-gm*ron/2.;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= \\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance of the NMOS driver= 0.69 mA/V^2\n",
+ "\n",
+ "\n",
+ "output resistances= 500.00 Kohm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= \n",
+ " -173.21 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg357"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.14\n",
+ "Kn1=500.*10**-3;\n",
+ "Kn2=200.*10**-3;\n",
+ "Vtn1=1.2;\n",
+ "Vtn2=Vtn1;\n",
+ "Idq1=0.2;\n",
+ "Idq2=0.5;\n",
+ "Vdsq1=6.;\n",
+ "Vdsq2=6.;\n",
+ "Ri=100.;\n",
+ "Rsi=4.;\n",
+ "Rs2=(10.-Vdsq2)/Idq2;\n",
+ "print\"%s %.2f %s\"%('\\nRs2= ',Rs2,' KOhm\\n')\n",
+ "Vgs2=math.sqrt(Idq2/Kn2)+Vtn2;\n",
+ "print\"%s %.2f %s\"%('\\ngate to source voltage for M2= ',Vgs2,' V\\n')\n",
+ "Vs2=-1.;\n",
+ "Vg2=Vs2+Vgs2;\n",
+ "print\"%s %.2f %s\"%('\\ngate voltage of M2= ',Vg2,' V\\n')\n",
+ "Vg1=Vg2;\n",
+ "Rd1=(5.-Vg1)/Idq1;\n",
+ "print\"%s %.2f %s\"%('\\nresistor Rd1= ',Rd1,' KOhm\\n')\n",
+ "Vs1=Vg1-Vdsq1;\n",
+ "print\"%s %.2f %s\"%('\\nsource voltage of M1= ',Vs1,' KOhm\\n')\n",
+ "Rs1=(Vs1+5.)/Idq1;\n",
+ "print\"%s %.2f %s\"%('\\nresistor Rs1= ',Rs1,' KOhm\\n')\n",
+ "Vgs1=math.sqrt(Idq1/Kn1)+Vtn1;\n",
+ "print\"%s %.2f %s\"%('\\ngate to source voltage for M1',Vgs1,' V\\n')\n",
+ "R1=Ri*10./(Vgs1+Idq1*Rs1);\n",
+ "print\"%s %.2f %s\"%('\\nR1= ',R1,' KOhm\\n')\n",
+ "##Ri=R1*R2/(R1+R2)\n",
+ "R2=Ri*R1/(R1-Ri);\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Rs2= 8.00 KOhm\n",
+ "\n",
+ "\n",
+ "gate to source voltage for M2= 2.78 V\n",
+ "\n",
+ "\n",
+ "gate voltage of M2= 1.78 V\n",
+ "\n",
+ "\n",
+ "resistor Rd1= 16.09 KOhm\n",
+ "\n",
+ "\n",
+ "source voltage of M1= -4.22 KOhm\n",
+ "\n",
+ "\n",
+ "resistor Rs1= 3.91 KOhm\n",
+ "\n",
+ "\n",
+ "gate to source voltage for M1 1.83 V\n",
+ "\n",
+ "\n",
+ "R1= 382.61 KOhm\n",
+ "\n",
+ "\n",
+ "R2= 135.38 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg358"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.15\n",
+ "Vtn1=1.2;\n",
+ "Vtn2=1.2;\n",
+ "Kn1=0.8;\n",
+ "Kn2=0.8;\n",
+ "##x=R1+R2+R3=300\n",
+ "x=300.;\n",
+ "Rs=10.;\n",
+ "Idq=0.4;\n",
+ "Vdsq1=2.5;\n",
+ "Vdsq2=2.5;\n",
+ "Vs1=Idq*Rs-5.;\n",
+ "print\"%s %.2f %s\"%('\\ndc voltage at source of M1= ',Vs1,' V\\n')\n",
+ "Vgs=math.sqrt(Idq/Kn1)+Vtn1;\n",
+ "print\"%s %.2f %s\"%('\\ngate to source voltage= ',Vgs,' V\\n')\n",
+ "R3=(Vgs+Vs1)*x/5.;\n",
+ "print\"%s %.2f %s\"%('\\nR3= ',R3,' KOhm\\n')\n",
+ "Vs2=Vdsq2+Vs1;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage at source of M2= ',Vs2,' V\\n')\n",
+ "##y=R2+R3\n",
+ "y=(Vgs+Vs2)*x/5.;\n",
+ "print\"%s %.2f %s\"%('\\nR2+R3= ',y,' KOhm\\n')\n",
+ "R2=150.;\n",
+ "R1=x-y;\n",
+ "print\"%s %.2f %s\"%('\\nR1=',R1,' KOhm\\n')\n",
+ "R3=y-R2;\n",
+ "print\"%s %.2f %s\"%('\\nR3= ',R3,' KOhm\\n')\n",
+ "Vd2=Vdsq2+Vs2;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage at drain of M2 = ',Vd2,'V\\n')\n",
+ "Rd=(5.-Vd2)/Idq;\n",
+ "print\"%s %.2f %s\"%('\\ndrain resistance= ',Rd,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "dc voltage at source of M1= -1.00 V\n",
+ "\n",
+ "\n",
+ "gate to source voltage= 1.91 V\n",
+ "\n",
+ "\n",
+ "R3= 54.43 KOhm\n",
+ "\n",
+ "\n",
+ "voltage at source of M2= 1.50 V\n",
+ "\n",
+ "\n",
+ "R2+R3= 204.43 KOhm\n",
+ "\n",
+ "\n",
+ "R1= 95.57 KOhm\n",
+ "\n",
+ "\n",
+ "R3= 54.43 KOhm\n",
+ "\n",
+ "\n",
+ "voltage at drain of M2 = 4.00 V\n",
+ "\n",
+ "\n",
+ "drain resistance= 2.50 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 6.17\n",
+ "Kn=0.8;\n",
+ "Vtn=1.2;\n",
+ "Vgs=1.91;\n",
+ "Rd=2.5;\n",
+ "gm=2.*Kn*(Vgs-Vtn);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "Av=-gm*Rd;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 1.14 mA/V\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= -2.84 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 6.18\n",
+ "##Determine the small signal voltage gain of a circuit in fig.6.55\n",
+ "Idss=12.;\n",
+ "Vp=-4.;\n",
+ "##lambda=y\n",
+ "y=0.008;\n",
+ "import numpy\n",
+ "from numpy.polynomial import Polynomial as P\n",
+ "p = P([26.4, 17.2, 2.025])\n",
+ "p.roots()\n",
+ "\n",
+ "\n",
+ "print('',p.roots(),' V\\n')\n",
+ "Vgsq=-2.01\n",
+ "Idq=Idss*(1.-Vgsq/Vp)**2;\n",
+ "print\"%s %.2f %s\"%('\\nquiescent drain current= ',Idq,' mA\\n')\n",
+ "gm=(-2*Idss/Vp)*(1.-Vgsq/Vp);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "ro=(1/(y*Idq));\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' KOhm\\n')\n",
+ "Rd=2.7;\n",
+ "Rl=4.;\n",
+ "x=Rd*Rl/(Rd+Rl);\n",
+ "Av=-gm*ro*x/(ro+x);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('', array([-6.48281115, -2.01101602]), ' V\\n')\n",
+ "\n",
+ "quiescent drain current= 2.97 mA\n",
+ "\n",
+ "\n",
+ "transconductance= 2.99 mA/V\n",
+ "\n",
+ "\n",
+ "output resistance= 42.09 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= -4.63 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg366"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 6.19\n",
+ "Idss=12.;\n",
+ "Vp=-4.;\n",
+ "Rl=10.;\n",
+ "##lambda=y\n",
+ "y=0.01;\n",
+ "Av=0.9;\n",
+ "##gm=(-2*Idss/Vp)*(1-Vgs/Vp)\n",
+ "gm=2.;\n",
+ "Vgs=(1.+gm*Vp/(2.*Idss))*Vp;\n",
+ "print\"%s %.2f %s\"%('\\ngate to source voltage= ',Vgs,' V\\n')\n",
+ "Idq=Idss*(1.-Vgs/Vp)**2;\n",
+ "print\"%s %.2f %s\"%('\\nquiescent drain current= ',Idq,' mA\\n')\n",
+ "Rs=(-Vgs+10.)/Idq;\n",
+ "print\"%s %.2f %s\"%('\\nRs= ',Rs,' KOhm\\n')\n",
+ "ro=(1./(y*Idq));\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance= ',ro,' KOhm\\n')\n",
+ "x=Rl*ro/(Rl+ro);\n",
+ "t=x*Rs/(x+Rs);\n",
+ "Av=gm*t/(1.+gm*t);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "gate to source voltage= -2.67 V\n",
+ "\n",
+ "\n",
+ "quiescent drain current= 1.33 mA\n",
+ "\n",
+ "\n",
+ "Rs= 9.50 KOhm\n",
+ "\n",
+ "\n",
+ "output resistance= 75.00 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= 0.90 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter7_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter7_1_2.ipynb new file mode 100644 index 00000000..ad03d9ee --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter7_1_2.ipynb @@ -0,0 +1,1012 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:dffc3a2403edde165f33b6308fbd20345cecf1da13bbb808db0bb00e10439bb7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter7-Frequency Response"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 7.1\n",
+ "Rs=1000.;\n",
+ "Rp=10000.;\n",
+ "Cs=1.*10**-6;\n",
+ "Cp=3.*10**-12;\n",
+ "Ts=(Rs+Rp)*Cs;\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Ts,' s\\n')\n",
+ "f=1/(2.*math.pi*Ts);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency= ',f,' Hz\\n')\n",
+ "x=20.*math.log10(Rp/(Rp+Rs));\n",
+ "print\"%s %.2f %s\"%('\\nmaximum magnitude = ',x,'dB\\n')\n",
+ "Rp=10.;##KOhm\n",
+ "Rs=1.;##Kohm\n",
+ "Cp=3.;##pF\n",
+ "Tp=Cp*Rs*Rp/(Rs+Rp);\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Tp,' ns\\n')\n",
+ "Tp=2.73*10**-3;##micro sec\n",
+ "f=1/(2.*math.pi*Tp);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency = ',f,'MHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "time constant= 0.01 s\n",
+ "\n",
+ "\n",
+ "corner frequency= 14.47 Hz\n",
+ "\n",
+ "\n",
+ "maximum magnitude = -0.83 dB\n",
+ "\n",
+ "\n",
+ "time constant= 2.73 ns\n",
+ "\n",
+ "\n",
+ "corner frequency = 58.30 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.2\n",
+ "Rs=1000.;\n",
+ "Rp=10000.;\n",
+ "Cs=1*10**-6;\n",
+ "Cp=3*10**-12;\n",
+ "Ts=(Rs+Rp)*Cs;\n",
+ "print\"%s %.2f %s\"%('\\nopen circuit time constant= ',Ts,' s\\n')\n",
+ "Rs=1.;##KOhm\n",
+ "Rp=10.;##KOhm\n",
+ "Cp=3.;##pF\n",
+ "Tp=Cp*Rs*Rp/(Rs+Rp);\n",
+ "print\"%s %.2f %s\"%('\\nshort circuit time constant= ',Tp,' ns\\n')\n",
+ "fL=1./(2.*math.pi*Ts);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency fL= ',fL,' Hz\\n')\n",
+ "Tp=2.73*10**-3;##microsec\n",
+ "fH=1/(2.*math.pi*Tp);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency fH= ',fH,' MHz\\n')\n",
+ "fL=14.5*10**-6;##MHz\n",
+ "fbw=fH-fL;\n",
+ "print\"%s %.2f %s\"%('\\nbandwidth = ',fbw,' MHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "open circuit time constant= 0.01 s\n",
+ "\n",
+ "\n",
+ "short circuit time constant= 2.73 ns\n",
+ "\n",
+ "\n",
+ "corner frequency fL= 14.47 Hz\n",
+ "\n",
+ "\n",
+ "corner frequency fH= 58.30 MHz\n",
+ "\n",
+ "\n",
+ "bandwidth = 58.30 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg400"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.3\n",
+ "R1=51.2;\n",
+ "R2=9.6;\n",
+ "Rc=2.;\n",
+ "Re=.4;\n",
+ "Rsi=.1;\n",
+ "Vt=0.026;\n",
+ "Cc=1.;\n",
+ "Vcc=10.;\n",
+ "Vbe=0.7;\n",
+ "b=100.;\n",
+ "Rb=8.08;\n",
+ "Icq=1.81;\n",
+ "gm=Icq/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "r=b*Vt/Icq;\n",
+ "print\"%s %.2f %s\"%('\\ndiffusion resistance= ',r,' KOhm\\n')\n",
+ "x=r+(1.+b)*Re;\n",
+ "y=x*R2/(x+R2);\n",
+ "Ri=y*R1/(R1+y);\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance= ',Ri,' KOhm\\n')\n",
+ "Ts=(Rsi+Ri)*Cc;\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Ts, 'ms')\n",
+ "Ts=6.87*10**-3;##Sec\n",
+ "fL=1/(2.*math.pi*Ts);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency fL= ',fL,' Hz\\n')\n",
+ "Rib=r+(1.+b)*Re;\n",
+ "print\"%s %.2f %s\"%('\\nRib= ',Rib,' KOhm\\n')\n",
+ "Av=(gm*r*Rc/(Rsi+Ri))*Rb/(Rb+Rib);\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal voltage gain= ',Av,'\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 69.62 mA/V\n",
+ "\n",
+ "\n",
+ "diffusion resistance= 1.44 KOhm\n",
+ "\n",
+ "\n",
+ "input resistance= 6.78 KOhm\n",
+ "\n",
+ "\n",
+ "time constant= 6.88 ms\n",
+ "\n",
+ "corner frequency fL= 23.17 Hz\n",
+ "\n",
+ "\n",
+ "Rib= 41.84 KOhm\n",
+ "\n",
+ "\n",
+ "small signal voltage gain= 4.71 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.4\n",
+ "fL=20.*10**-3;##KHz\n",
+ "Rd=6.7;\n",
+ "Rl=10;\n",
+ "Ts=1./(2.*math.pi*fL);\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Ts,' ms\\n')\n",
+ "Cc=Ts/(Rd+Rl);\n",
+ "print\"%s %.2f %s\"%('\\ncoupling capacitance= ',Cc,' microF\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "time constant= 7.96 ms\n",
+ "\n",
+ "\n",
+ "coupling capacitance= 0.48 microF\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.5\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "Rs=500.;\n",
+ "Rb=100000.;\n",
+ "Re=10000.;\n",
+ "Rl=10000.;\n",
+ "Va=120.;\n",
+ "Ccc2=1*10**-6;\n",
+ "Icq=0.838*0.001;\n",
+ "r=3100.;##small signal parameter\n",
+ "gm=32.2*0.001;\n",
+ "ro=143000.;\n",
+ "x=(r+Rs*Rb/(Rs+Rb))/(1+b);\n",
+ "y=ro*x/(ro+x);\n",
+ "Ro=Re*y/(Re+y);\n",
+ "print\"%s %.2f %s\"%('\\noutput resistance of emitter= ',Ro,' Ohm\\n')\n",
+ "Ts=(Ro+Rl)*Ccc2;\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Ts,' s\\n')\n",
+ "fL=1/(2.*math.pi*Ts);\n",
+ "print\"%s %.2f %s\"%('\\n3dB frequency= ',fL,' Hz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "output resistance of emitter= 35.48 Ohm\n",
+ "\n",
+ "\n",
+ "time constant= 0.01 s\n",
+ "\n",
+ "\n",
+ "3dB frequency= 15.86 Hz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.6\n",
+ "Rs=3.2;\n",
+ "Rd=10.;\n",
+ "Rl=20.;\n",
+ "Cl=10.;\n",
+ "Vtp=-2.;\n",
+ "Kp=0.25;\n",
+ "Idq=0.5;\n",
+ "Vsgq=3.41;\n",
+ "Vsdq=3.41;\n",
+ "gm=2.*Kp*(Vsgq+Vtp);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance = ',gm,'mA/V\\n')\n",
+ "Tp=Cl*Rd*Rl/(Rd+Rl);\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Tp,' ns\\n')\n",
+ "Tp=66.7*10**-3;##micro sec\n",
+ "fH=1./(2.*math.pi*Tp);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency= ',fH,' MHz\\n')\n",
+ "Av=(gm*Rd*Rl/(Rd+Rl))/(1+gm*Rs);\n",
+ "print\"%s %.2f %s\"%('\\nmaximum small signal voltage gain=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance = 0.71 mA/V\n",
+ "\n",
+ "\n",
+ "time constant= 66.67 ns\n",
+ "\n",
+ "\n",
+ "corner frequency= 2.39 MHz\n",
+ "\n",
+ "\n",
+ "maximum small signal voltage gain=\n",
+ " 1.44 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg409"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.7\n",
+ "Vbe=0.7;\n",
+ "b=100.;\n",
+ "Re=.5;\n",
+ "Rc=5.;\n",
+ "Rl=10.;\n",
+ "R1=40.;\n",
+ "Cc=10.;\n",
+ "R2=5.7;\n",
+ "Rs=.1;\n",
+ "Vt=0.026;\n",
+ "Icq=0.99;\n",
+ "gm=Icq/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "r=b*Vt/Icq;\n",
+ "print\"%s %.2f %s\"%('\\ndiffusion resistance= ',r,' KOhm\\n')\n",
+ "Ri=r+(1.+b)*Re;\n",
+ "print\"%s %.2f %s\"%('\\ninput resistance= ',Ri,' KOhm\\n')\n",
+ "x=Rc*Rl/(Rc+Rl);\n",
+ "y=R1*R2/(R1+R2);\n",
+ "t=y*Ri/(y+Ri);\n",
+ "Av=gm*r*x*(y/(y+Ri))*(1./(Rs+t));\n",
+ "print\"%s %.2f %s\"%('\\nmaximum small signal voltage gain=\\n',Av,'')\n",
+ "Ts=(Rs+t)*Cc;\n",
+ "print\"%s %.2f %s\"%('\\ntime constant=\\n',Ts,'ms')\n",
+ "Ts=46.6*0.001;##sec\n",
+ "Cl=15.;\n",
+ "Tp=x*Cl;\n",
+ "print\"%s %.2f %s\"%('\\ntime constant ',Tp,' ns\\n')\n",
+ "fL=1/(2.*math.pi*Ts);\n",
+ "print\"%s %.2f %s\"%('\\nlower corner frequency= ',fL,' Hz\\n')\n",
+ "Tp=50.*10**-3;##micro sec\n",
+ "fH=1/(2.*math.pi*Tp);\n",
+ "print\"%s %.2f %s\"%('\\nupper corner frequency= ',fH,' MHz\\n')\n",
+ "fL=3.4*10**-6;##MHz\n",
+ "fbw=fH-fL;\n",
+ "print\"%s %.2f %s\"%('\\nbandwidth = ',fbw,'MHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance= 38.08 mA/V\n",
+ "\n",
+ "\n",
+ "diffusion resistance= 2.63 KOhm\n",
+ "\n",
+ "\n",
+ "input resistance= 53.13 KOhm\n",
+ "\n",
+ "\n",
+ "maximum small signal voltage gain=\n",
+ " 6.14 \n",
+ "\n",
+ "time constant=\n",
+ " 46.61 ms\n",
+ "\n",
+ "time constant 50.00 ns\n",
+ "\n",
+ "\n",
+ "lower corner frequency= 3.42 Hz\n",
+ "\n",
+ "\n",
+ "upper corner frequency= 3.18 MHz\n",
+ "\n",
+ "\n",
+ "bandwidth = 3.18 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg412"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 7.8\n",
+ "Re=4.;\n",
+ "Rc=2.;\n",
+ "Rs=0.5;\n",
+ "Vt=0.026;\n",
+ "Ce=1*10**-3;\n",
+ "V1=5.;\n",
+ "Icq=1.06;\n",
+ "V2=-5.;\n",
+ "b=100.;\n",
+ "Vbe=0.7;\n",
+ "gm=Icq/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance = ',gm,'mA/V\\n')\n",
+ "r=b*Vt/Icq;\n",
+ "print\"%s %.2f %s\"%('\\ndiffusion resistance= ',r,' KOhm\\n')\n",
+ "Ta=Re*Ce;\n",
+ "print\"%s %.2f %s\"%('\\ntime constant Ta= ',Ta,'f s\\n')\n",
+ "Tb=(Re*Ce*(Rs+r))/(Rs+r+(1+b)*Re);\n",
+ "print\"%s %.2f %s\"%('\\ntime constant Tb= ',Tb,' s\\n')\n",
+ "fA=1/(2.*math.pi*Ta);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency = ',fA,'Hz\\n')\n",
+ "Tb=2.9*0.01;##msec\n",
+ "fB=1/(2.*math.pi*Tb);\n",
+ "print\"%s %.2f %s\"%('\\ncorner frequency =',fB,'khz')\n",
+ "Av=(gm*r*Rc)/(Rs+r+(1.+b)*Re);\n",
+ "print\"%s %.2f %s\"%('\\nlimiting low frequency horizontal asymptote= \\n',Av,'')\n",
+ "Av=gm*r*Rc/(Rs+r);\n",
+ "print\"%s %.2f %s\"%('\\nnlimiting high frequency horizontal asymptote=\\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance = 40.77 mA/V\n",
+ "\n",
+ "\n",
+ "diffusion resistance= 2.45 KOhm\n",
+ "\n",
+ "\n",
+ "time constant Ta= 0.00 f s\n",
+ "\n",
+ "\n",
+ "time constant Tb= 0.00 s\n",
+ "\n",
+ "\n",
+ "corner frequency = 39.79 Hz\n",
+ "\n",
+ "\n",
+ "corner frequency = 5.49 khz\n",
+ "\n",
+ "limiting low frequency horizontal asymptote= \n",
+ " 0.49 \n",
+ "\n",
+ "nlimiting high frequency horizontal asymptote=\n",
+ " 67.73 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg420"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.9\n",
+ "r=2600.;\n",
+ "C1=2.*10**-6;\n",
+ "C2=0.1*10**-6;\n",
+ "fB=1/(2.*math.pi*r*(C1+C2));\n",
+ "print\"%s %.2f %s\"%('\\n3dB frequency= ',fB,' MHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "3dB frequency= 29.15 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg421"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.10\n",
+ "fT=500.;\n",
+ "Ic=1.;\n",
+ "b=100.;\n",
+ "Vt=0.026;\n",
+ "C2=0.3*10**-12;\n",
+ "fB=fT/b;\n",
+ "print\"%s %.2f %s\"%('\\nbandwidth= ',fB,' MHz\\n')\n",
+ "gm=Ic/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,'mA/V\\n')\n",
+ "fT=500000000.;\n",
+ "gm=38.5*0.001;\n",
+ "C1=gm/(fT*2.*math.pi)-C2;\n",
+ "print\"%s %.2e %s\"%('\\ncapacitance = ',C1,'F\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "bandwidth= 5.00 MHz\n",
+ "\n",
+ "\n",
+ "transconductance= 38.46 mA/V\n",
+ "\n",
+ "\n",
+ "capacitance = 1.20e-11 F\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg430"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.12\n",
+ "Kn=0.25;\n",
+ "Vtn=1.;\n",
+ "Cgd=0.04*10**-3;\n",
+ "Cgs=0.2*10**-3;\n",
+ "Vgs=3.;\n",
+ "gm=2.*Kn*(Vgs-Vtn);\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance = ',gm,'mA/V\\n')\n",
+ "fT=gm/(2.*math.pi*(Cgd+Cgs));\n",
+ "print\"%s %.2f %s\"%('\\nunity gain bandwidth= ',fT,' MHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "transconductance = 1.00 mA/V\n",
+ "\n",
+ "\n",
+ "unity gain bandwidth= 663.15 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 7.13\n",
+ "gm=1.;\n",
+ "Cgd=0.04;\n",
+ "Rl=10.;\n",
+ "Cgs=0.2;\n",
+ "Cm=Cgd*(1.+gm*Rl);\n",
+ "print\"%s %.2f %s\"%('\\nMiller capacitance= ',Cm,' pF\\n')\n",
+ "Cm=0.44*0.001;##nF\n",
+ "Cgs=0.2*0.001;##nF\n",
+ "fT=gm/(2.*math.pi*(Cgs+Cm));\n",
+ "print\"%s %.2f %s\"%('\\ncutoff frequency= ',fT,' MHz\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Miller capacitance= 0.44 pF\n",
+ "\n",
+ "\n",
+ "cutoff frequency= 248.68 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.14\n",
+ "V1=5.;\n",
+ "V=-5.;\n",
+ "Rs=0.1;\n",
+ "Rb=40.;\n",
+ "R2=5.72;\n",
+ "Re=0.5;\n",
+ "Rc=5.;\n",
+ "Rl=10.;\n",
+ "b=150.;\n",
+ "Vbe=0.7;\n",
+ "C1=35*10**-3;\n",
+ "C2=4.;\n",
+ "Vt=0.026;\n",
+ "Icq=1.02;\n",
+ "r=b*Vt/Icq;\n",
+ "print\"%s %.2f %s\"%('\\nsmall signal parameter= ',r,' KOhm\\n')\n",
+ "gm=Icq/Vt;\n",
+ "print\"%s %.2f %s\"%('\\ntransconductance= ',gm,' mA/V\\n')\n",
+ "Cm=C2*(1+gm*Rc*Rl/(Rc+Rl));\n",
+ "print\"%s %.2f %s\"%('\\nMiller capacitance= ',Cm,' pF\\n')\n",
+ "Cm=527.*10**-3;\n",
+ "x=Rb*Rs/(Rb+Rs);\n",
+ "y=r*x/(r+x);\n",
+ "fH=1/(2.*math.pi*y*(C1+Cm));\n",
+ "print\"%s %.2f %s\"%('\\nupper corner frequency = ',fH,'MHz\\n')\n",
+ "t=Rb*r/(Rb+r);\n",
+ "p=Rc*Rl/(Rc+Rl);\n",
+ "Av=gm*p*t/(t+Rs);\n",
+ "print\"%s %.2f %s\"%('\\nmidband gain= ',Av,'\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "small signal parameter= 3.82 KOhm\n",
+ "\n",
+ "\n",
+ "transconductance= 39.23 mA/V\n",
+ "\n",
+ "\n",
+ "Miller capacitance= 527.08 pF\n",
+ "\n",
+ "\n",
+ "upper corner frequency = 2.91 MHz\n",
+ "\n",
+ "\n",
+ "midband gain= 127.13 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.15\n",
+ "V1=5.;\n",
+ "V=-5.;\n",
+ "Rs=0.1;\n",
+ "R1=40.;\n",
+ "R2=5.72;\n",
+ "Re=0.5;\n",
+ "Rc=5.;\n",
+ "Rl=10.;\n",
+ "b=150.;\n",
+ "Vbe=0.7;\n",
+ "C1=35.;\n",
+ "C2=4.;\n",
+ "Vt=0.026;\n",
+ "Icq=1.02;\n",
+ "gm=39.2;\n",
+ "r=3.82;\n",
+ "x=Re*Rs/(Re+Rs);\n",
+ "t=r/(1.+b);\n",
+ "y=t*x/(t+x);\n",
+ "Tp=y*C1;\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Tp,' ns\\n')\n",
+ "Tp=0.679*10**-3;##micro sec\n",
+ "f=1/(2.*math.pi*Tp);\n",
+ "print\"%s %.2f %s\"%('\\nupper frequency = ',f,'MHz\\n')\n",
+ "T=C2*Rc*Rl/(Rc+Rl);\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',T,' ns\\n')\n",
+ "T=13.3*10**-3;##micro sec\n",
+ "f=1/(2.*math.pi*T);\n",
+ "print\"%s %.2f %s\"%('\\nupper frequency= ',f,' MHz\\n')\n",
+ "x=Rc*Rl/(Rc+Rl);\n",
+ "y=Re*t/(Re+t);\n",
+ "Av=gm*x*(y/(y+Rs));\n",
+ "print\"%s %.2f %s\"%('\\nmidband voltage gain \\n',Av,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "time constant= 0.68 ns\n",
+ "\n",
+ "\n",
+ "upper frequency = 234.40 MHz\n",
+ "\n",
+ "\n",
+ "time constant= 13.33 ns\n",
+ "\n",
+ "\n",
+ "upper frequency= 11.97 MHz\n",
+ "\n",
+ "\n",
+ "midband voltage gain \n",
+ " 25.36 \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg443"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 7.16\n",
+ "V1=5.;\n",
+ "V=-5.;\n",
+ "Rs=0.1;\n",
+ "R1=42.5;\n",
+ "R2=20.5;\n",
+ "R3=28.3;\n",
+ "Re=5.4;\n",
+ "Rc=5.;\n",
+ "Rl=10.;\n",
+ "b=150.;\n",
+ "Vbe=0.7;\n",
+ "C1=35.;\n",
+ "C2=4.;\n",
+ "Vt=0.026;\n",
+ "Icq=1.02;\n",
+ "gm=39.2;\n",
+ "r=3.820;\n",
+ "Rb=R2*R3/(R2+R3);\n",
+ "x=Rb*r/(Rb+r);\n",
+ "y=Rs*x/(x+Rs);\n",
+ "Tp=y*(C1+2*C2);\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',Tp,' ns\\n')\n",
+ "Tp=Tp*10**-3;##micro sec\n",
+ "f=1/(2.*math.pi*Tp);\n",
+ "print\"%s %.2f %s\"%('\\n3dB frequency = ',f,'MHz\\n')\n",
+ "T=C2*Rc*Rl/(Rc+Rl);\n",
+ "print\"%s %.2f %s\"%('\\ntime constant= ',T,'ns\\n')\n",
+ "T=T*0.001;##micro sec\n",
+ "f=1/(2.*math.pi*T);\n",
+ "print\"%s %.2f %s\"%('\\nupper frequency= ',f,' MHz\\n')\n",
+ "x=Rc*Rl/(Rc+Rl);\n",
+ "y=Rb*r/(Rb+r);\n",
+ "Av=gm*x*(y/(y+Rs));\n",
+ "print\"%s %.2f %s\"%('\\nmidband voltage gain= ',Av,' \\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "time constant= 4.16 ns\n",
+ "\n",
+ "\n",
+ "3dB frequency = 38.29 MHz\n",
+ "\n",
+ "\n",
+ "time constant= 13.33 ns\n",
+ "\n",
+ "\n",
+ "upper frequency= 11.94 MHz\n",
+ "\n",
+ "\n",
+ "midband voltage gain= 126.30 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg447"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Example 7.17\n",
+ "V1=5.;\n",
+ "V=-5.;\n",
+ "Rs=0.1;\n",
+ "R1=40;\n",
+ "R2=5.720;\n",
+ "Re=0.5;\n",
+ "Rc=5.;\n",
+ "Rl=10.;\n",
+ "b=150.;\n",
+ "Vbe=0.7;\n",
+ "C1=35.;\n",
+ "C2=4.;\n",
+ "Vt=0.026;\n",
+ "Icq=1.02;\n",
+ "gm=39.2;\n",
+ "r=3.820;\n",
+ "t=r/(1.+b);\n",
+ "t=t*0.001;\n",
+ "f=1/(2.*math.pi*C1*t);\n",
+ "print'%s %.2f %s'%('\\nthe zero occurs at this frequency= ',f,' MHz\\n')\n",
+ "x=1+gm*Re*Rl/(Re+Rl);\n",
+ "Rb=R1*R2/(R1+R2)\n",
+ "d=x*r;\n",
+ "y=d*Rb/(d+Rb);\n",
+ "t=y*Rs/(y+Rs);\n",
+ "Tp=t*(C2+C1/x);\n",
+ "print'%s %.2f %s'%('\\ntime constant= ',Tp,' ns\\n')\n",
+ "Tp=Tp*10**-3;##micro sec\n",
+ "f=1/(2.*math.pi*Tp);\n",
+ "print'%s %.2f %s'%('\\n3dB frequency= ',f,' MHz\\n')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the zero occurs at this frequency= 179.75 MHz\n",
+ "\n",
+ "\n",
+ "time constant= 0.57 ns\n",
+ "\n",
+ "\n",
+ "3dB frequency= 281.24 MHz\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter8_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter8_1_2.ipynb new file mode 100644 index 00000000..c0c5ebd2 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter8_1_2.ipynb @@ -0,0 +1,447 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:cdd1f791d4e53fd374d40ced42151b47691b22d0170edabcbe64af78e36e0de3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-Ouput Stages and Power Amplifier "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg478"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 8.1\n",
+ "##let thermal resistance parameters be R\n",
+ "Rdcase=1.75;##degree celsius per watt\n",
+ "Rcsnk=1.;##degree celsius per watt\n",
+ "Rsamb=5.;##degree celsius per watt\n",
+ "Rcamb=50.;##degree celsius per watt\n",
+ "Tamb=30.;##ambient temperature \n",
+ "Tjmax=150.;##maximum junction temperature\n",
+ "Tdev=150.;##device temperature\n",
+ "##when no heat sink is used\n",
+ "P=(Tjmax-Tamb)/(Rdcase+Rcamb);\n",
+ "print\"%s %.2f %s\"%('\\nmaximum power dissipation= ',P,' W\\n')\n",
+ "##when heat sink is used\n",
+ "P=(Tjmax-Tamb)/(Rdcase+Rcsnk+Rsamb);\n",
+ "print\"%s %.2f %s\"%('\\nmaximum power dissipation= ',P,' W\\n')\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "maximum power dissipation= 2.32 W\n",
+ "\n",
+ "\n",
+ "maximum power dissipation= 15.48 W\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 8.3\n",
+ "Rcsnk=1.;##degree celsius per watt\n",
+ "Rsamb=5.;##degree celsius per watt\n",
+ "Tjmax=175.;##maximum junction temperature\n",
+ "Toc=25.;\n",
+ "Tamb=25.;\n",
+ "Pr=20.;##rated power W\n",
+ "Rdcase=(Tjmax-Toc)/Pr;\n",
+ "print\"%s %.2f %s\"%('\\ndevice to case thermal resistance= ',Rdcase,' C/W\\n')\n",
+ "P=(Tjmax-Tamb)/(Rdcase+Rcsnk+Rsamb);\n",
+ "print\"%s %.2f %s\"%('\\nmaximum power dissipation= ',P,' W\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "device to case thermal resistance= 7.50 C/W\n",
+ "\n",
+ "\n",
+ "maximum power dissipation= 11.11 W\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg492"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 8.7\n",
+ "Vdd=10.;\n",
+ "Rl=20.;\n",
+ "K=0.2;\n",
+ "Vt=1.;\n",
+ "vo=5.;\n",
+ "iL=vo/20.;\n",
+ "print\"%s %.2f %s\"%('\\niL= ',iL,' A\\n')\n",
+ "Idq=0.05;\n",
+ "##Idq=K*(Vbb/2-Vt)\n",
+ "Vbb=(math.sqrt(Idq/K)+1.)*2.;\n",
+ "print\"%s %.2f %s\"%('\\nVbb= ',Vbb,' V\\n')\n",
+ "iD=iL;\n",
+ "Vgsn=math.sqrt(iD/K)+Vt;\n",
+ "print\"%s %.2f %s\"%('\\nVgsn= ',Vgsn,' V\\n')\n",
+ "Vsgp=Vbb-Vgsn;\n",
+ "print\"%s %.2f %s\"%('\\nVsgp= ',Vsgp,' V\\n')\n",
+ "vi=vo+Vgsn-Vbb/2.;\n",
+ "print\"%s %.2f %s\"%('\\ninput voltage= ',vi,' V\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "iL= 0.25 A\n",
+ "\n",
+ "\n",
+ "Vbb= 3.00 V\n",
+ "\n",
+ "\n",
+ "Vgsn= 2.12 V\n",
+ "\n",
+ "\n",
+ "Vsgp= 0.88 V\n",
+ "\n",
+ "\n",
+ "input voltage= 5.62 V\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg498"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ " \n",
+ "##Example 8.8\n",
+ "Vcc=24.;\n",
+ "Rl=8.;\n",
+ "P=5.;\n",
+ "Vbe=0.7;\n",
+ "b=100.;\n",
+ "Vp=math.sqrt(2.*Rl*P);\n",
+ "print\"%s %.2f %s\"%('\\npeak output voltage= ',Vp,' V\\n')\n",
+ "Ip=Vp/Rl;\n",
+ "print\"%s %.2f %s\"%('\\npeak output current = ',Ip,'A\\n')\n",
+ "a=0.9*Vcc/Vp;\n",
+ "print\"%s %.2f %s\"%('\\na= ',a,'\\n')\n",
+ "Icq=Ip/(0.9*a);\n",
+ "print\"%s %.2f %s\"%('\\nIcq= ',Icq,' A\\n')\n",
+ "Pq=Vcc*Icq;\n",
+ "print\"%s %.2f %s\"%('\\nmaximum power dissipated in the transistor= ',Pq,' W\\n')\n",
+ "Ibq=Icq/b;\n",
+ "Ibq=Ibq*1000.;##mA\n",
+ "print\"%s %.2f %s\"%('\\nbase current Ibq= ',Ibq,' mA\\n')\n",
+ "Rth=2.500;\n",
+ "##Vth=Vcc*Rth/R1 and Vth=Ibq*Rth+Vbe\n",
+ "R1=Vcc*Rth/(Ibq*Rth+Vbe);\n",
+ "print\"%s %.2f %s\"%('\\nR1= ',R1,' KOhm\\n')\n",
+ "R2=Rth*R1/(R1-Rth);\n",
+ "print\"%s %.2f %s\"%('\\nR2= ',R2,' KOhm\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "peak output voltage= 8.94 V\n",
+ "\n",
+ "\n",
+ "peak output current = 1.12 A\n",
+ "\n",
+ "\n",
+ "a= 2.41 \n",
+ "\n",
+ "\n",
+ "Icq= 0.51 A\n",
+ "\n",
+ "\n",
+ "maximum power dissipated in the transistor= 12.35 W\n",
+ "\n",
+ "\n",
+ "base current Ibq= 5.14 mA\n",
+ "\n",
+ "\n",
+ "R1= 4.42 KOhm\n",
+ "\n",
+ "\n",
+ "R2= 5.75 KOhm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg500"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 8.9\n",
+ "Iso=3*10**-14;\n",
+ "Isq=10**-13;\n",
+ "b=75.;\n",
+ "Vt=0.026;\n",
+ "Rl=8.;\n",
+ "P=5.;\n",
+ "Vp=math.sqrt(2.*Rl*P);\n",
+ "print\"%s %.2f %s\"%('\\npeak voltage Vp= ',Vp,' V\\n')\n",
+ "Vcc=Vp/0.8;\n",
+ "print\"%s %.2f %s\"%('\\nsupply voltage= ',Vcc,' V\\n')\n",
+ "Ien=Vp/Rl;\n",
+ "print\"%s %.2f %s\"%('\\nemitter current= ',Ien,' A\\n')\n",
+ "Ibn=Ien/(1.+b);\n",
+ "Ibn=Ibn*1000.;##mA\n",
+ "print\"%s %.2f %s\"%('\\nbase current= ',Ibn,' mA\\n')\n",
+ "iD=0.020;\n",
+ "Vbb=2.*Vt*math.log(iD/Iso);\n",
+ "print\"%s %.2f %s\"%('\\nVbb= ',Vbb,' V\\n')\n",
+ "Icq=Isq*math.exp((Vbb/2.)/Vt);\n",
+ "Icq=Icq*1000.;##mA\n",
+ "print\"%s %.2f %s\"%('\\nquiescent collector current= ',Icq,' mA\\n')\n",
+ "Ibias=20.;##mA\n",
+ "iD=Ibias-Ibn;\n",
+ "print\"%s %.2f %s\"%('\\ndrain current= ',iD,' mA\\n')\n",
+ "iD=iD*0.001;##A\n",
+ "Vbb=2.*Vt*math.log(iD/Iso);\n",
+ "print\"%s %.2f %s\"%('\\nVbb= ',Vbb,' V\\n')\n",
+ "Icn=1.12;\n",
+ "Vben=Vt*math.log(Icn/Isq);\n",
+ "print\"%s %.2f %s\"%('\\nB-E voltage of Qn= ',Vben,' V\\n')\n",
+ "Vebp=Vbb-Vben;\n",
+ "print\"%s %.2f %s\"%('\\nemitter base voltage of Qp= ',Vebp,' V\\n')\n",
+ "Icp=Isq*math.exp(Vebp/Vt);\n",
+ "Icp=Icp*1000.;##mA\n",
+ "print\"%s %.2f %s\"%('\\nIcp ',Icp,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "peak voltage Vp= 8.94 V\n",
+ "\n",
+ "\n",
+ "supply voltage= 11.18 V\n",
+ "\n",
+ "\n",
+ "emitter current= 1.12 A\n",
+ "\n",
+ "\n",
+ "base current= 14.71 mA\n",
+ "\n",
+ "\n",
+ "Vbb= 1.42 V\n",
+ "\n",
+ "\n",
+ "quiescent collector current= 66.67 mA\n",
+ "\n",
+ "\n",
+ "drain current= 5.29 mA\n",
+ "\n",
+ "\n",
+ "Vbb= 1.35 V\n",
+ "\n",
+ "\n",
+ "B-E voltage of Qn= 0.78 V\n",
+ "\n",
+ "\n",
+ "emitter base voltage of Qp= 0.57 V\n",
+ "\n",
+ "\n",
+ "Icp 0.28 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 8.11\n",
+ "R1=2.;##KOhm\n",
+ "R2=R1;\n",
+ "Rl=.1;##KOhm\n",
+ "b=60.;\n",
+ "Vbe=0.6;\n",
+ "Veb=0.6;\n",
+ "V1=15.;\n",
+ "V2=V1;\n",
+ "iR1=(V1-Vbe)/R1;\n",
+ "##iR1=iR2=iE1=iE2\n",
+ "print\"%s %.2f %s\"%('\\niR1= ',iR1,' mA\\n')\n",
+ "vo=10.;\n",
+ "io=vo/Rl;\n",
+ "print\"%s %.2f %s\"%('\\noutput current= ',io,' mA\\n')\n",
+ "iB3=100./61.;\n",
+ "print\"%s %.2f %s\"%('\\niB3= ',iB3,'mA\\n')\n",
+ "iR1=(V1-(10.+Vbe))/R1;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent in R1= ',iR1,' mA\\n')\n",
+ "iE1=iR1-iB3;\n",
+ "print\"%s %.2f %s\"%('\\niE1= ',iE1,' mA\\n')\n",
+ "iB1=iE1/(1.+b);\n",
+ "iB1=iB1*1000.;##micro A\n",
+ "print\"%s %.2f %s\"%('\\niB1= ',iB1,' microA\\n')\n",
+ "iE2=(10-0.6+15.)/R1;\n",
+ "print\"%s %.2f %s\"%('\\niE2= ',iE2,' mA\\n')\n",
+ "iB2=iE2/(1.+b);\n",
+ "iB2=iB2*1000.;\n",
+ "print\"%s %.2f %s\"%('\\niB2= ',iB2,' microA\\n')\n",
+ "Ii=iB2-iB1;\n",
+ "print\"%s %.2f %s\"%('\\ninput current= ',Ii,' microA\\n')\n",
