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diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C10_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C10_1.ipynb new file mode 100644 index 00000000..1b13f377 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C10_1.ipynb @@ -0,0 +1,430 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10: AC Performance : Bandwidth, Slew rate and Noise"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.1 Page No 277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=0.35 #rise time\n",
+ "\n",
+ "#calculation\n",
+ "B=0.35/t #Bandwidth\n",
+ "\n",
+ "#result\n",
+ "print\"Bandwidth is\",B,\"MHz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bandwidth is 1.0 MHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.2 Page No 277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=0.35 #rise time\n",
+ "\n",
+ "#calculation\n",
+ "B=0.35/t #Bandwidth\n",
+ "gain=B\n",
+ "\n",
+ "#result\n",
+ "print\"Openloop Voltage gain is\",gain\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Openloop Voltage gain is 1.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.3 Page No 277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B1=10**6 #Hz\n",
+ "B2=100*10**3 #Hz\n",
+ "\n",
+ "#calculation\n",
+ "gain=B1/B2\n",
+ "\n",
+ "#Result\n",
+ "print\"As frenquency goes down by a factor 10, Gain rise sby same which is\",gain\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "As frenquency goes down by a factor 10, Gain rise sby same which is 10\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.4 Page No 278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=1.5 #Mhz\n",
+ "f=1 #KHz\n",
+ "\n",
+ "#Calculation\n",
+ "gain=B*1000/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Open loop Voltage gain is\",gain\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Open loop Voltage gain is 1500.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.5 Page No 279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Find the actual gain\n",
+ "\n",
+ "#Given\n",
+ "Acl=100.0 #Ideal Dc cloesd loop gain\n",
+ "Aol1=10000.0\n",
+ "Aol2=1000.0\n",
+ "Aol3=100.0\n",
+ "Aol4=10.0\n",
+ "Aol5=1.0\n",
+ "\n",
+ "#Calculation (a)\n",
+ "#For a noninverting amplifier\n",
+ "a1=Acl #ideal gain, a=(Rf+Ri)/Ri\n",
+ "actualAcl1=a1/((1+1*a1/Aol1))\n",
+ "#For the inverting amplifier\n",
+ "a2=101 #gain\n",
+ "actualAcl1_=-a1/(1+1*a2/Aol1)\n",
+ "\n",
+ "# (b)\n",
+ "#For a noninverting amplifier\n",
+ "a1=Acl #ideal gain, a=(Rf+Ri)/Ri\n",
+ "actualAcl2=a1/(1+1*a1/Aol2)\n",
+ "#For the inverting amplifier\n",
+ "a2=101 #gain\n",
+ "actualAcl2_=-a1/(1+1*a2/Aol2)\n",
+ "\n",
+ "#(c)\n",
+ "#For a noninverting amplifier\n",
+ "a1=Acl #ideal gain, a=(Rf+Ri)/Ri\n",
+ "actualAcl3=a1/(1+1*a1/Aol3)\n",
+ "#For the inverting amplifier\n",
+ "a2=101 #gain\n",
+ "actualAcl3_=-a1/(1+1*a2/Aol3)\n",
+ "\n",
+ "#(d)\n",
+ "#For a noninverting amplifier\n",
+ "a1=Acl #ideal gain, a=(Rf+Ri)/Ri\n",
+ "actualAcl4=a1/(1+1*a1/Aol4)\n",
+ "#For the inverting amplifier\n",
+ "a2=101 #gain\n",
+ "actualAcl4_=-a1/(1+1*a2/Aol4)\n",
+ "\n",
+ "#(e)\n",
+ "#For a noninverting amplifier\n",
+ "a1=Acl #ideal gain, a=(Rf+Ri)/Ri\n",
+ "actualAcl5=a1/(1+1*a1/Aol5)\n",
+ "#For the inverting amplifier\n",
+ "a2=101 #gain\n",
+ "actualAcl5_=-a1/(1+1*a2/Aol5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)Actual gain in noninverting amplifier is\",round(actualAcl5,2)\n",
+ "print\"Actual gain in inverting amplifier is\",round(actualAcl5_,2)\n",
+ "print\"(b)Actual gain in noninverting amplifier is\",round(actualAcl4,2)\n",
+ "print\"Actual gain in inverting amplifier is\",round(actualAcl4_,2)\n",
+ "print\"(c)Actual gain in noninverting amplifier is\",round(actualAcl3,2)\n",
+ "print\"Actual gain in inverting amplifier is\",round(actualAcl3_,2)\n",
+ "print\"(d)Actual gain in noninverting amplifier is\",round(actualAcl2,2)\n",
+ "print\"Actual gain in inverting amplifier is\",round(actualAcl2_,2)\n",
+ "print\"(e)Actual gain in noninverting amplifier is\",round(actualAcl1,0)\n",
+ "print\"Actual gain in inverting amplifier is\",round(actualAcl1_,2)\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Actual gain in noninverting amplifier is 0.99\n",
+ "Actual gain in inverting amplifier is -0.98\n",
+ "(b)Actual gain in noninverting amplifier is 9.09\n",
+ "Actual gain in inverting amplifier is -9.01\n",
+ "(c)Actual gain in noninverting amplifier is 50.0\n",
+ "Actual gain in inverting amplifier is -49.75\n",
+ "(d)Actual gain in noninverting amplifier is 90.91\n",
+ "Actual gain in inverting amplifier is -90.83\n",
+ "(e)Actual gain in noninverting amplifier is 99.0\n",
+ "Actual gain in inverting amplifier is -99.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.6 Page No 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ri=10.0 #Kohm\n",
+ "Rf=Ri\n",
+ "B=10**6 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "#(a)\n",
+ "Acl1=-Rf/Ri\n",
+ "fh1=B/((Ri+Rf)/Ri)\n",
+ "\n",
+ "#(b)\n",
+ "ACl2=(Ri+Rf)/Ri\n",
+ "\n",
+ "#(c)\n",
+ "Acl3=1 #Voltage follower gain\n",
+ "Rf2=0\n",
+ "Ri2=1\n",
+ "fh2=B/((Ri2+Rf2)/Ri2)\n",
+ "\n",
+ "#Resilt\n",
+ "print\"(a)The voltage gain is\",Acl1\n",
+ "print\"bandwidth is\",fh1/1000,\"khz\"\n",
+ "print\"(b)The voltage gain is\",ACl2\n",
+ "print\"(c)The voltage gain is\",Acl3\n",
+ "print\"bandwidth is\",fh2/1000000,\"Mhz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The voltage gain is -1.0\n",
+ "bandwidth is 500.0 khz\n",
+ "(b)The voltage gain is 2.0\n",
+ "(c)The voltage gain is 1\n",
+ "bandwidth is 1 Mhz\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.7 Page No 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=10 #V\n",
+ "slewrate=0.5 #per microsecond\n",