+ "Ii=Ii*0.001;##mA\n",
+ "Ai=io/Ii;\n",
+ "print\"%s %.2f %s\"%('\\ncurrent gain=\\n',Ai,'')\n",
+ "Ai=(1.+b)*R1/(2.*Rl);\n",
+ "print\"%s %.2f %s\"%('\\npredicted current gain=\\n',Ai,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "iR1= 7.20 mA\n",
+ "\n",
+ "\n",
+ "output current= 100.00 mA\n",
+ "\n",
+ "\n",
+ "iB3= 1.64 mA\n",
+ "\n",
+ "\n",
+ "current in R1= 2.20 mA\n",
+ "\n",
+ "\n",
+ "iE1= 0.56 mA\n",
+ "\n",
+ "\n",
+ "iB1= 9.19 microA\n",
+ "\n",
+ "\n",
+ "iE2= 12.20 mA\n",
+ "\n",
+ "\n",
+ "iB2= 200.00 microA\n",
+ "\n",
+ "\n",
+ "input current= 190.81 microA\n",
+ "\n",
+ "\n",
+ "current gain=\n",
+ " 524.08 \n",
+ "\n",
+ "predicted current gain=\n",
+ " 610.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter9_1_2.ipynb b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter9_1_2.ipynb new file mode 100644 index 00000000..01be9a77 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter9_1_2.ipynb @@ -0,0 +1,136 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:32f4ddc547fad5eac3ccd22992e1c23752ad6ef6cf432bf032a216c1338d002f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9-The Ideal Operational Amplifier"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg541"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "##Example 9.5\n",
+ "Zl=0.1;\n",
+ "R1=10.;\n",
+ "R2=1.;\n",
+ "R3=1.;\n",
+ "Rf=10.;\n",
+ "Vt=-5.;\n",
+ "iL=-Vt/R2;\n",
+ "print\"%s %.2f %s\"%('\\nload current= ',iL,' mA\\n')\n",
+ "vL=iL*Zl;\n",
+ "print\"%s %.2f %s\"%('\\nvoltage across the load= ',vL,' V\\n')\n",
+ "i4=vL/R2;\n",
+ "print\"%s %.2f %s\"%('\\ni4= ',i4,' mA\\n')\n",
+ "i3=i4+iL;\n",
+ "print\"%s %.2f %s\"%('\\ni3= ',i3,' mA\\n')\n",
+ "Vo=i3*R3+vL;\n",
+ "print\"%s %.2f %s\"%('\\noutput voltage= ',Vo,' V\\n')\n",
+ "i1=Vt/R1;\n",
+ "i2=i1;\n",
+ "print\"%s %.2f %s\"%('\\ni1= ',i1,' mA\\n')\n",
+ "print\"%s %.2f %s\"%('\\ni2= ',i2,' mA\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "load current= 5.00 mA\n",
+ "\n",
+ "\n",
+ "voltage across the load= 0.50 V\n",
+ "\n",
+ "\n",
+ "i4= 0.50 mA\n",
+ "\n",
+ "\n",
+ "i3= 5.50 mA\n",
+ "\n",
+ "\n",
+ "output voltage= 6.00 V\n",
+ "\n",
+ "\n",
+ "i1= -0.50 mA\n",
+ "\n",
+ "\n",
+ "i2= -0.50 mA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ " \n",
+ "##Example 9.9\n",
+ "##Vo=(-1/R1*C2)*integrate((-1)dt) \n",
+ "def fun(x):\n",
+ " y=-1\n",
+ " return y\n",
+ "Vo=10.;\n",
+ "I=scipy.integrate.quad(fun, 0, 1);\n",
+ "I1=I[0]\n",
+ "##let y=R1*C2\n",
+ "y1=I1/Vo;\n",
+ "print\"%s %.2f %s\"%('\\nR1C2= ',y1,' ms\\n')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "R1C2= -0.10 ms\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter3_2.png b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter3_2.png Binary files differnew file mode 100644 index 00000000..5143e4e5 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter3_2.png diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter4_2.png b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter4_2.png Binary files differnew file mode 100644 index 00000000..cd0b0135 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter4_2.png diff --git a/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter5_2.png b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter5_2.png Binary files differnew file mode 100644 index 00000000..e11eb074 --- /dev/null +++ b/Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter5_2.png diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch1.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch1.ipynb new file mode 100644 index 00000000..01e7a07d --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch1.ipynb @@ -0,0 +1,136 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-1 : Physical properties of elements" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No.4 Example 1.1." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "r1=0.53 (A.U)\n", + "r2=2.12 (meters)\n", + "r3=4.77 (meters)\n" + ] + } + ], + "source": [ + "from math import pi\n", + "epsilon=8.854*10**-12\n", + "h=6.62*10**-34 #planck's constant\n", + "m=9.1*10**-31 #mass of electron\n", + "q=1.6*10**-19 #charge of electron\n", + "for n in [1]:\n", + " r1=(epsilon*(h**2)*(n**2))/(pi*m*(q**2)) #radius of 1st orbit for hydrogen\n", + " x1=r1*10**10 # in A.U\n", + " print \"r1=%0.2f (A.U)\"%x1\n", + "\n", + "for n in [2]:\n", + " r2=(epsilon*(h**2)*(n**2))/(pi*m*(q**2)) #radius of 2st orbit for hydrogen\n", + " x2=r2*10**10 # in A.U\n", + " print \"r2=%0.2f (meters)\"%x2\n", + "\n", + "for n in [3]:\n", + " r3=(epsilon*(h**2)*(n**2))/(pi*m*(q**2)) #radius of 3st orbit for hydrogen\n", + " x3=r3*10**10 # in A.U\n", + " print \"r3=%0.2f (meters)\"%x3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 6 Example 1.2." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelenth of the emitted photon is =920.97 (Armstrong)\n" + ] + } + ], + "source": [ + "E1=-13.6# #energy of 10th state\n", + "E10=-13.6/10**2# #enery in the ground state\n", + "lamda=12400/(E10-E1)# #wavelength of emitted photon\n", + "print \"Wavelenth of the emitted photon is =%0.2f (Armstrong)\"%lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 6 Example 1.3." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the wavelength limit = 12400 / Einfinity-E2 =3647.06 (A.U)\n" + ] + } + ], + "source": [ + "Einfinity=0 #energy of electron at infinite orbit\n", + "E2=-13.6/2**2 #energy of electron at second orbit\n", + "wavelength=12400/(Einfinity-E2) #wavelength limit\n", + "print \"the wavelength limit = 12400 / Einfinity-E2 =%0.2f (A.U)\"%wavelength" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch10.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch10.ipynb new file mode 100644 index 00000000..0f3b8bbf --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch10.ipynb @@ -0,0 +1,239 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-10 : Multistage Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 316 Example 10.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For the second stage ZL = RE2 and the current gain of the second stage is\n", + " AI2 = -Ie2 / Ib2 = -hfc / (hoc*RE2) =55.45\n", + "For the second stage,\n", + " Ri2 = hic + (hrc*AI2*RE2) =223.42 kohm\n", + " AV2 = Vo/V2 = (AI2*Re2) / Ri2 = 0.99 \n", + "The First Stage :\n", + " RL1= RC1 || Ri2 =3.93 kohm\n", + "Current gain,\n", + " AI1 = -IC1/Ib1 = -hfe/(1+(hoe*RL1)) =-54.63\n", + "The input impedance of the first stage, which is also the input impedance of the cascaded amplifier is\n", + " Ri1 = hie + hre*AI1*RL1 =1.49 kohm\n", + "The voltage gain of the first stage is\n", + " AV1 = V2/V1 = (AI1*RL1) / Ri1 =-143.83\n", + "The output admittance of the first transistor Q1\n", + " Yo1(uA/V) = hoe - ((hfe*hre) / (hie+RS)) =11.36\n", + "The output impedance of the first stage\n", + " Ro1 = 1 / Yo1 =88.00 kohm\n", + "The output impedance taking RC1 into account is\n", + " Rot1(k-ohm) = Ro1 || RC1 =3.83 kohm\n", + "The output admittance of the second stage\n", + " Yo2 = hoc-((hfc*hrc) / (hic+Rot1)) =0.01 A/V\n", + "Output impedance,\n", + " RO2 = 1 / Yo2 =86.77 ohm\n", + "Hence, Ro2(ohm) = (RO2*RE2) / (RO2+RE2) =85.15 ohm\n", + " Ib2/Ic1 = -Rc1/ Rc1+Ri2 =-0.02\n", + " AI = -AI2*AI1*(Rc1 / Ri2+Rc1) =-53.29\n", + " AV = AV2*AV1 =-142.80\n", + "The overall voltage gain taking the source impedance into account,\n", + " AVs = Vo/Vs = Av(Ri1 / Ri1+Rs) =-101.86\n" + ] + } + ], + "source": [ + "hie=1600.\n", + "hfe=60.\n", + "hre=5*10**-4\n", + "hoe=25*10**-6\n", + "hic=1600.\n", + "hfc=-61.\n", + "hrc=1.\n", + "hoc=25*10**-6\n", + "print \"For the second stage ZL = RE2 and the current gain of the second stage is\"\n", + "RE2=4000.\n", + "AI2=-hfc/(1+(hoc*RE2))\n", + "print \" AI2 = -Ie2 / Ib2 = -hfc / (hoc*RE2) =%0.2f\"%AI2\n", + "print \"For the second stage,\"\n", + "Ri2 = hic + (hrc*AI2*RE2)\n", + "Ri22=Ri2*10**-3\n", + "print \" Ri2 = hic + (hrc*AI2*RE2) =%0.2f kohm\"%Ri22\n", + "Re2=4000.\n", + "AV2=(AI2*Re2)/Ri2\n", + "print \" AV2 = Vo/V2 = (AI2*Re2) / Ri2 = %0.2f \"%AV2\n", + "print \"The First Stage :\"\n", + "RC1=4000.\n", + "RL1=(RC1*Ri2)/(RC1+Ri2)\n", + "RL11=RL1*10**-3\n", + "print \" RL1= RC1 || Ri2 =%0.2f kohm\"%RL11\n", + "print \"Current gain,\"\n", + "AI1= -hfe/(1+(hoe*RL1))\n", + "print \" AI1 = -IC1/Ib1 = -hfe/(1+(hoe*RL1)) =%0.2f\"%AI1\n", + "print \"The input impedance of the first stage, which is also the input impedance of the cascaded amplifier is\"\n", + "Ri1=hie +(hre*AI1*RL1) # answer in textbook is wrong \n", + "Ri11=Ri1*10**-3\n", + "print \" Ri1 = hie + hre*AI1*RL1 =%0.2f kohm\"%Ri11\n", + "print \"The voltage gain of the first stage is\"\n", + "AV1=(AI1*RL1)/Ri1 # answer in textbook is wrong \n", + "print \" AV1 = V2/V1 = (AI1*RL1) / Ri1 =%0.2f\"%AV1\n", + "print \"The output admittance of the first transistor Q1\"\n", + "RS=600.\n", + "Yo1=hoe-((hfe*hre)/(hie+RS))\n", + "Yo0=Yo1*10**6\n", + "print \" Yo1(uA/V) = hoe - ((hfe*hre) / (hie+RS)) =%0.2f\"%Yo0\n", + "print \"The output impedance of the first stage\"\n", + "Ro1=1./Yo1\n", + "Ro0=Ro1*10**-3\n", + "print \" Ro1 = 1 / Yo1 =%0.2f kohm\"%Ro0\n", + "print \"The output impedance taking RC1 into account is\"\n", + "Rot1=(Ro1*RC1)/(Ro1+RC1)\n", + "Rott=Rot1*10**-3\n", + "print \" Rot1(k-ohm) = Ro1 || RC1 =%0.2f kohm\"%Rott\n", + "print \"The output admittance of the second stage\"\n", + "Yo2=hoc-((hfc*hrc)/(hic+Rot1))\n", + "print \" Yo2 = hoc-((hfc*hrc) / (hic+Rot1)) =%0.2f A/V\"%Yo2\n", + "print \"Output impedance,\"\n", + "RO2=1/(11.525*10**-3)\n", + "print \" RO2 = 1 / Yo2 =%0.2f ohm\"%RO2\n", + "Ro2=(87.*4000.)/(87+4000)\n", + "print \"Hence, Ro2(ohm) = (RO2*RE2) / (RO2+RE2) =%0.2f ohm\"%Ro2\n", + "Rc1=4000.\n", + "x=(-Rc1)/ (Rc1+Ri2)\n", + "print \" Ib2/Ic1 = -Rc1/ Rc1+Ri2 =%0.2f\"%x\n", + "AI=-AI2*x*AI1\n", + "print \" AI = -AI2*AI1*(Rc1 / Ri2+Rc1) =%0.2f\"%AI\n", + "AV=AV2*AV1\n", + "print \" AV = AV2*AV1 =%0.2f\"%AV # answer in textbook is wrong\n", + "print \"The overall voltage gain taking the source impedance into account,\"\n", + "AVs=AV*(Ri1/(Ri1+RS))\n", + "print \" AVs = Vo/Vs = Av(Ri1 / Ri1+Rs) =%0.2f\"%AVs # answer in textbook is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 325 Example 10.2." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ro = 1/hoe =33.33 kohm\n", + "RB = R1 || R2 =9.09 kohm\n", + "Ri = hie =1.20 kohm\n", + "RC'' = RC || Ro =4.35 kohm\n", + "Ri'' = RB || Ri =1.06 kohm\n", + "Rci'' = Rc'' || Ri'' =0.85 kohm\n", + "rbe = hfe / gm =1000.00 ohm\n", + "(a) Mid-band current gain,\n", + "AIm = (-hfe*R''C) / (RC''+Ri'') =-39.91\n", + "(b) Mid-band voltage gain,\n", + "AVm = (-hfe) * (Rcid/hie) =-36.25\n", + "(c) Lower 3dB frequency,\n", + "fL = 1 / (2*pi*CC*(R_C+R_i)) =4.87 Hz\n", + "Higher 3dB frequency,\n", + "fH = 1 / (2*pi*C*rbe) =994.72 kHz\n", + "(d) Voltage gain x bandwidth\n", + "|AVmfH| =36.06\n" + ] + } + ], + "source": [ + "from math import pi\n", + "hfe=50.\n", + "hie=1200.\n", + "hoe=30*10**-6\n", + "hre=2.5*10**-4\n", + "RC=5*10**3\n", + "C=160*10**-12\n", + "CC=6*10**-6\n", + "R1=100*10**3\n", + "R2=10*10**3\n", + "gm=50*10**-3\n", + "Ro=1./hoe\n", + "x1=(Ro*10**-3)\n", + "print \"Ro = 1/hoe =%0.2f kohm\"%x1\n", + "RB=(R1*R2)/(R1+R2)\n", + "x2=RB*10**-3\n", + "print \"RB = R1 || R2 =%0.2f kohm\"%x2\n", + "Ri=hie\n", + "x3=Ri*10**-3\n", + "print \"Ri = hie =%0.2f kohm\"%x3\n", + "R_C=(RC*Ro)/(RC+Ro)\n", + "x4=R_C*10**-3\n", + "print \"RC'' = RC || Ro =%0.2f kohm\"%x4\n", + "R_i=(RB*Ri)/(RB+Ri)\n", + "x6=R_i*10**-3\n", + "print \"Ri'' = RB || Ri =%0.2f kohm\"%x6\n", + "R_ci=(R_C*R_i)/(R_C+R_i)\n", + "x7=R_ci*10**-3\n", + "print \"Rci'' = Rc'' || Ri'' =%0.2f kohm\"%x7\n", + "rbe=hfe/gm\n", + "print \"rbe = hfe / gm =%0.2f ohm\"%rbe\n", + "print \"(a) Mid-band current gain,\"\n", + "AIm=(-50*4.35*10**3)/((4.35*10**3)+(1.1*10**3))\n", + "print \"AIm = (-hfe*R''C) / (RC''+Ri'') =%0.2f\"%AIm\n", + "print \"(b) Mid-band voltage gain,\"\n", + "AVm=(-50)*((0.87*10**3)/(1.2*10**3))\n", + "print \"AVm = (-hfe) * (Rcid/hie) =%0.2f\"%AVm\n", + "print \"(c) Lower 3dB frequency,\"\n", + "fL=1./(2*pi*6*10**-6*(5.45*10**3))\n", + "print \"fL = 1 / (2*pi*CC*(R_C+R_i)) =%0.2f Hz\"%fL\n", + "print \"Higher 3dB frequency,\"\n", + "fH=1/(2*pi*C*rbe)\n", + "x8=fH*10**-3\n", + "print \"fH = 1 / (2*pi*C*rbe) =%0.2f kHz\"%x8 # answer in textbook is wrong \n", + "print \"(d) Voltage gain x bandwidth\"\n", + "y=abs(AVm*fH)\n", + "x9=(y*10**-6)\n", + "print \"|AVmfH| =%0.2f\"%x9" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch11.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch11.ipynb new file mode 100644 index 00000000..0223a598 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch11.ipynb @@ -0,0 +1,230 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-11 : Frequency Response of Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 380 Example 11.2." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "BW = 0.35 / tr =35.00 MHz\n" + ] + } + ], + "source": [ + "tr=10*10**-9\n", + "BW=0.35/tr\n", + "x1=BW*10**-6\n", + "print \"BW = 0.35 / tr =%0.2f MHz\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 382 Example 11.3." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "AV = (-hfe*RL) / (RS + hie + ((hie*RS)/RB)) =-145.70 MF\n", + "Lower 3-dB point,\n", + "f1 = (1+hfe) / ((RS+hie)*2*pi*CE) =120.42\n" + ] + } + ], + "source": [ + "from math import pi\n", + "hfe=400.\n", + "hie=10*10**3\n", + "Rs=600.\n", + "RL=5*10**3\n", + "RE=1*10**3\n", + "VCC=12.\n", + "R1=15*10**3\n", + "R2=2.2*10**3\n", + "CE=50*10**-6\n", + "RB=(R1*R2)/(R1+R2)\n", + "Av=(-hfe*RL)/(Rs+hie+((hie*Rs)/RB))\n", + "print \"AV = (-hfe*RL) / (RS + hie + ((hie*RS)/RB)) =%0.2f MF\"%Av\n", + "print \"Lower 3-dB point,\"\n", + "f1=(1.+hfe)/((Rs+hie)*2*pi*CE)\n", + "print \"f1 = (1+hfe) / ((RS+hie)*2*pi*CE) =%0.2f\"%f1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 386 Example 11.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The lower 3 dB frequency, f1 = 1 / (2*pi*(RS+R1dash)*CC)\n", + "(a) R1'' = R1 || R2 || hie =500.00 ohm \n", + " CC = 1 / (2*pi*f1*(RS+R1'')) =1.16 uF\n", + "(b) R1''(ohm) = R1 || R2 || [hie+((1+hfe)*RCE)] =716.31 ohm\n", + " CC = 1 / (2*pi*f1*(RS+R1'')) =0.97 uF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "RS=600.\n", + "hie=1*10**3\n", + "hfe=60.\n", + "R1=5*10**3\n", + "R2=1.25*10**3\n", + "RCE=25.\n", + "f1=125.\n", + "print \"The lower 3 dB frequency, f1 = 1 / (2*pi*(RS+R1dash)*CC)\"\n", + "R1dash=(R1*R2*hie)/((R2*hie)+(R1*hie)+(R1*R2))\n", + "CC=1 / (2*pi*f1*(RS+R1dash))\n", + "x1=CC*10**6\n", + "print \"(a) R1'' = R1 || R2 || hie =%0.2f ohm \"%R1dash\n", + "print \" CC = 1 / (2*pi*f1*(RS+R1'')) =%0.2f uF\"%x1\n", + "x2=hie+((1.+hfe)*RCE)\n", + "R1dash=(R1*R2*x2)/((R2*x2)+(R1*x2)+(R1*R2))\n", + "CC=1 / (2*pi*f1*(RS+R1dash))\n", + "x3=CC*10**6\n", + "print \"(b) R1''(ohm) = R1 || R2 || [hie+((1+hfe)*RCE)] =%0.2f ohm\"%R1dash\n", + "print \" CC = 1 / (2*pi*f1*(RS+R1'')) =%0.2f uF\"%x3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 390 Example 11.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " gm = IC(mA)/26mV = 1/26 =38.46 m-mho\n", + " rb''e = hfe / gm =5.82 kohm\n", + " rbb'' = hie - rb''e = 6000-5824 =176.00 ohm\n", + " cb''e(pF) = gm/2*pi*fT - Cb''c =64.51 pF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "gm=1./26 #mho\n", + "x1=gm*10**3 #m-mho\n", + "print \" gm = IC(mA)/26mV = 1/26 =%0.2f m-mho\"%x1\n", + "rbe=224./(38.46*10**-3)\n", + "x2=rbe*10**-3 #k-ohm\n", + "print \" rb''e = hfe / gm =%0.2f kohm\"%x2\n", + "rbb=6000.-5824. #ohm\n", + "print \" rbb'' = hie - rb''e = 6000-5824 =%0.2f ohm\"%rbb\n", + "cbe=((38.46*10**-3)/(2*pi*(80*10**6)))-(12*10**-12) # farad\n", + "x3=cbe*10**12 #pF\n", + "print \" cb''e(pF) = gm/2*pi*fT - Cb''c =%0.2f pF\"%x3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 392 Example 11.6." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " f_alpha = hfe / 2*pi*rb''e*Cb''e =95.91 MHz\n", + " f_beta = 1 / 2*pi*rb''e*(Cb''e+Cb''c) =0.36 MHz\n", + " fT = gm / 2*pi*(Cb''e+Cb''c) =80.64 MHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "alpha=224./(2*pi*(5.9*10**3)*(63*10**-12)) #Hz\n", + "x1=alpha*10**-6 #MHz\n", + "print \" f_alpha = hfe / 2*pi*rb''e*Cb''e =%0.2f MHz\"%x1\n", + "beta=1/(2*pi*(5.9*10**3)*((63*10**-12)+(12*10**-12)))\n", + "x2=beta*10**-6\n", + "print \" f_beta = 1 / 2*pi*rb''e*(Cb''e+Cb''c) =%0.2f MHz\"%x2\n", + "fT=(38*10**-3)/(2*pi*((63*10**-12)+(12*10**-12)))\n", + "x3=fT*10**-6\n", + "print \" fT = gm / 2*pi*(Cb''e+Cb''c) =%0.2f MHz\"%x3" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch12.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch12.ipynb new file mode 100644 index 00000000..bd75f1dd --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch12.ipynb @@ -0,0 +1,256 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-12 : Large Signal Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 414 Example 12.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL'' = (N1/N2)**2 * RL =1.60 kohm\n" + ] + } + ], + "source": [ + "RL=16*10**2 #in ohm\n", + "x1=RL*10**-3 # in k-ohm\n", + "print \"RL'' = (N1/N2)**2 * RL =%0.2f kohm\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 414 Example 12.2." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(N1/N2)**2 = RL''/RL =900.00\n", + "N1/N2 =30.00\n", + "Hence, N1 : N2 = 30 : 1\n" + ] + } + ], + "source": [ + "x1=7200./8\n", + "print \"(N1/N2)**2 = RL''/RL =%0.2f\"%x1\n", + "x2=x1**0.5\n", + "print \"N1/N2 =%0.2f\"%x2\n", + "print \"Hence, N1 : N2 = 30 : 1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 415 Example 12.3." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Series-fed load\n", + "Overall efficiency, eta = 25(Vmax-Vmin / Vmax) =23.33 %\n", + "(ii) Transformer-coupled load\n", + "Overall efficiency, eta = 50*(Vmax-Vmin / Vmax+Vmin) =43.75 %\n" + ] + } + ], + "source": [ + "print \"(i) Series-fed load\"\n", + "eta=(25.*14)/15. #in percentage\n", + "print \"Overall efficiency, eta = 25(Vmax-Vmin / Vmax) =%0.2f %%\"%eta\n", + "print \"(ii) Transformer-coupled load\"\n", + "eta=50.*(14./16) #in percentage\n", + "print \"Overall efficiency, eta = 50*(Vmax-Vmin / Vmax+Vmin) =%0.2f %%\"%eta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 415 Example 12.4." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector circuity efficiency,\n", + " eta = (pi/4)*(1-(VCE/VCC))*100 =68.07 %\n" + ] + } + ], + "source": [ + "from math import pi\n", + "VCE=2.\n", + "VCC=15.\n", + "eta=(pi/4.)*(1-(VCE/VCC))*100.\n", + "print \"Collector circuity efficiency,\"\n", + "print \" eta = (pi/4)*(1-(VCE/VCC))*100 =%0.2f %%\"%eta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 416 Example 12.5." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that, TJ = TA + theta*PD\n", + "Therefore, TJ = 27 degree C + (8 degree C/W)*3W =51.00 degree C\n" + ] + } + ], + "source": [ + "theta=8.\n", + "TA=27.\n", + "PD=3.\n", + "TJ=TA+(theta*PD)\n", + "print \"We know that, TJ = TA + theta*PD\"\n", + "print \"Therefore, TJ = 27 degree C + (8 degree C/W)*3W =%0.2f degree C\"%TJ" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 417 Example 12.6." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "PD = (TJ-TA)/thetaJ-A = (160-40)/80 =1.50 W\n" + ] + } + ], + "source": [ + "TJ=160.\n", + "TA=40.\n", + "theta=80.\n", + "PD=(TJ-TA)/theta\n", + "print \"PD = (TJ-TA)/thetaJ-A = (160-40)/80 =%0.2f W\"%PD" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 418 Example 12.7." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " theta_J-A = theta_J-C + theta_C-A || theta_HS-A =12.31 degree C/w\n", + " PD = TJ-TA / theta_J-A =9.75 W\n" + ] + } + ], + "source": [ + "thetaH=8.\n", + "TA=40.\n", + "TJ=160.\n", + "thetaJ=5.\n", + "thetaC=85.\n", + "x1=(thetaC*thetaH)/(thetaC+thetaH)\n", + "theta=thetaJ+x1\n", + "print \" theta_J-A = theta_J-C + theta_C-A || theta_HS-A =%0.2f degree C/w\"%theta\n", + "PD=(TJ-TA)/theta\n", + "print \" PD = TJ-TA / theta_J-A =%0.2f W\"%PD" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch14.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch14.ipynb new file mode 100644 index 00000000..8f68f296 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch14.ipynb @@ -0,0 +1,374 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-14 : Feedback Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 452 Example 14.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change in gain of the amplifier with feedback is\n", + " dAf/Af = dA/A * 1/(1+A*beta) = 0.24 %\n" + ] + } + ], + "source": [ + "A=1000.\n", + "beta=0.04\n", + "dA=10.\n", + "print \"The percentage change in gain of the amplifier with feedback is\"\n", + "dAf=dA*(1/(1+(A*beta)))\n", + "print \" dAf/Af = dA/A * 1/(1+A*beta) = %0.2f %%\"%dAf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 452 Example 14.2." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(1 + A*beta) =10.00\n", + "Also, the gain with feedback is\n", + " Af = A / (1+A*beta)\n", + "Therefore, A =1000.00\n", + " 1 + A*beta = 10# i.e. A*beta = 9\n", + "Therefore, beta =0.01\n" + ] + } + ], + "source": [ + "Af=100.\n", + "dAf=0.02\n", + "dA=0.2\n", + "Ab=dA/dAf\n", + "print \"(1 + A*beta) =%0.2f\"%Ab\n", + "print \"Also, the gain with feedback is\"\n", + "print \" Af = A / (1+A*beta)\"\n", + "A=Af*Ab\n", + "print \"Therefore, A =%0.2f\"%A\n", + "print \" 1 + A*beta = 10# i.e. A*beta = 9\"\n", + "beta=9./A\n", + "print \"Therefore, beta =%0.2f\"%beta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 455 Example 14.3." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) We have BWf = (1 + A*beta) * BW\n", + " BWf =1.50 MHz\n", + "Gain with feedback, Af = A / (1+ A*beta) =20.83\n", + "(b) BWf = (1 + A*beta'') * BW\n", + "1*10**6 = (1 + 125*beta'')*250*10**3\n", + "Therefore, beta =0.02\n", + "i.e. beta (in %) =2.40\n" + ] + } + ], + "source": [ + "A=125.\n", + "BW=250*10**3\n", + "beta=0.04\n", + "print \"(a) We have BWf = (1 + A*beta) * BW\"\n", + "BWf = (1 + (A*beta))*BW\n", + "x1=BWf*10**-6\n", + "print \" BWf =%0.2f MHz\"%x1\n", + "Af=A/(1+(A*beta))\n", + "print \"Gain with feedback, Af = A / (1+ A*beta) =%0.2f\"%Af\n", + "print \"(b) BWf = (1 + A*beta'') * BW\"\n", + "print \"1*10**6 = (1 + 125*beta'')*250*10**3\"\n", + "Bd=3./125\n", + "print \"Therefore, beta =%0.2f\"%Bd\n", + "Bd1=Bd*100\n", + "print \"i.e. beta (in %%) =%0.2f\"%Bd1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 456 Example 14.4." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain with feedback\n", + " Af = A / (1 + A*beta) =80.00\n", + "New lower 3dB frequency,\n", + " f_1f = f1 / 1+A*beta =10.00 Hz\n", + "New upper 3dB frequency,\n", + " f2f = (1+A*beta)*f2 =1.00 MHz\n", + "Distortion with feedback,\n", + " Df = D / 1+A*beta =2.00 %\n" + ] + } + ], + "source": [ + "A=400.\n", + "f1=50.\n", + "f2=200*10**3\n", + "D=10.\n", + "beta=0.01\n", + "print \"The voltage gain with feedback\"\n", + "Af=A/(1+(A*beta))\n", + "print \" Af = A / (1 + A*beta) =%0.2f\"%Af\n", + "print \"New lower 3dB frequency,\"\n", + "f1f=f1/(1+(A*beta))\n", + "print \" f_1f = f1 / 1+A*beta =%0.2f Hz\"%f1f\n", + "print \"New upper 3dB frequency,\"\n", + "f2f=(1+(A*beta))*f2\n", + "x2=f2f*10**-6\n", + "print \" f2f = (1+A*beta)*f2 =%0.2f MHz\"%x2\n", + "print \"Distortion with feedback,\"\n", + "Df=D/(1+(A*beta))\n", + "print \" Df = D / 1+A*beta =%0.2f %%\"%Df" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 458 Example 14.5" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain, Af = A / (1+A*beta) =83.33\n", + "Input resistance, Rif = (1+(A*beta))*Ri =18.00 kohm\n", + "Output resistance, Rof = Ro / (1+A*beta) =3.33 kohm\n" + ] + } + ], + "source": [ + "A=500.\n", + "Ri=3*10**3\n", + "Ro=20*10**3\n", + "beta=0.01\n", + "Af=A/(1+(A*beta))\n", + "print \"Voltage gain, Af = A / (1+A*beta) =%0.2f\"%Af\n", + "Rif=(1+(A*beta))*Ri\n", + "x1=Rif*10**-3\n", + "print \"Input resistance, Rif = (1+(A*beta))*Ri =%0.2f kohm\"%x1\n", + "Rof=Ro/(1+(A*beta))\n", + "x2=Rof*10**-3\n", + "print \"Output resistance, Rof = Ro / (1+A*beta) =%0.2f kohm\"%x2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 460 Example 14.6." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Ai = 1 + hfe =81.00\n", + " Ri = hie + (1+hfe)*RL =167.00 kohm\n", + " Av = Ai*RL / Ri =0.00\n", + " Ro = hie+Rs / 1+hfe =69.14 ohm\n", + " Rof = Ro || RL =66.82 ohm\n" + ] + } + ], + "source": [ + "Ai=1+80\n", + "print \" Ai = 1 + hfe =%0.2f\"%Ai\n", + "Ri=(5*10**3)+((1+80)*(2*10**3)) #in ohm\n", + "x1=Ri*10**-3 #in k-ohm\n", + "print \" Ri = hie + (1+hfe)*RL =%0.2f kohm\"%x1\n", + "Av=(81*2*10**3)/(167*10**3)\n", + "print \" Av = Ai*RL / Ri =%0.2f\"%Av\n", + "Ro=(5000.+600)/(1.+80) # in ohm\n", + "print \" Ro = hie+Rs / 1+hfe =%0.2f ohm\"%Ro\n", + "Rof=(69.13*2000)/(2069.13) #in ohm\n", + "print \" Rof = Ro || RL =%0.2f ohm\"%Rof" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 461 Example 14.7." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " R''L = RB || RL =1000.00 ohm\n", + " Av = -hfe*R''L / hie =-30.48\n", + " Rif = hie || (RB / 1-Av) =1013.17 ohm\n", + " Avf = Vo/Vs = Av*Rif / RS+Rif =-19.14\n", + " Rof = (RB / RS) * (RS+hie / hfe) =4.67 kohm\n", + " R''of = Rof || RL =1.40 kohm\n" + ] + } + ], + "source": [ + "RL=((40*2)/42)*10**3 #in ohm\n", + "print \" R''L = RB || RL =%0.2f ohm\"%RL\n", + "Av=(-80*1905.)/5000.\n", + "print \" Av = -hfe*R''L / hie =%0.2f\"%Av\n", + "x1=(40000.)/(1+30.48)\n", + "Rif=(x1*5000.)/(x1+5000) #in ohm\n", + "print \" Rif = hie || (RB / 1-Av) =%0.2f ohm\"%Rif\n", + "Avf=(-30.48*1013.172)/(600+1013.172)\n", + "print \" Avf = Vo/Vs = Av*Rif / RS+Rif =%0.2f\"%Avf\n", + "Rof=(40000/600.)*(5600./80) #in ohm\n", + "x2=Rof*10**-3 #in k-ohm\n", + "print \" Rof = (RB / RS) * (RS+hie / hfe) =%0.2f kohm\"%x2\n", + "Roff=(4.666*2)/(6.666) #in k-ohm\n", + "print \" R''of = Rof || RL =%0.2f kohm\"%Roff," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 462 Example 14.8." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) A = -hfe*RL / hie =-40.00\n", + " Ri = hie = 2 k-ohm\n", + "(b) beta = Re / RL =0.10\n", + "(c) Rif = hie + (1+hfe)*Re =10.10 kohm\n", + "(d) Af = -hfe*RL / Rif =-7.92\n", + "(e) Loop gain, beta = -40*0.1 = -4 i.e. 20log4 =12.04 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "R1=20.*10**3\n", + "R2=20.*10**3\n", + "hie=2.*10**3\n", + "RL=1.0*10**3\n", + "Re=100.\n", + "hfe=80.\n", + "A=(-hfe*RL)/hie\n", + "print \"(a) A = -hfe*RL / hie =%0.2f\"%A\n", + "print \" Ri = hie = 2 k-ohm\"\n", + "beta=Re/RL\n", + "print \"(b) beta = Re / RL =%0.2f\"%beta\n", + "Rif=hie+((1+hfe)*Re)\n", + "x1=Rif*10**-3\n", + "print \"(c) Rif = hie + (1+hfe)*Re =%0.2f kohm\"%x1\n", + "Af=(-hfe*RL)/Rif\n", + "print \"(d) Af = -hfe*RL / Rif =%0.2f\"%Af\n", + "lg=20.*log10(4)\n", + "print \"(e) Loop gain, beta = -40*0.1 = -4 i.e. 20log4 =%0.2f dB\"%lg" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch15.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch15.ipynb new file mode 100644 index 00000000..3e5b1387 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch15.ipynb @@ -0,0 +1,462 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-15 : Oscillators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 475 Example 15.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L1 = (1 / 4*pi**2*fo**2*C) - L1 =0.04 mH\n" + ] + } + ], + "source": [ + "from math import pi\n", + "L1=(1./(4*(pi**2)*((120*10**3)**2)*0.004*10**-6))-(0.4*10**-3) #in henry\n", + "x1=L1*10**3 #in mH\n", + "print \"L1 = (1 / 4*pi**2*fo**2*C) - L1 =%0.2f mH\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 475 Example 15.2." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When fo = 950 kHz\n", + " C =13.89 pF\n", + "When fo = 2050 kHz\n", + " C = 2.98 pF\n", + "Hence, the range of capacitance is from 2.98 pF to 13.89 pF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "print \"When fo = 950 kHz\"\n", + "C=1./(4*(pi**2)*((2*10**-3)+(20*10**-6))*((950*10**3)**2)) #farady\n", + "x1=C*10**12 #pF\n", + "print \" C =%0.2f pF\"%x1\n", + "print \"When fo = 2050 kHz\"\n", + "C=1./(4*(pi**2)*((2*10**-3)+(20*10**-6))*((2050*10**3)**2)) #farady\n", + "x1=C*10**12 #pF\n", + "print \" C = %0.2f pF\"%x1\n", + "print \"Hence, the range of capacitance is from 2.98 pF to 13.89 pF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 476 Example 15.3" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fo = 1 / 2*pi*sqrt(50*10**-6*500*10**-12) =1.01 MHz\n", + "Feedback factor, beta = L1 / L2 =3.17\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "L1=38.*10**-6\n", + "L2=12.*10**-6\n", + "C=500.*10**-12\n", + "L=L1+L2\n", + "fo = 1. / (2*pi*sqrt(L*C))\n", + "x1=fo*10**-6\n", + "print \"fo = 1 / 2*pi*sqrt(50*10**-6*500*10**-12) =%0.2f MHz\"%x1\n", + "beta=L1/L2\n", + "print \"Feedback factor, beta = L1 / L2 =%0.2f\"% beta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 478 Example 15.4." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L = (C1+C2) / (4*pi**2*fo**2*C1*C2) =13.93 mH\n", + "The voltage gain required to produce oscillation is\n", + " Av > C1/C2 =10.00\n" + ] + } + ], + "source": [ + "from math import pi\n", + "C1=0.2*10**-6\n", + "C2=0.02*10**-6\n", + "fo=10.*10**3\n", + "L=(C1+C2)/(4*pi**2*fo**2*C1*C2)\n", + "x1=L*10**3\n", + "print \"L = (C1+C2) / (4*pi**2*fo**2*C1*C2) =%0.2f mH\"%x1\n", + "print \"The voltage gain required to produce oscillation is\"\n", + "x2=C1/C2\n", + "print \" Av > C1/C2 =%0.2f\"%x2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 480 Example 15.5." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) In a Colpitts oscillator, a series combination of C1 and C2 which is in parallel with inductance L and frequency of oscillations is\n", + " fo = 1 / 2pi*sqrt(LCeq) = 1 / 2pi*sqrt(L*C1*C2/C1+C2) =87.17 kHz\n", + "Vf = Vo*C1 / C2 =2.00 V\n", + " Gain = 500*10**-12 / 100*10**-12 =5.00\n", + "C1 = C2 / 10 =50.00 pF\n", + "(v) The frequncy of oscillation is\n", + "fo = 118.03 kHz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "L=40.*10**-3\n", + "C1=100.*10**-12\n", + "C2=500.*10**-12\n", + "Vo=10.\n", + "print \"(i) In a Colpitts oscillator, a series combination of C1 and C2 which is in parallel with inductance L and frequency of oscillations is\"\n", + "fo=1./ (2*pi*sqrt((L*C1*C2)/(C1+C2)))\n", + "x1=fo*10**-3\n", + "print \" fo = 1 / 2pi*sqrt(LCeq) = 1 / 2pi*sqrt(L*C1*C2/C1+C2) =%0.2f kHz\"%x1\n", + "Vf=(Vo*C1)/C2\n", + "print \"Vf = Vo*C1 / C2 =%0.2f V\"%Vf\n", + "gain=C2/C1\n", + "print \" Gain = 500*10**-12 / 100*10**-12 =%0.2f\"%gain\n", + "x2=C2/10\n", + "x3=x2*10**12\n", + "print \"C1 = C2 / 10 =%0.2f pF\"%x3\n", + "print \"(v) The frequncy of oscillation is\"\n", + "fo=1./ (2*pi*sqrt((40*50*500*10**-27)/((50*10**-12)+(500*10**-12))))\n", + "x4=fo*10**-3\n", + "print \"fo = %0.2f kHz\"% x4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 481 Example 15.6." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When fo = 400 kHz, Cmax(pF) =2638.57\n", + "When fo = 1200 kHz, Cmin = 293.17 pF\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "fo1=400.*10**3\n", + "fo2=1200.*10**3\n", + "Lp=60.*10**-6\n", + "C = 1 / (4*pi**2*fo1**2*Lp)\n", + "x1=C*10**12\n", + "print \"When fo = 400 kHz, Cmax(pF) =%0.2f\"%x1 # answer in textbook is wrong\n", + "C = 1. / (4*pi**2*fo2**2*Lp)\n", + "x2=C*10**12\n", + "print \"When fo = 1200 kHz, Cmin = %0.2f pF\"%x2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 482 Example 15.7." + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When fo = 540 kHz, Cmax = 86.87 pF\n", + "When fo = 1650 kHz, Cmin = 9.30 pF\n", + "Hence, the capacitor range required is 9.3-86.87 pF\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "fo1=540.*10**3\n", + "fo2=1650.*10**3\n", + "L=1*10.**-3\n", + "Cmax = 1. / (4*pi**2*fo1**2*L)\n", + "x1=Cmax*10**12\n", + "print \"When fo = 540 kHz, Cmax = %0.2f pF\"%x1\n", + "Cmin = 1. / (4*pi**2*fo2**2*L)\n", + "x2=Cmin*10**12\n", + "print \"When fo = 1650 kHz, Cmin = %0.2f pF\"%x2\n", + "print \"Hence, the capacitor range required is 9.3-86.87 pF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 483 Example 15.8." + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " fo =3.25 kHz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "fo=1./(2*pi*(200*10**3)*(100*10**-12)*sqrt(6)) #in Hz\n", + "x1=fo*10**-3 #in kHz\n", + "print \" fo =%0.2f kHz\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 486 Example 15.9." + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fo = 606.69 Hz\n", + "beta =155.70\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "fo=1./(2*3.142*10000*(0.01*10**-6)*sqrt(6+((4*2.2*10**3)/(10000)))) #in Hz\n", + "print \"fo = %0.2f Hz\"%fo\n", + "beta=23.+(29.*(10/2.2))+(4*(2.2/10))\n", + "print \"beta =%0.2f\"%beta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 487 Example 15.10." + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " C=0.42 nF\n", + " hfe >=50.68\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "fo=1./(2*pi*(10*10**3)*(7.1*10**3)*sqrt(6+((4*40*10**3)/(7.1*10**3)))) # in Farady\n", + "x1=fo*10**9 # in nF\n", + "print \" C=%0.2f nF\"%x1\n", + "h=23.+(29.*(7.1/40))+(4*(40/7.1))\n", + "print \" hfe >=%0.2f\"% h" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 488 Example 15.11." + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore, C = 1 / 2*pi*R*fo =159.15 pF\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "C=1./(2*pi*100000*10000) # in farady\n", + "x1=C*10**12 #in pF\n", + "print \"Therefore, C = 1 / 2*pi*R*fo =%0.2f pF\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 489 Example 15.12." + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The series resonant frequencies of the crystal is\n", + " fs = 1 / 2*pi*sqrt(L*Cs) = 918.88 kHz\n", + "Q factor of the crystal at fs = omegaS*L / R = 2*pi*fs*L / R =577.36\n", + "(b) The parallel resonant frequency of the crystal is\n", + " fp = 1/2pi * sqrt((Cs+Cp)/(L*Cs*Cp)) =946.05 kHz\n", + "Q factor of the crystal at fp = omegaS*L / R = 2*pi*fs*L / R =594.39\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "print \"(a) The series resonant frequencies of the crystal is\"\n", + "fs=1./(2*pi*sqrt(0.5*0.06*10**-12)) #in Hz\n", + "x1=fs*10**-3 #in kHz\n", + "print \" fs = 1 / 2*pi*sqrt(L*Cs) = %0.2f kHz\"%x1\n", + "fs=(2.*pi*(918.9*10**3)*0.5)/(5*10**3)\n", + "print \"Q factor of the crystal at fs = omegaS*L / R = 2*pi*fs*L / R =%0.2f\"%fs\n", + "print \"(b) The parallel resonant frequency of the crystal is\"\n", + "fp=(1./(2*pi))*sqrt((1.06*10**-12)/(0.5*(0.06*10**-12)*(1*10**-12))) # in Hz\n", + "x1=fp*10**-3\n", + "print \" fp = 1/2pi * sqrt((Cs+Cp)/(L*Cs*Cp)) =%0.2f kHz\"%x1\n", + "fp=(2.*pi*(946.*10**3)*0.5)/(5.*10**3)\n", + "print \"Q factor of the crystal at fp = omegaS*L / R = 2*pi*fs*L / R =%0.2f\"%fp" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch16.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch16.ipynb new file mode 100644 index 00000000..e4790b69 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch16.ipynb @@ -0,0 +1,766 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-16 : Wave Shaping and Multivibrator Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 512 Example 16.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given tr = 35 ns\n", + "We know that, tr = 0.35 / BW\n", + "Therefore, BW = 0.35 / tr =10.00 MHz\n" + ] + } + ], + "source": [ + "print \"Given tr = 35 ns\"\n", + "bw=0.35/(35*10**-9) # in Hz\n", + "x1=bw*10**-6 #in MHz\n", + "print \"We know that, tr = 0.35 / BW\"\n", + "print \"Therefore, BW = 0.35 / tr =%0.2f MHz\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 514 Example 16.2." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given ton = 70 ns\n", + " C = ton / 0.1*Rs =1166.67 pF\n", + " tre = 2.3*RB*C =15.46 u-seconds\n", + " f = 1/2T = 1/2tre =33.33 kHz\n" + ] + } + ], + "source": [ + "print \"Given ton = 70 ns\"\n", + "C=(70*10**-9)/(0.1*600) # in faraday\n", + "x1=C*10**12 # in pF\n", + "print \" C = ton / 0.1*Rs =%0.2f pF\"%x1 # approximately 1200 pF\n", + "tre=2.3*(5.6*10**3)*(1200*10**-12) # in seconds\n", + "x2=tre*10**6 #in us\n", + "print \" tre = 2.3*RB*C =%0.2f u-seconds\"%x2\n", + "f=1./(2*(15*10**-6)) #in Hz\n", + "x3=f*10**-3 #in kHz\n", + "print \" f = 1/2T = 1/2tre =%0.2f kHz\"%x3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 514 Example 16.3." + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "transition voltage: 3.00\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f0239952290>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "from numpy import arange,sin,pi\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,subplot,title,xlabel,show,ylabel\n", + "amp = 15.\n", + "vi_t=3.# # transition voltage\n", + "t=arange(0,2*pi+0.1,0.1)\n", + "vi=[]\n", + "for x in t:\n", + " vi.append(amp*sin(x))\n", + "vo=[]\n", + "for x in vi:\n", + " vo.append(x+3)# # output voltage\n", + "print 'transition voltage: %0.2f'%vi_t\n", + "for i in range(0,len(t)):\n", + " if(vo[(i)]<=0):\n", + " vo[(i)]=0\n", + "\n", + "subplot(2,1,1)\n", + "plot(t,vo,2,'011','',[0,0,7,18])\n", + "title('Ouptut voltage in sin wave')\n", + "xlabel('t')\n", + "ylabel('vo')\n", + "show()\n", + "\n", + "\n", + "t=arange(0,20+0.1,0.1)\n", + "vo=[]\n", + "for i in range(0,int(len(t)/2)):\n", + " vo.append(15+3)\n", + "\n", + "for i in range(int(len(t)/2-1),len(t)-1):\n", + " vo.append(0)\n", + "subplot(3,1,2)\n", + "\n", + "plot(t,vo,2,'011','',[0,-5,21,20])#\n", + "title('Ouptut voltage in square wave')\n", + "xlabel('t')\n", + "ylabel('vo')#\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 515 Example 16.4." + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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wF7An8AJwvLu/FarGLKnw87wA6CQaCnJgNXBqyXi2DMDMPgMsBpazecjnfKJd\nGKr+jKYuBEREpH3SNhwkIiJtpBAQESkwhYCISIEpBERECkwhICJSYAoBEZECUwiIVMnMOszsf4Su\nQ6SZFAIi1dsB+J+hixBpJoWASPUuAcbHFz+5NHQxIs2gFcMiVTKzvYBfufsnQ9ci0izqBESqp4uf\nSO4oBERECkwhIFK9vwDbhS5CpJkUAiJVcvce4P+Z2QpNDEteaGJYRKTA1AmIiBSYQkBEpMAUAiIi\nBaYQEBEpMIWAiEiBKQRERApMISAiUmAKARGRAvv/vixeuz53rIgAAAAASUVORK5CYII=\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f0239638290>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "from numpy import arange\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,subplot,title,xlabel,show,ylabel\n", + "\n", + "t= arange(0,20+0.1,0.1)\n", + "x=[]\n", + "for i in range(0,len(t)):\n", + " if(t[(i)]<=5):\n", + " x.append((15.0/5)*t[(i)])\n", + " elif(t[(i)]>=5 and t[(i)]<=15):\n", + " x.append(-3.2*t[(i)]+30)\n", + " elif(t[(i)]>=15 and t[(i)]<=20):\n", + " x.append((15./5)*t[(i)]-60)\n", + "y=[]\n", + "for i in range(0,len(t)):\n", + " if(x[(i)]>3):\n", + " y.append(x[(i)])\n", + " elif(x[i]<=3):\n", + " y.append(3)\n", + "plot(t,y,2,'011','',[0,0,20,16])#\n", + "\n", + "title('output voltage')\n", + "xlabel('t')\n", + "ylabel('Vo')\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 516 Example 16.5." + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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F4UsoblGAp1/zidlPcPTCUaZ3nB7yV5RUKjM5cv4Itb+uzYdNPgzZ/Y3huEUR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\nPe7uQYEcBeyOphxEC4VSeLq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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa57b0ed350>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max output voltage is 5V\n", + "min output voltage is -3V\n" + ] + } + ], + "source": [ + "from numpy import arange\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,subplot,title,xlabel,show,ylabel,legend\n", + "\n", + "\n", + "#let input wave be V_in=V_p_in*sin(2*pi*f*t) \n", + "f=1.# #Frequency is 1Hz\n", + "T=1./f#\n", + "V_p_in=10# #Peak input voltage\n", + "V_th=0.7# #knee voltage of diode\n", + "#let n be double the number of cycles of output shown in graph\n", + "for n in range(0,2):\n", + " t=arange(T*n/2.,T*(n+1)/2.+0.0005,0.0005) #time for each half cycle\n", + " V_in=[]\n", + " for tt in t:\n", + " V_in.append(V_p_in*sin(2*pi*f*tt))\n", + " Vout=V_in#\n", + " if (n%2)==0: #positive half,D1 conducts till V_in=5V\n", + " a=(Vout<5)# \n", + " b=(Vout>5)# \n", + " y=[]\n", + " for vv in Vout:\n", + " y.append(a*vv+5*b)# #output follows input till 5V then is constant at 5V\n", + " else: #negative half, D2 conducts till V_in=-3V\n", + " a=(Vout<-3)# \n", + " b=(Vout>-3)#\n", + " for vv in Vout:\n", + " y.append(-3*a+b*vv) #output follows input till -3V then stays constant at -3V\n", + " \n", + " plot(t,y[0:1001])\n", + " plot(t,V_in)\n", + "legend(['output','input'])\n", + "title('Positive and Negative diode limiter')\n", + "xlabel('t')\n", + "ylabel('Vo')\n", + "show()\n", + "print 'max output voltage is 5V'\n", + "print 'min output voltage is -3V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 518 Example 16.8." + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V_DC = 10.00 V\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f0239c14410>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "from numpy import arange,pi,sin\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,subplot,title,xlabel,show,ylabel\n", + "#Positive Clamping circuit\n", + "#let input voltage be V_in=V_p_in*sin(2*pi*f*t)\n", + "V_p_in=10.#\n", + "V_DC=(V_p_in)# #DC level added to output\n", + "print 'V_DC = %0.2f V'%V_DC\n", + "for n in range(0,2):\n", + " t=arange(n/2,(n+1)/2+0.0005,0.0005)\n", + " V_in=[]\n", + " for tt in t:\n", + " V_in.append(V_p_in*sin(2*pi*tt))\n", + " Vout=[]\n", + " for vv in V_in:\n", + " Vout.append(V_DC+vv)\n", + " plot(t,Vout)\n", + "\n", + "title('Positive clipper graph')\n", + "xlabel('t')\n", + "ylabel('Vo')\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 519 Example 16.9." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V_DC = -12.00 V\n" + ] + }, + { + "data": { + "image/png": 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YWMnL/Gn0aFi8GBYu9DuSxDUnew5tJrWh90m9md1ztiUTY7AeSsJatAgGDHA3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\n+oD3fmlgCnAasA+4RVUzg+z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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa57b361650>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "from numpy import arange,pi,sin\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,subplot,title,xlabel,show,ylabel\n", + "#Negative Clamping circuit\n", + "#let input voltage be V_in=V_p_in*sin(2*pi*f*t)\n", + "V_p_in=12#\n", + "V_DC=-(V_p_in)# #DC level added to output\n", + "print 'V_DC = %0.2f V'%V_DC\n", + "#t=[]\n", + "for n in range(0,2):\n", + " t=(n/2.,(n+1)/2.+0.0005,0.0005)\n", + " V_in=[]\n", + " for tt in t:\n", + " V_in.append(V_p_in*sin(2*pi*tt))\n", + " Vout=[]\n", + " for vv in V_in:\n", + " Vout.append(V_DC+vv)\n", + " plot(t,Vout)\n", + "\n", + "title('Negative clipper graph')\n", + "xlabel('t')\n", + "ylabel('Vo')\n", + "show()\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 520 Example 16.10." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of a symmetrical astable multivibrator is\n", + " f = 1/1.386RC =36.08 kHz\n" + ] + } + ], + "source": [ + "f=1./(1.386*(20*10**3)*(1000*10**-12)) #in Hz\n", + "x1=f*10**-3 # in kHz\n", + "print \"The frequency of a symmetrical astable multivibrator is\"\n", + "print \" f = 1/1.386RC =%0.2f kHz\"%x1 # answer in textbook is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 521 Example 16.11." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The period of oscillation for an asymmetrical astable multivibrator is,\n", + " T = 0.693(R1C1+R2C2) = 360.36 us\n", + "Therefore, the frequency of oscillation, f = 1/T =2.78 kHz\n" + ] + } + ], + "source": [ + "print \"The period of oscillation for an asymmetrical astable multivibrator is,\"\n", + "t=0.693*(((2*10**3)*0.01*10**-6)+((10*10**3)*(0.05*10**-6))) # seconds\n", + "x1=t*10**6 # in us\n", + "print \" T = 0.693(R1C1+R2C2) = %0.2f us\"%x1\n", + "f=1./(360.36*10**-6) # in Hz\n", + "x2=f*10**-3 # in kHz\n", + "print \"Therefore, the frequency of oscillation, f = 1/T =%0.2f kHz\"%x2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 522 Example 16.12." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The period of oscillation is, T = 1/f = 10.00 us\n", + " T1 = 2us (given)\n", + "Hence, T2 = T - T1 =8.00 us\n", + " T1 = 0.693*R1C1\n", + "Therefore, C1 = T1 / 0.693R1 =144.30 pF\n", + " T2 = 0.693*R2*C2\n", + "Therefore, C2 = T2 / 0.693R2 =577.20 pF\n" + ] + } + ], + "source": [ + "t=1./(100*10**3) # in seconds\n", + "x1=t*10**6 # in us\n", + "print \"The period of oscillation is, T = 1/f = %0.2f us\"%x1\n", + "print \" T1 = 2us (given)\"\n", + "t2=10-2 # in us\n", + "print \"Hence, T2 = T - T1 =%0.2f us\"%t2\n", + "print \" T1 = 0.693*R1C1\"\n", + "c1=(2*10**-6)/(0.693*(20*10**3)) # in faraday\n", + "x1=c1*10**12 # in pF\n", + "print \"Therefore, C1 = T1 / 0.693R1 =%0.2f pF\"%x1 #answer in textbook is wrong\n", + "c2=(8*10**-6)/(0.693*(20*10**3)) # in faraday\n", + "x1=c2*10**12 # in pF\n", + "print \" T2 = 0.693*R2*C2\" #answer in textbook is wrong\n", + "print \"Therefore, C2 = T2 / 0.693R2 =%0.2f pF\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 523 Example 16.13." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " RC = 12-0.2/1*10**-3 = 11.80 kohm\n", + " R <= hfe*RC\n", + " R <=1.18 Mohm\n", + "Hence, let us assume that R = R1 = R2 = 1 M-ohm\n", + " Toff = 0.693*R*C1\n", + "Therefore, C1 =28.86 pF\n", + " Ton = 0.693*R*C2\n", + "Therefore, C2 = 14.43 pF\n" + ] + } + ], + "source": [ + "rc=(12-0.2)/(1*10**-3) # in ohm\n", + "x1=rc*10**-3 # in k-ohm\n", + "print \" RC = 12-0.2/1*10**-3 = %0.2f kohm\"%x1\n", + "r=100.*11.8*10**3 # in ohm\n", + "x1=r*10**-6 # in M-ohm\n", + "print \" R <= hfe*RC\"\n", + "print \" R <=%0.2f Mohm\"%x1\n", + "print \"Hence, let us assume that R = R1 = R2 = 1 M-ohm\"\n", + "print \" Toff = 0.693*R*C1\"\n", + "c1=(20*10**-6)/(0.693*10**6) # in faraday\n", + "x1=c1*10**12 # in pF\n", + "print \"Therefore, C1 =%0.2f pF\"%x1\n", + "print \" Ton = 0.693*R*C2\"\n", + "c1=(10*10**-6)/(0.693*10**6) # in faraday\n", + "x1=c1*10**12 # in pF\n", + "print \"Therefore, C2 = %0.2f pF\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 523 Example 16.14." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At stable state, Q2 is ON and Q2 is OFF:\n", + " RC2(ohm) = RC1(ohm) = VCC-VCE(sat) / IC(sat) =950.00 ohm\n", + "IB2(sat) = IC(sat) / hfe(min) =0.30 mA\n", + "Also, IB1(sat) = 0.3 mA\n", + " R = VCC-VBE(sat) / IB2(sat) = 17.67 kohm\n", + " [because, VBE(sat) = 0.7 V for Si transistor]\n", + "At quasi-stable state, Q1 is ON and Q2 is OFF\n", + " T = 0.693*R*C\n", + "Therefore, C= T / 0.693*R =0.01 uF\n", + "Assume, IB1(sat) = IR2\n", + "Therefore, IR1 = IB1(sat)+IR2 =0.60 mA\n", + " VCC = VBE(sat) + IR1(RC2+R1)\n", + "Therefore, R1 = (VCC-VBE(sat) / IR1) - RC2 =7.88 kohm\n", + " R2 = VBE(sat)-(-VBB) / IR2 =7.33 kohm\n", + "The speed up capacitor C1 is chosen such that R1C1 = 1 us and hence,\n", + " C1 = 127.67 pF\n" + ] + } + ], + "source": [ + "print \"At stable state, Q2 is ON and Q2 is OFF:\"\n", + "rc2=(6.-0.3)/(6*10**-3) # in ohm\n", + "print \" RC2(ohm) = RC1(ohm) = VCC-VCE(sat) / IC(sat) =%0.2f ohm\"%rc2\n", + "ib2=(6.*10**-3)/20 # in ampere\n", + "x1=ib2*10**3 # in mA\n", + "print \"IB2(sat) = IC(sat) / hfe(min) =%0.2f mA\"%x1\n", + "print \"Also, IB1(sat) = 0.3 mA\"\n", + "r=(6-0.7)/(0.3*10**-3) # in ohm\n", + "x1=r*10**-3 # in k-ohm\n", + "print \" R = VCC-VBE(sat) / IB2(sat) = %0.2f kohm\"%x1\n", + "print \" [because, VBE(sat) = 0.7 V for Si transistor]\"\n", + "print \"At quasi-stable state, Q1 is ON and Q2 is OFF\"\n", + "print \" T = 0.693*R*C\"\n", + "c=(140.*10**-6)/(0.693*17.67*10**3) # in F\n", + "x1=c*10**6 # in uF\n", + "print \"Therefore, C= T / 0.693*R =%0.2f uF\"%x1\n", + "print \"Assume, IB1(sat) = IR2\"\n", + "ir2=0.3+0.3 # in mA\n", + "print \"Therefore, IR1 = IB1(sat)+IR2 =%0.2f mA\"%ir2\n", + "r1=((6-0.7)/(0.6*10**-3))-950 # in ohm\n", + "x1=r1*10**-3 # in k-ohm\n", + "print \" VCC = VBE(sat) + IR1(RC2+R1)\"\n", + "print \"Therefore, R1 = (VCC-VBE(sat) / IR1) - RC2 =%0.2f kohm\"%x1\n", + "r2=(0.7+1.5)/(0.3*10**-3) # in ohm\n", + "x1=r2*10**-3 # in k-ohm\n", + "print \" R2 = VBE(sat)-(-VBB) / IR2 =%0.2f kohm\"%x1\n", + "print \"The speed up capacitor C1 is chosen such that R1C1 = 1 us and hence,\"\n", + "c1=(1.0*10**-6)/(7.833*10**3) # in F\n", + "x1=c1*10**12 # in pF\n", + "print \" C1 = %0.2f pF\" %x1 # answer in textbook is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 524 Example 16.15." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " VB1 = -VBB*R2 / R2+R3 = -1.57 V\n", + " IC2 = [VCC-VC2(sat) / RC2] - [VC2(sat)-(-VBB) / R2+R3] =5.21 mA\n", + " IB2 > IC2 / hfe(min) > 0.27\n", + "Therefore, I6 = 0.13 mA\n", + " I3 =0.63 mA\n", + " VC1 = 10.62 V\n" + ] + } + ], + "source": [ + "vb1=(-12.*15*10**3)/(115.*10**3) # in volts\n", + "print \" VB1 = -VBB*R2 / R2+R3 = %0.2f V\"%vb1\n", + "ic2=((12-0.3)/(2.2*10**3))-((0.3+12)/(115*10**3)) # in A\n", + "x1=ic2*10**3 # in mA (Since Q2 is ON VC2(sat) = 0.3 V)\n", + "print \" IC2 = [VCC-VC2(sat) / RC2] - [VC2(sat)-(-VBB) / R2+R3] =%0.2f mA\"%x1 # answer in textbook is wrong\n", + "ib2=(5.35*10**-3)/20 # in A\n", + "x1=ib2*10**3 # in mA\n", + "print \" IB2 > IC2 / hfe(min) > %0.2f\"%x1 # approximately 0.5 mA\n", + "i6=(0.7+12)/(100) # in mA\n", + "print \"Therefore, I6 = %0.2f mA\"%i6\n", + "i3=0.5+0.127 # in mA\n", + "print \" I3 =%0.2f mA\"%i3\n", + "vc1=12-((0.627*10**-3)*(2.2*10**3))\n", + "print \" VC1 = %0.2f V\"%vc1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 525 Example 16.16." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage across RE is VE = VB2 - VBE =4.30 V\n", + " RE = VE / IE =2.15 kohm\n", + " IC*RC2 = VCC - VE - VCE(sat) = 7.50 V\n", + " RC2 =3.75 kohm\n", + " I2 = 0.1*IC2 =0.20 mA\n", + " R2 = VB2 / I2 =25.00 kohm\n", + " IB2 = IC2 / hfe(min) = 0.00 uA\n", + "RC1 + R1 =31.82\n", + " I1 = VB2 / R2 =0.12 mA\n", + " IC1 = IE = VB1-VBE / RE =1.07 mA\n", + "Therefore, RC1 =4.84 kohm\n", + " R1 = 26.96 kohm\n", + " RB < hfe*RE\n", + " RB = hfe*RE / 10 =21.50 kohm\n" + ] + } + ], + "source": [ + "ve=5-0.7 # in volts\n", + "print \"Voltage across RE is VE = VB2 - VBE =%0.2f V\"%ve\n", + "re=4.3/2 # in k-ohm\n", + "print \" RE = VE / IE =%0.2f kohm\"%re\n", + "x=12-4.3-0.2 # in volts\n", + "print \" IC*RC2 = VCC - VE - VCE(sat) = %0.2f V\"%x\n", + "rc2=7.5/(2) # in k-ohm\n", + "print \" RC2 =%0.2f kohm\"%rc2\n", + "i2=0.1*2 # in mA\n", + "print \" I2 = 0.1*IC2 =%0.2f mA\"%i2\n", + "r2=5/0.2 # in k-ohm\n", + "print \" R2 = VB2 / I2 =%0.2f kohm\"%r2\n", + "ib2=(210**-3)/100 # in A\n", + "x1=ib2*10**6 # in uA\n", + "print \" IB2 = IC2 / hfe(min) = %0.2f uA\"%x1\n", + "x=7/(0.22) # in k-ohm\n", + "print \"RC1 + R1 =%0.2f\"%x\n", + "i1=3./25 # in mA\n", + "print \" I1 = VB2 / R2 =%0.2f mA\"%i1\n", + "ic1=(3-0.7)/2.15 # in mA\n", + "print \" IC1 = IE = VB1-VBE / RE =%0.2f mA\"%ic1\n", + "rc1=(12-((0.12*10**-3)*(56.8*10**3)))/(1.07*10**-3) # in ohm\n", + "x1=rc1*10**-3 # in k-ohm\n", + "print \"Therefore, RC1 =%0.2f kohm\"%x1\n", + "r1=31.8-4.84\n", + "print \" R1 = %0.2f kohm\"%r1\n", + "rb=(100*2.15)/10\n", + "print \" RB < hfe*RE\"\n", + "print \" RB = hfe*RE / 10 =%0.2f kohm\"%rb" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch17.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch17.ipynb new file mode 100644 index 00000000..6ff93046 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch17.ipynb @@ -0,0 +1,73 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-17 : Blocking Oscillators and Time Based Generators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 536 Example 17.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RE < VBB-VP/IP, i.e. RE(k-ohm) < 20-2.9/1.6*10**-3 =10.69 kohm\n", + "RE > VBB-VV/IV, i.e. RE(k-ohm) < 20-1.118/3.5*10**-3 =5.39 kohm\n", + "Therefore, CE = 0.24 uF\n", + "Therefore, R1 = VR1/IE =20.00 ohm\n", + " R2(ohm) = 11*10**3/250 =44.00 ohm\n" + ] + } + ], + "source": [ + "RE=(20-2.9)/(1.6) # in k-ohm\n", + "print \"RE < VBB-VP/IP, i.e. RE(k-ohm) < 20-2.9/1.6*10**-3 =%0.2f kohm\"%RE\n", + "RE=(20-1.118)/(3.5) # in k-ohm\n", + "print \"RE > VBB-VV/IV, i.e. RE(k-ohm) < 20-1.118/3.5*10**-3 =%0.2f kohm\"%RE\n", + "# answer in textbook is wrong\n", + "CE=1./(500*(2.303*10**4)*0.36) # in farady\n", + "x1=CE*10**6 # in uF\n", + "print \"Therefore, CE = %0.2f uF\"%x1\n", + "R1=5./(250*10**-3) #in ohm\n", + "print \"Therefore, R1 = VR1/IE =%0.2f ohm\"%R1\n", + "R2=11000./250\n", + "print \" R2(ohm) = 11*10**3/250 =%0.2f ohm\"%R2" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch18.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch18.ipynb new file mode 100644 index 00000000..125917ee --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch18.ipynb @@ -0,0 +1,916 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-18 : Rectifiers and Power Supplies" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 548 Example 18.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Peak value of current, Im = Vm / rf+RL = 295.45 mA\n", + " Average current, Id.c. = Im / pi = 94.04 mA\n", + " RMS value of current, Irms = Im / 2 = 147.72 mA\n", + "(b) D.C. power output, Pd.c. = (Id.c.)**2 * RL = 8.84 W\n", + "(c) AC input power, Pac = (Irms)**2 * (rf+RL) = 24.00 \n", + "(d) Efficiency of rectification, eta = Pdc / Pac = 36.85 %\n" + ] + } + ], + "source": [ + "from math import pi\n", + "im=325./(100+1000) # in A\n", + "x1=im*10**3 # in mA\n", + "print \"(a) Peak value of current, Im = Vm / rf+RL = %0.2f mA\"%x1\n", + "idc=295.45/pi # in mA\n", + "print \" Average current, Id.c. = Im / pi = %0.2f mA\"%idc\n", + "irms=295.45/2 # in mA\n", + "print \" RMS value of current, Irms = Im / 2 = %0.2f mA\"%irms\n", + "pdc=((94.046*10**-3)**2)*1000 # in W\n", + "print \"(b) D.C. power output, Pd.c. = (Id.c.)**2 * RL = %0.2f W\"%pdc\n", + "pac=((147.725*10**-3)**2)*1100 # in W\n", + "print \"(c) AC input power, Pac = (Irms)**2 * (rf+RL) = %0.2f \"%pac\n", + "eta=(8.845/24)*100 # in percentage\n", + "print \"(d) Efficiency of rectification, eta = Pdc / Pac = %0.2f %%\"%eta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 548 Example 18.2." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average value of load current, Id.c.= Vdc / RL = 48.00 mA\n", + "Maximum value of load current, Im= pi * Idc = 150.80 mA\n", + "Therefore, maximum ac voltage required at the input,\n", + " Vm = Im * (rf+RL) = 82.94 V\n" + ] + } + ], + "source": [ + "from math import pi\n", + "icd=(24./500)*10**3 # in mA\n", + "print \"Average value of load current, Id.c.= Vdc / RL = %0.2f mA\"%icd\n", + "im=pi*48 # in mA\n", + "print \"Maximum value of load current, Im= pi * Idc = %0.2f mA\"%im\n", + "print \"Therefore, maximum ac voltage required at the input,\"\n", + "vm=550*150.8*10**-3 # in V\n", + "print \" Vm = Im * (rf+RL) = %0.2f V\"%vm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 549 Example 18.3." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The transformer secondary voltage = 46.00 V\n", + " Maximum value of secondary voltage, Vm = 65.05 V\n", + " Therefore, d.c. output voltage, Vd.c. = Vm / pi = 20.69 V\n", + "(b) PIV of a diode = Vm = 65 V\n", + "(c) Maximum value of load current, Im= Vm / RL = 0.22 A\n", + " Therefore, maximum value of power delivered to the load,\n", + " Pm = Im**2 * RL = 14.13 W\n", + "(d) The average value of load current, Id.c.(A) = Vdc / RL = 0.07 A\n", + " Therefore, average value of power delivered to the load,\n", + " Pd.c. = (Idc)**2 * RL = 1.43 W\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "x1=230./5 # in V\n", + "vm=sqrt(2) * 46 # in V\n", + "vdc=65./pi # in V\n", + "im=65./300 # in A\n", + "pm=0.217**2 * 300 # in W\n", + "idc=20.7/300 # in A\n", + "pdc=(0.069**2)*300 # in W\n", + "print \"(a) The transformer secondary voltage = %0.2f V\"%x1\n", + "print \" Maximum value of secondary voltage, Vm = %0.2f V\"%vm\n", + "print \" Therefore, d.c. output voltage, Vd.c. = Vm / pi = %0.2f V\"%vdc\n", + "print \"(b) PIV of a diode = Vm = 65 V\"\n", + "print \"(c) Maximum value of load current, Im= Vm / RL = %0.2f A\"%im\n", + "print \" Therefore, maximum value of power delivered to the load,\"\n", + "print \" Pm = Im**2 * RL = %0.2f W\"%pm\n", + "print \"(d) The average value of load current, Id.c.(A) = Vdc / RL = %0.2f A\"%idc\n", + "print \" Therefore, average value of power delivered to the load,\"\n", + "print \" Pd.c. = (Idc)**2 * RL = %0.2f W\"%pdc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 550 Example 18.4." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage across the two ends of secondary(in V) = 230 / 5 = 46.00 V\n", + "Voltage from center tapping to one end, Vrms = 23.00 V\n", + "(a) d.c. voltage across the load, Vdc = 2Vm / pi = 20.71 V\n", + "(b) d.c. current flowing through the load,\n", + " Idc = Vdc / (rs+rf+RL) = 20.70 mA\n", + "(c) d.c. power delivered to the load,\n", + " Pdc = (Idc)**2 * RL = 0.39 W\n", + "(d) PIV across each diode = 2Vm = 65.05 W\n", + "(e) Ripple voltage, Vr,rms = sqrt(Vrms**2 - Vdc**2) = 10.03 V\n", + " Frequency of ripple voltage = 120.00 Hz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "x1=230./5 # in V\n", + "vrms=46./2 # in V\n", + "vdc=(2.*23*sqrt(2))/pi # in V\n", + "idc=(20.7/1000)*10**3 # in mA\n", + "pdc=((20.7*10**-3)**2)*900 # in W\n", + "piv=2*23*sqrt(2) # in V\n", + "vrrms=sqrt(23**2 - 20.7**2) # in V\n", + "f=2*60 # in Hz\n", + "print \"The voltage across the two ends of secondary(in V) = 230 / 5 = %0.2f V\"%x1\n", + "print \"Voltage from center tapping to one end, Vrms = %0.2f V\"%vrms\n", + "print \"(a) d.c. voltage across the load, Vdc = 2Vm / pi = %0.2f V\"% vdc\n", + "print \"(b) d.c. current flowing through the load,\"\n", + "print \" Idc = Vdc / (rs+rf+RL) = %0.2f mA\"%idc\n", + "print \"(c) d.c. power delivered to the load,\"\n", + "print \" Pdc = (Idc)**2 * RL = %0.2f W\"%pdc\n", + "print \"(d) PIV across each diode = 2Vm = %0.2f W\"%piv\n", + "print \"(e) Ripple voltage, Vr,rms = sqrt(Vrms**2 - Vdc**2) = %0.2f V\"%vrrms\n", + "print \" Frequency of ripple voltage = %0.2f Hz\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 552 Example 18.5." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore, Imax = 320.00 mA\n", + "The maximum value of the secondary voltage,\n", + " Vm = 141.42 V\n", + "Therefore, the value of load resistor that gives the largest d.c. power output\n", + " RL = Vm / Imax = 441.88 ohm\n", + "(b) D.C.(load) voltage, Vdc(V) = (2*141.4)/pi = 90.02 V\n", + " D.C. load current, Idc = Vdc / RL = 0.20 A\n", + "(c) PIV of each diode = 2Vm = 282.8 V\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "imax=0.8*400 # in mA\n", + "print \"Therefore, Imax = %0.2f mA\"%imax \n", + "print \"The maximum value of the secondary voltage,\"\n", + "vm=sqrt(2)*100 # in V\n", + "print \" Vm = %0.2f V\"%vm\n", + "print \"Therefore, the value of load resistor that gives the largest d.c. power output\"\n", + "RL=141.4/(320*10**-3)\n", + "print \" RL = Vm / Imax = %0.2f ohm\"%RL\n", + "vdc=(2*141.4)/pi\n", + "print \"(b) D.C.(load) voltage, Vdc(V) = (2*141.4)/pi = %0.2f V\"%vdc\n", + "idc=90./442\n", + "print \" D.C. load current, Idc = Vdc / RL = %0.2f A\"%idc\n", + "print \"(c) PIV of each diode = 2Vm = 282.8 V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 553 Example 18.6." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "D.C. power delivered to the load,\n", + " Pdc = Vdc**2 / RL\n", + "Therefore, Vdc = sqrt(Pdc*RL) = 100.00 V\n", + "The ripple factor, gamma = Vac / Vdc\n", + "i.e. 0.01 = Vac / 100\n", + "Therefore, the ac ripple voltage across the load, Vac = 1 V\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "print \"D.C. power delivered to the load,\"\n", + "print \" Pdc = Vdc**2 / RL\"\n", + "vdc=sqrt(50.*200)\n", + "print \"Therefore, Vdc = sqrt(Pdc*RL) = %0.2f V\"%vdc\n", + "print \"The ripple factor, gamma = Vac / Vdc\"\n", + "print \"i.e. 0.01 = Vac / 100\"\n", + "print \"Therefore, the ac ripple voltage across the load, Vac = 1 V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 554 Example 18.7. " + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The rms value of the transformer secondary voltage,\n", + " Vrms = 57.50 V\n", + " The maximum value of the secondary voltage\n", + " Vm = 81.32 V Therefore, d.c. output voltage, Vdc = 2Vm / pi = 51.76 V\n", + "(b) D.C. power delivered to the load,\n", + " Pd.c. = (Vdc)**2 / RL = 2.70 W\n", + "(c) PIV across each diode = Vm = 81.3 V\n", + "(d) Output frequency = 2 x 50 = 100 Hz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "Vrms=230./4 # in V\n", + "vm=sqrt(2)*57.5 # in V\n", + "vdc=(2*81.3)/pi # in V\n", + "pdc=52.**2/1000 # in W\n", + "print \"(a) The rms value of the transformer secondary voltage,\"\n", + "print \" Vrms = %0.2f V\"%Vrms\n", + "print \" The maximum value of the secondary voltage\"\n", + "print \" Vm = %0.2f V\"%vm,\n", + "print \"Therefore, d.c. output voltage, Vdc = 2Vm / pi = %0.2f V\"%vdc\n", + "print \"(b) D.C. power delivered to the load,\"\n", + "print \" Pd.c. = (Vdc)**2 / RL = %0.2f W\"%pdc\n", + "print \"(c) PIV across each diode = Vm = 81.3 V\"\n", + "print \"(d) Output frequency = 2 x 50 = 100 Hz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 556 Example 18.8." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that the ripple factor for inductor filter is gamma = RL / 3*sqrt(2)*omega*L\n", + "Therefore, L = 1.56 Henry\n" + ] + } + ], + "source": [ + "L=0.0625/0.04 # in H\n", + "print \"We know that the ripple factor for inductor filter is gamma = RL / 3*sqrt(2)*omega*L\"\n", + "print \"Therefore, L = %0.2f Henry\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 558 Example 18.9." + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that the ripple factor for capacitor filter is\n", + " gamma = 1 / 4*sqrt(3)*f*C*RL\n", + "Therefore, C = 72.20 pF\n" + ] + } + ], + "source": [ + "print \"We know that the ripple factor for capacitor filter is\"\n", + "print \" gamma = 1 / 4*sqrt(3)*f*C*RL\"\n", + "c=(0.722)/0.01 # in pF\n", + "print \"Therefore, C = %0.2f pF\"%c" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 560 Example 18.10" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The effective load resistance RL = 50.00 ohm\n", + "We know that the ripple factor, gamma = 1.194 / LC\n", + "i.e. LC = 59.70 \n", + "Critical value of L(mH) = RL / 3*omega = 50 / 3*2*pi*f = 53mH\n", + "Taking L = 60 mH (about 20% higher), C will be about 1000 uF\n" + ] + } + ], + "source": [ + "rl=10./(200*10**-3) # in ohm\n", + "lc=1.194/0.02 \n", + "print \"The effective load resistance RL = %0.2f ohm\"%rl\n", + "print \"We know that the ripple factor, gamma = 1.194 / LC\"\n", + "print \"i.e. LC = %0.2f \"%lc\n", + "print \"Critical value of L(mH) = RL / 3*omega = 50 / 3*2*pi*f = 53mH\"\n", + "print \"Taking L = 60 mH (about 20% higher), C will be about 1000 uF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 561 Example 18.11" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " RL = 50.00 ohm\n", + " 0.02 = 5700 / L*C1*C2*50 = 114 / L*C1*C2\n", + "If we assume L = 10 mH and C1 = C2 = C, we have\n", + " 0.02 = 114 / L*C**2 = 11.4 / C**2\n", + " C**2 = 570.00 \n", + "therefore, C = 23.87 uF\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "rl=(10./(200*10**-3)) # in ohm\n", + "c2=11.4/0.02\n", + "c=sqrt(570.) # in uF\n", + "print \" RL = %0.2f ohm\"%rl\n", + "print \" 0.02 = 5700 / L*C1*C2*50 = 114 / L*C1*C2\"\n", + "print \"If we assume L = 10 mH and C1 = C2 = C, we have\"\n", + "print \" 0.02 = 114 / L*C**2 = 11.4 / C**2\"\n", + "print \" C**2 = %0.2f \"%c2\n", + "print \"therefore, C = %0.2f uF\"%c" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 562 Example 18.12." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Pz = Vz * Iz_max = 0.40 W\n", + "Hence a 0.5Z 10 zener can be selected\n", + "Value of load resistance, RL\n", + " RL_min = Vo / IL_max = 200.00 ohm\n", + " RL_max = Vo / IL_min = 333.33 ohm\n", + "Value of input resistance, R\n", + " Rmax = Vin(max)-Vo / ILmin+IZmax = 285.71 ohm\n", + "142.857142857 Rmax(ohm) = Vin(min)-Vo / ILmax+IZmin =\n", + " R = Rmax+Rmin / 2 = 214.50 ohm\n" + ] + } + ], + "source": [ + "pz=10.*40*10**-3 # in W\n", + "print \" Pz = Vz * Iz_max = %0.2f W\"%pz\n", + "print \"Hence a 0.5Z 10 zener can be selected\"\n", + "print \"Value of load resistance, RL\"\n", + "rlmin=10./(50*10**-3) # in ohm\n", + "print \" RL_min = Vo / IL_max = %0.2f ohm\"%rlmin\n", + "rlmax=10./(30*10**-3) # in ohm\n", + "print \" RL_max = Vo / IL_min = %0.2f ohm\"%rlmax\n", + "print \"Value of input resistance, R\"\n", + "rmax=(30.-10)/((30+40)*10**-3) # in ohm\n", + "print \" Rmax = Vin(max)-Vo / ILmin+IZmax = %0.2f ohm\"%rmax\n", + "rmin=(20.-10)/((50+20)*10**-3) # in ohm\n", + "print rmin,\" Rmax(ohm) = Vin(min)-Vo / ILmax+IZmin =\"\n", + "r=(286+143.)/2\n", + "print \" R = Rmax+Rmin / 2 = %0.2f ohm\"%r # answer in textbook is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 563 Example 18.13." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " RL = Vo / IL = 250.00 ohm\n", + "Hence, the series resistance R(ohm) = Vi(min)-Vo / IZ(min)+IL = 120.00 ohm\n", + "The various values are given in the Zener regulator shown in Fig. 18.19\n" + ] + } + ], + "source": [ + "rl=5./(20*10**-3) # in ohm\n", + "print \" RL = Vo / IL = %0.2f ohm\"%rl\n", + "r=(8.-5)/((5.+20)*10**-3) # in ohm\n", + "print \"Hence, the series resistance R(ohm) = Vi(min)-Vo / IZ(min)+IL = %0.2f ohm\"%r\n", + "print \"The various values are given in the Zener regulator shown in Fig. 18.19\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 564 Example 18.14." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Let IZ = IZ(min) and IL = 0\n", + " The total current I = IL + IZ = 10 mA\n", + " Therefore, R = 1000.00 ohm\n", + "(ii) For IZ = IZ(max) = 100 mA and IL = 20 mA\n", + " I = IL + IZ = 120.00 mA\n", + " Therefore, R = 83.33 ohm\n", + "(iii) The range of R varies from 83.33 ohm to 1000 ohm\n" + ] + } + ], + "source": [ + "print \"(i) Let IZ = IZ(min) and IL = 0\"\n", + "print \" The total current I = IL + IZ = 10 mA\"\n", + "r=10./(10*10**-3) # in ohm\n", + "print \" Therefore, R = %0.2f ohm\"%r\n", + "print \"(ii) For IZ = IZ(max) = 100 mA and IL = 20 mA\"\n", + "i=20+100. # in mA\n", + "print \" I = IL + IZ = %0.2f mA\"%i\n", + "r=10./(120*10**-3)\n", + "print \" Therefore, R = %0.2f ohm\"%r\n", + "print \"(iii) The range of R varies from 83.33 ohm to 1000 ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 565 Example 18.15." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Here, load resistance is RL = Vo / IL = 500.00 ohm\n", + "Maximum Zener Current Iz_max = 80.00 mA\n", + "The minimum input voltage required will be when Iz = 0. Under this condition,\n", + " I = IL = 10 mA\n", + "Minimum input voltage Vi_min = Vo + IR\n", + "Hence, Vi_min = 8.00 V\n", + "or 8 = 5 + (10*10**-3)R\n", + "Therefore, Rmax = 300.00 ohm\n", + "Now, maximum input voltage, Vi_max = 5 + [(80+10)10**-3]R\n", + " Rmin = 77.78 ohm\n", + "The value of R is chosen between 77.77 ohm and 300 ohm\n" + ] + } + ], + "source": [ + "rl=5./(10*10**-3) # in ohm\n", + "print \"Here, load resistance is RL = Vo / IL = %0.2f ohm\"%rl\n", + "iz=400./5 # in mA\n", + "print \"Maximum Zener Current Iz_max = %0.2f mA\"%iz\n", + "print \"The minimum input voltage required will be when Iz = 0. Under this condition,\"\n", + "print \" I = IL = 10 mA\"\n", + "print \"Minimum input voltage Vi_min = Vo + IR\"\n", + "vi=10.-2 # in V\n", + "print \"Hence, Vi_min = %0.2f V\"%vi\n", + "print \"or 8 = 5 + (10*10**-3)R\"\n", + "rmax=3./(10*10**-3) # in ohm\n", + "print \"Therefore, Rmax = %0.2f ohm\"%rmax\n", + "print \"Now, maximum input voltage, Vi_max = 5 + [(80+10)10**-3]R\"\n", + "rmin=7./(90*10**-3) # in ohm\n", + "print \" Rmin = %0.2f ohm\"%rmin\n", + "print \"The value of R is chosen between 77.77 ohm and 300 ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 566 Example 18.16." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The load current, IL( = Vo / RL = 20.00 mA\n", + "Max. Zener current, Iz_max = 25.00 mA\n", + " Rmax(ohm) = Vi-Vo / IL_min+IZ_max = 177.78 ohm\n" + ] + } + ], + "source": [ + "il=(24./1200)*10**3 # in mA\n", + "print \"The load current, IL( = Vo / RL = %0.2f mA\"%il\n", + "iz=600./24 # in mA\n", + "print \"Max. Zener current, Iz_max = %0.2f mA\"%iz\n", + "rmax=(32.-24)/((20+25)*10**-3) # in ohm\n", + "print \" Rmax(ohm) = Vi-Vo / IL_min+IZ_max = %0.2f ohm\"%rmax" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 567 Example 18.17." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Refer to fig.18.24. We know that\n", + " Vi_min(V) = Vo + 3V = 18.00 V\n", + "Assuming the ripple voltage Vr = 2V(max), the input voltage is\n", + " Vi = Vi(min) + Vr/2 = 19.00 V\n", + "Then Vz = Vi /2 = 9.50 V (use the zener diode 1N758 for 10V)\n", + "Therefore, Vz = 10 V\n", + " Iz = 20 mA\n", + " R1 = Vi-Vz / Iz = 450.00 ohm\n", + "Let I2 = IB(max) = 50 uA\n", + " R2 = Vo-Vz / I2 = 100.00 kohm\n", + " R3 = Vz / I2 = 200.00 kohm\n", + "Select C1 = 50 uF\n", + "Specification of transistor Q1\n", + " VCE_max = Vi_max(V) = Vi + Vr/2 = 20.00 V\n", + " IE = IL = 50 mA\n", + " P = VCE*IL = (Vi-Vo) * IL = 200.00 mW\n", + "Use the transistor 2N718 for Q1\n" + ] + } + ], + "source": [ + "vi=15.+3 # in V\n", + "print \"Refer to fig.18.24. We know that\"\n", + "print \" Vi_min(V) = Vo + 3V = %0.2f V\"%vi\n", + "vi=18.+1 # in V\n", + "print \"Assuming the ripple voltage Vr = 2V(max), the input voltage is\"\n", + "print \" Vi = Vi(min) + Vr/2 = %0.2f V\"%vi\n", + "vz=19./2 # in V\n", + "print \"Then Vz = Vi /2 = %0.2f V (use the zener diode 1N758 for 10V)\"%vz\n", + "print \"Therefore, Vz = 10 V\"\n", + "print \" Iz = 20 mA\"\n", + "r1=(19.-10)/(20*10**-3) # in ohm\n", + "print \" R1 = Vi-Vz / Iz = %0.2f ohm\"%r1\n", + "print \"Let I2 = IB(max) = 50 uA\"\n", + "r2=((15.-10)/(50*10**-6))*10**-3 # in k-ohm\n", + "print \" R2 = Vo-Vz / I2 = %0.2f kohm\"%r2\n", + "r3=(10./(50*10**-6))*10**-3 # in k-ohm\n", + "print \" R3 = Vz / I2 = %0.2f kohm\"%r3\n", + "print \"Select C1 = 50 uF\"\n", + "print \"Specification of transistor Q1\"\n", + "vce=19.+1 # in V\n", + "print \" VCE_max = Vi_max(V) = Vi + Vr/2 = %0.2f V\"%vce\n", + "print \" IE = IL = 50 mA\"\n", + "p=((19.-15)*50) # in mW\n", + "print \" P = VCE*IL = (Vi-Vo) * IL = %0.2f mW\"%p\n", + "print \"Use the transistor 2N718 for Q1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 568 Example 18.18." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Selection of Zener diode\n", + " RLmin = Vo / ILmax = 400.00 ohm\n", + " Vz = Vo / 2 = 10.00 V\n", + "The current flowing through the Zener,\n", + " Iz = IE2 + IR1 = 20.00 mA\n", + " Pz = Vz*Iz = 0.20 W\n", + "Selection of transistor Q1\n", + " IE1 = IR1 + IR2 + IL = 70.00 mA\n", + " Vi(max) - Vo = 30 -20 = 10 V\n", + "For transistor SL100, the rating are\n", + " IC(max) = 500 mA\n", + " VCE(max) = 50 V\n", + " hre = 50 - 280\n", + "Hence, SL100 can be chosen for Q1\n", + "\n", + "Selection of transistor Q2\n", + " Therefore, VCE2_max = (Vo + VBE1) - Vz = 10.60 V\n", + "For transistor BC107, the rating are\n", + " VCEO(max) = 45 V\n", + " IC(max) = 200 mA\n", + " hFE = 125 - 300\n", + "Hence, transistor BC107 is selected for Q2\n", + "Selection of resistor R1, R2 and R3\n", + " VR1 = Vo - Vz = 10.00 V\n", + " R1 = VR1 / IR1 = 1.00 kohm\n", + " VR2 = Vo - VR3 = 9.40 V\n", + " R2 = VR2 / IR2 = 940.00 ohm\n", + " VR3 = Vz + VBE2(sat) = 10.60 V\n", + " R3 = VR3 / IR3 = 1060.00 ohm\n", + "Selection of resistor R4\n", + " VB1 = VC2(V) = Vo + VBE1 = 20.60 V\n", + " IB1 = IC1 / beta = 1.40 mA\n", + " IR4 = IB1 + IC2 = 11.40 mA\n", + " R4_max = VR4(max) / IR4 = Vi(max)-VB1 / IR4 = 824.56 ohm\n", + " R4_min = VR4(min) / IR4 = Vi(min)-VB1 / IR4 = 122.81 ohm\n", + " R4 = R4(max)+R4(min) / 2 = 474.00 ohm\n" + ] + } + ], + "source": [ + "rlmin=20./(50*10**-3) # in ohm\n", + "print \"Selection of Zener diode\"\n", + "print \" RLmin = Vo / ILmax = %0.2f ohm\"%rlmin\n", + "vz=20./2 # in V\n", + "print \" Vz = Vo / 2 = %0.2f V\"%vz\n", + "print \"The current flowing through the Zener,\"\n", + "iz=10.+10 # in mA\n", + "print \" Iz = IE2 + IR1 = %0.2f mA\"%iz\n", + "pz=10.*20*10**-3 # in W\n", + "print \" Pz = Vz*Iz = %0.2f W\"%pz # > 0.5 W\n", + "print \"Selection of transistor Q1\"\n", + "ie1=10.+10+50 # in mA\n", + "print \" IE1 = IR1 + IR2 + IL = %0.2f mA\"%ie1\n", + "print \" Vi(max) - Vo = 30 -20 = 10 V\"\n", + "print \"For transistor SL100, the rating are\"\n", + "print \" IC(max) = 500 mA\"\n", + "print \" VCE(max) = 50 V\"\n", + "print \" hre = 50 - 280\"\n", + "print \"Hence, SL100 can be chosen for Q1\"\n", + "print \"\"\n", + "print \"Selection of transistor Q2\"\n", + "vce2=20.6-10 # in V\n", + "print \" Therefore, VCE2_max = (Vo + VBE1) - Vz = %0.2f V\"%vce2\n", + "print \"For transistor BC107, the rating are\"\n", + "print \" VCEO(max) = 45 V\"\n", + "print \" IC(max) = 200 mA\"\n", + "print \" hFE = 125 - 300\"\n", + "print \"Hence, transistor BC107 is selected for Q2\"\n", + "print \"Selection of resistor R1, R2 and R3\"\n", + "vr1=20-10. # in V\n", + "print \" VR1 = Vo - Vz = %0.2f V\"%vr1\n", + "r1=10./(10) # in k-ohm\n", + "print \" R1 = VR1 / IR1 = %0.2f kohm\"%r1\n", + "vr2=20.-10.6 # in V\n", + "print \" VR2 = Vo - VR3 = %0.2f V\"%vr2\n", + "r2=9.4/(10*10**-3) # in ohm\n", + "print \" R2 = VR2 / IR2 = %0.2f ohm\"%r2\n", + "vr3=10+0.6 # in V\n", + "print \" VR3 = Vz + VBE2(sat) = %0.2f V\"%vr3\n", + "r3=10.6/(10*10**-3) # in ohm\n", + "print \" R3 = VR3 / IR3 = %0.2f ohm\"%r3\n", + "print \"Selection of resistor R4\"\n", + "vb1=20+0.6 # in V\n", + "print \" VB1 = VC2(V) = Vo + VBE1 = %0.2f V\"%vb1\n", + "ib1=70./50 # in mA\n", + "print \" IB1 = IC1 / beta = %0.2f mA\"%ib1\n", + "ir4=11.4 # in mA\n", + "print \" IR4 = IB1 + IC2 = %0.2f mA\"%ir4\n", + "r4max=(30-20.6)/(11.4*10**-3) # in ohm\n", + "print \" R4_max = VR4(max) / IR4 = Vi(max)-VB1 / IR4 = %0.2f ohm\"%r4max\n", + "r4min=(22-20.6)/(11.4*10**-3) # in ohm\n", + "print \" R4_min = VR4(min) / IR4 = Vi(min)-VB1 / IR4 = %0.2f ohm\"%r4min\n", + "r4=(825.+123)/2 # in ohm\n", + "print \" R4 = R4(max)+R4(min) / 2 = %0.2f ohm\"%r4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 569 Example 18.19." + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A bridge rectifier or full wave rectifier is used to get the pulsating d.c. output.\n", + " RL = Vdc / TL = 90.00 ohm\n", + "A capacitor filter is used to remove the ripple and get a smooth output.\n", + " Ripple factor gamma = 1 / 4*sqrt(3)*f*C*RL\n", + "Assume the ripple factor to be 0.03\n", + " C = 1069.17 uF\n", + "The short circuit resistance Rsc connected with the series pass transistor is\n", + " Rsc = VBE / Ilim_it = 4.67 ohm\n", + "Assume 7.6 V Zener diode in series with 1.5 k-ohm\n", + "The designed circuit is shown in fig.18.32.\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "print \"A bridge rectifier or full wave rectifier is used to get the pulsating d.c. output.\"\n", + "rl=9./(100*10**-3) # in ohm\n", + "print \" RL = Vdc / TL = %0.2f ohm\"%rl\n", + "print \"A capacitor filter is used to remove the ripple and get a smooth output.\"\n", + "print \" Ripple factor gamma = 1 / 4*sqrt(3)*f*C*RL\"\n", + "print \"Assume the ripple factor to be 0.03\"\n", + "c=(1./(4*sqrt(3)*50*0.03*90))*10**6 # in uF\n", + "print \" C = %0.2f uF\"%c # = 1000 uF\n", + "print \"The short circuit resistance Rsc connected with the series pass transistor is\"\n", + "rsc=0.7/(150*10**-3) # in ohm\n", + "print \" Rsc = VBE / Ilim_it = %0.2f ohm\"%rsc\n", + "print \"Assume 7.6 V Zener diode in series with 1.5 k-ohm\"\n", + "print \"The designed circuit is shown in fig.18.32.\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch20.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch20.ipynb new file mode 100644 index 00000000..e1b955eb --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch20.ipynb @@ -0,0 +1,322 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-20 : Operational Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 620 Example 20.1." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " CMRR = Ad / Acm = 10**5\n", + " Therefore, the common-mode gain, Acm = Ad / CMRR =1.00\n" + ] + } + ], + "source": [ + "print \" CMRR = Ad / Acm = 10**5\"\n", + "acm=(1.0*10**5)/(10**5)\n", + "print \" Therefore, the common-mode gain, Acm = Ad / CMRR =%0.2f\"%acm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 620 Example 20.2." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The slew rate, SR = dVo / dt\n", + " SR =5.00 V/us\n" + ] + } + ], + "source": [ + "sr=20./(4) # in V/us\n", + "print \" The slew rate, SR = dVo / dt\"\n", + "print \" SR =%0.2f V/us\"%sr" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 621 Example 20.3." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,\n", + " SR = 2*pi*f*Vm / 10**6 = 0.5 V/us\n", + "The maximum output voltage, Vm = A*Vid =1.00 V\n", + "The maximum frequency of the input for which undistorted output is obtained is given by,\n", + " fmax = SR*10**6 / 2*pi*Vm =79.58\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "print \"The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,\"\n", + "print \" SR = 2*pi*f*Vm / 10**6 = 0.5 V/us\"\n", + "vm=50.*(20*10**-3) # in volts\n", + "print \"The maximum output voltage, Vm = A*Vid =%0.2f V\"%vm\n", + "print \"The maximum frequency of the input for which undistorted output is obtained is given by,\"\n", + "f=(0.5*10**6)/(2*pi*1) # in kHz\n", + "x1=f*10**-3\n", + "print \" fmax = SR*10**6 / 2*pi*Vm =%0.2f\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 622 Example 20.4." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,\n", + " SR = 2*pi*f*Vm / 10**6 = 0.5 V/us\n", + "The maximum output voltage, Vm(V peak-to-peak) = SR*10**6 / 2*pi*f =1.99 = 3.98 V peak-to-peak\n", + "The maximum peak-to-peak input voltage for undistorted output is,\n", + " Vid(V peak-to-peak) = Vm/A = 0.40\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "print \"The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,\"\n", + "print \" SR = 2*pi*f*Vm / 10**6 = 0.5 V/us\"\n", + "vm=(0.5*10**6)/(2*pi*(40*10**3)) # in volts\n", + "print \"The maximum output voltage, Vm(V peak-to-peak) = SR*10**6 / 2*pi*f =%0.2f\"%vm,\" = 3.98 V peak-to-peak\"\n", + "print \"The maximum peak-to-peak input voltage for undistorted output is,\"\n", + "vid=3.98/10 # in volts\n", + "print \" Vid(V peak-to-peak) = Vm/A = %0.2f\"%vid" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 623 Example 20.5." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The closed-loop voltage gain Af = -RF / R1 =-10.00\n" + ] + } + ], + "source": [ + "af=-10./1\n", + "print \" The closed-loop voltage gain Af = -RF / R1 =%0.2f\"%af" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 624 Example 20.6. " + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The closed-loop voltage gain, AF = 1 + RF/R1 =11.00\n", + " The feedback factor, beta = R1 / R1+RF =0.09\n" + ] + } + ], + "source": [ + "af=1.+(10./1)\n", + "print \" The closed-loop voltage gain, AF = 1 + RF/R1 =%0.2f\"%af\n", + "beta=1/(1+10.)\n", + "print \" The feedback factor, beta = R1 / R1+RF =%0.2f\"%beta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 625 Example 20.7." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage is given by,\n", + " Vo = -Rf/R * (V1+V2+...+Vn) =-9.00 V\n" + ] + } + ], + "source": [ + "v=-(2+3+4) # in volts\n", + "print \"The output voltage is given by,\"\n", + "print \" Vo = -Rf/R * (V1+V2+...+Vn) =%0.2f V\"%v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 626 Example 20.8." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1. Given: fL = 1 kHz\n", + "2. Since R and C values are not given, let assume C = 0.01 uF\n", + "3. Therefore, R(k-ohm) = 1 / 2*pi*fL*C =15.92 kohm\n", + "4. Given pass band gain A = 1 + Rf/Ri = 2 i.e. the value of Rf = Ri\n", + "Let Rf = Ri = 10 k-ohm. The high pass circuit values are shown in Fig.20.31\n" + ] + } + ], + "source": [ + "from math import pi\n", + "print \"1. Given: fL = 1 kHz\"\n", + "print \"2. Since R and C values are not given, let assume C = 0.01 uF\"\n", + "r=1/(2*pi*(10**3)*(0.01*10**-6))\n", + "x1=r*10**-3 # in k-ohm\n", + "print \"3. Therefore, R(k-ohm) = 1 / 2*pi*fL*C =%0.2f kohm\"%x1\n", + "print \"4. Given pass band gain A = 1 + Rf/Ri = 2 i.e. the value of Rf = Ri\"\n", + "print \"Let Rf = Ri = 10 k-ohm. The high pass circuit values are shown in Fig.20.31\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 627 Example 20.9." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(b) With C = 0.01 uF, R = 0.5*10**-3/0.01*10**-6 =50.00 kohm\n", + "(c) Maximum value of differential input voltage is\n", + " 2*Vsat*(R2 / R1+R2) =0.00\n", + "Therefore, the peak values for the differential input voltage just exceed +-2 x 6.48 V\n" + ] + } + ], + "source": [ + "r=(0.5)/0.01 # in k-ohm\n", + "print \"(b) With C = 0.01 uF, R = 0.5*10**-3/0.01*10**-6 =%0.2f kohm\"%r\n", + "print \"(c) Maximum value of differential input voltage is\"\n", + "x=2.*14*(100/(100+116))\n", + "print \" 2*Vsat*(R2 / R1+R2) =%0.2f\"%x\n", + "print \"Therefore, the peak values for the differential input voltage just exceed +-2 x 6.48 V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch21.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch21.ipynb new file mode 100644 index 00000000..50d5965c --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch21.ipynb @@ -0,0 +1,119 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-21 : Transducers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 632 Example 21.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The electron mobility, un = sigma*RH =2000.00 cm**2/V-s\n" + ] + } + ], + "source": [ + "u=10.*200 # in cm**2/V-s\n", + "print \"The electron mobility, un = sigma*RH =%0.2f cm**2/V-s\"%u" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 633 Example 21.2." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that the electron mobilty, un = sigma/nq\n", + "Therefore, the electron concentration,\n", + " n = sigma / uq =1.25e+22 m**-3\n" + ] + } + ], + "source": [ + "n=10./((50*10**-4)*(1.6*10**-19)) # m**-3\n", + "print \"We know that the electron mobilty, un = sigma/nq\"\n", + "print \"Therefore, the electron concentration,\"\n", + "print \" n = sigma / uq =%0.2e m**-3\"%n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 634 Example 21.3." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that the number of conduction electrons, i.e. electron density,\n", + " n = B*I/VH*q*w =5.00e+21 m**3\n" + ] + } + ], + "source": [ + "n=(1.2*20)/(60*(1.6*10**-19)*(0.5*10**-3)) # in m**3\n", + "print \"We know that the number of conduction electrons, i.e. electron density,\"\n", + "print \" n = B*I/VH*q*w =%0.2e m**3\"%n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch24.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch24.ipynb new file mode 100644 index 00000000..ce7b87dc --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch24.ipynb @@ -0,0 +1,293 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-24 : Digital Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 645 Example 24.1." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The procedure is as follows.\n", + "12 divided by 8 = quotient 1 with a remainder of 4\n", + "1 divided by 8 = quotient 0 with a remainder of 1\n", + "Therefore, decimal 12 = octal 14\n" + ] + } + ], + "source": [ + "#convert decimal 12 to an octal number\n", + "def base10toN(num, base):\n", + " \"\"\"Change ``num'' to given base\n", + " Upto base 36 is supported.\"\"\"\n", + "\n", + " converted_string, modstring = \"\", \"\"\n", + " currentnum = num\n", + " if not 1 < base < 37:\n", + " raise ValueError(\"base must be between 2 and 36\")\n", + " if not num:\n", + " return '0'\n", + " while currentnum:\n", + " mod = currentnum % base\n", + " currentnum = currentnum // base\n", + " converted_string = chr(48 + mod + 7*(mod > 10)) + converted_string\n", + " return converted_string\n", + "\n", + "o=base10toN(12,8)\n", + "print \"The procedure is as follows.\"\n", + "print \"12 divided by 8 = quotient 1 with a remainder of 4\"\n", + "print \"1 divided by 8 = quotient 0 with a remainder of 1\"\n", + "print \"Therefore, decimal 12 = octal\",o" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 646 Example 24.2." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) octal 144 = decimal 100\n", + "(ii) octal 237 = decimal 159\n", + "(iii) octal 120 = decimal 80\n" + ] + } + ], + "source": [ + "# convert octal number to decimal.\n", + "d=int('144',8)\n", + "print \"(i) octal 144 = decimal\",d\n", + "d1=int(\"237\",8)\n", + "print \"(ii) octal 237 = decimal\",d1\n", + "d2=int('120',8)\n", + "print \"(iii) octal 120 = decimal\",d2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 647 Example 24.3." + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The procedure is as follows,\n", + "(i) 112 divided by 16 = quotient 7 with a remainder of 0\n", + " 7 divided by 16 = quotient 0 with a remainder of 7\n", + "decimal 112 = hex 70\n", + "(ii) 253 divided by 16 = quotient 7 with a remainder of 13 i.e. D\n", + " 15 divided by 16 = quotient 0 with a remainder of 15 i.e. F\n", + "decimal 253 = hex FD\n" + ] + } + ], + "source": [ + "#convert decimal to hexadecimal number\n", + "def base10toN(num, base):\n", + " \"\"\"Change ``num'' to given base\n", + " Upto base 36 is supported.\"\"\"\n", + "\n", + " converted_string, modstring = \"\", \"\"\n", + " currentnum = num\n", + " if not 1 < base < 37:\n", + " raise ValueError(\"base must be between 2 and 36\")\n", + " if not num:\n", + " return '0'\n", + " while currentnum:\n", + " mod = currentnum % base\n", + " currentnum = currentnum // base\n", + " converted_string = chr(48 + mod + 7*(mod > 10)) + converted_string\n", + " return converted_string\n", + "\n", + "h=base10toN(112,16)\n", + "print \"The procedure is as follows,\"\n", + "print \"(i) 112 divided by 16 = quotient 7 with a remainder of 0\"\n", + "print \" 7 divided by 16 = quotient 0 with a remainder of 7\"\n", + "print \"decimal 112 = hex\",h\n", + "print \"(ii) 253 divided by 16 = quotient 7 with a remainder of 13 i.e. D\"\n", + "print \" 15 divided by 16 = quotient 0 with a remainder of 15 i.e. F\"\n", + "h=base10toN(253,16)\n", + "print \"decimal 253 = hex\",h" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 648 Example 24.4." + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) hex 4AB = decimal 1195.0\n", + "(ii) hex 23F = decimal 575.0\n" + ] + } + ], + "source": [ + "#convert hexadecimal number to decimal\n", + "h=float.fromhex('4AB')\n", + "print \"(i) hex 4AB = decimal\",h\n", + "h=float.fromhex('23F')\n", + "print \"(ii) hex 23F = decimal\",h" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 650 Example 24.5." + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) 1101 x 1100 = 10011100\n", + "(ii) 1000 x 101 = 101000\n", + "(iii) 1111 x 1001 = 10000111\n" + ] + } + ], + "source": [ + "# multiply binary numbers\n", + "h=int('1101',2)\n", + "o=int('1100',2)\n", + "p=h*o\n", + "z=bin(p)[2:]\n", + "print \"(i) 1101 x 1100 =\",z\n", + "h=int('1000',2)\n", + "o=int('101',2)\n", + "p=h*o\n", + "z=bin(p)[2:]\n", + "print \"(ii) 1000 x 101 =\",z\n", + "h=int('1111',2)\n", + "o=int('1001',2)\n", + "p=h*o\n", + "z=bin(p)[2:]\n", + "print \"(iii) 1111 x 1001 =\",z" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 651 Example 24.6." + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) 110 / 10\n", + " = binary 11\n", + " = decimal 3\n", + "(ii) 1111 / 110\n", + " = binary 10\n", + " = decimal) 2\n" + ] + } + ], + "source": [ + "# perform the binary divisions\n", + "\n", + "x=int('110',2)\n", + "x1=int('10',2)\n", + "x2=x/x1\n", + "x3=bin(x2)[2:]\n", + "print \"(i) 110 / 10\"\n", + "print \" = binary\",x3\n", + "print \" = decimal\",x2\n", + "x=int('1111',2)\n", + "x1=int('110',2)\n", + "x2=x/x1\n", + "x3=bin(x2)[2:]\n", + "print \"(ii) 1111 / 110\"\n", + "print \" = binary\",x3\n", + "print \" = decimal)\",x2" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch3.