+ "\n",
+ "#Calculation\n",
+ "t=V/slewrate\n",
+ "\n",
+ "#Result\n",
+ "print\"The time taken is\",t,\"microsecond\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The time taken is 20.0 microsecond\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.8 Page No 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "slewrate=0.5 #per microsecond\n",
+ "V1=10.0 #Peak voltage\n",
+ "V2=1\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "fmax1=slewrate*1000/(6.28*V1)\n",
+ "fmax2=slewrate*1000/(6.28*V2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)maximum frenquency is\",round(fmax1,1),\"KHz\"\n",
+ "print\"(b)maximum frenquency is\",round(fmax2,0),\"KHz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)maximum frenquency is 8.0 KHz\n",
+ "(b)maximum frenquency is 80.0 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.9 Page No 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B1=500 #Khz, bandwidth\n",
+ "gain1=-1 #Gain for inverting amplifier\n",
+ "gain2=2 #Gain for non-inverting amplifier\n",
+ "slewrate=0.5 #per micros\n",
+ "V=10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Vopmax=slewrate*10**6/(6.28*B1*10**3)\n",
+ "fmax=slewrate*1000/(6.28*V)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum operating voltage is\",round(Vopmax*1000,1),\"mV\"\n",
+ "print\"maximum frenquency is\",round(fmax,0),\"Khz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum operating voltage is 159.2 mV\n",
+ "maximum frenquency is 8.0 Khz\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C11_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C11_1.ipynb new file mode 100644 index 00000000..057dd53b --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C11_1.ipynb @@ -0,0 +1,694 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11: Active Filters"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.1 Page No 298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10 #Kohm\n",
+ "C=0.001 #microF\n",
+ "\n",
+ "#Calculation\n",
+ "wc=1/(R*10**3*C*10**-6)\n",
+ "fc=wc/(6.28)\n",
+ "\n",
+ "#Result\n",
+ "print\"The cut off frequency is\",round(fc/1000,2),\"Khz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The cut off frequency is 15.92 Khz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.2 Page No 298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=2 #Frequency, khz\n",
+ "C=0.005 #MicroF\n",
+ "\n",
+ "#Calculation\n",
+ "R=1/(6.28*2*10**3*C*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of R is\",round(R/1000,1),\"Kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R is 15.9 Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.3 Page No 298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=30 #rad/s, frequency\n",
+ "C=0.01 #microF\n",
+ "R=1/(f*10**3*C*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of R is\",round(R/1000,1),\"Kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R is 3.3 Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.4 Page No 301"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=1 #Khz, cut off frequency\n",
+ "C1=0.01 #MicroF\n",
+ "\n",
+ "#Calculation\n",
+ "C2=2*C1\n",
+ "R=0.707/(6.28*f*10**3*C1*10**-6)\n",
+ "Rf=2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of R is\",round(R,0),\"ohm\"\n",
+ "print\"The value of Rf is\",round(Rf,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R is 11258.0 ohm\n",
+ "The value of Rf is 22516.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.5 Page No 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "C3=0.01 #microF, from fig 11.5 (a)\n",
+ "f=1.0 #Khz, cut off frequency\n",
+ "\n",
+ "#Calculation\n",
+ "C1=C3/2.0\n",
+ "C2=2*C3\n",
+ "R=1/(6.28*f*10**3*C3*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of C1=\",C1,\"microF\"\n",
+ "print\"The value of C2=\",C2,\"microF\"\n",
+ "print\"The value of R=\",round(R),\"ohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of C1= 0.005 microF\n",
+ "The value of C2= 0.02 microF\n",
+ "The value of R= 15924.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.6 Page No 307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=0.002 #microF, capacitance\n",
+ "fc=10 #Khz, frequency\n",
+ "\n",
+ "#Calculation\n",
+ "R=1/(6.28*f*10**3*C*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of R=\",round(R/10000),\"kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R= 8.0 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.7 Page No 307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=22.0 #kohm\n",
+ "C=0.001 #microF\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "wc=1/(R*10**3*C*10**-6)\n",
+ "fc=wc/(2*math.pi)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of wc is\",round(wc/10000,2),\"krad/s\"\n",
+ "print\"The value of fc is\",round(fc/10,1),\"Hz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of wc is 4.55 krad/s\n",
+ "The value of fc is 723.4 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.8 Page No 309"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=0.01 #microF\n",
+ "C2=C1\n",
+ "f=1 #KHz, frequency\n",
+ "\n",
+ "#Calculation\n",
+ "#From eq 11.9\n",
+ "R1=1.414/(6.28*f*10**3*C1*10**-6)\n",
+ "R2=R1/2.0\n",
+ "\n",
+ "#result\n",
+ "print\"The value of R1 is\",round(R1/1000,1),\"kohm\"\n",
+ "print\"The value of R2 is\",round(R2/1000,1),\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R1 is 22.5 kohm\n",
+ "The value of R2 is 11.3 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.9 Page No 309"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=125 #pF\n",
+ "C2=C1 #KHz,cut off frequency\n",
+ "f=80 #Krad/s cutoff frequency\n",
+ "\n",
+ "#Calculation\n",
+ "#From eq 11.9\n",
+ "R1=1.414/(f*10**3*C1*10**-12)\n",
+ "R2=R1/2.0\n",
+ "\n",
+ "#result\n",
+ "print\"The value of R1 is\",round(R1/1000,0),\"kohm\"\n",