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch3.ipynb new file mode 100644 index 00000000..adbb80a5 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch3.ipynb @@ -0,0 +1,691 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-3 : Electron Ballistics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 : Page 48" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Speed of the electron, v =sqrt(2*q*V/m) = 4.19e+07 m/s\n", + "The kinetic energy = q x V = 5000 eV\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "q=1.6*10**-19 #charge of electron\n", + "V=5000 #potential difference\n", + "m=9.1*10**-31 #mass of electron\n", + "v=sqrt(2*q*V/m) #speed of electron\n", + "print \"Speed of the electron, v =sqrt(2*q*V/m) = %0.2e m/s\"% v\n", + "ke=(q*V)/(1.6*10**-9) #kinetic energyin eV\n", + "x1=ke*10**10\n", + "print \"The kinetic energy = q x V = %0.f eV\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 Page 48" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass of the charged particle = 1000 times the mass of an electron = 9.10e-28 kg\n", + "The charge of the partical = 1.6*10**-19 C\n", + "Therefore, The velocity, v = sqrt(2*q*V/me) = 5.93e+05 m/s\n", + "Kinetic energy = q x V = 1000.00 eV\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "me=1000*9.1*10**-31\n", + "print \"Mass of the charged particle = 1000 times the mass of an electron = %0.2e kg\"%me\n", + "print \"The charge of the partical = 1.6*10**-19 C\"\n", + "q=1.6*10**-19 #charge of the particle\n", + "V=1000 #potential difference\n", + "v=sqrt(2*q*V/me)\n", + "print \"Therefore, The velocity, v = sqrt(2*q*V/me) = %0.2e m/s\"%v\n", + "ke=(q*V)/(1.6*10**-19) # in eV\n", + "print \"Kinetic energy = q x V = %0.2f eV\"%ke" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 : Page 50" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore, E = V / d = 5.83e+04 \n", + " ax = qE / m = 1.03e+16 m/s**2\n", + "We know that,\n", + " x = vox*t + 0.5*a*t**2\n", + " vx = vox + ax*t\n", + "(i) Consider x = 3*10**-3 m\n", + "3*10**-3 = 3*10**6*t + 5.13*10**15*t**2\n", + "Solving this equation,\n", + "t = 5.26e-10 seconds \n", + "vx = 8.40e+06 m/s \n", + "(ii) Consider x = 6*10**-6 m\n", + "t**2+(5.85*10**-10)*t-(1.17*10**-18) = 0\n", + "Solving this equation,\n", + "t = 8.28e-10 seconds \n", + "vx = 1.15e+07 m/s\n" + ] + } + ], + "source": [ + "from sympy import symbols, solve\n", + "from math import sqrt\n", + "d=6*10**-3\n", + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "vax=3*10**6\n", + "E=350/d\n", + "print \"Therefore, E = V / d = %0.2e \"%E\n", + "ax=q*E/m\n", + "print \" ax = qE / m = %0.2e m/s**2\"%ax\n", + "print \"We know that,\"\n", + "print \" x = vox*t + 0.5*a*t**2\"\n", + "print \" vx = vox + ax*t\"\n", + "print \"(i) Consider x = 3*10**-3 m\"\n", + "print \"3*10**-3 = 3*10**6*t + 5.13*10**15*t**2\"\n", + "print \"Solving this equation,\"\n", + "t=symbols('t')\n", + "p1=(5.13*10**15)*t**2+(3*10**6)*t-3*10**-3\n", + "t1=solve(p1,t)\n", + "ans1=t1[1]\n", + "print \"t = %0.2e seconds \"%ans1\n", + "vx=(3*10**6)+((1.026*10**16)*(5.264*10**-10))\n", + "print \"vx = %0.2e m/s \"%vx\n", + "print \"(ii) Consider x = 6*10**-6 m\"\n", + "print \"t**2+(5.85*10**-10)*t-(1.17*10**-18) = 0\"\n", + "print \"Solving this equation,\"\n", + "t=symbols('t')\n", + "p2=t**2+(5.85*10**-10)*t-1.17*10**-18\n", + "t2=solve(p2, t)\n", + "ans2=t2[1]\n", + "print \"t = %0.2e seconds \"%ans2\n", + "vx1=(3*10**6)+((8.28*10**-10)*(1.026*10**16))\n", + "print \"vx = %0.2e m/s\"%vx1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 : Page 51" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)The electron starts from rest at plate A, therefore, the initial velocity is zero. The velocity of electron on reaching plate B is\n", + "v = sqrt(2*q*V/m) = 8.39e+06 m/s\n", + "(ii)Time taken by the electron to travel from plate A to plate B can be calculated from the average velocity of the electron in transit.The average velocity is,\n", + "vaverage = (Initial velocity + Final velocity) / 2 = 4.19e+06 m/s\n", + "Therefore, time taken for travel is,\n", + "Time = Separation between the plates / Average velocity = 7.16e-10 seconds\n", + "(iii)Kinetic energy of the electron on reaching the plate B is\n", + "Kinetic energy = q V = 3.20e-17 Joules\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "V=200\n", + "m=9.1*10**-31\n", + "v=sqrt(2*q*V/m)\n", + "print \"(i)The electron starts from rest at plate A, therefore, the initial velocity is zero. The velocity of electron on reaching plate B is\"\n", + "print \"v = sqrt(2*q*V/m) = %0.2e m/s\"%v\n", + "iv=0 #initial velocity\n", + "fv=8.38*10**6 #final velocity\n", + "va=(iv+fv)/2 #average velocity of electron in transit\n", + "print \"(ii)Time taken by the electron to travel from plate A to plate B can be calculated from the average velocity of the electron in transit.The average velocity is,\"\n", + "print \"vaverage = (Initial velocity + Final velocity) / 2 = %0.2e m/s\"%va\n", + "sp=3*10**-3 #separation between the plates\n", + "time=sp/va\n", + "print \"Therefore, time taken for travel is,\"\n", + "print \"Time = Separation between the plates / Average velocity = %0.2e seconds\"%time\n", + "ke=q*V\n", + "print \"(iii)Kinetic energy of the electron on reaching the plate B is\"\n", + "print \"Kinetic energy = q V = %0.2e Joules\"%ke" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 : Page 51" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed acquired by electron due to the applied voltage is\n", + "v = sqrt(vinitial**2+(2*q*V/m)) = 1.03e+07 m/s\n", + "The average velocity,\n", + "vaverage = (vinitial + vfinal) / 2 = 5.66e+06 m/s\n", + "Therefore, time for travel = seperation between plates / vaverage = 1.41e-09 seconds\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "vinitial=1*10**6\n", + "q=1.6*10**-19\n", + "V=300\n", + "m=9.1*10**-31\n", + "vfinal=10.33*10**6\n", + "sp=8*10**-3 #separation between plates\n", + "v=sqrt(vinitial**2+(2*q*V/m))\n", + "print \"The speed acquired by electron due to the applied voltage is\"\n", + "print \"v = sqrt(vinitial**2+(2*q*V/m)) = %0.2e m/s\"%v\n", + "va=(vinitial+vfinal)/2\n", + "print \"The average velocity,\"\n", + "print \"vaverage = (vinitial + vfinal) / 2 = %0.2e m/s\"%va\n", + "time=sp/va\n", + "print \"Therefore, time for travel = seperation between plates / vaverage = %0.2e seconds\"%time" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 : Page 52" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The electric field intensity,\n", + "E = -5t / d*10*-9 = -5t / 10**-9*1*10**-2 = 5*10**11*t (for 0 < t < t1)\n", + " = 0 (for t1 < t < infinity)\n", + "(i) The position of the electron after 1ns,\n", + " d(um) = (5*10**11)*(1.76*10**11)*((1*10**-9)**3/6) = 14.67 um\n", + "(ii) The rest of the distance to be covered by the electron = 0.8cm - 14.7 um = 0.80\n", + "Since, the potential difference drops to zero volt, after 1ns, the electron will travel the distance of 0.799 cm with a constant velocity of\n", + "vx = (5*10**11)*(q/m)*(t**2/2) = 4.40e+04 m/s\n", + "Therefore, the time t2 = d / vx = 1.81e-07 seconds\n", + "The total time of transit of electron from cathode to anode = 1.82e-07 seconds\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "d=(5*10**11*1.76*10**11)*(((1*10**-9)**3)/6)\n", + "x1=d*10**6\n", + "print \"The electric field intensity,\"\n", + "print \"E = -5t / d*10*-9 = -5t / 10**-9*1*10**-2 = 5*10**11*t (for 0 < t < t1)\"\n", + "print \" = 0 (for t1 < t < infinity)\"\n", + "print \"(i) The position of the electron after 1ns,\"\n", + "print \" d(um) = (5*10**11)*(1.76*10**11)*((1*10**-9)**3/6) = %0.2f um\"%x1\n", + "x2=0.8-(d*10**2)\n", + "print \"(ii) The rest of the distance to be covered by the electron = 0.8cm - 14.7 um = %0.2f\"%x2\n", + "print \"Since, the potential difference drops to zero volt, after 1ns, the electron will travel the distance of 0.799 cm with a constant velocity of\"\n", + "vx=(5*10**11*1.76*10**11)*(((1*10**-9)**2)/2)\n", + "print \"vx = (5*10**11)*(q/m)*(t**2/2) = %0.2e m/s\"%vx\n", + "x3=(x2/vx)*10**-2\n", + "print \"Therefore, the time t2 = d / vx = %0.2e seconds\"%x3\n", + "x4=(1*10**-9)+x3\n", + "print \"The total time of transit of electron from cathode to anode = %0.2e seconds\"%x4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7 : Page 56" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of the electron is = sqrt(2qVa/m) = 3.75e+06 m/s\n", + "The time taken for one revolution is T = 2*pi*m / B*q = 3.93e-11 seconds\n", + "The pitch = T*v*cos(theta) = 1.28e-04 meters\n", + "Thus, the electron has travelled = 1.28e-04 meters\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "q=1.6*10**-19\n", + "Va=40\n", + "m=9.1*10**-31\n", + "B=0.91\n", + "ve=sqrt(2*q*Va/m)\n", + "print \"The velocity of the electron is = sqrt(2qVa/m) = %0.2e m/s\"%ve\n", + "tt=(2*pi*m)/(B*q)\n", + "print \"The time taken for one revolution is T = 2*pi*m / B*q = %0.2e seconds\"%tt\n", + "p=tt*ve*(sqrt(3)/2) #cos(30)=sqrt(3)/2\n", + "print \"The pitch = T*v*cos(theta) = %0.2e meters\"%p\n", + "print \"Thus, the electron has travelled = %0.2e meters\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.8 : Page 56" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The velocity of the charged particle before entering the field is,\n", + "v = sqrt(2aV/m) * sqrt(2(3q)V/2m) = sqrt(6qV/2m) = 5.14e+06 m/s\n", + "(ii) The radius of the helical path is\n", + "r = Mvsine(theta) / QB = 2mvsine(theta) / 3qB = 0.41 mm\n", + "(iii) Time for one revolution,\n", + "T = 2*pi*M / B*Q = 2*pi*(2m) / B(3q) = 1.19e-09 seconds\n" + ] + } + ], + "source": [ + "from math import radians as rdn, sin,pi,sqrt\n", + "radians=rdn(25)\n", + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "V=50\n", + "Q=3*q\n", + "M=2*m\n", + "v=sqrt(2*Q*V/M)\n", + "print \"(i) The velocity of the charged particle before entering the field is,\"\n", + "print \"v = sqrt(2aV/m) * sqrt(2(3q)V/2m) = sqrt(6qV/2m) = %0.2e m/s\"%v\n", + "B=0.02\n", + "r=(M*v*sin(radians))/(Q*B)\n", + "r1=r*10**3\n", + "print \"(ii) The radius of the helical path is\"\n", + "print \"r = Mvsine(theta) / QB = 2mvsine(theta) / 3qB = %0.2f mm\"%r1\n", + "T=(2*pi*M)/(B*Q)\n", + "print \"(iii) Time for one revolution,\"\n", + "print \"T = 2*pi*M / B*Q = 2*pi*(2m) / B(3q) = %0.2e seconds\"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9 : Page 58" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given, T = 35.5/B *10**-12 s, B = 0.01 Wb/m**3, Va = 900V\n", + "Therefore, T = 3.55*10**-9 s\n", + "Velocity, v(m/s) = sqrt(2qVa/m) = 1.78e+07 m/s\n", + "Radius, r(mm) = mv/qB = v/(q/m)B = 10.11 mm\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "print \"Given, T = 35.5/B *10**-12 s, B = 0.01 Wb/m**3, Va = 900V\"\n", + "print \"Therefore, T = 3.55*10**-9 s\"\n", + "T = 3.55*10**-9\n", + "Va=900\n", + "v=sqrt(2*(1.76*10**11)*900)\n", + "print \"Velocity, v(m/s) = sqrt(2qVa/m) = %0.2e m/s\"%v\n", + "r=(17.799*10**6)/(0.01*1.76*10**11)\n", + "x1=r*10**3\n", + "print \"Radius, r(mm) = mv/qB = v/(q/m)B = %0.2f mm\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.10 : Page 60" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The velocity of the electron, v = 1.45e+07 m/s\n", + "(ii) ma = qE\n", + "Thus, acceleration, a(m/s)= qE / m = (q/m)(Vd/d) = 4.40e+14 m/s\n", + "(iii) The deflection on the screen, D(cm)= ILVd / 2Vad = 1.46 cm\n", + "(iv) Deflection sensitivity(cm/V)= D / Vd = 0.07 cm/V\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "Va=600\n", + "l=3.5\n", + "d=0.8\n", + "L=20\n", + "Vd=20\n", + "format(9)\n", + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "v=sqrt(2*q*Va/m)\n", + "print \"(i) The velocity of the electron, v = %0.2e m/s\"%v\n", + "a=(q/m)*(Vd/d)\n", + "a1=a*10**2\n", + "print \"(ii) ma = qE\"\n", + "print \"Thus, acceleration, a(m/s)= qE / m = (q/m)(Vd/d) = %0.2e m/s\"%a1\n", + "D=(l*L*Vd)/(2*Va*d)\n", + "print\"(iii) The deflection on the screen, D(cm)= ILVd / 2Vad = %0.2f cm\"% D\n", + "Ds=D/Vd\n", + "print \"(iv) Deflection sensitivity(cm/V)= D / Vd = %0.2f cm/V\"%Ds" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.11 : Page 61" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The velocity of the beam, v = sqrt(2qVa / m) = 1.68e+07 m/s\n", + "(ii) The deflection of the beam, D = lLVd / 2dVa\n", + "Therefore, the voltage that must be applied to the plates, Vd = 20.00 V\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "Va=800\n", + "l=2\n", + "d=0.5\n", + "L=20\n", + "D=1\n", + "v=sqrt(2*q*Va/m)\n", + "print \"(i) The velocity of the beam, v = sqrt(2qVa / m) = %0.2e m/s\"%v\n", + "Vd=(D*2*d*Va)/(l*L)\n", + "print \"(ii) The deflection of the beam, D = lLVd / 2dVa\"\n", + "print \"Therefore, the voltage that must be applied to the plates, Vd = %0.2f V\"%Vd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.12 : Page 61" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Velocity of beam, v = sqrt(2qVa/m) = 1.88e+07 m/s\n", + "(ii) Deflection sensitivity = D/Vd\n", + "where D = l*L*Vd / 2*Va*d = 0.01 cm\n", + "Therefore, the deflection sensitivity = 4.00e-04 cm/V\n", + "(iii) To find the angle of deflection, theta :\n", + " tan(theta) = D/L-l\n", + "Therefore, theta = tan**-1(D/L-l) = 0.032 degrees\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import degrees, atan\n", + "v=sqrt((2*(1.6*10**-19)*1000)/(9.1*10**-31))\n", + "print \"(i) Velocity of beam, v = sqrt(2qVa/m) = %0.2e m/s\"%v\n", + "D=((2*10**-2)*(20*10**-2)*25)/(2*1000*(0.5*10**-2))\n", + "print \"(ii) Deflection sensitivity = D/Vd\"\n", + "print \"where D = l*L*Vd / 2*Va*d = %0.2f cm\"%D\n", + "ds=D/25\n", + "print \"Therefore, the deflection sensitivity = %0.2e cm/V\"%ds\n", + "theta=degrees(atan(1/1800))\n", + "print \"(iii) To find the angle of deflection, theta :\"\n", + "print \" tan(theta) = D/L-l\"\n", + "print \"Therefore, theta = tan**-1(D/L-l) = %0.3f degrees\"%theta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.13 : Page 62" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The electron starts moving in the +y direction, but, since acceleration is along the -y direction, its velocity isreduced to zero at time t=t''\n", + "v0y = v0 * cos(theta) = 1.50e+05 m/s\n", + "ay = qE / m = 1.60e+14 m/s**2\n", + "t'' = v0y / ay = 0.94 ns\n" + ] + } + ], + "source": [ + "from math import cos,pi\n", + "v0=3*10**5\n", + "E=910\n", + "theta=60\n", + "m=9.109*10**-31\n", + "q=1.6*10**-19\n", + "print \"The electron starts moving in the +y direction, but, since acceleration is along the -y direction, its velocity isreduced to zero at time t=t''\"\n", + "v0y=v0*cos(theta*pi/180)\n", + "print \"v0y = v0 * cos(theta) = %0.2e m/s\"%v0y\n", + "ay=(q*E)/m\n", + "print \"ay = qE / m = %0.2e m/s**2\"%ay\n", + "tdash=v0y/ay\n", + "x1=tdash*10**9\n", + "print \"t'' = v0y / ay = %0.2f ns\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.14 : Page 62" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The deflection of the spot,\n", + "D = (IBL/sqrt(Va))*sqrt(q/2m) = 0.42 cm\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "D=(((2*10**-2)*(1*10**-4)*(20*10**-2))/sqrt(800))*sqrt((1.6*10**-19)/(2*9.1*10**-31))\n", + "x1=D*10**2\n", + "print \"The deflection of the spot,\"\n", + "print \"D = (IBL/sqrt(Va))*sqrt(q/2m) = %0.2f cm\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.15 : Page 62" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnetostatic deflection, D = (IBL/sqrt(Va))*sqrt(q/2m)\n", + "The electrostatic deflection, D = lLVd / 2dVa\n", + "For returning the beam back to the centre, the electrostatic deflection and the magnetostatic deflection must be equal, i.e.,\n", + "(IBL/sqrt(Va))*sqrt(q/2m) = lLVd / 2dVa\n", + "Therefore,\n", + "Vd = dB*sqrt(2*Va*q/m) = 33.55 V\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "print \"The magnetostatic deflection, D = (IBL/sqrt(Va))*sqrt(q/2m)\"\n", + "print \"The electrostatic deflection, D = lLVd / 2dVa\"\n", + "print \"For returning the beam back to the centre, the electrostatic deflection and the magnetostatic deflection must be equal, i.e.,\"\n", + "print \"(IBL/sqrt(Va))*sqrt(q/2m) = lLVd / 2dVa\"\n", + "print \"Therefore,\"\n", + "Vd=(1*10**-2*2*10**-4)*sqrt((2*800*1.6*10**-19)/(9.1*10**-31))\n", + "print \"Vd = dB*sqrt(2*Va*q/m) = %0.2f V\"%Vd" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch4.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch4.ipynb new file mode 100644 index 00000000..c7f2ac66 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch4.ipynb @@ -0,0 +1,600 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-4 : Semiconductor Diodes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 78 Example 4.1." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The intrinsic conductivity for germanium,\n", + "sigmai(S/cm) = q*ni*(un+up) = 0.02\n", + "(ii) The intrinsic conductivity for silicon,\n", + "sigmai(S/cm)= q*ni*(un+np) =4.32e-06\n" + ] + } + ], + "source": [ + "un1=3800 #mobility of free electrons in pure germanium\n", + "up1=1800 #mobility of free holes in pure germanium\n", + "un2=1300 #mobility of free electrons in pure silicon\n", + "up2=500 #mobility of free holes in pure silicon\n", + "q=1.6*10**-19\n", + "nig=2.5*10**13\n", + "nis=1.5*10**10\n", + "sigma1=q*nig*(un1+up1)\n", + "print \"(i) The intrinsic conductivity for germanium,\"\n", + "print \"sigmai(S/cm) = q*ni*(un+up) = %0.2f\"%sigma1\n", + "sigma2=q*nis*(un2+up2)\n", + "print \"(ii) The intrinsic conductivity for silicon,\"\n", + "print \"sigmai(S/cm)= q*ni*(un+np) =%0.2e\"%sigma2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 79 Example 4.6." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) In intrensic condition, n=p=ni\n", + "Hence, sigma_i = q*ni*(un+up)\n", + "sigma_i = 4.32e-06 S/cm\n", + "(b) Number of silicon atoms/cm**3 = 5*10**22\n", + "Hence, ND = 5.00e+14 cm**-3\n", + "Further, n = ND\n", + "Therefore, p = ni**2/n = ni**2/ND\n", + "p = 4.50e+05 cm**-3\n", + "Thus p << n. Hence p may be neglected while calculating the conductivity.\n", + "Hence, sigma = n*q*un = ND*q*un\n", + "sigma = 0.10 S/cm\n", + "(c) NA = 1.00e+15 cm**-3\n", + "Further, p = NA\n", + "Hence, n = ni**2/p = ni**2/NA\n", + "n = 2.25e+05 cm**-3\n", + "Thus p >> n. Hence n may be neglected while calculating the conductivity.\n", + "Hence, sigma = p*q*up = NA*q*up\n", + "sigma = 0.08 S/cm\n", + "(d) With both types of impurities present simultaneously, the net acceptor impurity density is,\n", + "Na = 5.00e+14 cm**-3\n", + "Hence, sigma = Na*q*up\n", + "sigma = 0.04 S/cm\n" + ] + } + ], + "source": [ + "\n", + "ni=1.5*10**10\n", + "un=1300\n", + "up=500\n", + "q=1.6*10**-19\n", + "nos=5*10**22\n", + "print \"(a) In intrensic condition, n=p=ni\"\n", + "print \"Hence, sigma_i = q*ni*(un+up)\"\n", + "sigma_i = q*ni*(un+up)\n", + "print \"sigma_i = %0.2e S/cm\"%sigma_i\n", + "print \"(b) Number of silicon atoms/cm**3 = 5*10**22\"\n", + "ND=5*10**22/10**8\n", + "print \"Hence, ND = %0.2e cm**-3\"%ND\n", + "print \"Further, n = ND\"\n", + "print \"Therefore, p = ni**2/n = ni**2/ND\"\n", + "p=ni**2/ND\n", + "print \"p = %0.2e cm**-3\"%p # wrong answer in textbook\n", + "print \"Thus p << n. Hence p may be neglected while calculating the conductivity.\"\n", + "print \"Hence, sigma = n*q*un = ND*q*un\"\n", + "sigma=ND*q*un\n", + "print \"sigma = %0.2f S/cm\"%sigma\n", + "NA=(5*10**22)/(5*10**7)\n", + "print \"(c) NA = %0.2e cm**-3\"%NA\n", + "print \"Further, p = NA\"\n", + "print \"Hence, n = ni**2/p = ni**2/NA\"\n", + "n=ni**2/NA\n", + "print \"n = %0.2e cm**-3\"%n\n", + "print \"Thus p >> n. Hence n may be neglected while calculating the conductivity.\"\n", + "print \"Hence, sigma = p*q*up = NA*q*up\"\n", + "sigma1=NA*q*up\n", + "print \"sigma = %0.2f S/cm\"%sigma1\n", + "print \"(d) With both types of impurities present simultaneously, the net acceptor impurity density is,\"\n", + "Na=NA-ND\n", + "print \"Na = %0.2e cm**-3\"%Na\n", + "print \"Hence, sigma = Na*q*up\"\n", + "sigma2=Na*q*up\n", + "print \"sigma = %0.2f S/cm\"%sigma2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 83 Example 4.7." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) n = p = ni = 2.5*10**13 cm**-3\n", + "Therefore, conductivity, sigma = q*ni*(un+np) =0.02 S/cm\n", + "Hence, resistivity rho = 1 / sigma =44.64 (ohm-cm)\n", + "\n", + "(b) ND = 4.40e+15 cm**-3\n", + "Also, n = ND\n", + "Therefore, p = ni**2 / n = ni**2 / ND =142045454545.45 (holes/cm**3)\n", + "Here, as n >> p, p can be neglected.\n", + "Therefore, conductivity, sigma = n*q*un = ND*q*un =2.68 (S/cm)\n", + "Hence, resistivity, rho = 1 / sigma =0.37 (ohm-cm)\n", + "\n", + "(c) NA = 4.40e+14 (cm**-3)\n", + "Also, p = NA\n", + "Therefore, n = ni**2 / p = ni**2 / NA =1420454545454.55 (electrons/cm**3)\n", + "Here, as p >> n, n may be neglected. Then,\n", + "Conductivity, sigma = p*q*up = NA*q*up =0.13 (S/cm)\n", + "Hence, resistivity, rho = 1 / sigma = 7.89 (ohm-cm)\n", + "\n", + "(d) with both p and n type impurities present,\n", + " ND = 4.4*10**15 cm**-3 and NA = 4.4*10**14 cm**-3\n", + "Therefore, the net donor density ND'' is\n", + "ND'' = (ND - NA) =3960000000000000.00 (cm**-3)\n", + "Therefore, effective n = ND'' = 3.96*10**15 cm**-3\n", + "p = ni**2 / N''D =157828282828.28 (cm**-3)\n", + "Here again p(= ni**2 / N''D) is very small compared with N''D and may be neglected in calculating the effective conductivity.\n", + "Therefore, conductivity, sigma = ND''*q*un =2.41 (S/cm)\n", + "Hence, resistivity, rho = 1 / sigma =0.42 (ohm-cm)\n" + ] + } + ], + "source": [ + "ni=2.5*10**13\n", + "un=3800\n", + "up=1800\n", + "nog=4.4*10**22\n", + "q=1.6*10**-19\n", + "sigma=q*ni*(un+up)\n", + "print \"(a) n = p = ni = 2.5*10**13 cm**-3\"\n", + "print \"Therefore, conductivity, sigma = q*ni*(un+np) =%0.2f S/cm\"%sigma\n", + "rho=1/sigma\n", + "print \"Hence, resistivity rho = 1 / sigma =%0.2f (ohm-cm)\"%rho\n", + "ND=(4.4*10**22)/10**7\n", + "print \"\\n(b) ND = %0.2e cm**-3\"%ND\n", + "p=ni**2/ND\n", + "print \"Also, n = ND\"\n", + "print \"Therefore, p = ni**2 / n = ni**2 / ND =%0.2f (holes/cm**3)\"%p\n", + "print \"Here, as n >> p, p can be neglected.\"\n", + "sigma1=ND*q*un\n", + "print \"Therefore, conductivity, sigma = n*q*un = ND*q*un =%0.2f (S/cm)\"%sigma1\n", + "rho1=1/sigma1\n", + "print \"Hence, resistivity, rho = 1 / sigma =%0.2f (ohm-cm)\"%rho1\n", + "\n", + "NA=(4.4*10**22)/10**8\n", + "print \"\\n(c) NA = %0.2e (cm**-3)\"%NA\n", + "print \"Also, p = NA\"\n", + "n=ni**2/NA\n", + "print \"Therefore, n = ni**2 / p = ni**2 / NA =%0.2f (electrons/cm**3)\"%n\n", + "sigma2=NA*q*up\n", + "print \"Here, as p >> n, n may be neglected. Then,\"\n", + "print \"Conductivity, sigma = p*q*up = NA*q*up =%0.2f (S/cm)\"%sigma2\n", + "rho2=1/sigma2\n", + "print \"Hence, resistivity, rho = 1 / sigma = %0.2f (ohm-cm)\"%rho2\n", + "\n", + "print \"\\n(d) with both p and n type impurities present,\"\n", + "print \" ND = 4.4*10**15 cm**-3 and NA = 4.4*10**14 cm**-3\"\n", + "print \"Therefore, the net donor density ND'' is\"\n", + "Nd=ND-NA\n", + "print \"ND'' = (ND - NA) =%0.2f (cm**-3)\"%Nd\n", + "print \"Therefore, effective n = ND'' = 3.96*10**15 cm**-3\"\n", + "p1=ni**2/Nd\n", + "print \"p = ni**2 / N''D =%0.2f (cm**-3)\"%p1\n", + "print \"Here again p(= ni**2 / N''D) is very small compared with N''D and may be neglected in calculating the effective conductivity.\"\n", + "sigma3=Nd*q*un\n", + "print \"Therefore, conductivity, sigma = ND''*q*un =%0.2f (S/cm)\"%sigma3\n", + "rho3=1/sigma3\n", + "print \"Hence, resistivity, rho = 1 / sigma =%0.2f (ohm-cm)\"%rho3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 87 Example 4.8" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "sigma_i = qni(un+up) = 1 / 25*10**4\n", + "\n", + "Therefore, ni = sigma_i / q(un+up) =14492753623.19\n", + "\n", + "Net donor density, ND(= n) = 3.00e+10 (cm**-3)\n", + "Hence, p = ni**2 / ND =7001330252.75 (cm**-3)\n", + "Hence, sigma = q*(n*un + p*up) =0.00\n", + "Therefore, total conduction current density, J = sigma*E =0.00 A/cm**2\n" + ] + } + ], + "source": [ + "un=1250\n", + "up=475\n", + "q=1.6*10**-19\n", + "sigma_i=1/(25*10**4)\n", + "format(9)\n", + "ni=1/((25*10**4)*(1.6*10**-19)*(1250+475))\n", + "print \"sigma_i = qni(un+up) = 1 / 25*10**4\"\n", + "print \"\\nTherefore, ni = sigma_i / q(un+up) =%0.2f\"%ni\n", + "\n", + "ND=(4*10**10)-10**10\n", + "print \"\\nNet donor density, ND(= n) = %0.2e (cm**-3)\"%ND\n", + "p=ni**2/ND\n", + "print \"Hence, p = ni**2 / ND =%0.2f (cm**-3)\"%p\n", + "sigma=(1.6*10**-19)*((1250*3*10**10)+(475*0.7*10**10))\n", + "print \"Hence, sigma = q*(n*un + p*up) =%0.2f\"%sigma\n", + "\n", + "J=6.532*4*10**-6\n", + "print \"Therefore, total conduction current density, J = sigma*E =%0.2f A/cm**2\"%J" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 92 Example 4.9." + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Concentration in N-type silicon\n", + "The conductivity of an N-type Silicon is sigma = q*n*un\n", + "Concentratoin of electrons, n = sigma / q*un =1.44e+18 (cm**-3)\n", + "Hence, concentration of holes, p = ni**2 / n =1.56e+02 (cm**-3)\n", + "(b) Concentration in P-type silicon\n", + "The conductivity of a P-type Silicon is sigma = q*p*up\n", + "Hence, concentratoin of holes, p = sigma / q*up =3.75e+18 (cm**-3)\n", + "and concentration of electrons, n = ni**2 / p = 60.00 (cm**-3)\n" + ] + } + ], + "source": [ + "ni=1.5*10**10\n", + "un=1300\n", + "up=500\n", + "q=1.6*10**-19\n", + "sigma=300\n", + "print \"(a) Concentration in N-type silicon\"\n", + "format(10)\n", + "n=sigma/(q*un)\n", + "print \"The conductivity of an N-type Silicon is sigma = q*n*un\"\n", + "print \"Concentratoin of electrons, n = sigma / q*un =%0.2e (cm**-3)\"%n\n", + "p=ni**2/n\n", + "print \"Hence, concentration of holes, p = ni**2 / n =%0.2e (cm**-3)\"%p\n", + "print \"(b) Concentration in P-type silicon\"\n", + "p=sigma/(q*up)\n", + "print \"The conductivity of a P-type Silicon is sigma = q*p*up\"\n", + "print \"Hence, concentratoin of holes, p = sigma / q*up =%0.2e (cm**-3)\"%p\n", + "n=ni**2/p\n", + "print \"and concentration of electrons, n = ni**2 / p = %0.2f (cm**-3)\"%n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 93 Example 4.10." + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Density of added impurity atoms is, ND = 4.20e+22 (atoms/m**3)\n", + "Also, n = ND\n", + "Therefore, p = ni**2 / n = ni**2 / ND =1.49e+16 (m**-3)\n", + "Here, as p << n, p may be neglected.\n", + "Therefore, sigma = q*ND*un =2553.60 (S/m)\n", + "Therefore, resistivity, rho = 1 / sigma =0.00 ohm-m\n", + "Resistance, R = rho*L / A =78.32 kohm\n", + "Voltage drop, V = RI =78.32 mV\n" + ] + } + ], + "source": [ + "ND=(4.2*10**28)/10**6\n", + "print \"Density of added impurity atoms is, ND = %0.2e (atoms/m**3)\"%ND\n", + "ni=2.5*10**19\n", + "p=ni**2/ND\n", + "print \"Also, n = ND\"\n", + "print \"Therefore, p = ni**2 / n = ni**2 / ND =%0.2e (m**-3)\"%p\n", + "print \"Here, as p << n, p may be neglected.