+ "print\"The value of R2 is\",round(R2/1000,1),\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R1 is 141.0 kohm\n",
+ "The value of R2 is 70.7 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.10 Page No 311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=159.0 #Hz, cut off frequency\n",
+ "C=0.1 #microF\n",
+ "C1=C\n",
+ "C2=C\n",
+ "C3=C\n",
+ "wc=1\n",
+ "\n",
+ "#calculation\n",
+ "R3=1/(wc*10**3*C*10**-6)\n",
+ "R1=2*R3\n",
+ "R2=R3/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The value of R3 is\",round(R3/1000),\"kohm\"\n",
+ "print\"(b)The value of R1 is\",round(R1/1000),\"kohm\"\n",
+ "print\"(c)The value of R2 is\",round(R2/1000,0),\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The value of R3 is 10.0 kohm\n",
+ "(b)The value of R1 is 20.0 kohm\n",
+ "(c)The value of R2 is 5.0 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.11 Page No 311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=60 #kHz, cut off frequency\n",
+ "C=220 #pF\n",
+ "C1=C\n",
+ "C2=C\n",
+ "C3=C\n",
+ "\n",
+ "#calculation\n",
+ "R3=1/(6.28*f*10**3*C*10**-12)\n",
+ "R1=2*R3\n",
+ "R2=R3/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The value of R3 is\",round(R3/1000,0),\"kohm\"\n",
+ "print\"(b)The value of R1 is\",round(R1/1000),\"kohm\"\n",
+ "print\"(c)The value of R2 is\",round(R2/1000),\"kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The value of R3 is 12.0 kohm\n",
+ "(b)The value of R1 is 24.0 kohm\n",
+ "(c)The value of R2 is 6.0 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.12 Page No 313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=300 # Hz, lower cutoff frequency\n",
+ "f2=3000 #Hz, upper cutoff frequency\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=f2-f1\n",
+ "f=math.sqrt(f1*f2)\n",
+ "\n",
+ "#Result\n",
+ "print\"The bandwidth is\",B,\"Hz\"\n",
+ "print\"The frequency is\",round(f,1),\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The bandwidth is 2700 Hz\n",
+ "The frequency is 948.7 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.13 Page No 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fr=950 #Hz, resonant frequency\n",
+ "B=2700 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fl=math.sqrt(B**2/4.0+fr**2)-B/2.0\n",
+ "fh=fl+B\n",
+ "\n",
+ "#Result\n",
+ "print\"The lower cutoff frequency is\",round(fl,1),\"Hz\"\n",
+ "print\"The upper cutoff frequency is\",round(fh,1),\"Hz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lower cutoff frequency is 300.8 Hz\n",
+ "The upper cutoff frequency is 3000.8 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.14 Page No 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=2700.0 #Hz\n",
+ "fr=950 #Hz resonant frequency\n",
+ "\n",
+ "#calculation\n",
+ "Q=fr/B\n",
+ "\n",
+ "#Result\n",
+ "print\"The quality factor\",round(Q,2)\n",
+ "\n",
+ "if Q<0.5:\n",
+ " print\"This filter is classified\"\n",
+ "else:\n",
+ " print\"This filter is not classified\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The quality factor 0.35\n",
+ "This filter is classified\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.15 Page No 318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fr=1000 #Hz, resonant frequency\n",
+ "fl=80 #From eq 11.16\n",
+ "fh=1280 #upper frequency\n",
+ "C=0.015 #microF\n",
+ "Q=2.0 #Quality factor\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=fr/Q\n",
+ "fl=math.sqrt(B**2/4.0+fr**2)-B/2.0\n",
+ "fh=fl+B\n",
+ "\n",
+ "R=0.1591/(B*C*10**-6)\n",
+ "Rr=R/(2*Q**2-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Rr is\",round(Rr/1000,2),\"kohm\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Rr is 3.03 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.16 Page No 319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=21.21*10**3 #ohm, resiatance\n",
+ "Rr=3.03*10**3 \n",
+ "C=0.015*10**-6 #F, capacitance\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "#From eq 11.12\n",
+ "fr=0.1125*math.sqrt(1+R/Rr)/(R*C)\n",
+ "B=0.1591/(R*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"The resonant frequency is\",round(fr,0),\"Hz\"\n",
+ "print\"Bandwidth is\",round(B,0),\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resonant frequency is 1000.0 Hz\n",
+ "Bandwidth is 500.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.17 Page No 321"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fr=120 #Hz resonant frequency\n",
+ "B=12 #Hz\n",
+ "Q=10 #Quality factor\n",
+ "\n",
+ "#Calculation\n",
+ "C=0.33 #MicroF\n",
+ "R=0.1591/(B*C*10**-6)\n",
+ "Rr=R/(2*Q**2-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\",round(R/1000,1),\"kohm\"\n",
+ "print\"The value of Rr is\",round(Rr,1),\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 40.2 kohm\n",
+ "The value of Rr is 201.9 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C12_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C12_1.ipynb new file mode 100644 index 00000000..ca52f7c0 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C12_1.ipynb @@ -0,0 +1,358 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Modulating,Demodulating and frequency Changing with the multiplier"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.1 Page No 332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#(a)\n",
+ "V=10.0\n",
+ "x1=10 #Polarity combination of voltages\n",
+ "y1=10 #V\n",
+ "#(b)\n",
+ "x2=-10 #V, 2nd combination\n",
+ "y2=10 #V\n",
+ "#(c)\n",
+ "x3=10 #V\n",
+ "y3=-10 #V\n",
+ "#(d)\n",
+ "x4=-10 #V\n",
+ "y4=-10 #V\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "Vo1=x1*y1/V\n",
+ "Vo2=x2*y2/V\n",
+ "Vo3=x3*y3/V\n",
+ "Vo4=x4*y4/V\n",
+ "\n",
+ "#Result\n",
+ "print\"The values of Vo are\",Vo1,\"V ,\",Vo2,\"V ,\",Vo3,\"V ,\",Vo4,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The values of Vo are 10.0 V , -10.0 V , -10.0 V , 10.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.2 Page No 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei1=10 #V\n",
+ "Ei2=-10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Vo1=Ei1**2/Ei1\n",
+ "Vo2=Ei2**2/Ei1\n",
+ "\n",
+ "#Reult\n",
+ "print\"(a)The value of Vo is\",Vo1,\"V\"\n",
+ "print\"(b)The value of Vo is\",Vo2,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The value of Vo is 10 V\n",
+ "(b)The value of Vo is 10 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.4 Page No 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei1=10 #V\n",
+ "f1=1 #KHz\n",
+ "Ei2=2 #V\n",
+ "f1=2.5 #KHz\n",
+ "\n",
+ "#Calculation\n",
+ "dcvalue=Ei1**2/20.0\n",
+ "pac=Ei1**2/20.0\n",
+ "\n",
+ "\n",
+ "#(b)\n",
+ "dcvalue_=Ei2**2/20.0\n",
+ "pac_=Ei2**2/20.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The DC and AC output voltage are\",dcvalue,\"V,\",pac,\"V at 2 KHz\"\n",
+ "print\"(b)The DC and AC output voltage are\",dcvalue_,\"V,\",pac_,\"V at 5 KHz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The DC and AC output voltage are 5.0 V, 5.0 V at 2 KHz\n",
+ "(b)The DC and AC output voltage are 0.2 V, 0.2 V at 5 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.5 Page No 339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Exp=5 #V\n",
+ "Eyp=5 #V\n",
+ "Vodc=1.25 #DC component of Vo\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "theta=math.acos(20*Vodc/(Exp*Eyp))\n",
+ "\n",
+ "#Result\n",
+ "print\"The angle is\",theta,\"degree\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle is 0.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.6 Page No 339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vo=0 #AVerage value of DC component\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "theta=math.acos(Vo)\n",
+ "\n",
+ "#Result\n",
+ "print\"The angle is\",round(theta*180/3.14,0),\"degree\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle is 90.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.7 Page No 339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "theta1=30 #degree\n",
+ "theta2=45 #degree\n",
+ "theta3=60 #degree\n",
+ "theta4=0 #Degree from example 12.5 \n",
+ "theta5=90 #Degree from example 12.6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vodc1=math.cos(theta1*3.14/180.0)\n",
+ "Vodc2=math.cos(theta2*3.14/180.0)\n",
+ "Vodc3=math.cos(theta3*3.14/180.0)\n",
+ "Vodc4=math.cos(theta4*3.14/180.0)\n",
+ "Vodc5=math.cos(theta5*3.14/180.0)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The Output Dc voltages are\",round(Vodc1,3),\"V ,\",round(Vodc2,3),\"V ,\",round(Vodc3,3),\"V ,\",round(Vodc4,3),\"V ,\",round(Vodc5,3),\"V ,\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Output Dc voltages are 0.866 V , 0.707 V , 0.5 V , 1.0 V , 0.001 V ,\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.8 Page No 340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#GIven\n",
+ "theta1=90 #degree\n",
+ "theta2=120 #degree\n",
+ "theta3=135 #degree\n",
+ "theta4=150 #degree\n",
+ "theta5=180 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vodc1=math.cos(theta1*3.14/180.0)\n",
+ "Vodc2=math.cos(theta2)\n",
+ "Vodc3=math.cos(theta3)\n",
+ "Vodc4=math.cos(theta4)\n",
+ "Vodc5=math.cos(theta5)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Voltages are\",round(Vodc1,3),\"V ,\",round(Vodc2,3),\"V \",round(Vodc3,3),\"V \",round(Vodc4,3),\"V \",round(Vodc5,3),\"V \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Voltages are 0.001 V , 0.814 V -0.996 V 0.699 V -0.598 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 12.10 Page No 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ecp=5 #V, Voltage\n",
+ "fc=10000 #Hz frequency\n",
+ "Em=5 #V peak voltage\n",
+ "fm=1000 #Hz frequency\n",
+ "\n",
+ "#Calculation\n",
+ "Vc=Em*Ecp/(10.0) #The carrier term peak voltage\n",
+ "Vp=Vc #product term Peak voltage\n",
+ "Vs=Em*Ecp/(20.0) #The side term peak voltagec\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The carrier term peak voltage is\",Vc,\"V\"\n",
+ "print\"The product term peak voltage is\",Vp,\"V\"\n",
+ "print\"The side term peak voltage is\",Vs,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The carrier term peak voltage is 2.5 V\n",
+ "The product term peak voltage is 2.5 V\n",
+ "The side term peak voltage is 1.25 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C13_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C13_1.ipynb new file mode 100644 index 00000000..19a9a8b1 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C13_1.ipynb @@ -0,0 +1,496 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13: Integrated circuit Timers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.1 Page No 371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From fig 13.7 (a)\n",
+ "Ra=6.8 #kohm, resistance\n",
+ "Rb=3.3 #Kohm\n",
+ "C=0.1 #microF, capacitance\n",
+ "\n",
+ "#Calculation\n",
+ "thigh=0.695*(Ra+Rb)*C\n",
+ "tlow=0.695*Rb*C\n",
+ "f=1.44/((Ra+2*Rb)*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of high time is\",round(thigh,1),\"ms\"\n",
+ "print\"The value of low time is\",round(tlow,2),\"ms\"\n",
+ "print\"The value of frequency is\",round(f,2),\"khz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of high time is 0.7 ms\n",
+ "The value of low time is 0.23 ms\n",
+ "The value of frequency is 1.07 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.2 Page No 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From fig 13.7 (a)\n",
+ "Ra=6.8 #kohm, resistance\n",
+ "Rb=3.3 #Kohm\n",
+ "\n",
+ "#Calculation\n",