\"\n", + "q=1.6*10**-19\n", + "un=0.38\n", + "sigma=q*ND*un\n", + "print \"Therefore, sigma = q*ND*un =%0.2f (S/m)\"%sigma\n", + "\n", + "rho=1/sigma\n", + "print \"Therefore, resistivity, rho = 1 / sigma =%0.2f ohm-m\"%rho\n", + "\n", + "L=5*10**-3\n", + "A=5*10**-6\n", + "R=(rho*L)/A**2\n", + "R1=R*10**-3\n", + "print \"Resistance, R = rho*L / A =%0.2f kohm\"%R1\n", + "I=10**-6\n", + "V=R*I\n", + "V1=V*10**3\n", + "print \"Voltage drop, V = RI =%0.2f mV\"%V1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 94 Example 4.11." + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Resistivity, rho = 1 / sigma = 1 / NA*q*up = 6 ohm-cm\n", + "Therefore, NA = 1 / 6*q*up =5.79e+14 (1/cm**3)\n", + "Similarly, ND(1/cm**3) = 1 / 4*q*un =4.11e+14 (1/cm**3)\n", + "Therefore, Va = VT*ln(ND*NA / ni**2) = 0.15 V\n", + "Hence, Eo = 0.15 eV\n", + "\n", + "(b) Vo = 0.026*ln(2*ND*2*NA / ni**2) =0.19 V\n", + "Therefore, Eo(eV) = 0.19 eV\n" + ] + } + ], + "source": [ + "from math import log\n", + "q=1.6*10**-19\n", + "ni=2.5*10**13\n", + "up=1800\n", + "un=3800\n", + "VT=0.026\n", + "rho=6\n", + "format(9)\n", + "NA=1/(6*q*up)\n", + "print \"(a) Resistivity, rho = 1 / sigma = 1 / NA*q*up = 6 ohm-cm\"\n", + "print \"Therefore, NA = 1 / 6*q*up =%0.2e (1/cm**3)\"%NA\n", + "ND=1/(4*q*un)\n", + "print \"Similarly, ND(1/cm**3) = 1 / 4*q*un =%0.2e (1/cm**3)\"%ND\n", + "Va=VT*log((ND*NA)/ni**2)\n", + "print \"Therefore, Va = VT*ln(ND*NA / ni**2) = %0.2f V\"%Va\n", + "print \"Hence, Eo = %0.2f eV\"%Va\n", + "Va1=0.026*log((2*ND*2*NA)/ni**2)\n", + "print \"\\n(b) Vo = 0.026*ln(2*ND*2*NA / ni**2) =%0.2f V\"%Va1\n", + "print \"Therefore, Eo(eV) = %0.2f eV\"%Va1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 97 Example 4.12." + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current flowing through the PN diode under forward bias is,\n", + "I = Io*(e**40*VF - 1) =120.73 uA\n" + ] + } + ], + "source": [ + "from math import exp\n", + "Ia=0.3*10**-6\n", + "VF=0.15\n", + "I=Ia*((exp(40*VF))-1)\n", + "I1=I*10**6\n", + "print \"The current flowing through the PN diode under forward bias is,\"\n", + "print \"I = Io*(e**40*VF - 1) =%0.2f uA\"%I1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 97 Example 4.13." + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volt-equivalent of the temperature(T) is,\n", + "VT(V) = T / 11600 = 0.03\n", + "Therefore, the diode current, I = Io*e**((VF/eta*VT)-1) =1.18 A\n" + ] + } + ], + "source": [ + "from math import exp\n", + "VF=0.6\n", + "T=298\n", + "Io=10**-5\n", + "eta=2\n", + "VT=T/11600.0\n", + "print \"The volt-equivalent of the temperature(T) is,\"\n", + "print \"VT(V) = T / 11600 = %0.2f\"%VT\n", + "I=Io*((exp((VF/(eta*VT))))-1)\n", + "print \"Therefore, the diode current, I = Io*e**((VF/eta*VT)-1) =%0.2f A\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 98 Example 4.16." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Forward resistance of a PN junction diode, rf = (eta*VT)/I where VT = T/11600 and eta = 2 for silicon\n", + "Therefore, rf = 2*(T/11600) / 5*10**-3\n", + "rf = 10.34 ohm\n" + ] + } + ], + "source": [ + "I=5*10**-3\n", + "T=300\n", + "print \"Forward resistance of a PN junction diode, rf = (eta*VT)/I where VT = T/11600 and eta = 2 for silicon\"\n", + "print \"Therefore, rf = 2*(T/11600) / 5*10**-3\"\n", + "eta=2 #for silicon\n", + "rf=600/(11600*5*10**-3)\n", + "print \"rf = %0.2f ohm\"%rf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 100 Example 4.17." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The saturation current at 400 K is,\n", + "Io2 = Io1 * 2**((T2-T1)/10)\n", + " = 7.5*10**-6 * 2**(127-27/10)\n", + "Io2 = 7.68 mA\n" + ] + } + ], + "source": [ + "Io1=7.5*10**-6\n", + "T1=27\n", + "T2=127\n", + "print \"The saturation current at 400 K is,\"\n", + "print \"Io2 = Io1 * 2**((T2-T1)/10)\"\n", + "print \" = 7.5*10**-6 * 2**(127-27/10)\"\n", + "Io2=Io1*(2**((T2-T1)/10))\n", + "I=Io2*10**3\n", + "print \"Io2 = %0.2f mA\"%I" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch5.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch5.ipynb new file mode 100644 index 00000000..bcea886e --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch5.ipynb @@ -0,0 +1,70 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-5 : Special Diodes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 132 Exmaple 5.1." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The barrier height for N-type material is,\n", + " Theta_BN = Theta_M - Chi = 0.25 V\n", + "The built-in potential is given by,\n", + " Theta_IN = Theta_BN - (kT/q)*ln(NC/ND) =-0.22 V\n" + ] + } + ], + "source": [ + "from math import log\n", + "thetaM=4.26 #work function\n", + "chi=4.01 #electron affinity\n", + "thetaBN=thetaM-chi\n", + "print \"The barrier height for N-type material is,\"\n", + "print \" Theta_BN = Theta_M - Chi = %0.2f V\"%thetaBN\n", + "thetaIN=thetaBN-((((1.38*10**-23)*300)/(1.6*10**-19)))*log((2.8*10**25)/(4*10**17))\n", + "print \"The built-in potential is given by,\"\n", + "print \" Theta_IN = Theta_BN - (kT/q)*ln(NC/ND) =%0.2f V\"%thetaIN \n", + "# answer in the textbook is wrong, even if we take log10 we get a answer 0.047." + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch6.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch6.ipynb new file mode 100644 index 00000000..03eb7a71 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch6.ipynb @@ -0,0 +1,1581 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-6 : Bipolar Junction Transistor " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 151 Example 6.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter current is,\n", + "IE = IB + IC\n", + "10 = IB + 9.8\n", + "Therefore, IB(mA) =0.20\n" + ] + } + ], + "source": [ + "IE=10\n", + "IC=9.8\n", + "print \"The emitter current is,\"\n", + "print \"IE = IB + IC\"\n", + "print \"10 = IB + 9.8\"\n", + "IB=IE-IC\n", + "print \"Therefore, IB(mA) =%0.2f\"%IB" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 152 Example 6.2." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common-base d.c. current gain,\n", + "alpha = IC/IE = 0.9873\n" + ] + } + ], + "source": [ + "IE=6.28\n", + "IC=6.20\n", + "print \"The common-base d.c. current gain,\"\n", + "alpha=IC/IE\n", + "print \"alpha = IC/IE = %0.4f\"%alpha" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 155 Example 6.3." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common-base d.c. current gain (alpha) is,\n", + "alpha = 0.967 = IC/IE = IC/10\n", + "\n", + "Therefore, IC = 9.67 mA\n", + "The emitter current, IE = IB + IC\n", + "Therefore, IB =0.33 mA\n" + ] + } + ], + "source": [ + "alpha=0.967\n", + "IE=10\n", + "print \"The common-base d.c. current gain (alpha) is,\"\n", + "print \"alpha = 0.967 = IC/IE = IC/10\"\n", + "IC=alpha*IE\n", + "print \"\\nTherefore, IC = %0.2f mA\"%IC\n", + "print \"The emitter current, IE = IB + IC\"\n", + "IB=IE-IC\n", + "print \"Therefore, IB =%0.2f mA\"%IB" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 156 Example 6.4." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common-base d.c. current gain, alpha = IC/IE\n", + "Therefore, IC =9.80 mA\n", + "The emitter current, IE = IB + IC\n", + "Therefore, IB =0.20 mA\n" + ] + } + ], + "source": [ + "IE=10\n", + "alpha=0.98\n", + "print \"The common-base d.c. current gain, alpha = IC/IE\"\n", + "IC=alpha*IE\n", + "print \"Therefore, IC =%0.2f mA\"%IC\n", + "print \"The emitter current, IE = IB + IC\"\n", + "IB=IE-IC\n", + "print \"Therefore, IB =%0.2f mA\"%IB" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 158 Example 6.5." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "If alpha=0.97, beta = alpha/(1 - alpha)\n", + "beta = 32.3333\n", + "If beta=200, alpha = beta/(beta + 1)\n", + "alpha =0.0000\n" + ] + } + ], + "source": [ + "alpha=0.97\n", + "print \"If alpha=0.97, beta = alpha/(1 - alpha)\"\n", + "beta=alpha/(1-alpha)\n", + "print \"beta = %0.4f\"%beta\n", + "beta1=200\n", + "print \"If beta=200, alpha = beta/(beta + 1)\"\n", + "alpha1 =beta1/(beta1+1)\n", + "print \"alpha =%0.4f\"%alpha1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 158 Example 6.6." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "beta = 100 = IC / IB\n", + "Therefore, IB =0.40 mA\n", + "IE = IB + IC\n", + "IE =40.40 mA\n" + ] + } + ], + "source": [ + "beta=100.0\n", + "IC=40\n", + "print \"beta = 100 = IC / IB\"\n", + "IB=IC/beta\n", + "print \"Therefore, IB =%0.2f mA\"%IB\n", + "print \"IE = IB + IC\"\n", + "IE=IB+IC\n", + "print \"IE =%0.2f mA\"%IE" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 159 Example 6.7." + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common-base current gain, alpha = beta / (beta + 1) =0.9934\n", + "Also, alpha = IC / IE\n", + "Therefore, IC = 9.93 mA\n", + "the emitter current, IE = IB + IC\n", + "Therefore, IB =0.07 mA\n" + ] + } + ], + "source": [ + "beta=150.\n", + "IE=10\n", + "alpha=beta/(beta+1)\n", + "print \"The common-base current gain, alpha = beta / (beta + 1) =%0.4f\"%alpha\n", + "print \"Also, alpha = IC / IE\"\n", + "IC=alpha*IE\n", + "print \"Therefore, IC = %0.2f mA\"%IC\n", + "print \"the emitter current, IE = IB + IC\"\n", + "IB=IE-IC\n", + "print \"Therefore, IB =%0.2f mA\"%IB" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 160 Example 6.8." + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that (beta), beta = 170 = IC / IB\n", + "Therefore, IB =0.47 mA\n", + "and IE = IB + IC =80.47 mA\n" + ] + } + ], + "source": [ + "beta=170.\n", + "IC=80\n", + "print \"We know that (beta), beta = 170 = IC / IB\"\n", + "IB=IC/beta\n", + "print \"Therefore, IB =%0.2f mA\"%IB\n", + "IE=IB+IC\n", + "print \"and IE = IB + IC =%0.2f mA\"%IE" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 161 Example 6.9." + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "beta = 200 = IC / IB\n", + "Therefore, IC = 25.00 mA\n", + "and IE = IB + IC =25.12 mA\n" + ] + } + ], + "source": [ + "IB=0.125\n", + "beta=200\n", + "print \"beta = 200 = IC / IB\"\n", + "IC=beta*IB\n", + "print \"Therefore, IC = %0.2f mA\"%IC\n", + "IE=IB+IC\n", + "print \"and IE = IB + IC =%0.2f mA\"%IE" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 162 Example 6.10" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that base current, IB = IE / (1 + beta) =0.12 mA\n", + "and collector current, IC = IE - IB =11.88 mA\n" + ] + } + ], + "source": [ + "IE=12.\n", + "beta=100\n", + "IB=IE/(1+beta)\n", + "print \"We know that base current, IB = IE / (1 + beta) =%0.2f mA\"%IB\n", + "IC=IE-IB\n", + "print \"and collector current, IC = IE - IB =%0.2f mA\"%IC" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 162 Example 6.11" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) To find beta of the transistor \n", + "beta = IC / IB = 20.0000\n", + "(b) To find alpha of the transistor\n", + "alpha = beta / (1+beta) = 0.9524\n", + "(c) To find emitter current, IE\n", + "IE = IB + IC = 2.10 mA\n", + "(d) To find the new value of beta when delta_IB = 25uA and delta_IC = 0.6mA\n", + "Therefore, IB = 125.00 uA\n", + " IC = 2.60 mA\n", + "New value of beta of the transistor,\n", + "beta = IC / IB =20.8000\n" + ] + } + ], + "source": [ + "IB=100*10**-6\n", + "IC=2*10**-3\n", + "beta=IC/IB\n", + "print \"(a) To find beta of the transistor \"\n", + "print \"beta = IC / IB = %0.4f\"%beta\n", + "alpha=beta/(beta+1)\n", + "print \"(b) To find alpha of the transistor\"\n", + "print \"alpha = beta / (1+beta) = %0.4f\"%alpha\n", + "IE=IB+IC\n", + "IE1=IE*10**3\n", + "print \"(c) To find emitter current, IE\"\n", + "print \"IE = IB + IC = %0.2f mA\"%IE1\n", + "# answer in the textbook is wrong\n", + "print \"(d) To find the new value of beta when delta_IB = 25uA and delta_IC = 0.6mA\"\n", + "delta_IB=25*10**-6\n", + "delta_IC=0.6*10**-3\n", + "IB1=IB+delta_IB\n", + "IB11=IB1*10**6\n", + "IC1=IC+delta_IC\n", + "IC11=IC1*10**3\n", + "print \"Therefore, IB = %0.2f uA\"%IB11\n", + "print \" IC = %0.2f mA\"%IC11\n", + "beta1=IC1/IB1\n", + "print \"New value of beta of the transistor,\"\n", + "print \"beta = IC / IB =%0.4f\"%beta1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 163 Example 6.12." + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The collector current is, IC = ((alpha*IB)/(1-alpha))+(ICO/(1-alpha)) = 5.15 mA\n", + "The emitter current is, IE = IB + IC =5.25 mA\n" + ] + } + ], + "source": [ + "alpha=0.98\n", + "ICO=5*10**-6\n", + "ICBO=ICO\n", + "IB=100*10**-6\n", + "IC=((alpha*IB)/(1-alpha))+(ICO/(1-alpha))\n", + "IC1=IC*10**3\n", + "print \"The collector current is, IC = ((alpha*IB)/(1-alpha))+(ICO/(1-alpha)) = %0.2f mA\"%IC1\n", + "IE=IB+IC\n", + "IE1=IE*10**3\n", + "print \"The emitter current is, IE = IB + IC =%0.2f mA\"%IE1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 164 Example 6.13." + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) To find the value of collector current when IB = 0.25mA\n", + "IC = (beta*IB) + ((1+beta)*ICBO) = 13.01 mA\n", + "(b) To find the value of new collector current if temperature rises to 50 C\n", + "I''CBO(beta=50) = ICBO*(2**((T2-T1)/10)) =49.25 uA\n", + "Therefore, the collector current at 50 C is\n", + "IC = (beta*IB) + ((1+beta)*I''CBO) =15.01 mA\n" + ] + } + ], + "source": [ + "ICBO=10*10**-6\n", + "hFE=50\n", + "beta=hFE\n", + "IB=0.25*10**-3\n", + "IC=(beta*IB)+((1+beta)*ICBO)\n", + "IC1=IC*10**3\n", + "print \"(a) To find the value of collector current when IB = 0.25mA\"\n", + "print \"IC = (beta*IB) + ((1+beta)*ICBO) = %0.2f mA\"%IC1\n", + "T1=27.\n", + "T2=50.\n", + "I_CBO = ICBO * (2**((T2-T1)/10))\n", + "I_CBO1=I_CBO*10**6\n", + "print \"(b) To find the value of new collector current if temperature rises to 50 C\"\n", + "print \"I''CBO(beta=50) = ICBO*(2**((T2-T1)/10)) =%0.2f uA\"%I_CBO1\n", + "IC2=(beta*IB)+((1+beta)*I_CBO)\n", + "IC3=IC2*10**3\n", + "print \"Therefore, the collector current at 50 C is\"\n", + "print \"IC = (beta*IB) + ((1+beta)*I''CBO) =%0.2f mA\"%IC3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 165 Example 6.14." + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current gain of the transistor is alpha = delta_IC/delta_IE = 0.9900\n" + ] + } + ], + "source": [ + "delta_IC=0.99*10**-3\n", + "delta_IE=1*10**-3\n", + "alpha=delta_IC/delta_IE\n", + "print \"The current gain of the transistor is alpha = delta_IC/delta_IE = %0.4f\"%alpha" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 165 Example 6.15" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The d.c. current gain of the transistor in CB mode is, alpha_dc = beta_dc/(1+beta_dc) =0.99\n" + ] + } + ], + "source": [ + "beta_dc=100.\n", + "alpha_dc=beta_dc/(1+beta_dc)\n", + "print \"The d.c. current gain of the transistor in CB mode is, alpha_dc = beta_dc/(1+beta_dc) =%0.2f\"%alpha_dc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 165 Example 6.16." + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Common base current gain is, alpha = delta_IC/delta_IE =0.99\n", + "Common-emitter current gain is beta = alpha / (1-alpha) =199.00\n" + ] + } + ], + "source": [ + "delta_IC=0.995*10**-3\n", + "delta_IE=1*10**-3\n", + "alpha=delta_IC/delta_IE\n", + "print \"Common base current gain is, alpha = delta_IC/delta_IE =%0.2f\"%alpha\n", + "beta=alpha/(1-alpha)\n", + "print \"Common-emitter current gain is beta = alpha / (1-alpha) =%0.2f\"%beta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 166 Example 6.17." + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that, alpha = beta/(1+beta)\n", + "Therefore, the common base current gain is, alpha =0.98\n", + "We also know that, alpha = IC / IE\n", + "Therefore, IC = alpha * IE = 2.94 mA\n" + ] + } + ], + "source": [ + "beta=49.\n", + "alpha=beta/(1+beta)\n", + "print \"We know that, alpha = beta/(1+beta)\"\n", + "print \"Therefore, the common base current gain is, alpha =%0.2f\"%alpha\n", + "print \"We also know that, alpha = IC / IE\"\n", + "IE=3*10**-3\n", + "IC=alpha*IE\n", + "IC1=IC*10**3\n", + "print \"Therefore, IC = alpha * IE = %0.2f mA\"%IC1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 166 Example 6.18." + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The collector current, IC(mA) = beta * IB =2.25 mA\n", + "The emitter current, IE(mA) = IC + IB =2.26 mA\n", + "Common-base current gain, alpha = beta/(1+beta) = 0.99\n" + ] + } + ], + "source": [ + "IB=15*10**-6\n", + "beta=150.\n", + "IC=beta*IB\n", + "IC1=IC*10**3\n", + "print \"The collector current, IC(mA) = beta * IB =%0.2f mA\"%IC1\n", + "IE=IC+IB\n", + "IE1=IE*10**3\n", + "print \"The emitter current, IE(mA) = IC + IB =%0.2f mA\"%IE1\n", + "alpha=beta/(1+beta)\n", + "print \"Common-base current gain, alpha = beta/(1+beta) = %0.2f\"%alpha" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 167 Example 6.19." + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Referring to fig.6.18, the base current is,\n", + "IB = (VBB - VBE) / RB =16.50 uA\n", + "The collector current is, IC = beta*IB =3.30 mA\n", + "The emitter current is, IE = IC + IB =3.32 mA\n", + "Therefore, VCE = VCC - IC*RC =3.40 V\n" + ] + } + ], + "source": [ + "print \"Referring to fig.6.18, the base current is,\"\n", + "VBB=4\n", + "VBE=0.7\n", + "RB=200*10**3\n", + "IB=(VBB-VBE)/RB\n", + "IB1=IB*10**6\n", + "print \"IB = (VBB - VBE) / RB =%0.2f uA\"%IB1\n", + "beta=200\n", + "IC=beta*IB\n", + "IC1=IC*10**3\n", + "print \"The collector current is, IC = beta*IB =%0.2f mA\"%IC1\n", + "IE=IC+IB\n", + "IE1=IE*10**3\n", + "print \"The emitter current is, IE = IC + IB =%0.2f mA\"%IE1\n", + "VCC=10\n", + "RC=2*10**3\n", + "VCE=VCC-(IC*RC)\n", + "print \"Therefore, VCE = VCC - IC*RC =%0.2f V\"%VCE" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 167 Example 6.20." + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "2.48 IC = ((alpha_dc*IB)/(1-alpha_dc)) + (ICBO/(1-alpha_dc)) =2.48 mA\n", + "Therefore, IE = IB + IC =2.50 mA\n" + ] + } + ], + "source": [ + "alpha_dc=0.99\n", + "ICBO=5*10**-6\n", + "IB=20*10**-6\n", + "IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))\n", + "IC1=IC*10**3\n", + "print IC1,\"IC = ((alpha_dc*IB)/(1-alpha_dc)) + (ICBO/(1-alpha_dc)) =%0.2f mA\"%IC1\n", + "IE=IB+IC\n", + "IE1=IE*10**3\n", + "print \"Therefore, IE = IB + IC =%0.2f mA\"%IE1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 168 Example 6.21." + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The leakage current ICBO = 0.2 uA\n", + " ICEO = 18 uA\n", + "Assume that IB = 30 mA\n", + "IE = IB + IC\n", + "IC = IE - IB = (beta*IB)+((1+beta)*ICBO)\n", + "We know that, ICEO = ICBO/(1-alpha) = (1+beta)*ICBO\n", + "beta = (ICEO / ICBO)-1 =89.00\n", + "IC = (beta*IB) + ((1+beta)*ICBO) =2.67 A\n", + "alpha_dc = 1 - (ICBO / ICEO) =0.99\n", + "beta_dc = (IC-ICBO) / (IB-ICEO) =89.05\n" + ] + } + ], + "source": [ + "ICBO=0.2*10**-6\n", + "ICEO=18*10**-6\n", + "IB=30*10**-3\n", + "print \"The leakage current ICBO = 0.2 uA\"\n", + "print \" ICEO = 18 uA\"\n", + "print \"Assume that IB = 30 mA\"\n", + "print \"IE = IB + IC\"\n", + "print \"IC = IE - IB = (beta*IB)+((1+beta)*ICBO)\"\n", + "print \"We know that, ICEO = ICBO/(1-alpha) = (1+beta)*ICBO\"\n", + "beta=(ICEO/ICBO)-1\n", + "print \"beta = (ICEO / ICBO)-1 =%0.2f\"%beta\n", + "IC=(beta*IB)+((1+beta)*ICBO)\n", + "print \"IC = (beta*IB) + ((1+beta)*ICBO) =%0.2f A\"%IC\n", + "alpha_dc=1-(ICBO/ICEO)\n", + "print \"alpha_dc = 1 - (ICBO / ICEO) =%0.2f\"%alpha_dc\n", + "beta_dc=(IC-ICBO)/(IB-ICEO)\n", + "print \"beta_dc = (IC-ICBO) / (IB-ICEO) =%0.2f\"%beta_dc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 168 Example 6.22." + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Assume that, IB = 1 mA\n", + "IC = ((alpha_dc*IB) / (1-alpha_dc)) + (ICBO/(1-alpha_dc)) = 104.00 mA\n", + "IE = IC + IB = 105.00 mA\n" + ] + } + ], + "source": [ + "alpha_dc=0.99\n", + "ICBO=50*10**-6\n", + "IB=1*10**-3\n", + "IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))\n", + "IC1=IC*10**3\n", + "print \"Assume that, IB = 1 mA\"\n", + "print \"IC = ((alpha_dc*IB) / (1-alpha_dc)) + (ICBO/(1-alpha_dc)) = %0.2f mA\"%IC1\n", + "IE=IC+IB\n", + "IE1=IE*10**3\n", + "print \"IE = IC + IB = %0.2f mA\"%IE1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 169 Example 6.23." + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) DC load line:\n", + "Maximum VCE = VCC = 24V\n", + "Maximum IC = VCC / RC = 0.00 mA\n", + "(ii) For fixing the optimum operating point Q, mark the middle of the d.c. load line AB and the corresponding VCE and IC values can be found\n", + "Here, VCEQ(V) = VCC / 2 = 12.00 V\n", + " ICQ = 1.5 mA\n", + "\n", + "(iii) AC load line\n", + "AC load, R_a.c. = RC || RL = 6.00 kohm\n", + "Therefore, maximum VCE(V) = VCEQ + ICQ*R_a.c. = 21.00 \n", + "This locates the point D(OD = 21V) on the VCE axis\n", + "Maximum IC = ICQ + VCEQ/R_a.c. = 1.50 mA\n", + "This locates the point C(OC = 3.5mA) on the IC axis. By joining points C and D a.c. load line CD is constructed. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1e72637b90>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend\n", + "print \"(i) DC load line:\"\n", + "print \"Maximum VCE = VCC = 24V\"\n", + "IC=24/(8*10**3) #in Ampere\n", + "x1=IC*10**3 #in mA\n", + "print \"Maximum IC = VCC / RC = %0.2f mA\"%x1\n", + "print \"(ii) For fixing the optimum operating point Q, mark the middle of the d.c. load line AB and the corresponding VCE and IC values can be found\"\n", + "VCEQ=24./2\n", + "print \"Here, VCEQ(V) = VCC / 2 = %0.2f V\"%VCEQ #in volts\n", + "print \" ICQ = 1.5 mA\"\n", + "print \"\"\n", + "print \"(iii) AC load line\"\n", + "Rac=(8*24.)/(8+24) #in k-ohm\n", + "print \"AC load, R_a.c. = RC || RL = %0.2f kohm\"%Rac\n", + "VCE=12+((1.5*10**-3)*(6*10**3)) #in Volts\n", + "print \"Therefore, maximum VCE(V) = VCEQ + ICQ*R_a.c. = %0.2f \"%VCE\n", + "print \"This locates the point D(OD = 21V) on the VCE axis\"\n", + "IC=(1.5*10**-3)+(12/(6*10**3)) #in Ampere\n", + "x3=IC*10**3 #in mA\n", + "print \"Maximum IC = ICQ + VCEQ/R_a.c. = %0.2f mA\"%x3\n", + "print \"This locates the point C(OC = 3.5mA) on the IC axis. By joining points C and D a.c. load line CD is constructed. \"\n", + "x=[24,0]\n", + "y=[0,3]\n", + "plot(x,y)\n", + "x1=[21,0]\n", + "y1=[0,3.5]\n", + "plot(x1,y1)\n", + "title(\"Fig.6.22(b)\")\n", + "xlabel(\"VCE(V)\")\n", + "ylabel(\"IC(mA)\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 170 Example 6.24." + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) DC load line:\n", + "Maximum VCE = VCC = 12V, which locates the point B(OB = 12V) of the d.c. load line\n", + "Maximum IC = VCC / (RC+RE) = 0.00 mA\n", + "(ii) Operating point Q\n", + "Therefore, V2 = 0.00 V\n", + " V2 = VBE + IE*RE\n", + "Therefore, IE = V2-VBE / RE = 3.30 mA\n", + " IC = IE =3.30 mA\n", + "VCE = VCC - IC(RC+RE) = 5.40 V\n", + "(iii) AC load line\n", + "AC load, Ra.c.(k-ohm) = RC || RL = 0.60 kohm\n", + "Therefore, maximum VCE = VCEQ + ICQ*Ra.c. = 7.38 V\n", + "Maximum IC(mA) = ICQ + VCEQ/Ra.c. = 12.30 mA\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1e723cb6d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend\n", + "print \"(i) DC load line:\"\n", + "print \"Maximum VCE = VCC = 12V, which locates the point B(OB = 12V) of the d.c. load line\"\n", + "IC=12/(2*10**3) #in Ampere\n", + "x1=IC*10**3 #in mA\n", + "print \"Maximum IC = VCC / (RC+RE) = %0.2f mA\"%x1\n", + "\n", + "print \"(ii) Operating point Q\"\n", + "V2=((4*10**3)/(12*10**3))*12 #in V\n", + "print \"Therefore, V2 = %0.2f V\"%V2\n", + "print \" V2 = VBE + IE*RE\"\n", + "IE=(4-0.7)/(1*10**3) #in Ampere\n", + "x2=IE*10**3 #in mA\n", + "print \"Therefore, IE = V2-VBE / RE = %0.2f mA\"%x2\n", + "IC=x2 #in mA\n", + "print \" IC = IE =%0.2f mA\"%IC\n", + "VCE=12-((3.3*10**-3)*(2*10**3)) #in volts\n", + "print \"VCE = VCC - IC(RC+RE) = %0.2f V\"%VCE\n", + "print \"(iii) AC load line\"\n", + "Rac=1.5/2.5 #in k-ohm\n", + "print \"AC load, Ra.c.(k-ohm) = RC || RL = %0.2f kohm\"%Rac\n", + "VCE=5.4+((3.3*10**-3)*(0.6*10**3)) #in Volts\n", + "print \"Therefore, maximum VCE = VCEQ + ICQ*Ra.c. = %0.2f V\"%VCE\n", + "IC=(3.3*10**-3)+(5.4/(0.6*10**3)) #in Ampere\n", + "x3=IC*10**3 #in mA\n", + "print \"Maximum IC(mA) = ICQ + VCEQ/Ra.c. = %0.2f mA\"%x3\n", + "x=[7.38,0]\n", + "y=[0,12.3]\n", + "plot(x,y)\n", + "x1=[12,0]\n", + "y1=[0,6]\n", + "plot(x1,y1)\n", + "title(\"Fig.6.23(b)\")\n", + "xlabel(\"VCE(V) -->\")\n", + "ylabel(\"IC(mA) -->\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 171 Example 6.25." + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The collector resistance is, RC = (VCC - VCEQ) / ICQ =4.00 kohm\n", + "The base current is, IBQ = ICQ / beta =10.00 uA\n", + "The base resistance is, RB = (VCC - VBE(on)) / IBQ =0.93 Mohm\n" + ] + } + ], + "source": [ + "ICQ=1*10**-3\n", + "VCEQ=6.\n", + "VCC=10.\n", + "beta=100.\n", + "VBE=0.7\n", + "RC=(VCC-VCEQ)/ICQ\n", + "RC1=RC*10**-3\n", + "RC2=round(RC1)\n", + "print \"The collector resistance is, RC = (VCC - VCEQ) / ICQ =%0.2f kohm\"%RC2\n", + "IBQ=ICQ/beta\n", + "IBQ1=IBQ*10**6\n", + "print \"The base current is, IBQ = ICQ / beta =%0.2f uA\"%IBQ1\n", + "RB=(VCC-VBE)/IBQ\n", + "RB1=RB*10**-6\n", + "print \"The base resistance is, RB = (VCC - VBE(on)) / IBQ =%0.2f Mohm\"%RB1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 171 Example 6.26." + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VBB = IB*RB + VBE(on) + IE*RE\n", + "Also, IE = IB + IC = IB + beta*IB = (1 + beta)*IB\n", + "The base current, IB = (VBB - VBE(on)) / (RB + ((1+beta)*RE)) =53.35 uA\n", + "Therefore, IC = beta*IB =5.33 mA\n", + "IE = IC + IB = 5.39 mA\n", + "VCE = VCC - (IC*RC) - (IE*RE) =4.63 V\n", + "The Q point is at\n", + "VCEQ = 4.63 V\n", + "and ICQ(mA) = 5.33 mA\n" + ] + } + ], + "source": [ + "beta=100\n", + "VBE=0.7\n", + "VCC=10\n", + "RB=20*10**3\n", + "RC=0.4*10**3\n", + "RE=0.6*10**3\n", + "VBB=5\n", + "print \"VBB = IB*RB + VBE(on) + IE*RE\"\n", + "print \"Also, IE = IB + IC = IB + beta*IB = (1 + beta)*IB\"\n", + "IB=(VBB-VBE)/(RB+((1+beta)*RE))\n", + "IB1=IB*10**6\n", + "print \"The base current, IB = (VBB - VBE(on)) / (RB + ((1+beta)*RE)) =%0.2f uA\"%IB1\n", + "IC=beta*IB\n", + "IC1=IC*10**3\n", + "print \"Therefore, IC = beta*IB =%0.2f mA\"%IC1\n", + "IE=IC+IB\n", + "IE1=IE*10**3\n", + "print \"IE = IC + IB = %0.2f mA\"%IE1\n", + "VCE=VCC-(IC*RC)-(IE*RE)\n", + "print \"VCE = VCC - (IC*RC) - (IE*RE) =%0.2f V\"%VCE\n", + "print \"The Q point is at\"\n", + "print \"VCEQ = %0.2f V\"%VCE\n", + "print \"and ICQ(mA) = %0.2f mA\"%IC1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 172 Example 6.27. " + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) DC load line:\n", + "When VCE = 0, IC = VCC/RC = 0.00 mA\n", + "\n", + "(ii) Operating point Q:\n", + "Therefore, IB = VCC-VBE / RB = 10.00 uA\n", + "Therefore, IC(mA) = beta*IB = 1.00 mA\n", + " VCE = VCC - IC*RC = 03 V\n", + "Therefore operating point is VCEQ = 3 V and ICQ = 1 mA\n", + "\n", + "(iii) Stability factor: S = 1 + beta = 1 + 100 = 101\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1e72288a10>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend\n", + "print \"(i) DC load line:\"\n", + "IC=6/(3*10**3) #in Ampere\n", + "x1=IC*10**3 #in mA\n", + "print \"When VCE = 0, IC = VCC/RC = %0.2f mA\"%x1\n", + "print \"\"\n", + "print \"(ii) Operating point Q:\"\n", + "IB=(6-0.7)/(530*10**3)\n", + "x2=IB*10**6\n", + "print \"Therefore, IB = VCC-VBE / RB = %0.2f uA\"%x2\n", + "IC=100*10*10**-6 # in Ampere\n", + "x3=IC*10**3 # in mA\n", + "print \"Therefore, IC(mA) = beta*IB = %0.2f mA\"%x3\n", + "VCE=6-((1*10**-3)*(3*10**3)) # in volts\n", + "print \" VCE = VCC - IC*RC = %02.f V\"%VCE\n", + "print \"Therefore operating point is VCEQ = 3 V and ICQ = 1 mA\"\n", + "print \"\"\n", + "print \"(iii) Stability factor: S = 1 + beta = 1 + 100 = 101\"\n", + "x=[6,0]\n", + "y=[0,2]\n", + "plot(x,y)\n", + "title(\"DC load line\")\n", + "xlabel(\"VCE (V) --->\")\n", + "ylabel(\"IC (mA) --->\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 172 Example 6.28." + ] + }, + { + "cell_type": "code", + "execution_count": 55, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) To determine RB :\n", + "RC = (VCC - VCE) / IC =5.00 kohm\n", + "IB = IC / beta = 10.00 uA\n", + "RB = (VCC - VBE - (IC*RC)) / IB = 630.00 kohm\n", + "(b) Stability factor, S =(1 + beta) / (1 + (beta*(RC / (RC+RB)))) =56.51\n", + "(c) VCC = (beta*IB*RC) + (IB*RB) + VBE\n", + " = IB * ((beta*RC) + RB) + VBE\n", + "IB = (VCC-VBE) / ((beta*RC)+RB) =12.84 uA\n", + "IC = beta*IB =0.64 mA\n", + "VCE = VCC - (IC*RC) =8.79 V\n", + "Therefore the coordinates of new operating point are :\n", + "VCEQ(V) =8.79 V\n", + "ICQ =0.64 mA\n" + ] + } + ], + "source": [ + "VCC=12.\n", + "beta=100.\n", + "VBE=0.7\n", + "print \"(a) To determine RB :\"\n", + "VCE=7\n", + "IC=1*10**-3\n", + "RC=(VCC-VCE)/IC\n", + "RC1=RC*10**-3\n", + "print \"RC = (VCC - VCE) / IC =%0.2f kohm\"%RC1\n", + "IB=IC/beta\n", + "IB1=IB*10**6\n", + "print \"IB = IC / beta = %0.2f uA\"%IB1\n", + "RB=(VCC-VBE-(IC*RC))/IB\n", + "RB1=RB*10**-3\n", + "print \"RB = (VCC - VBE - (IC*RC)) / IB = %0.2f kohm\"%RB1\n", + "S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n", + "print \"(b) Stability factor, S =(1 + beta) / (1 + (beta*(RC / (RC+RB)))) =%0.2f\"%S\n", + "beta1=50\n", + "print \"(c) VCC = (beta*IB*RC) + (IB*RB) + VBE\"\n", + "print \" = IB * ((beta*RC) + RB) + VBE\"\n", + "IB=(VCC-VBE)/((beta1*RC)+RB)\n", + "IB1=IB*10**6\n", + "print \"IB = (VCC-VBE) / ((beta*RC)+RB) =%0.2f uA\"%IB1\n", + "IC=beta1*IB\n", + "IC1=IC*10**3\n", + "print \"IC = beta*IB =%0.2f mA\"%IC1\n", + "VCE=VCC-(IC*RC)\n", + "print \"VCE = VCC - (IC*RC) =%0.2f V\"%VCE\n", + "print \"Therefore the coordinates of new operating point are :\"\n", + "print \"VCEQ(V) =%0.2f V\"%VCE\n", + "print \"ICQ =%0.2f mA\"%IC1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 173 Example 6.29." + ] + }, + { + "cell_type": "code", + "execution_count": 56, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RB = VCEQ / IB =32.00 kohm\n", + "Stability factor, S = (1+beta) / 1 + (beta*(RC/RC+RB)) =56.90\n" + ] + } + ], + "source": [ + "VCC=12.\n", + "RC=250.\n", + "IB=0.25*10**-3\n", + "beta=100.\n", + "VCEQ=8.\n", + "RB=VCEQ/IB\n", + "RB1=RB*10**-3\n", + "print \"RB = VCEQ / IB =%0.2f kohm\"%RB1\n", + "S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n", + "print \"Stability factor, S = (1+beta) / 1 + (beta*(RC/RC+RB)) =%0.2f\"%S" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 173 Example 6.30." + ] + }, + { + "cell_type": "code", + "execution_count": 59, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For a germanium transistor, VBE=0.3V. As alpha=0.985\n", + "beta = alpha / ( 1 - alpha) =66.00\n", + "(a) To find the coordinates of the operating point\n", + "Referring to fig. 6.29,\n", + "Thevenin voltage, VT = (R2 / (R1+R2)) * VCC =0.00 V\n", + "Thevenin resistance, RB = (R1 * R2) / (R1 + R2) = 14.74 kohm\n", + "Therefore, IC =-0.13 mA\n", + "Since IB is very small, IC ~ IE = 1.73 mA\n", + "Therefore, VCE = VCC - (IC*RC) - (IE*RE) =16.67 V\n", + "Therefore, the coordinates of the operating point are :\n", + "IC = -0.13 mA\n", + "VCE =16.67 V\n", + "(b) To find the stability factor S\n", + "S =7.24\n" + ] + } + ], + "source": [ + "VCC=16\n", + "RC=3*10**3\n", + "RE=2*10**3\n", + "R1=56*10**3\n", + "R2=20*10**3\n", + "alpha=0.985\n", + "VBE=0.3\n", + "print \"For a germanium transistor, VBE=0.3V. As alpha=0.985\"\n", + "beta=alpha/(1-alpha)\n", + "beta1=round(beta)\n", + "print \"beta = alpha / ( 1 - alpha) =%0.2f\"%beta1\n", + "print \"(a) To find the coordinates of the operating point\"\n", + "print \"Referring to fig. 6.29,\"\n", + "VT=(R2/(R1+R2))*VCC\n", + "print \"Thevenin voltage, VT = (R2 / (R1+R2)) * VCC =%0.2f V\"%VT\n", + "RB=(R1*R2)/(R1+R2)\n", + "RB1=RB*10**-3\n", + "print \"Thevenin resistance, RB = (R1 * R2) / (R1 + R2) = %0.2f kohm\"%RB1\n", + "IC=(VT-VBE)/((RB/beta)+(RE/beta)+RE)\n", + "IC1=IC*10**3\n", + "print \"Therefore, IC =%0.2f mA\"%IC1\n", + "print \"Since IB is very small, IC ~ IE = 1.73 mA\"\n", + "IE=IC\n", + "VCE=VCC-(IC*RC)-(IE*RE)\n", + "print \"Therefore, VCE = VCC - (IC*RC) - (IE*RE) =%0.2f V\"%VCE\n", + "print \"Therefore, the coordinates of the operating point are :\"\n", + "print \"IC = %0.2f mA\"%IC1\n", + "print \"VCE =%0.2f V\"%VCE\n", + "print \"(b) To find the stability factor S\"\n", + "S = (1+beta)*((1+(RB/RE))/(1+beta+(RB/RE)))\n", + "print \"S =%0.2f\"%S" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 174 Example 6.31." + ] + }, + { + "cell_type": "code", + "execution_count": 60, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) To determine RE,\n", + "VCE = VCC - (IC*RC) - (IE*RE)\n", + "Therefore, RE = 1.30 kohm\n", + "(b) To determine R1 and R2,\n", + "Therefore, RB(k-ohm) = ((RE*beta) / (((1+beta)/S)-1)) - RE = 5.92 kohm\n", + "Therefore, R2 = 6.50 kohm\n", + "Therefore, R1 = R2 / ((R2/RB)-1)= 64.00 kohm\n" + ] + } + ], + "source": [ + "VCE=12\n", + "IC=2*10**-3\n", + "VCC=24\n", + "VBE=0.7\n", + "beta=50\n", + "RC=4.7*10**3\n", + "S=5.1\n", + "print \"(a) To determine RE,\"\n", + "print \"VCE = VCC - (IC*RC) - (IE*RE)\"\n", + "RE = (VCC - (IC*RC) - VCE)/IC #IC=IE\n", + "RE1=RE*10**-3\n", + "print \"Therefore, RE = %0.2f kohm\"%RE1\n", + "\n", + "print \"(b) To determine R1 and R2,\"\n", + "RB=((RE*beta)/(((1+beta)/S)-1))-RE\n", + "RB1=(RB*10**-3)\n", + "print \"Therefore, RB(k-ohm) = ((RE*beta) / (((1+beta)/S)-1)) - RE = %0.2f kohm\"%RB1\n", + "R2=0.1*beta*RE\n", + "R2_1=R2*10**-3\n", + "print \"Therefore, R2 = %0.2f kohm\"%R2_1\n", + "R1=(5.9*10**3*R2)/(R2-(5.9*10**3)) #RB=5.9\n", + "R1_1=round(R1*10**-3)\n", + "print \"Therefore, R1 = R2 / ((R2/RB)-1)= %0.2f kohm\"%R1_1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 174 Example 6.32." + ] + }, + { + "cell_type": "code", + "execution_count": 61, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RTH = R1 || R2 = 10.00 kohm\n", + "VTH = (R2 / (R1+R2)) * VCC = 1.79 V\n", + "IBQ = (VTH-VBE(on)) / (RTH + ((1+beta)*RE)) =15.46 uA\n", + "Therefore, ICQ = beta * IBQ =2.32 mA\n", + "IEQ = IBQ + ICQ = 2.33 mA\n", + "VCEQ = VCC - (ICQ*RC) - (IEQ*RE) = 4.43 V\n", + "The Q point is at :\n", + "VCEQ = 4.43 V\n", + "ICQ = 2.32 mA\n" + ] + } + ], + "source": [ + "R1=56*10**3\n", + "R2=12.2*10**3\n", + "RC=2*10**3\n", + "RE=400\n", + "VCC=10\n", + "VBE=0.7\n", + "beta=150\n", + "RTH=(R1*R2)/(R1+R2)\n", + "RTH1=round(RTH*10**-3)\n", + "print \"RTH = R1 || R2 = %0.2f kohm\"%RTH1\n", + "VTH=(R2/(R1+R2))*VCC\n", + "print \"VTH = (R2 / (R1+R2)) * VCC = %0.2f V\"%VTH\n", + "IBQ=(VTH-VBE)/(RTH+((1+beta)*RE))\n", + "IBQ1=IBQ*10**6\n", + "print \"IBQ = (VTH-VBE(on)) / (RTH + ((1+beta)*RE)) =%0.2f uA\"%IBQ1\n", + "ICQ=beta*IBQ\n", + "ICQ1=ICQ*10**3\n", + "print \"Therefore, ICQ = beta * IBQ =%0.2f mA\"%ICQ1\n", + "IEQ=IBQ+ICQ\n", + "IEQ1=IEQ*10**3\n", + "print \"IEQ = IBQ + ICQ = %0.2f mA\"%IEQ1\n", + "VCEQ=VCC-(ICQ*RC)-(IEQ*RE)\n", + "print \"VCEQ = VCC - (ICQ*RC) - (IEQ*RE) = %0.2f V\"%VCEQ\n", + "print \"The Q point is at :\"\n", + "print \"VCEQ = %0.2f V\"%VCEQ\n", + "print \"ICQ = %0.2f mA\"%ICQ1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 174 Example 6.33. " + ] + }, + { + "cell_type": "code", + "execution_count": 64, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IB = 128.90 uA\n", + "IC = 7.73 mA \n", + "VCE = 5.75 V\n", + "To find stability factor, (S):\n", + "Stability factor for voltage divider bias is\n", + "S =(1+beta)/(1+(beta*(RE/(RE+RB)))) = where RB = R1 || R2 = 61\n" + ] + } + ], + "source": [ + "VCC=22\n", + "RC=2*10**3\n", + "beta=60\n", + "VBE=0.6\n", + "R1=100*10**3\n", + "R2=5*10**3\n", + "RE=100\n", + "a=VCC-VBE-((R1*VCC)/(R1+R2))\n", + "c=(((R1*R2)/(R1+R2))+((1+beta)*RE))\n", + "IB=a/c\n", + "IB1=IB*10**6\n", + "print \"IB = %0.2f uA\"%IB1\n", + "IC=beta*IB\n", + "IC1=IC*10**3\n", + "print \"IC = %0.2f mA \"%IC1\n", + "VCE = VCC - (IC*RC) - ((1+beta)*IB*RE)\n", + "print \"VCE = %0.2f V\"%VCE\n", + "print \"To find stability factor, (S):\"\n", + "print \"Stability factor for voltage divider bias is\"\n", + "RB=(R1*R2)/(R1+R2)\n", + "S=(1+beta)/(1+(beta*(RE/(RE+RB))))\n", + "print \"S =(1+beta)/(1+(beta*(RE/(RE+RB)))) = where RB = R1 || R2 = %0.f\"%S" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 175 Example 6.34." + ] + }, + { + "cell_type": "code", + "execution_count": 68, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IB = 46.50 uA\n", + "Hence, IC = beta * IB =2.33 mA\n", + "VCE = VCC - IC*RC = 5.35 V\n", + "Therefore,the co-ordinates of the new operating point are:\n", + "VCEQ = 5.35 V\n", + "ICQ = 2.33 mA\n", + "To find the stability factor S\n", + "S = (1+beta) / (1 + (beta*[RC/(RC+RB)])) =51.00\n" + ] + } + ], + "source": [ + "VCC=10\n", + "RC=2*10**3\n", + "beta=50\n", + "RB=100*10**3\n", + "VBE=0.7 #collector to base resistor\n", + "IB=(VCC-VBE)/(RB+(beta*RC))\n", + "IB1=IB*10**6\n", + "print \"IB = %0.2f uA\"%IB1\n", + "IC=beta*IB\n", + "IC1=IC*10**3\n", + "print \"Hence, IC = beta * IB =%0.2f mA\"%IC1\n", + "VCE=VCC-(IC*RC)\n", + "print \"VCE = VCC - IC*RC = %0.2f V\"%VCE\n", + "print \"Therefore,the co-ordinates of the new operating point are:\"\n", + "print \"VCEQ = %0.2f V\"%VCE\n", + "print \"ICQ = %0.2f mA\"%IC1\n", + "print \"To find the stability factor S\"\n", + "S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n", + "print \"S = (1+beta) / (1 + (beta*[RC/(RC+RB)])) =%0.2f\"%S" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch7.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch7.ipynb new file mode 100644 index 00000000..6c83db1c --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch7.ipynb @@ -0,0 +1,502 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-7: Field Effect Transistor " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 180 Example 7.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VGS = 12 V, IG = 10**-9 A\n", + "Therefore, gate-to-source resistance = VGS / IG =12000.00 Mohm\n" + ] + } + ], + "source": [ + "VGS=12.\n", + "IG=10**-9\n", + "GSR=VGS/IG\n", + "GSR1=GSR*10**-6\n", + "print \"VGS = 12 V, IG = 10**-9 A\"\n", + "print \"Therefore, gate-to-source resistance = VGS / IG =%0.2f Mohm\"%GSR1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 181 Example 7.2." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "delta_VGS = 4 - 3.9 = 0.1 V\n", + "delta_ID = 1.6 - 1.3 = 0.3 mA\n", + "Therefore, transconductance, gm = delta_ID / delta_VGS =3.00 m-mho\n" + ] + } + ], + "source": [ + "delta_VGS=0.1\n", + "delta_ID=0.3*10**-3\n", + "print \"delta_VGS = 4 - 3.9 = 0.1 V\"\n", + "print \"delta_ID = 1.6 - 1.3 = 0.3 mA\"\n", + "gm=delta_ID/delta_VGS\n", + "gm1=gm*10**3\n", + "print \"Therefore, transconductance, gm = delta_ID / delta_VGS =%0.2f m-mho\"%gm1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 184 Example 7.3" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ID = IDSS*[1 - (VGS/VGS_off)]**2\n", + "Therefore, VGS =-6.00 V\n", + "VP = |VGS_off| = 6.00 V\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "VGSoff=-6\n", + "IDSS=8\n", + "ID=4\n", + "print \"ID = IDSS*[1 - (VGS/VGS_off)]**2\"\n", + "VGS=(1-sqrt(ID/IDSS))*VGSoff\n", + "print \"Therefore, VGS =%0.2f V\"%VGS\n", + "VP=abs(VGSoff)\n", + "print \"VP = |VGS_off| = %0.2f V\"%VP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 184 Example 7.4." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimum value of VDS for pinch-off to occur for VGS = -2 V is\n", + "VDSmin = VGS - VP =3.00 V\n", + "IDS = IDSS * [1-(VGS/VP)]**2 =8.00 mA\n" + ] + } + ], + "source": [ + "VGS=-2\n", + "VP=-5\n", + "IDSS=8*10**-3\n", + "print \"The minimum value of VDS for pinch-off to occur for VGS = -2 V is\"\n", + "VDSmin=VGS-VP\n", + "print \"VDSmin = VGS - VP =%0.2f V\"%VDSmin\n", + "IDS=IDSS*(1-(VGS/VP))**2\n", + "IDS1=IDS*10**3\n", + "print \"IDS = IDSS * [1-(VGS/VP)]**2 =%0.2f mA\"%IDS1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 186 Example 7.5." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of drain current at Q-point,\n", + "IDQ = IDSS / 2 =5.00 mA\n", + "and the value of drain-to-source at Q-point,\n", + "VDSQ = VDD / 2 = 10.00 V\n", + "Therefore, the operating point is at:\n", + "VDS =10.00 V\n", + "ID(mA) =5.00 mA\n", + "Therefore, RD = 2.50 kohm\n", + "The source voltage or voltage across the source resistor RS is\n", + " VS = -VGS = -3 V\n", + "Also,VS = ID*RS \n", + "Therefore, RS = 750.00 ohm\n" + ] + } + ], + "source": [ + "IDSS=10*10**-3\n", + "VGS=-3\n", + "ID=4*10**-3\n", + "VDD=20.\n", + "print \"The value of drain current at Q-point,\"\n", + "IDQ=IDSS/2\n", + "IDQ1=IDQ*10**3\n", + "print \"IDQ = IDSS / 2 =%0.2f mA\"%IDQ1\n", + "print \"and the value of drain-to-source at Q-point,\"\n", + "VDSQ=VDD/2.\n", + "print \"VDSQ = VDD / 2 = %0.2f V\"%VDSQ\n", + "print \"Therefore, the operating point is at:\"\n", + "print \"VDS =%0.2f V\"%VDSQ\n", + "print \"ID(mA) =%0.2f mA\"%IDQ1\n", + "RD=(VDD-VDSQ)/ID\n", + "RD1=RD*10**-3\n", + "print \"Therefore, RD = %0.2f kohm\"%RD1\n", + "print \"The source voltage or voltage across the source resistor RS is\"\n", + "VS=-VGS\n", + "print \" VS = -VGS = -3 V\"\n", + "print \"Also,VS = ID*RS \"\n", + "RS=VS/ID\n", + "print \"Therefore, RS = %0.2f ohm\"%RS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 186 Example 7.6." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that, ID = IDSS * [1 - (VGS/VP)]**2\n", + "Substituting the given values, we get\n", + " ID =40.00 mA\n", + "Therefore, RS = |VGSQ / ID| =125.00 ohm\n" + ] + } + ], + "source": [ + "IDSS=40*10**-3\n", + "VP=-10\n", + "VGSQ=-5\n", + "print \"We know that, ID = IDSS * [1 - (VGS/VP)]**2\"\n", + "print \"Substituting the given values, we get\"\n", + "ID = IDSS*(1-(VGSQ/VP))**2\n", + "ID1=ID*10**3\n", + "print \" ID =%0.2f mA\"%ID1\n", + "RS=abs(VGSQ/ID)\n", + "print \"Therefore, RS = |VGSQ / ID| =%0.2f ohm\"%RS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 187 Example 7.7. " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From fig.7.13.,\n", + " VGG = VDD*(R2 / (R1+R2)) = 10.00 V\n", + "Therefore, ID=3.39 or 4.72 mA\n", + "As ID = 4.72mA > 4mA = IDSS, this value is inappropriate. So, IDQ=3.39 mA is selected.\n", + "Therefore,\n", + " VGSQ = VGG - (IDQ*RS) =-0.17 V\n", + "and VDSQ = VDD - (IDQ*(RD+RS)) =10.75 V\n", + "Then, VDGQ = VDSQ - VGSQ = 10.92 V\n", + "which is grater than |VP| = 2 V. Hence, the FET is in the pinch-off region.\n" + ] + } + ], + "source": [ + "from sympy import symbols,solve\n", + "VDD=24.\n", + "R2=8.57*10**6\n", + "R1=12*10**6\n", + "VP=-2\n", + "IDSS=4*10**-3\n", + "RD=910.\n", + "RS=3*10**3\n", + "print \"From fig.7.13.,\"\n", + "VGG=round(VDD*(R2/(R1+R2)))\n", + "print \" VGG = VDD*(R2 / (R1+R2)) = %0.2f V\"%VGG\n", + "x=symbols('x')\n", + "p1=solve(9*x**2-73*x+144,x)\n", + "ans1=p1[0]\n", + "ans2=p1[1]\n", + "print \"Therefore, ID=%0.2f or %0.2f mA\"%(ans1,ans2)\n", + "print \"As ID = 4.72mA > 4mA = IDSS, this value is inappropriate. So, IDQ=3.39 mA is selected.\"\n", + "print \"Therefore,\"\n", + "IDQ=3.39*10**-3\n", + "VGSQ=VGG-(IDQ*RS)\n", + "print \" VGSQ = VGG - (IDQ*RS) =%0.2f V\"%VGSQ\n", + "VDSQ=VDD-(IDQ*(RD+RS))\n", + "print \"and VDSQ = VDD - (IDQ*(RD+RS)) =%0.2f V\"%VDSQ\n", + "VDGQ = VDSQ - VGSQ\n", + "print \"Then, VDGQ = VDSQ - VGSQ = %0.2f V\"%VDGQ\n", + "print \"which is grater than |VP| = 2 V. Hence, the FET is in the pinch-off region.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 190 Example 7.8." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Assume that the JFET is biased in the saturation region. Then the dc drain current is given by\n", + " ID = IDSS*(1-(VGS/VP))**2\n", + "Therefore, VGS = -1.03 V\n", + "The voltage at the source terminal is\n", + " VS = (ID*RS) - 5 = -2.50 V\n", + "The gate voltage is\n", + " VG = VGS + VS = -3.53 V\n", + "The gate voltage is\n", + " VG = ((R2 / (R1 + R2))*10) - 5\n", + "Therefore, R2 = 17.70 kohm\n", + "and R1(k-ohm) = 102.30 kohm\n", + "The drain-to-source voltage is\n", + "VDS = 5 - ID*RD - ID*RS - (-5)\n", + " RD = 0.50 kohm\n", + "VGS - VP = 2.47 \n", + "Here, since VDS > (VGS-VP), the JFET is biased in the saturation region, which satisfies the initial assumption\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "IDSS=10*10**-3\n", + "VP=-3.5\n", + "Rth=120*10**3 #R1+R2=120 k-ohm\n", + "ID=5*10**-3\n", + "VDS=5\n", + "RS=0.5*10**3\n", + "print \"Assume that the JFET is biased in the saturation region. Then the dc drain current is given by\"\n", + "print \" ID = IDSS*(1-(VGS/VP))**2\"\n", + "VGS=VP*(1-(sqrt(ID/IDSS)))\n", + "print \"Therefore, VGS = %0.2f V\"%VGS\n", + "# textbook answer is wrong\n", + "print \"The voltage at the source terminal is\"\n", + "VS=(ID*RS)-5\n", + "print \" VS = (ID*RS) - 5 = %0.2f V\"%VS\n", + "print \"The gate voltage is\"\n", + "VG=VGS+VS\n", + "print \" VG = VGS + VS = %0.2f V\"%VG\n", + "print \"The gate voltage is\"\n", + "print \" VG = ((R2 / (R1 + R2))*10) - 5\"\n", + "R2=(Rth*(VG+5))/10\n", + "R2_1=R2*10**-3\n", + "print \"Therefore, R2 = %0.2f kohm\"%R2_1\n", + "# textbook answer is wrong\n", + "R1=Rth-R2\n", + "R1_1=R1*10**-3\n", + "print \"and R1(k-ohm) = %0.2f kohm\"%R1_1\n", + "# textbook answer is wrong\n", + "print \"The drain-to-source voltage is\"\n", + "print \"VDS = 5 - ID*RD - ID*RS - (-5)\"\n", + "RD=(10-VDS-(ID*RS))/ID\n", + "RD1=RD*10**-3\n", + "print \" RD = %0.2f kohm\"%RD1\n", + "x=VGS-VP\n", + "print \"VGS - VP = %0.2f \"%x # textbook has taken a different value hence the wrong answer in textbook\n", + "print \"Here, since VDS > (VGS-VP), the JFET is biased in the saturation region, which satisfies the initial assumption\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 191 Example 7.9." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VGSt = 3.50 V\n", + "Therefore,\n", + " VDSt = VGSt - VTN = 2.00 V\n", + "Therefore, RD(k-ohm) = (VDD - VDSQ) / IDQ = 3.33 kohm\n", + "Then, IDQ = KN*(VGSQ-VTN)**2\n", + "Therefore, VGSQ = 2.72 V\n", + " R1 = 439.56 kohm\n", + " R2 = 129.45 kohm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "KN=1*10**-3\n", + "lamda=0.01\n", + "Ri=100*10**3\n", + "IDt=4*10**-3\n", + "IDQ=1.5*10**-3\n", + "VTN=1.5\n", + "VDD=12.\n", + "VDSQ=7.\n", + "VGSt=sqrt(IDt/KN)+VTN\n", + "print \"VGSt = %0.2f V\"%VGSt\n", + "print \"Therefore,\"\n", + "VDSt=VGSt-VTN\n", + "print \" VDSt = VGSt - VTN = %0.2f V\"%VDSt\n", + "RD=(VDD-VDSQ)/IDQ\n", + "RD1=RD*10**-3\n", + "print \"Therefore, RD(k-ohm) = (VDD - VDSQ) / IDQ = %0.2f kohm\"%RD1\n", + "print \"Then, IDQ = KN*(VGSQ-VTN)**2\"\n", + "VGSQ=(sqrt(IDQ/KN))+VTN\n", + "print \"Therefore, VGSQ = %0.2f V\"%VGSQ\n", + "R1=1200/2.73\n", + "print \" R1 = %0.2f kohm\"%R1\n", + "R2=R1/((12/2.73)-1)\n", + "print \" R2 = %0.2f kohm\"%R2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 192 Example 7.10." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ID = 0.40 mA\n", + "The d.c. drain-to-source voltage is\n", + " VDS = VDD - ID*RS = 3.00 V\n", + "Then, VDSsat = VGS - VTN = 2.00 V\n", + "Since VDS > VDSsat, the MOSFET is biased in the saturation region\n" + ] + } + ], + "source": [ + "\n", + "VTN=-2\n", + "KN=0.1*10**-3\n", + "VDD=5\n", + "RS=5*10**3\n", + "VGS=0\n", + "ID=KN*(-VTN)**2\n", + "ID1=ID*10**3\n", + "print \"ID = %0.2f mA\"%ID1\n", + "print \"The d.c. drain-to-source voltage is\"\n", + "VDS=VDD-(ID*RS)\n", + "print \" VDS = VDD - ID*RS = %0.2f V\"%VDS\n", + "VDSsat=VGS-VTN\n", + "print \"Then, VDSsat = VGS - VTN = %0.2f V\"%VDSsat\n", + "print \"Since VDS > VDSsat, the MOSFET is biased in the saturation region\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch8.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch8.ipynb new file mode 100644 index 00000000..2b4eb01f --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch8.ipynb @@ -0,0 +1,174 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-8 : Thyristors " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 210 Example 8.1." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We have,\n", + " V1 = Vm*sin(theta)\n", + "Therefore,\n", + " Firing angel, theta =0.01\n", + " Conduction angle = 180 - theta = 179.99\n", + "Average voltage, Vav = (Vm/2pi) * (1+cos(theta))\n", + " Vav = 70.03 V\n", + "Average current, Iav = Vav / RL =0.70 A\n", + "Power output = Vav*Iav = 49.04 W\n", + "The time during which the SCR remains OFF is\n", + " t = 1.67 ms\n" + ] + } + ], + "source": [ + "from math import asin,pi,cos\n", + "Vm=220.\n", + "V1=110.\n", + "RL=100.\n", + "print \"We have,\"\n", + "print \" V1 = Vm*sin(theta)\"\n", + "print \"Therefore,\"\n", + "x=asin(V1/Vm)*pi/180\n", + "print \" Firing angel, theta =%0.2f\"%x\n", + "ca=180-x\n", + "print \" Conduction angle = 180 - theta = %0.2f\"%ca\n", + "print \"Average voltage, Vav = (Vm/2pi) * (1+cos(theta))\"\n", + "Vav = (Vm/(2*pi))*(1+cos(x*pi/180))\n", + "print \" Vav = %0.2f V\"%Vav\n", + "Iav=Vav/RL\n", + "print \"Average current, Iav = Vav / RL =%0.2f A\"%Iav\n", + "po=Vav*Iav\n", + "print \"Power output = Vav*Iav = %0.2f W\"%po\n", + "print \"The time during which the SCR remains OFF is\"\n", + "t=1./(2*6*50)\n", + "t1=t*10**3\n", + "print \" t = %0.2f ms\"%t1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 211 Example 8.2." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For an SCR full wave rectifier,\n", + " Vdc = (Vm/pi)*(1+cos(theta))\n", + "Therefore, theta =0.02\n", + "For 50Hz, T = 20 ms for 360\n", + "Therefore t = (20*10**3/360)*63.34 = 1.07e-03 ms\n", + "Load current, Iav = Vav / RL = 15.00 A\n" + ] + } + ], + "source": [ + "from math import sqrt,pi,acos\n", + "Vdc=150.\n", + "Vm=230*sqrt(2)\n", + "RL=10.\n", + "print \"For an SCR full wave rectifier,\"\n", + "print \" Vdc = (Vm/pi)*(1+cos(theta))\"\n", + "x=acos(((Vdc*pi)/Vm)-1)*pi/180\n", + "print \"Therefore, theta =%0.2f\"%x\n", + "print \"For 50Hz, T = 20 ms for 360\"\n", + "t = (20./360)*x\n", + "print \"Therefore t = (20*10**3/360)*63.34 = %0.2e ms\"%t\n", + "Iav=Vdc/RL\n", + "print \"Load current, Iav = Vav / RL = %0.2f A\"%Iav" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 212 Example 8.3." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "As the supply voltage is 400 sin 314t, Vm = 400 V\n", + "Peak inverse voltage(PIV) = sqrt(3)*Vm =692.82 V\n", + "RMS value of current = 20 V\n", + "Average value of current, Iav = RMS value/form factor =18.00 A\n", + "Power rating of the SCR(kW) = PIV * Iav =12.47 kW\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "Vm=400\n", + "PIV=sqrt(3)*Vm\n", + "print \"As the supply voltage is 400 sin 314t, Vm = 400 V\"\n", + "print \"Peak inverse voltage(PIV) = sqrt(3)*Vm =%0.2f V\"%PIV\n", + "RMS=20\n", + "ff=1.11\n", + "Iav=round(RMS/ff)\n", + "print \"RMS value of current = 20 V\"\n", + "print \"Average value of current, Iav = RMS value/form factor =%0.2f A\"%Iav\n", + "pr=PIV*Iav\n", + "pr1=pr*10**-3\n", + "print \"Power rating of the SCR(kW) = PIV * Iav =%0.2f kW\"%pr1" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch9.ipynb b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch9.ipynb new file mode 100644 index 00000000..0ce15525 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/Ch9.ipynb @@ -0,0 +1,1478 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-9 : Midband Analysis of Small Signal Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 221 Example 9.1." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Exact analysis :\n", + "Current gain, AI = -hfe / 1+hoe*RL = -48.78\n", + "Input resistance, Ri = hie - (hfe*hre / hoe+(1/RL)) = 600.00 ohm\n", + "Voltage gain, Av = AI*(RL/Ri) = -49.26 \n", + " Yo = hoe - (hfe*hre / hie+Rs) = 0.00 mho\n", + " Ro(k-ohm) = 1/Yo = 51.43 kohm\n", + " Approximate analysis\n", + " AI = -hfe = -50\n", + " Ri = hie = 1 k-ohm\n", + " Av = - hfe*RL / hie = -50.00\n", + " Ro = infinity\n" + ] + } + ], + "source": [ + "print \" Exact analysis :\"\n", + "AI=(-50)/(1+((25*10**-6)*(10**3)))\n", + "print \"Current gain, AI = -hfe / 1+hoe*RL = %0.2f\"%AI\n", + "Ri=1000-((50*2*10**-4)/((25*10**-6)+(1/1000))) #in ohm\n", + "print \"Input resistance, Ri = hie - (hfe*hre / hoe+(1/RL)) = %0.2f ohm\"%Ri\n", + "Av=(-48.78)*(1000/990.24)\n", + "print \"Voltage gain, Av = AI*(RL/Ri) = %0.2f \"%Av\n", + "Yo=(25*10**-6)-((100*10**-4)/(1000+800)) #in mho\n", + "print \" Yo = hoe - (hfe*hre / hie+Rs) = %0.2f mho\"%Yo\n", + "Ro=1/Yo #in ohm\n", + "x1=Ro*10**-3\n", + "print \" Ro(k-ohm) = 1/Yo = %0.2f kohm\"%x1\n", + "print \" Approximate analysis\"\n", + "print \" AI = -hfe = -50\"\n", + "print \" Ri = hie = 1 k-ohm\"\n", + "Av=-(50.*1000)/1000\n", + "print \" Av = - hfe*RL / hie = %0.2f\"%Av\n", + "print \" Ro = infinity\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 223 Example 9.2." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) For RE = 200 ohm,\n", + " hoe*(RE + RC) = 0.05 \n", + "Since hoe*(RE+RC) < 0.1, the approximate model is permissible.\n", + " AI = -hfe = -55\n", + " Ri = hie + (1+hfe)*RE = 12.50 kohm\n", + " Av = AI * (RC/Ri) = 0.00 \n", + "Output resistance, Ro = infinity\n", + "Output terminal resistance, RoT = Ro || RC = 2 k-ohm\n", + "(ii) For RE = 400 ohm\n", + " hoe*(RE + RC) = 0.05 \n", + "Since hoe*(RE+RC) < 0.1, the approximate model is permissible.\n", + " AI = -hfe = -55\n", + " Ri(k-ohm) = hie + (1+hfe)*RE = 23.70 kohm\n", + " Av = AI * (RC/Ri) = -4.64\n", + "Output resistance, Ro = infinity\n", + "Output terminal resistance, RoT = Ro || RC = 2 k-ohm\n", + "(iii) For RE = 1000 ohm\n", + "Since hoe*(RE+RC) < 0.1, the approximate model is permissible.\n", + " AI = -hfe = -55\n", + " Ri(k-ohm) = hie + (1+hfe)*RE = 57.30 kohm\n", + " Av = AI * (RC/Ri) = 0.00 \n", + "Output resistance, Ro = infinity\n", + "Output terminal resistance, RoT = Ro || RC = 2 k-ohm\n" + ] + } + ], + "source": [ + "RC=2*10**3\n", + "hie=1300\n", + "hre=2*10**-4\n", + "hfe=55\n", + "hoe=22*10**-6\n", + "print \"(i) For RE = 200 ohm,\"\n", + "RE=200\n", + "x=hoe*(RE+RC)\n", + "print \" hoe*(RE + RC) = %0.2f \"%x\n", + "print \"Since hoe*(RE+RC) < 0.1, the approximate model is permissible.\"\n", + "AI=-hfe\n", + "print \" AI = -hfe = -55\"\n", + "Ri=hie+((1+hfe)*RE)\n", + "x1=Ri*10**-3\n", + "print \" Ri = hie + (1+hfe)*RE = %0.2f kohm\"%x1\n", + "Av=AI*(RC/Ri)\n", + "print \" Av = AI * (RC/Ri) = %0.2f \"%Av\n", + "print \"Output resistance, Ro = infinity\"\n", + "print \"Output terminal resistance, RoT = Ro || RC = 2 k-ohm\"\n", + "print \"(ii) For RE = 400 ohm\"\n", + "RE=400.\n", + "x2=hoe*(RE+RC)\n", + "print \" hoe*(RE + RC) = %0.2f \"%x2\n", + "print \"Since hoe*(RE+RC) < 0.1, the approximate model is permissible.\"\n", + "AI=-hfe\n", + "print \" AI = -hfe = -55\"\n", + "Ri=hie+((1+hfe)*RE)\n", + "x3=Ri*10**-3\n", + "print \" Ri(k-ohm) = hie + (1+hfe)*RE = %0.2f kohm\"%x3\n", + "Av=AI*(RC/Ri)\n", + "print \" Av = AI * (RC/Ri) = %0.2f\"%Av\n", + "print \"Output resistance, Ro = infinity\"\n", + "print \"Output terminal resistance, RoT = Ro || RC = 2 k-ohm\"\n", + "print \"(iii) For RE = 1000 ohm\"\n", + "print \"Since hoe*(RE+RC) < 0.1, the approximate model is permissible.\"\n", + "AI=-hfe\n", + "print \" AI = -hfe = -55\"\n", + "Ri=1300+((1+55)*1000)\n", + "x3=Ri*10**-3\n", + "print \" Ri(k-ohm) = hie + (1+hfe)*RE = %0.2f kohm\"%x3\n", + "Av=AI*(RC/Ri)\n", + "print \" Av = AI * (RC/Ri) = %0.2f \"%Av\n", + "print \"Output resistance, Ro = infinity\"\n", + "print \"Output terminal resistance, RoT = Ro || RC = 2 k-ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 225 Example 9.3." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Conversion formulae :\n", + " hic = hie = 1200 ohm,\n", + " hfc = -(1+hfe) = -61.00 \n", + "hre = 1, hoc = hoe = 25 uA/V\n", + "Exact analysis :\n", + "Current gain, AI = -hfe / (1 + (hoc*RL)) = 58.10 \n", + "Input impedance, Ri(k-ohm) = hic + hrc*AI*RL = 117.39 kohm\n", + "Voltage gain, Av = AI*RL / Ri = 0.99 \n", + "Output resistance, Ro :\n", + " Yo = 1/Ro = hoc - (hfc*hrc/hic+Rs) =0.03 mho\n", + " Ro = 34.40 ohm\n", + "Approximate analysis :\n", + "Current gain, AI = 1 + hfe = 61.00 \n", + "Input impedance, Ri = hie + (1+hfe)RL = 123.20 kohm\n", + "Voltage gain, Av = 1 - hie/Ri = 0.99 \n", + "Output resistance, Ro:\n", + " Yo(mho) = (1+hfe) / (hie+RS) = 0.03 mho\n", + " Ro = 34.43 ohm \n" + ] + } + ], + "source": [ + "RS=900.\n", + "RL=2000.\n", + "hie=1200.\n", + "hre=2*10**-4\n", + "hfe=60.\n", + "hoe=25*10**-6\n", + "print \"Conversion formulae :\"\n", + "hic=hie\n", + "print \" hic = hie = 1200 ohm,\"\n", + "hfc=-(1+hfe)\n", + "print \" hfc = -(1+hfe) = %0.2f \"%hfc\n", + "print \"hre = 1, hoc = hoe = 25 uA/V\"\n", + "hoc=hoe\n", + "hre=1\n", + "print \"Exact analysis :\"\n", + "format(7)\n", + "AI=-hfc/(1+(hoc*RL))\n", + "print \"Current gain, AI = -hfe / (1 + (hoc*RL)) = %0.2f \"%AI\n", + "Ri=hic + (hre*AI*RL)\n", + "x1=Ri*10**-3\n", + "print \"Input impedance, Ri(k-ohm) = hic + hrc*AI*RL = %0.2f kohm\"%x1\n", + "Av=(AI*RL)/Ri\n", + "print \"Voltage gain, Av = AI*RL / Ri = %0.2f \"%Av\n", + "Yo=hoc-((hfc*hre)/(hic+RS))\n", + "print \"Output resistance, Ro :\"\n", + "print \" Yo = 1/Ro = hoc - (hfc*hrc/hic+Rs) =%0.2f mho\"%Yo\n", + "Ro=1./Yo\n", + "print \" Ro = %0.2f ohm\"%Ro\n", + "print \"Approximate analysis :\"\n", + "AI=1+hfe\n", + "print \"Current gain, AI = 1 + hfe = %0.2f \"%AI\n", + "Ri=hie+((1+hfe)*RL)\n", + "x2=Ri*10**-3\n", + "print \"Input impedance, Ri = hie + (1+hfe)RL = %0.2f kohm\"%x2\n", + "Av=1-(hie/Ri)\n", + "print \"Voltage gain, Av = 1 - hie/Ri = %0.2f \"%Av\n", + "print \"Output resistance, Ro:\"\n", + "Yo=(1+hfe)/(hie+RS)\n", + "print \" Yo(mho) = (1+hfe) / (hie+RS) = %0.2f mho\"%Yo\n", + "Ro=1./Yo\n", + "print \" Ro = %0.2f ohm \"%Ro" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 228 Example 9.4." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current gain, AI = -hfc / 1+hoc*RL''\n", + "where, RL'' = RE || RL = 0.05 kohm\n", + "Therefore, AI = 86.21 \n", + "Input resistance, Ri = hic + hrc*AI*RL'' = 691.06 kohm\n", + "Voltage gain, Av = AI*RL'' / Ri = 1.00 \n", + "Output resistance, Ro = 1 / Yo\n", + " Yo = hoc - (hfc*hrc)/(hic+RS'')\n", + "where, RS'' = RS || R1 || R2 = 0.91 kohm\n", + " Yo = 0.04 \n", + " Ro = 22.99 ohm\n", + " Ro'' = Ro || RLdash = 22.92 ohm\n" + ] + } + ], + "source": [ + "hic=1.4*10**3\n", + "hfc=-100\n", + "hrc=1\n", + "hoc=20*10**-6\n", + "R1=20*10**3\n", + "RS=1*10**3\n", + "R2=20*10**3\n", + "RE=10*10**3\n", + "RL=40*10**3\n", + "print \"Current gain, AI = -hfc / 1+hoc*RL''\"\n", + "RLd=(RE*RL)/(RE+RL)\n", + "x1=RLd*10**-3\n", + "print \"where, RL'' = RE || RL = %0.2f kohm\"%x\n", + "AI = -hfc / (1+(hoc*RLd))\n", + "print \"Therefore, AI = %0.2f \"%AI \n", + "Ri=hic+(hrc*AI*RLd)\n", + "x2=Ri*10**-3\n", + "print \"Input resistance, Ri = hic + hrc*AI*RL'' = %0.2f kohm\"%x2\n", + "Av=(AI*RLd)/Ri\n", + "print \"Voltage gain, Av = AI*RL'' / Ri = %0.2f \"%Av\n", + "print \"Output resistance, Ro = 1 / Yo\"\n", + "print \" Yo = hoc - (hfc*hrc)/(hic+RS'')\"\n", + "RSd=(RS*R1*R2)/((R1*R2)+(RS*R2)+(RS*R1))\n", + "x3=RSd*10**-3\n", + "print \"where, RS'' = RS || R1 || R2 = %0.2f kohm\"%x3\n", + "Yo = hoc - ((hfc*hrc)/(hic+RSd))\n", + "print \" Yo = %0.2f \"%Yo\n", + "# answer in textbook is wrong\n", + "Ro=1/0.0435\n", + "print \" Ro = %0.2f ohm\"%Ro\n", + "Rod=(Ro*RLd)/(Ro+RLd)\n", + "print \" Ro'' = Ro || RLdash = %0.2f ohm\"%Rod" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 229 Example 9.5." + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Exact analysis\n", + "Current gain, AI = -hfb / (1 + hob*RL) = 0.98 \n", + "Input impedance, Ri(ohm) = hib + hrb*AI*RL = 22.29 ohm\n", + "Voltage gain, Av = AI*RL / Ri = 43.94 \n", + "Overall current gain, Avc = Av*Ri / Ri+Rs = 0.80 \n", + "Overall current gain, AIS = AI*Rs / Ri+Rs = 0.96 \n", + "Output admittance, Yo(u-mho) = hob * (hfb*hrb / hib+Rs) = 0.74 \n", + " Ro(M-ohm) = 1 / Yo = 1.35\n", + "Power gain, AP = Av*AI = 43.04 \n", + "Approximate analysis\n", + "Current gain, AI = -hfb = 0.98 \n", + "Input impedance, Ri = hib = 22.00 ohm\n", + "Reaaranging this equation, hfe = -hfb / 1+hfb = 49.00\n", + "From Table 9.3, hib = hie / 1+hfe\n", + " hie = hib(1+hfe) = 1100.00 ohm\n", + " Av = 44.55\n", + "Output impedance, Ro = infinity\n", + "Overall voltage gain, Avs = Av*Ri / Ri+Rs = 0.80 \n", + "Overall current gain, AIS = AI*Rs / Ri+Rs = 0.96 \n", + "Power gain, AP = Av*AI = 43.65\n" + ] + } + ], + "source": [ + "Rs=1200.