+ "D=Rb/(Ra+2*Rb)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Duty Cycle is\",round(D,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Duty Cycle is 0.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.3 Page No 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#GIven\n",
+ "V=5 #V, Voltage\n",
+ "E1=0 #V\n",
+ "E2=1\n",
+ "Re=3 #Kohm, resistance (fig 13-10)\n",
+ "C=1 #microF\n",
+ "\n",
+ "#Calculation\n",
+ "I=(V-E1)/(3.0) #Current\n",
+ "fc=3/(Re*C)\n",
+ "df=0.2*fc*E2\n",
+ "fout=fc+df\n",
+ "E3=-1 #V\n",
+ "df_=0.2*fc*E3\n",
+ "fout_=fc+df_\n",
+ "Elower=-4\n",
+ "Eupper=4\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The charge current is\",round(I,2),\"mA\"\n",
+ "print\"(b)The current frequency is\",fc,\"KHz\"\n",
+ "print\"(c)df=\",df*1000,\"Hz\",\"\\nand The frequency shift is\",(df-df_)*1000,\"Hz\"\n",
+ "print\"fout was initially\",fout*1000,\"Hz . then changes to\",fout_*1000,\"Hz\"\n",
+ "print\"(d)The positive nd negative limits are\",Eupper,\"V and\",Elower,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The charge current is 1.67 mA\n",
+ "(b)The current frequency is 1 KHz\n",
+ "(c)df= 200.0 Hz \n",
+ "and The frequency shift is 400.0 Hz\n",
+ "fout was initially 1200.0 Hz . then changes to 800.0 Hz\n",
+ "(d)The positive nd negative limits are 4 V and -4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.4 Page No 380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#GIven\n",
+ "Ra=9.1 #Kohm\n",
+ "thigh=1 #ms\n",
+ "\n",
+ "#Calculation\n",
+ "C=thigh*10**-3/(1.1*Ra)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of C is\",round(C*10**3,1),\"microF\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of C is 0.1 microF\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.5 Page No 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ra=10 #Kohm\n",
+ "C=0.2 #microF\n",
+ "Ci=0.001 #microF\n",
+ "\n",
+ "#calculation\n",
+ "thigh=1.1*(Ra*10**3*C*10**-6)\n",
+ "t=Ra*10**3*Ci\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The value of high time is\",thigh*1000,\"ms\"\n",
+ "print\"(b)Time constant is\",t/1000,\"ms\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The value of high time is 2.2 ms\n",
+ "(b)Time constant is 0.01 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.6 Page No 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ra=10 #Kohm\n",
+ "C=0.1 #microF\n",
+ "f=1 #kHz\n",
+ "\n",
+ "#Calculation\n",
+ "thigh=1.1*(Ra*10**3*C*10**-6)\n",
+ "#thigh should exceed two periods and be less than 3 periods.\n",
+ "thigh_=2.2 #ms, chose a value\n",
+ "Ra=thigh_/(1.1*C*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The time interval is\",thigh*1000,\"ms\"\n",
+ "print\"Ra is\",Ra/10**6,\"Kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The time interval is 1.1 ms\n",
+ "Ra is 20.0 Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.7 Page No 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=100 #Kohm\n",
+ "C=0.01 #microF\n",
+ "p1=3 #pin\n",
+ "p2=4\n",
+ "p3=7\n",
+ "p4=8\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "T=R*10**3*C*10**-6\n",
+ "tlow1=4*T #Using Table 13-2, pin=3 , time= 4*T\n",
+ "tlow2=8*T\n",
+ "tlow3=64*T\n",
+ "tlow4=128*T\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The required time is\",tlow1*1000,\"ms\"\n",
+ "print\"(b)The required time is\",tlow2*1000,\"ms\"\n",
+ "print\"(c)The required time is\",tlow3*1000,\"ms\"\n",
+ "print\"(d)The required time is\",tlow4*1000,\"ms\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The required time is 4.0 ms\n",
+ "(b)The required time is 8.0 ms\n",
+ "(c)The required time is 64.0 ms\n",
+ "(d)The required time is 128.0 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.8 Page No 389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=1 #s\n",
+ "\n",
+ "#Calculation\n",
+ "#Pins 3,6,7 are jumpered\n",
+ "Tsum1=8*T+16*T\n",
+ "Tsum2=4*T+32*T+64*T\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The timing cycle is\",Tsum1,\"s\"\n",
+ "print\"(b)The timing cycle is\",Tsum2,\"s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The timing cycle is 24 s\n",
+ "(b)The timing cycle is 100 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.9 Page No 389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=1 #microF\n",
+ "R=5 #Mohm\n",
+ "T=5 #s\n",
+ "\n",
+ "#Calculation\n",
+ "Tsum1=4*T+8*T\n",
+ "t=1*T\n",
+ "Tsum2=T+2*T+4*T+8*T+16*T+32*T+64*T+128*T\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The timing cycle is\",Tsum1/60,\"min\"\n",
+ "print\"(b)Minimum programmeble timing cycle is\",t,\"s\"\n",
+ "print\"(c)maximum programmable timing cycle is\",round(Tsum2,2),\"s or 21 min 15 s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The timing cycle is 1 min\n",
+ "(b)Minimum programmeble timing cycle is 5 s\n",
+ "(c)maximum programmable timing cycle is 1275.0 s or 21 min 15 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.10 Page No 391"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=2.5 #ms\n",
+ "\n",
+ "#pin from fig 13-19\n",
+ "p1=1\n",
+ "p2=2\n",
+ "p3=3\n",
+ "p4=4\n",
+ "\n",
+ "#Calculation\n",
+ "#Time base rating\n",
+ "t1=T\n",
+ "t2=2*T\n",
+ "t3=4*T\n",
+ "t4=8*T\n",
+ "\n",
+ "#Period\n",
+ "P1=2*T\n",
+ "P2=4*T\n",
+ "P3=8*T\n",
+ "P4=16*T\n",
+ "\n",
+ "#Frequency\n",
+ "f1=1/(P1)\n",
+ "f2=1/(P2)\n",
+ "f3=1/(P3)\n",
+ "f4=1/(P4)\n",
+ "\n",
+ "#Result\n",
+ "print\"The frequencies are\",f1*1000,\"Hz,\",f2*1000,\"Hz,\",f3*1000,\"Hz,\",f4*1000,\"Hz\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequencies are 200.0 Hz, 100.0 Hz, 50.0 Hz, 25.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.11 Page No 393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From fig 13.11 (a)\n",
+ "T=1 #ms\n",
+ "\n",
+ "#Calculation\n",
+ "Tsum=T+4*T\n",
+ "period=Tsum+T\n",
+ "f=1/(period*10**-3)\n",
+ "\n",
+ "#Result\n",
+ "print\"The frequency is\",round(f,1),\"Hz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency is 166.7 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C14_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C14_1.ipynb new file mode 100644 index 00000000..5eaa01c3 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C14_1.ipynb @@ -0,0 +1,509 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14:Digital to analog Converters"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.1 Page no 403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=8 #An 8-bit DAC wire\n",
+ "V1=0 #Voltage range\n",
+ "V2=5.12\n",
+ "\n",
+ "#Calculation\n",
+ "R=2**n\n",
+ "dvo=V2/R\n",
+ "Vfs=V2*(1-1/R)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The resolution is\",R\n",
+ "print\"(b)The output change per bit is\",dvo*1000,\"mV/bit\"\n",
+ "print\"(c)the ideal full scale output voltage is\",Vfs,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The resolution is 256\n",
+ "(b)The output change per bit is 20.0 mV/bit\n",
+ "(c)the ideal full scale output voltage is 5.12 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.2 Page no 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Res=20 #mV/bit\n",
+ "#From the given code 00010110 and 10000000\n",
+ "#The binary weighted code for unipolar operation\n",
+ "b7=0\n",
+ "b6=0\n",
+ "b5=0\n",
+ "b4=1\n",
+ "b3=0\n",
+ "b2=1\n",
+ "b1=1\n",
+ "b0=0\n",
+ "#For part (b)\n",
+ "b7_=1\n",
+ "b6_=0\n",
+ "b5_=0\n",
+ "b4_=0\n",
+ "b3_=0\n",
+ "b2_=0\n",
+ "b1_=0\n",
+ "b0_=0\n",
+ "\n",
+ "#Calculation\n",
+ "D=b7*2**7+b6*2**6+b5*2**5+b4*2**4+b3*2**3+b2*2**2+b1*2**1+b0*2**0\n",
+ "Vo=D*Res*10**-3\n",
+ "D_=b7_*2**7+b6_*2**6+b5_*2**5+b4_*2**4+b3_*2**3+b2_*2**2+b1_*2**1+b0_*2**0\n",
+ "Vo_=D_*Res*10**-3\n",
+ "\n",
+ "#Resul\n",
+ "print\"(a)The output voltage is\",Vo,\"V\"\n",
+ "print\"(b)The output voltage is\",Vo_,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The output voltage is 0.44 V\n",
+ "(b)The output voltage is 2.56 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.3 Page no 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From Example 14.1\n",
+ "n=8 #An 8-bit DAC wire\n",
+ "V1=0 #Voltage range\n",
+ "V2=5.12\n",
+ "\n",
+ "#Calculation\n",
+ "dvo=2*V2/2**n \n",
+ "\n",
+ "#Result\n",
+ "print\"The full scale output voltage range\",dvo*1000,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The full scale output voltage range 40.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.4 Page no 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vref=10.24 #V, input ref voltage\n",
+ "p=0.05 #Percentage error\n",
+ "n=12 #12 bit unipolar DAC\n",
+ "\n",
+ "#Calculation\n",
+ "Voff=0.05/100*Vref\n",
+ "dvo=Vref/(2**n)\n",
+ "a=Voff/dvo\n",
+ "\n",
+ "#REsult\n",
+ "print\"The offset voltage is\",Voff*1000,\"mV\"\n",
+ "print\"Interms of least significant bit(s) it is\",dvo*1000,\"mV/bit\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The offset voltage is 5.12 mV\n",
+ "Interms of least significant bit(s) it is 2.5 mV/bit\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.5 Page no 407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=8.0 #8-bit DAC\n",
+ "error=0.2 #percent\n",
+ "Vref=5.12 #V\n",
+ "Vos=0\n",
+ "\n",
+ "#Calculation\n",
+ "V11=(((-error/100.0))+1)*(Vref)+Vos\n",
+ "\n",
+ "#REsult\n",
+ "print\"The minimum output voltage is\",round(V11,1),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum output voltage is 5.1 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.6 Page no 411"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=4.0 #4 bit resistance ladder\n",
+ "R=10.0 #kohm\n",
+ "Vref=10.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Res=Vref/(2**n*R*10**3)\n",
+ "\n",
+ "#From dicimal value of binary 1111\n",
+ "D=15\n",
+ "Iout=Res*D\n",
+ "\n",
+ "#Result\n",
+ "print\"The resolution of 1 LSB is\",Res*10**6,\"microA\"\n",
+ "print\"output current is\",Iout*1000,\"microA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resolution of 1 LSB is 62.5 microA\n",
+ "output current is 0.9375 microA\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.7 Page no 412"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Io=62.5 #microA, from ex. 14.6\n",
+ "Rf=10 #kohm , from fig.14.7\n",
+ "Vref=10 #V, reference voltage\n",
+ "R=Rf\n",
+ "n=4 #no of bits\n",
+ "\n",
+ "#Calculation\n",
+ "VR=Io*Rf*10**-3\n",
+ "D=15 #For a digital input of 1111\n",
+ "Vo=-VR*D\n",
+ "\n",
+ "#REsult\n",
+ "print\"The voltage resolution of 1 LSB is\",VR,\"V\"\n",
+ "print\"When digital input is 1111 , Vo is\",Vo,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The voltage resolution of 1 LSB is 0.625 V\n",
+ "When digital input is 1111 , Vo is -9.375 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.8 Page no 416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vref=10.0 #V, from fig. 14.8. reference voltage\n",
+ "Rref=5.0 #kohm\n",
+ "n=8.0 #no of bits\n",
+ "\n",
+ "#Calculation\n",
+ "Iref=Vref/Rref\n",
+ "I=Vref/(Rref*2**n)\n",
+ "\n",
+ "#Result\n",
+ "print\"The ladder input current is\",Iref,\"mA\"\n",
+ "print\"Current value of 1 LSB is\",round(I*1000,3),\"microA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ladder input current is 2.0 mA\n",
+ "Current value of 1 LSB is 7.813 microA\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.9 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=7.812 #Current resolution from ex 14.8\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "Ifs=I*10**-3*255 #From eq. 14-15(a)\n",
+ "#The value of D for part (a),(b),(c)\n",
+ "D1=1 #for digital input 00000001\n",
+ "D2=128\t\t#for digital input 10000000\n",
+ "D3=255\t\t#for digital input 11111111\n",
+ "Iout1=I*D1\n",