\n", + "RL=1000.\n", + "hib=22.\n", + "hrb=3*10**-4\n", + "hfb=-0.98\n", + "hob=0.5*10**-6\n", + "print \" Exact analysis\"\n", + "AI=-hfb/(1+(hob*RL))\n", + "print \"Current gain, AI = -hfb / (1 + hob*RL) = %0.2f \"%AI\n", + "Ri=hib+(hrb*AI*RL)\n", + "print \"Input impedance, Ri(ohm) = hib + hrb*AI*RL = %0.2f ohm\"%Ri\n", + "Av=(AI*RL)/Ri\n", + "print \"Voltage gain, Av = AI*RL / Ri = %0.2f \"%Av\n", + "Avs=(Av*Ri)/(Ri+Rs)\n", + "print \"Overall current gain, Avc = Av*Ri / Ri+Rs = %0.2f \"%Avs\n", + "AIS=(AI*Rs)/(Ri+Rs)\n", + "print \"Overall current gain, AIS = AI*Rs / Ri+Rs = %0.2f \"%AIS\n", + "Yo=hob-((hfb*hrb)/(hib+Rs))\n", + "x1=Yo*10**6\n", + "print \"Output admittance, Yo(u-mho) = hob * (hfb*hrb / hib+Rs) = %0.2f \"%x1\n", + "Ro=1/Yo\n", + "x2=Ro*10**-6\n", + "print \" Ro(M-ohm) = 1 / Yo = %0.2f\"%x2\n", + "AP=Av*AI\n", + "print \"Power gain, AP = Av*AI = %0.2f \"%AP\n", + "print \"Approximate analysis\"\n", + "AI=-hfb\n", + "print \"Current gain, AI = -hfb = %0.2f \"%AI\n", + "Ri=hib\n", + "print \"Input impedance, Ri = hib = %0.2f ohm\"%Ri\n", + "hfe = -hfb / (1+hfb)\n", + "print \"Reaaranging this equation, hfe = -hfb / 1+hfb = %0.2f\"%hfe\n", + "print \"From Table 9.3, hib = hie / 1+hfe\"\n", + "hie=hib*(1+hfe)\n", + "print \" hie = hib(1+hfe) = %0.2f ohm\"%hie\n", + "Av=hfe*RL / hie\n", + "print \" Av = %0.2f\"%Av\n", + "print \"Output impedance, Ro = infinity\"\n", + "Avs=(Av*Ri)/(Ri+Rs)\n", + "print \"Overall voltage gain, Avs = Av*Ri / Ri+Rs = %0.2f \"%Avs\n", + "AIS=(AI*Rs)/(Ri+Rs)\n", + "print \"Overall current gain, AIS = AI*Rs / Ri+Rs = %0.2f \"%AIS\n", + "AP=Av*AI\n", + "print \"Power gain, AP = Av*AI = %0.2f\"%AP" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 230 Example 9.6." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current gain, AI = -hfb / 1+hob*RL''\n", + "where, RL'' = RC || RL = 6.46 kohm\n", + " AI = 0.98 \n", + "Input impedance Ri :\n", + " Ri = hib + hrb*AI*RL'' = 25.83 ohm\n", + "Voltage gain Av :\n", + " Av = (AI*RL'') / Ri = 244.36 \n", + "Output Resistance Ro :\n", + "The output admittance\n", + " Yo(u-mho) = 1 / Ro = hob - (hfb*hrb / hib+RS'') = where RS'' = RS || RE = 0.99 u-mho\n", + " Ro = 1 / Yo = 1.01 M-ohm \n", + " RS'' = Ro || RL'' = 6.42 kohm\n" + ] + } + ], + "source": [ + "hib=24.\n", + "hfb=-0.98\n", + "hob=0.49*10**-6\n", + "hrb=2.9*10**-4\n", + "RS=600.\n", + "RE=6*10**3\n", + "RC=12*10**3\n", + "RL=14*10**3\n", + "print \"Current gain, AI = -hfb / 1+hob*RL''\"\n", + "RLd=(RC*RL)/(RC+RL)\n", + "x1=RLd*10**-3\n", + "print \"where, RL'' = RC || RL = %0.2f kohm\"%x1\n", + "AI=-hfb / (1+hob*RLd)\n", + "print \" AI = %0.2f \"%AI\n", + "print \"Input impedance Ri :\"\n", + "Ri=hib+(hrb*AI*RLd)\n", + "print \" Ri = hib + hrb*AI*RL'' = %0.2f ohm\"%Ri\n", + "print \"Voltage gain Av :\"\n", + "Av=(AI*RLd)/Ri\n", + "print \" Av = (AI*RL'') / Ri = %0.2f \"%Av\n", + "print \"Output Resistance Ro :\"\n", + "print \"The output admittance\"\n", + "RSd=(RS*RE)/(RS+RE)\n", + "Yo=hob-((hfb*hrb)/(hib+RSd))\n", + "x4=Yo*10**6\n", + "print \" Yo(u-mho) = 1 / Ro = hob - (hfb*hrb / hib+RS'') = where RS'' = RS || RE = %0.2f u-mho\"%x4\n", + "Ro=1./Yo\n", + "x2=Ro*10**-6\n", + "print \" Ro = 1 / Yo = %0.2f M-ohm \"%x2\n", + "RSd=(Ro*RLd)/(Ro+RLd)\n", + "x3=RSd*10**-3\n", + "print \" RS'' = Ro || RL'' = %0.2f kohm\"%x3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 232 Example 9.7." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From h-parameter model\n", + " Zi = hie = 500 ohm\n", + " Zo = RC = 5.1 k-ohm\n", + " Av = (-hfe*RC) / hie = -612.00 \n", + " AI = -hfe = -60\n", + "From re model\n", + " Zi = beta*re where re = 26mV / Ie\n", + "From the circuit, Ib = (VCC - VBE) / RB = 51.82 uA\n", + " Ie = Ic = beta*Ib = 3.11 mA\n", + " re = 26mV / Ie = 8.37 ohm\n", + " Zi = beta*re = 502.20 ohm\n", + " Zo = RC = 5.1 k-ohm\n", + " Av = -RC / re = -609.00 \n", + " AI = -beta = -60\n" + ] + } + ], + "source": [ + "hfe=60.\n", + "hie=500.\n", + "IC=3*10**-3\n", + "RB=220*10**3\n", + "RC=5.1*10**3\n", + "VCC=12.\n", + "VBE=0.6\n", + "print \"From h-parameter model\"\n", + "beta=hfe\n", + "Zo=RC\n", + "Av=(-hfe*RC)/hie\n", + "print \" Zi = hie = 500 ohm\"\n", + "print \" Zo = RC = 5.1 k-ohm\"\n", + "print \" Av = (-hfe*RC) / hie = %0.2f \"%Av\n", + "print \" AI = -hfe = -60\"\n", + "print \"From re model\"\n", + "print \" Zi = beta*re where re = 26mV / Ie\"\n", + "Ib=(VCC - VBE)/RB\n", + "x1=Ib*10**6\n", + "print \"From the circuit, Ib = (VCC - VBE) / RB = %0.2f uA\"%x1\n", + "Ie=beta*(51.8*10**-6)\n", + "x2=Ie*10**3\n", + "print \" Ie = Ic = beta*Ib = %0.2f mA\"%x2\n", + "re = (26) / (3.108)\n", + "print \" re = 26mV / Ie = %0.2f ohm\"%re\n", + "Zi = beta*8.37\n", + "print \" Zi = beta*re = %0.2f ohm\"%Zi\n", + "print \" Zo = RC = 5.1 k-ohm\"\n", + "Av=int(-RC/re)\n", + "print \" Av = -RC / re = %0.2f \"%Av\n", + "print \" AI = -beta = -60\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 233 Example 9.8." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "h-parameter analysis :\n", + "Zi = RB || hie\n", + " RB = R1 || R2 = 40 k-ohm || 4.7 k-ohm = 4.21 \n", + " Zi = 4.2 k-ohm || 3.2 k-ohm = 1.82\n", + " Zo = RC = 4 k-ohm\n", + " Av = -hfe*RC / hie = -125.00 \n", + " AI = -hfe*RB / RB+hie = -56.79 \n", + "Using r model :\n", + "To find IB,\n", + " VB = R2*VCC / R1+R2 = 1.68 \n", + "Using Thevenin equivalent for input part,\n", + "IB = (VB-VBE) / (RB+((1+beta)*RE)) = 8.63 uA\n", + " IC = beta*IB = 0.86 mA\n", + " IE ~ IC = 0.86 mA\n", + "30.2325581395 re = 26mV / IE = 30.23 ohm\n", + " Zi = RB || beta*re = 1.76 kohm\n", + " Zo = RC = 4 k-ohm\n", + " Av = -RC / re = -132.31\n", + " AI = (-beta*RB) / (RB+(beta*re)) = -58.15\n" + ] + } + ], + "source": [ + "hie=3.2*10**3\n", + "hfe=100.\n", + "R1=40*10**3\n", + "R2=4.7*10**3\n", + "RC=4*10**3\n", + "VCC=16.\n", + "VBE=0.6\n", + "RE=1.2*10**3\n", + "beta=100.\n", + "print \"h-parameter analysis :\"\n", + "print \"Zi = RB || hie\"\n", + "RB=(R1*R2)/(R1+R2)\n", + "x1=RB*10**-3\n", + "print \" RB = R1 || R2 = 40 k-ohm || 4.7 k-ohm = %0.2f \"%x1\n", + "Zi=(RB*hie)/(RB+hie)\n", + "x2=Zi*10**-3\n", + "print \" Zi = 4.2 k-ohm || 3.2 k-ohm = %0.2f\"%x2\n", + "print \" Zo = RC = 4 k-ohm\"\n", + "Av=(-hfe*RC)/hie\n", + "print \" Av = -hfe*RC / hie = %0.2f \"%Av\n", + "AI=(-hfe*RB)/(RB+hie)\n", + "print \" AI = -hfe*RB / RB+hie = %0.2f \"%AI\n", + "print \"Using r model :\"\n", + "print \"To find IB,\"\n", + "VB=(R2*VCC)/(R1+R2)\n", + "print \" VB = R2*VCC / R1+R2 = %0.2f \"%VB\n", + "print \"Using Thevenin equivalent for input part,\"\n", + "IB=1.082/(125.4*10**3)\n", + "x3=IB*10**6\n", + "print \"IB = (VB-VBE) / (RB+((1+beta)*RE)) = %0.2f uA\"%x3\n", + "IC=beta*IB\n", + "x4=IC*10**3\n", + "print \" IC = beta*IB = %0.2f mA\"%x4\n", + "print \" IE ~ IC = %0.2f mA\"%x4\n", + "IE = IC\n", + "re=(26*10**-3)/(0.86*10**-3)\n", + "print re,\" re = 26mV / IE = %0.2f ohm\"%re\n", + "Zi=(RB*beta*re)/(RB+(beta*re))\n", + "x5=Zi*10**-3\n", + "print \" Zi = RB || beta*re = %0.2f kohm\"%x5\n", + "print \" Zo = RC = 4 k-ohm\"\n", + "Av=-RC/re\n", + "print \" Av = -RC / re = %0.2f\"%Av\n", + "AI=(-100*(4.2*10**3))/((4.2*10**3)+(100*30.23))\n", + "print \" AI = (-beta*RB) / (RB+(beta*re)) = %0.2f\"%AI" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 235 Example 9.9." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " |IE| = VEE-VBE / RE = 1.85 mA\n", + " re(ohm) = 26mV / IE = 14.05 ohm\n", + " Zi = RE || re = 14.00 ohm\n", + " Zo = RC = 3.00 kohm\n", + " Av = RC / re = 213.52 \n", + " AI = 1\n" + ] + } + ], + "source": [ + "VBE=0.6\n", + "VEE=8.\n", + "VCC=10.\n", + "RE=4*10**3\n", + "RC=3*10**3\n", + "IE=(VEE-VBE)/RE\n", + "x1=IE*10**3\n", + "print \" |IE| = VEE-VBE / RE = %0.2f mA\"%x1\n", + "re=(26*10**-3)/IE\n", + "print \" re(ohm) = 26mV / IE = %0.2f ohm\"%re\n", + "Zi=(RE*re)/(RE+re)\n", + "print \" Zi = RE || re = %0.2f ohm\"%Zi\n", + "Zo=RC*10**-3\n", + "print \" Zo = RC = %0.2f kohm\"%Zo\n", + "Av=3000/14.05\n", + "print \" Av = RC / re = %0.2f \"%Av\n", + "print \" AI = 1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 238 Example 9.10." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that IB = VCC-VBE / RB+(1+beta)*RE\n", + "Therefore, IB = 85.67 uA\n", + "IE = (1+beta)*IB = 8.57 mA\n", + "The dynamic resistance is\n", + " re = 3.03 ohm\n", + "The input impedance of the amplifier is\n", + " Zb = (1+beta)(re+RE) = 92.22 k-ohm \n", + "The input impedance of the amplifier stage is\n", + " Zi = RB || Zb = 41.36 kohm\n", + "The voltage gain of the amplifier is\n", + "Av = RE / re+RE = 1.00 \n" + ] + } + ], + "source": [ + "print \"We know that IB = VCC-VBE / RB+(1+beta)*RE\"\n", + "IB=((15-0.7)/((75*10**3)+(101*910)))*10**6\n", + "print \"Therefore, IB = %0.2f uA\"%IB # in uA\n", + "print \"IE = (1+beta)*IB = 8.57 mA\"\n", + "print \"The dynamic resistance is\"\n", + "re=0.026/(8.57*10**-3)\n", + "print \" re = %0.2f ohm\"% re # in ohm\n", + "print \"The input impedance of the amplifier is\"\n", + "zb=(101*(3.03+910))*10**-3 # in k-ohm\n", + "print \" Zb = (1+beta)(re+RE) = %0.2f k-ohm \"%zb\n", + "print \"The input impedance of the amplifier stage is\"\n", + "Zi=((75*92.2*10**6)/((75*10**3)+(92.2*10**3)))*10**-3 # in k-ohm\n", + "print \" Zi = RB || Zb = %0.2f kohm\"%Zi\n", + "print \"The voltage gain of the amplifier is\"\n", + "av=910./(3.03+910)\n", + "print \"Av = RE / re+RE = %0.2f \"%av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 240 Example 9.11." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From fig.9.55, IB = (VCC-VBE) / (RB + (1+beta)*RE) = 11.58 uA\n", + " IE = (1+beta)*IB = 1.17 mA\n", + "The load resistance of the emitter follower is rL = RE || RL = 49.25 ohm \n", + " Zi = RB || (1+beta)(re+rL) = 7.13 kohm\n", + " VL / VS = (rL/re+rL)(Zi/Rs+Zi) = 0.59 \n" + ] + } + ], + "source": [ + "\n", + "VCC=10.\n", + "RB=470*10**3\n", + "RE=3.3*10**3\n", + "beta=100.\n", + "RS=1*10**3\n", + "RL=50.\n", + "re=22.4\n", + "VBE=0.7\n", + "IB = (VCC-VBE) / (RB + ((1+beta)*RE))\n", + "x1=IB*10**6\n", + "print \"From fig.9.55, IB = (VCC-VBE) / (RB + (1+beta)*RE) = %0.2f uA\"%x1\n", + "IE=(1+beta)*IB\n", + "x2=IE*10**3\n", + "print \" IE = (1+beta)*IB = %0.2f mA\"%x2\n", + "rL=(RE*RL)/(RE+RL)\n", + "print \"The load resistance of the emitter follower is rL = RE || RL = %0.2f ohm \"%rL # answer in textbook is wrong\n", + "x=(1+beta)*(re+rL)\n", + "Zi=(RB*x)/(RB+x)\n", + "x3=Zi*10**-3\n", + "print \" Zi = RB || (1+beta)(re+rL) = %0.2f kohm\"%x3\n", + "y=(50/(22.4+50))*((7.13*10**3)/((1*10**3)+(7.3*10**3))) # answer in textbook is wrong\n", + "print \" VL / VS = (rL/re+rL)(Zi/Rs+Zi) = %0.2f \"%y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 241 Example 9.12. " + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that, IE = VEE-VBE / RE\n", + "Therefore, IE = 2.65 mA\n", + " Zb = re(ohm) = 9.81 ohm\n", + " Zi(ohm) = re || RE = 9.76 ohm\n", + " Av = RC / re = 101.92 \n", + " VL / VS = Av*(re/re+RS)*(RL/RL+RS) = 13.38\n", + " VL(in mV (rms)) = Av*VS = 133.75 \n", + " iL( in uA (rms)) = VL / RL = 33.44 \n", + " iL / iS = alpha*(RS/RS+re)*(RC/RC+RL) = 0.17 \n" + ] + } + ], + "source": [ + "RS=50.\n", + "RE=2*10**3\n", + "Ro=1*10**3\n", + "RL=4*10**3\n", + "VEE=6.\n", + "VBE=0.7\n", + "RC=1000.\n", + "VS=10*10**-3\n", + "IE=(VEE-VBE)/RE\n", + "x1=IE*10**3\n", + "print \"We know that, IE = VEE-VBE / RE\"\n", + "print \"Therefore, IE = %0.2f mA\"%x1\n", + "re=0.026/IE\n", + "print \" Zb = re(ohm) = %0.2f ohm\"%re\n", + "Zi=(re*RE)/(re+RE)\n", + "print \" Zi(ohm) = re || RE = %0.2f ohm\"%Zi\n", + "Av=RC/re\n", + "print \" Av = RC / re = %0.2f \"%Av\n", + "x=Av*(re/(re+RS))*(RL/(RL+RC))\n", + "print \" VL / VS = Av*(re/re+RS)*(RL/RL+RS) = %0.2f\"%x\n", + "VL=x*VS\n", + "x2=VL*10**3\n", + "print \" VL(in mV (rms)) = Av*VS = %0.2f \"%x2\n", + "iL=VL/RL\n", + "x3=iL*10**6\n", + "print \" iL( in uA (rms)) = VL / RL = %0.2f \"%x3\n", + "alpha=1.\n", + "y=alpha*(RS/(RS+re))*(RC/(RC+RL))\n", + "print \" iL / iS = alpha*(RS/RS+re)*(RC/RC+RL) = %0.2f \"%y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 243 Example 9.13." + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter current of the common base amplifier is\n", + " IE = VEE-VBE / RE = 0.00 A\n", + " re = 0.026 / IE = 24.14 ohm\n", + " Av = RC /re = 497.20 \n", + " VL/VS = Av*(re/re+RS)*(RL/RL+RC) = 195.23 \n", + " iL/iS = Ai*(RS/RS+re)*(RC/RC+RL) = 0.44\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "RC=12*10**3\n", + "RL=15*10**3\n", + "RS=10.\n", + "RE=22*10**3\n", + "VEE=24.\n", + "VBE=0.3\n", + "print \"The emitter current of the common base amplifier is\"\n", + "IE=(VEE-VBE)/RE\n", + "print \" IE = VEE-VBE / RE = %0.2f A\"%IE\n", + "re=0.026/IE\n", + "print \" re = 0.026 / IE = %0.2f ohm\"%re\n", + "Av=RC/re\n", + "print \" Av = RC /re = %0.2f \"%Av\n", + "x=497*(24.14/(24.14+10))*((15*10**3)/((12*10**3)+(15*10**3)))\n", + "print \" VL/VS = Av*(re/re+RS)*(RL/RL+RC) = %0.2f \"%x\n", + "Ai=3.413\n", + "y=Ai*(RS/(RS+re))*(RC/(RC+RL))\n", + "print \" iL/iS = Ai*(RS/RS+re)*(RC/RC+RL) = %0.2f\"%y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 244 Example 9.14." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that, IE(mA) = VEE-VBE / RE = 1.77 mA\n", + " re = 0.026 / IE = 14.72 ohm\n", + " Zi = RE || re = 14.68 ohm\n", + " Zo = RC || re = 2.20 kohm\n", + " Av = Zo/Zi = RC||rc/RE||re = 149.68 \n", + " VL/VS = Av*(Zi/RS+Zi)*(RL/RL+Zo) = 51.94 \n", + " VL = Av*VS = 149.68 rms\n" + ] + } + ], + "source": [ + "rc=1.5*10**6\n", + "RE=4.7*10**3\n", + "Ro=2.2*10**3\n", + "RS=20\n", + "RL=10*10**3\n", + "VS=20*10**-3\n", + "VEE=9\n", + "VBE=0.7\n", + "IE=(VEE-VBE)/RE\n", + "x1=IE*10**3\n", + "print \"We know that, IE(mA) = VEE-VBE / RE = %0.2f mA\"%x1\n", + "re=0.026/IE\n", + "print \" re = 0.026 / IE = %0.2f ohm\"%re\n", + "Zi=(RE*re)/(RE+re)\n", + "print \" Zi = RE || re = %0.2f ohm\"%Zi\n", + "Zo=(Ro*rc)/(Ro+rc)\n", + "x2=Zo*10**-3\n", + "print \" Zo = RC || re = %0.2f kohm\"%x2\n", + "Av=Zo/Zi\n", + "print \" Av = Zo/Zi = RC||rc/RE||re = %0.2f \"%Av\n", + "x=Av*(Zi/(RS+Zi))*(RL/(RL+Zo))\n", + "print \" VL/VS = Av*(Zi/RS+Zi)*(RL/RL+Zo) = %0.2f \"%x\n", + "y=x*VS\n", + "print \" VL = Av*VS = %0.2f rms\"%Av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 245 Example 9.15." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " VB = (R2 / R1+R2)*VCC = 1.48 V\n", + " VE = 1.39 - 0.7 = 0.69 V\n", + " IE(mA) = VE / RE = 1.01 mA\n", + " re = 0.026/IE = 25.62 ohm\n", + " Zi = R1 || R2 || beta*(re+RE) = 4.00 kohm\n", + "The overall voltage gain is VL/VS = (-RC/RE+re)*(Zi/RS+Zi)*(RL/RC+RL) = -3.33 \n", + " Zi = R1 || R2 || betare = 1.56 kohm\n", + " VL/VS = (-RC/re)*(Zi/RS+Zi)*(RL/RC+RL) = -76.27\n" + ] + } + ], + "source": [ + "beta=100.\n", + "VCC=10.\n", + "R2=4.7*10**3\n", + "R1=27*10**3\n", + "RE=680.\n", + "RC=3.3*10**3\n", + "RS=600.\n", + "RL=15*10**3\n", + "VB=(10*4.7*10**3)/((27*10**3)+(4.7*10**3))\n", + "print \" VB = (R2 / R1+R2)*VCC = %0.2f V\"%VB\n", + "# answer in textbook is wrong\n", + "VE=1.39-0.7\n", + "print \" VE = 1.39 - 0.7 = %0.2f V\"%VE\n", + "IE=VE/RE\n", + "x1=IE*10**3\n", + "print \" IE(mA) = VE / RE = %0.2f mA\"%x1\n", + "re=0.026/IE\n", + "print \" re = 0.026/IE = %0.2f ohm\"%re\n", + "x=beta*(re+RE)\n", + "Zi=(R1*R2*x)/((R2*x)+(R1*x)+(R1+R2)) # answer in textbook is wrong\n", + "x2=Zi*10**-3\n", + "print \" Zi = R1 || R2 || beta*(re+RE) = %0.2f kohm\"%x2\n", + "y=(-RC/(RE+re))*(Zi/(RS+Zi))*(RL/(RC+RL))\n", + "print \"The overall voltage gain is VL/VS = (-RC/RE+re)*(Zi/RS+Zi)*(RL/RC+RL) = %0.2f \"%y\n", + "u=beta*re\n", + "Zi=(R1*R2*u)/((R2*u)+(R1*u)+(R1*R2))\n", + "x3=Zi*10**-3\n", + "print \" Zi = R1 || R2 || betare = %0.2f kohm\"%x3\n", + "z=(-RC/re)*(Zi/(RS+Zi))*(RL/(RC+RL)) # answer in textbook is wrong\n", + "print \" VL/VS = (-RC/re)*(Zi/RS+Zi)*(RL/RC+RL) = %0.2f\"%z" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 246 Example 9.16." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " VB1 = (RB3*VCC)/(RB3+RB2+RB1) = 3.38 V\n", + " IE1 = VE1/RE = (VB1-VBE1)/RE = 2.06 mA\n", + " re1 = 26mV/IE1 = 12.64 ohm\n", + " re2 = 12.64 ohm (since IE2 = IE1)\n", + "Voltage gain of the second stage,\n", + " Av2 = RC / re2 = 174.11\n", + "Overall voltage gain,\n", + " Av = Av1*Av2 = -174.11\n" + ] + } + ], + "source": [ + "RB1=7.5*10**3\n", + "RB2=6.8*10**3\n", + "RB3=3.3*10**3\n", + "RE=1.3*10**3\n", + "RC=2.2*10**3\n", + "beta1=120.\n", + "beta2=120.\n", + "VCC=18.\n", + "VBE1=0.7\n", + "VB1=(RB3*VCC)/(RB3+RB2+RB1)\n", + "print \" VB1 = (RB3*VCC)/(RB3+RB2+RB1) = %0.2f V\"%VB1\n", + "IE1=(VB1-VBE1)/RE\n", + "x1=IE1*10**3\n", + "print \" IE1 = VE1/RE = (VB1-VBE1)/RE = %0.2f mA\"%x1\n", + "re1=(26*10**-3)/IE1\n", + "print \" re1 = 26mV/IE1 = %0.2f ohm\"%re1\n", + "re2=re1\n", + "print \" re2 = %0.2f ohm (since IE2 = IE1)\"%re2\n", + "print \"Voltage gain of the second stage,\"\n", + "Av2=RC/re2\n", + "print \" Av2 = RC / re2 = %0.2f\"%Av2\n", + "print \"Overall voltage gain,\"\n", + "Av1=-1\n", + "Av=Av1*Av2\n", + "print \" Av = Av1*Av2 = %0.2f\"%Av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 248 Example 9.17. " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain,\n", + " Av = Vo/Vi = -u*RD / RD+rd = -6.25\n", + "The minus sign indicates a 180 degree phase shift between Vi and Vo\n", + "Input impedance Zi(M-ohm) = RG = 10.00 \n", + "Output impedance Zo(k-ohm) = RD = 5.00 \n" + ] + } + ], + "source": [ + "RD=5*10**3\n", + "RG=10*10**6\n", + "u=50.\n", + "rd=35*10**3\n", + "print \"The voltage gain,\"\n", + "Av=(-u*RD)/(RD+rd)\n", + "print \" Av = Vo/Vi = -u*RD / RD+rd = %0.2f\"%Av\n", + "print \"The minus sign indicates a 180 degree phase shift between Vi and Vo\"\n", + "Zi=RG*10**-6\n", + "print \"Input impedance Zi(M-ohm) = RG = %0.2f \"%Zi\n", + "Zo=RD*10**-3\n", + "print \"Output impedance Zo(k-ohm) = RD = %0.2f \"%Zo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 252 Example 9.18." + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain,\n", + " Av = Vo/Vi = u*RS / (u+1)*RS+rd = 0.84 \n", + "Output impedance, Zo(ohm) = 1/gm || RS = (rd/u) || RS = 595.74 ohm\n" + ] + } + ], + "source": [ + "RS=4*10**3\n", + "RG=10*10**6\n", + "u=50.\n", + "rd=35*10**3\n", + "print \"The voltage gain,\"\n", + "Av=(u*RS)/(((1+u)*RS)+rd)\n", + "print \" Av = Vo/Vi = u*RS / (u+1)*RS+rd = %0.2f \"%Av\n", + "x=rd/u\n", + "Zo=(x*RS)/(RS+x)\n", + "print \"Output impedance, Zo(ohm) = 1/gm || RS = (rd/u) || RS = %0.2f ohm\"%Zo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 254 Example 9.19." + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain,\n", + " Av = Vo/Vi = (gm*rd + 1)*RD / (RD+rd) = 2.76\n", + "Input impedance, Zi(k-ohm) = RS || 1/gm = 0.41 k0hm\n", + "Output impedance, Zo ~ RD = 2 k-ohm\n" + ] + } + ], + "source": [ + "RD=2*10**3\n", + "RS=1*10**3\n", + "gm=1.43*10**-3\n", + "rd=35*10**3\n", + "print \"The voltage gain,\"\n", + "Av=(((gm*rd)+1)*RD)/(RD+rd)\n", + "print \" Av = Vo/Vi = (gm*rd + 1)*RD / (RD+rd) = %0.2f\"%Av\n", + "x=1./gm\n", + "Zi=(RS*x)/(RS+x)\n", + "x1=Zi*10**-3\n", + "print \"Input impedance, Zi(k-ohm) = RS || 1/gm = %0.2f k0hm\"%x1\n", + "print \"Output impedance, Zo ~ RD = 2 k-ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 256 Example 9.20." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " In the first set,\n", + " Vid = Vd(uV) = V1 = V2 = 200.00 uV\n", + " Vc(uV) = 1/2(V1+V2) = 0.00\n", + " In the second set,\n", + " Vd = V1 = V2 = 200.00 uV\n", + " Vc = 1/2(V1+V2) = 1000.00 uV\n" + ] + } + ], + "source": [ + "print \" In the first set,\"\n", + "Vid=100-(-100) #in uV\n", + "print \" Vid = Vd(uV) = V1 = V2 = %0.2f uV\"%Vid\n", + "Vc=(1/2)*(100+(-100)) # in uV\n", + "print \" Vc(uV) = 1/2(V1+V2) = %0.2f\"%Vc\n", + "print \" In the second set,\"\n", + "Vd=1100-900 # in uV\n", + "print \" Vd = V1 = V2 = %0.2f uV\"%Vd\n", + "Vc=(1./2)*(1100+900)\n", + "print \" Vc = 1/2(V1+V2) = %0.2f uV\"%Vc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 258 Example 9.21." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " IE = (VEE - VBE)/2*REE = 110.00 uA\n", + " IC = alpha_F*IE = 108.91 uA\n", + " IB = IC / beta_F = 1.09 uA\n", + " VC = VCC - IC*RC = 7.92 V\n", + " VCE = VC - VE = 8.62 V\n", + " IE = VEE / 2*REE = 115.38 uA\n" + ] + } + ], + "source": [ + "VEE=15.\n", + "VBE=0.7\n", + "REE=65*10**3\n", + "IE = (VEE - VBE)/(2*REE)\n", + "IE1=IE*10**6\n", + "print \" IE = (VEE - VBE)/2*REE = %0.2f uA\"%IE1\n", + "alphaF=100./101.\n", + "IC=(alphaF*IE)\n", + "IC1=IC*10**6\n", + "print \" IC = alpha_F*IE = %0.2f uA\"%IC1\n", + "betaF=100.\n", + "IB=IC/betaF\n", + "IB1=IB*10**6\n", + "print \" IB = IC / beta_F = %0.2f uA\"%IB1\n", + "VCC=VEE\n", + "RC=REE\n", + "VC=VCC-(IC*RC)\n", + "print \" VC = VCC - IC*RC = %0.2f V\"%VC\n", + "VE=-0.7\n", + "VCE=VC - VE\n", + "print \" VCE = VC - VE = %0.2f V\"%VCE\n", + "IE=(VEE/(2*REE))*10**6\n", + "print \" IE = VEE / 2*REE = %0.2f uA\"%IE" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 262 Example 9.22." + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " IDS = ISS / 2 = 87.50 uA\n", + " VGS = VTH + sqrt(ISS/Kn) = 1.24 V\n", + " VDS = VDD - (IDS*RD) + VGS = 7.55 V\n", + "Checking for saturation,\n", + " VGS - VTN = 0.24 \n", + "and VDS >= 0.2. Thus, both transistors in the differential amplifier are baised at Q-point of :\n", + "87.50\n", + "7.55\n", + " VIC= 7.31 V\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "VDD=12.\n", + "VSS=VDD\n", + "ISS=175*10**-6\n", + "RD=65*10**3\n", + "Kn=3*10**-3\n", + "VTN=1.\n", + "IDS=ISS/2.\n", + "IDS1=IDS*10**6\n", + "print \" IDS = ISS / 2 = %0.2f uA\"%IDS1\n", + "VGS=VTN+sqrt(ISS/Kn)\n", + "print \" VGS = VTH + sqrt(ISS/Kn) = %0.2f V\"%VGS\n", + "VDS=VDD-(IDS*RD)+VGS\n", + "print \" VDS = VDD - (IDS*RD) + VGS = %0.2f V\"%VDS\n", + "print \"Checking for saturation,\"\n", + "x=VGS-VTN\n", + "print \" VGS - VTN = %0.2f \"%x\n", + "print \"and VDS >= 0.2. Thus, both transistors in the differential amplifier are baised at Q-point of :\"\n", + "print \"%0.2f\" %IDS1\n", + "print \"%0.2f\"%(VDS)\n", + "VIC = VDD - IDS*RD + VTN\n", + "print \" VIC= %0.2f V\"%VIC" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 263 Example 9.23." + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "beta = hfe = 100\n", + " IE = (VEE-VBE) / ((2*RE)+(RS/beta)) = 1.01 mA\n", + "IC ~ IE = 1.009 mA\n", + " Therefore ICQ = 1.01 mA\n", + " VCE = VCC + VBE - IC*RC = 12.70 V\n", + "and VCEQ = 12.70 V\n", + "The differential gain is : \n", + " Ad = hfe*RC / RS+hie = 135.54 \n", + "Common mode gain is : \n", + " AC = (hfe*Re) / (((2*RE)*(1+hfe)) + RS + hie) = 0.40 \n", + "CMRR = Ad / AC = 341.72 \n", + "CMRR = 20log|Ad/AC| = 50.67 dB\n", + "The output voltage is Vo = Ad*Vd + AC*VC. Here,\n", + " Ad [mV(peak-peak)] = VS1 - VS2 = 20.00 \n", + "Then, VC [mV(peak-peak)]= (VS1+VS2) / 2 = 50.00 \n", + "Therefore, Vo [V(peak-peak)] = 2.73 \n" + ] + } + ], + "source": [ + "from math import log10\n", + "VS1=60*10**-3\n", + "VS2=40*10**-3\n", + "hie=3.2*10**3\n", + "hfe=100.\n", + "VEE=12.\n", + "VCC=VEE\n", + "VBE=0.7\n", + "beta=hfe\n", + "RE=5.6*10**3\n", + "RS=120.\n", + "RC=4.5*10**3\n", + "Rc=4.5*10**-5\n", + "IE=(VEE-VBE)/((2*RE)+(RS/beta))\n", + "IE1=IE*10**3\n", + "print \"beta = hfe = 100\"\n", + "print \" IE = (VEE-VBE) / ((2*RE)+(RS/beta)) = %0.2f mA\"%IE1\n", + "IC=IE\n", + "print \"IC ~ IE = 1.009 mA\"\n", + "print \" Therefore ICQ = %0.2f mA\"%IE1\n", + "VCE=VCC+VBE-(IC*Rc)\n", + "print \" VCE = VCC + VBE - IC*RC = %0.2f V\"%VCE\n", + "# answer in textbook is wrong\n", + "print \"and VCEQ = %0.2f V\"%VCE # answer in textbook is wrong\n", + "print \"The differential gain is : \"\n", + "Ad=(hfe*RC)/(RS+hie)\n", + "print \" Ad = hfe*RC / RS+hie = %0.2f \"%Ad\n", + "print \"Common mode gain is : \"\n", + "AC=(hfe*RC)/(((2*RE)*(1+hfe))+RS+hie)\n", + "print \" AC = (hfe*Re) / (((2*RE)*(1+hfe)) + RS + hie) = %0.2f \"%AC\n", + "CMRR = Ad / AC\n", + "print \"CMRR = Ad / AC = %0.2f \"%CMRR\n", + "CMRR1=20*log10(135.54/0.3966)\n", + "print \"CMRR = 20log|Ad/AC| = %0.2f dB\"%CMRR1\n", + "print \"The output voltage is Vo = Ad*Vd + AC*VC. Here,\"\n", + "Vd=VS1-VS2\n", + "Vd1=Vd*10**3\n", + "print \" Ad [mV(peak-peak)] = VS1 - VS2 = %0.2f \"%Vd1\n", + "VC=(VS1+VS2)/2\n", + "VC1=VC*10**3\n", + "print \"Then, VC [mV(peak-peak)]= (VS1+VS2) / 2 = %0.2f \"%VC1\n", + "Vo = Ad*Vd + AC*VC\n", + "print \"Therefore, Vo [V(peak-peak)] = %0.2f \"%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No. 264 Example 9.24." + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the circuit 9.90(a),\n", + " RL = hoe*(RL || RC) = 0.05 \n", + "For equivalent circuit refer fig.9.90(b).\n", + " Input resistance, Ri = hie || 100k = 398.41 \n", + " Output resistance, Ro = 50k || 3k || 5k = 1807.23 \n", + "Therefore, Vo/Vi = -hfe*Ro / hie = -180.72 \n", + " Vi/VS = Ri/(Ri+RS) = 0.28 \n", + "Hence, Avs = Vo/VS = (Vo/Vi)*(Vi/VS) = 51.49 \n" + ] + } + ], + "source": [ + "hie=400.\n", + "hre=2.1*10**-4\n", + "hfe=40.\n", + "hoe=25*10**-6\n", + "RL=5*10**3\n", + "RC=3*10**3\n", + "print \"From the circuit 9.90(a),\"\n", + "Rth=(RL*RC)/(RL+RC)\n", + "RLd=hoe*(Rth)\n", + "print \" RL = hoe*(RL || RC) = %0.2f \"%RLd\n", + "print \"For equivalent circuit refer fig.9.90(b).\"\n", + "Ri=(hie*100*10**3)/(hie+(100*10**3))\n", + "print \" Input resistance, Ri = hie || 100k = %0.2f \"%Ri\n", + "R1=50.0*10**3\n", + "Ro=(R1*RC*RL)/((RC*RL)+(R1*RL)+(R1*RC))\n", + "print \" Output resistance, Ro = 50k || 3k || 5k = %0.2f \"%Ro\n", + "x=(-hfe*Ro)/hie\n", + "print \"Therefore, Vo/Vi = -hfe*Ro / hie = %0.2f \"%x\n", + "RS=1*10**3\n", + "y=Ri/(Ri+RS)\n", + "print \" Vi/VS = Ri/(Ri+RS) = %0.2f \"%y\n", + "Avs=abs(x*y)\n", + "print \"Hence, Avs = Vo/VS = (Vo/Vi)*(Vi/VS) = %0.2f \"%Avs" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/KenrgyCh3.png b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/KenrgyCh3.png Binary files differnew file mode 100644 index 00000000..c0680ae6 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/KenrgyCh3.png diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/OpV_ch16.png b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/OpV_ch16.png Binary files differnew file mode 100644 index 00000000..f4fd0cd5 --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/OpV_ch16.png diff --git a/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/nVeClipper16.png b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/nVeClipper16.png Binary files differnew file mode 100644 index 00000000..6bfa71eb --- /dev/null +++ b/Electronics_Devices_And_Circuits_by_S._Salivahanan,_N._S._Kumar_And_A._Vallavaraj/screenshots/nVeClipper16.png diff --git a/Machine_Design_by_U.C._Jindal/README.txt b/Machine_Design_by_U.C._Jindal/README.txt new file mode 100644 index 00000000..b15f5d6b --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/README.txt @@ -0,0 +1,10 @@ +Contributed By: Preeti Rani +Course: btech +College/Institute/Organization: Techwords Institute +Department/Designation: ME +Book Title: Machine Design +Author: U.C. Jindal +Publisher: Dorling Kindersley (India) +Year of publication: 2010 +Isbn: 978-81-317-1659-5 +Edition: 1
\ No newline at end of file diff --git a/Thyristors_Theory_And_Applications_by_R._K._Sugandhi_And_K._K._Sugandhi/README.txt b/Thyristors_Theory_And_Applications_by_R._K._Sugandhi_And_K._K._Sugandhi/README.txt new file mode 100644 index 00000000..9ab4af48 --- /dev/null +++ b/Thyristors_Theory_And_Applications_by_R._K._Sugandhi_And_K._K._Sugandhi/README.txt @@ -0,0 +1,10 @@ +Contributed By: Meena Chandrupatla +Course: others +College/Institute/Organization: Vijaya Bank +Department/Designation: Accounts +Book Title: Thyristors Theory And Applications +Author: R. K. Sugandhi And K. K. Sugandhi +Publisher: Wiley Eastern Limited, New Delhi +Year of publication: 1983 +Isbn: 978-0-85226-852-0 +Edition: 2
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