+ "Iout2=I*D2\n",
+ "Iout3=I*D3\n",
+ "Iout_1=Ifs-Iout1*10**-3\n",
+ "Iout_2=Ifs-Iout2*10**-3\n",
+ "Iout_3=Ifs-Iout3*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The Iout is\",Iout1,\"mA and Iout_ is\",round(Iout_1,3),\"mA\"\n",
+ "print\"(b) The Iout is\",Iout2,\"mA and Iout_ is\",round(Iout_2,3),\"mA\"\n",
+ "print\"(c) The Iout is\",Iout3,\"mA and Iout_ is\",Iout_3,\"mA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The Iout is 7.812 mA and Iout_ is 1.984 mA\n",
+ "(b) The Iout is 999.936 mA and Iout_ is 0.992 mA\n",
+ "(c) The Iout is 1992.06 mA and Iout_ is 0.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.10 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From fig 14.8 (a)\n",
+ "Vref=10 #V\n",
+ "Rf=5\t\t#kohm\n",
+ "Rref=5.0\t\t#kohm\n",
+ "n=8\t\t#\n",
+ "\n",
+ "#Calculation\n",
+ "VR=Vref*Rf/(Rref*2**n)\n",
+ "D1=1\t\t#Value of D for 00000001\n",
+ "D2=255\t\t#Value of D for 11111111\n",
+ "Vo1=VR*1000*D1\n",
+ "Vo2=VR*1000*D2\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Vo is \",round(Vo1,0),\"V\"\n",
+ "print\"(b) Vo is \",round(Vo2/1000,3),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Vo is 39.0 V\n",
+ "(b) Vo is 9.961 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 14.11 Page no 418"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From fig 14.9(a)\n",
+ "I=8 #microA, value of 1 LSB\n",
+ "Rf=5\t\t# kohm\n",
+ "\n",
+ "#Calculation\n",
+ "#from eq 14-15(a)\n",
+ "Ifs=I*10**-3*255\n",
+ "D1=0 \t\t#Value of D for 00000000\n",
+ "Iout1=I*D1\n",
+ "Iout_1=Ifs-Iout1\n",
+ "Vo1=(Iout1-Iout_1)*Rf\n",
+ "\n",
+ "D2=127 #From digital value(b)\n",
+ "Iout2=I*D2\n",
+ "Iout_2=Ifs-Iout2/1000.0\n",
+ "Vo2=(Iout2/1000.0-Iout_2)*Rf\n",
+ "\n",
+ "D3=128 \t\t#Value of D for 00000000\n",
+ "Iout3=I*D3\n",
+ "Iout_3=Ifs-Iout3/1000.0\n",
+ "Vo3=(Iout3/1000.0-Iout_3)*Rf\n",
+ "\n",
+ "D4=255 \t\t#Value of D for 00000000\n",
+ "Iout4=I*D4\n",
+ "Iout_4=Ifs-Iout4/1000\n",
+ "Vo4=(Iout4/1000-Iout_4)*Rf\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)Vo is \",Vo1,\"V . (b)Vo is \",round(Vo2,3),\"V. (c)Vo is \",Vo3,\"V .(d)Vo is \",round(Vo4,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " 2.04\n",
+ "(a)Vo is -10.2 V . (b)Vo is -0.04 V. (c)Vo is 0.04 V .(d)Vo is 10.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C15_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C15_1.ipynb new file mode 100644 index 00000000..27c99a7a --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C15_1.ipynb @@ -0,0 +1,321 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 15: Analog to digital converters"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.1 Page No 432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=8 \t\t#8 bit ADC\n",
+ "V1=0\t\t#V, input unipolar voltage\n",
+ "V2=10\t\t#V\n",
+ "Vi=5.2 \t#V, applied input voltage\n",
+ "\n",
+ "#calculation\n",
+ "LSB=V2*1000/(2**n)\n",
+ "Vifs=V2-LSB*10**-3\n",
+ "D=Vi/LSB*10**3\n",
+ "binary=10000101\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The minimum input voltage is\",LSB,\"mV\"\n",
+ "print\"(b)Input voltage that will cause all 1s at ADC's output\",Vifs,\"V\"\n",
+ "print\"(c)The No of LSB's are \",round(D,0),\"So The digital output code is\",binary\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The minimum input voltage is 39 mV\n",
+ "(b)Input voltage that will cause all 1s at ADC's output 9.961 V\n",
+ "(c)The No of LSB's are 133.0 So The digital output code is 10000101\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.2 Page No 434"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given (from example 15.1)\n",
+ "n=8 \t\t#8 bit ADC\n",
+ "FSR=10\t\t#V\n",
+ "LSB=39.1 #mA, from ex. 15.1\n",
+ "\n",
+ "#Calculation\n",
+ "Vifs=FSR*10**3-LSB+LSB/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"The voltahe is\",Vifs/1000,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The voltahe is 9.98045 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.3 Page No 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "error=0.5 #percent\n",
+ "FSR=10\t\t#V\n",
+ "Vifs=9.961 #From ex 15.1 (c)\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "V=FSR*error/100.0\n",
+ "Vin=Vifs-V*10**-3\n",
+ "\n",
+ "#result\n",
+ "print\"The required voltage is\",round(Vin,2),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required voltage is 9.96 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.4 Page No 438"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vin1=100 #mV, input voltage\n",
+ "Vin2=200 #mV, input voltage\n",
+ "\n",
+ "#Calculation\n",
+ "T2=0.833*Vin1 #From eq 15.6(b)\n",
+ "T2_=0.833*Vin2\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The required tome is\",T2,\"ms\"\n",
+ "print\"(b)The required tome is\",T2_,\"ms\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The required tome is 83.3 ms\n",
+ "(b)The required tome is 166.6 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.5 Page No 439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vin=100 #mV\n",
+ "\n",
+ "#Calculation\n",
+ "Vo=10*Vin #From eq 15.7(c)\n",
+ "\n",
+ "#Result\n",
+ "print\"The digital output is\",Vo,\"counts \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The digital output is 1000 counts \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.6 Page No 442"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "n=8 \t\t#Bit\n",
+ "t=1 #micros, time for 1 clock pulse\n",
+ "\n",
+ "#Calcultion\n",
+ "Tc=t*(n+1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The required time is\",Tc,\"microsecond\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required time is 9 microsecond\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.7 Page No 450"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=8 \t\t#8 bit ADC\n",
+ "t=10 \t\t#micros, conversion time\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fmax=1/(2*math.pi*2**n*t*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\"The maximum frequency is\",round(fmax,0),\"Hz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum frequency is 62.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 15.8 Page No 451"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=50 \t\t#ns, aperture time\n",
+ "n=8\n",
+ "error=1 #LSB\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fmax=1/(2*math.pi*2**n*t*10**-9)\n",
+ "\n",
+ "#Result\n",
+ "print\"The maximum frequency is\",round(fmax/1000,1),\"Hz\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum frequency is 12.4 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C16_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C16_1.ipynb new file mode 100644 index 00000000..6ba96200 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C16_1.ipynb @@ -0,0 +1,344 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16:Power Supplies"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 16.1 Page No 455"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#from fig. 16.1\n",
+ "Erms=24 #V, rms voltage\n",
+ "\n",
+ "#Calculation\n",
+ "Em=1.4*Erms\n",
+ "\n",
+ "#Result\n",
+ "print\"The maximum instantaneous voltage is\",round(Em,0),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum instantaneous voltage is 34.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 16.2 Page No 460"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From fig 16.4 and 16.5\n",
+ "#at no load\n",
+ "Ilnl=0 #A, load current\t\t\n",
+ "Vdcnl=34\t#V\n",
+ "Ilfl=1 #A, at full load\n",
+ "Vdcfl=24 #V\t\n",
+ "Il=0.5 \n",
+ "\n",
+ "#Calculation\n",
+ "Ro=(Vdcnl-Vdcfl)/(Ilfl)\n",
+ "Vdc=Vdcnl-Il*Ro\n",
+ "\n",
+ "#result\n",
+ "print\"Th output resistance is\",Ro,\"ohm\"\n",
+ "print\"The dc output voltage is\",Vdc,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Th output resistance is 10 ohm\n",
+ "The dc output voltage is 29.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 16.3 Page No 461"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#GIven\n",
+ "Vdcnl=34\t#V, at no load\n",
+ "Ilfl=1 #A, at full load\n",
+ "Vdcfl=24 #V, at full load\n",
+ "\n",
+ "#Calculation\n",
+ "regu=(Vdcnl-Vdcfl)*100/Vdcfl\n",
+ "\n",
+ "#Result\n",
+ "print\"Thepercent regulation is\",round(regu,1),\"percent\"\t\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thepercent regulation is 41.0 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 16.4 Page No 463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "If=1 #A, full load current\n",
+ "Vf=24 #V full load voltage\n",
+ "C=1000 \t\t#microF , capacitance\n",
+ "\n",
+ "#Calculation\n",
+ "dvo=If/(200*C**10**-6)\n",
+ "Vrms=dvo/(3.5)\n",
+ "Vlmin=Vf-(round(dvo*1000,0)/2.0)\n",
+ "\n",
+ "#result\n",
+ "print\"(a)peak to peak and rms value of ripple voltage is\",round(dvo*1000,0),\"V and\",round(Vrms*1000,2),\"V\"\n",
+ "print\"(b) The minimum instanttaneous output voltage is\",Vlmin,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)peak to peak and rms value of ripple voltage is 5.0 V and 1.43 V\n",
+ "(b) The minimum instanttaneous output voltage is 21.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 16.5 Page No 463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#from example 16.4\n",
+ "Vf=24.0 #V\n",
+ "Vrms=1.43 #V\n",
+ "\n",
+ "#Calculation\n",
+ "ripple=Vrms/Vf*100\n",
+ "\n",
+ "#result\n",
+ "print\"The percentage ripple is\",round(ripple,0),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage ripple is 6.0 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 16.6 Page No 464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vdcfl=12.0 #V, voltage\n",
+ "Ilfl=1.0 #A current\n",
+ "ripp=10.0 #percent, ripple\n",
+ "Ro=10.0 #ohm, guess value\n",
+ "\n",
+ "\n",
+ "#calculation\n",
+ "#DC design procedure\n",
+ "#from eq 16.3(a)\n",
+ "Vdcnl=Vdcfl+Ilfl*Ro\n",
+ "Em=Vdcnl\n",
+ "Erms=Em/1.4\n",
+ "#For a FWB rectifier\n",
+ "Isec=1.8*Ilfl\n",
+ "PIV=Vdcnl+(20/100.0)*Vdcnl\n",
+ "\n",
+ "#AC design procedure\n",
+ "Vrms=(ripp/100.0)*Vdcfl\n",
+ "dvo=3.5*Vrms #From eq 16.5 (b)\n",
+ "C=Ilfl/(200*dvo)\n",
+ "WVDC=Vdcnl+(20/100.0)*Vdcnl\n",
+ "\n",
+ "#result\n",
+ "print\" Transformer at :115V/18V at \",round(Isec,0),\"A\"\n",
+ "print\"Four diodes :each 1 A at PIV\",PIV,\"V (or more)\"\n",
+ "print\"THree capacitors: 500 muF connected in parallel,WVDC=\",WVDC,\"V (or more)\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Transformer at :115V/18V at 2.0 A\n",
+ "Four diodes :each 1 A at PIV 26.4 V (or more)\n",
+ "THree capacitors: 500 muF connected in parallel,WVDC= 26.4 V (or more)\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Analysis Example 16.7 Page No 466"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ro=7\t\t#ohm\n",
+ "Ilfl=1\t\t#A, \n",
+ "C=1000 #microF\n",
+ "\n",
+ "#Calculation\n",
+ "Erms=1.8*Vrms\n",
+ "Em=1.4*Erms\n",
+ "Vdcnl=Em\n",
+ "Vdcfl=Vdcnl-Ilfl*Ro\n",
+ "reg=(Vdcnl-Vdcfl)*100/Vdcfl \n",
+ "\n",
+ "dvo=(Ilfl/200*C*10**-6)\n",
+ "Vrms=dvo/3.5\n",
+ "rip=Vrms*100/Vdcfl\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)Vdcfl is\",Vdcfl,\"V and percentag eregulation is\",round(reg,0)\n",
+ "print\"(b)percentage ripple is\",round(rip,1)\n",
+ "\n",
+ "#Plot\n",
+ "I=[0,5,10]\n",
+ "VdcNL=[26,22,18.5]\n",
+ "a1=plot(I,VdcNL)\n",
+ "show(a1)\n",
+ "\n",
+ "t1=[0,16.7]\n",
+ "Vdcnl=[26,26]\n",
+ "t2=[0,2,8.34,10.34,16.7]\n",
+ "Vdcfl=[16,21,16,21.5,16]\n",
+ "a2=plot(t1,Vdcnl)\n",
+ "a3=plot(t2,Vdcfl)\n",
+ "show(a2)\n",
+ "show(a3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Vdcfl is -7.0 V and percentag eregulation is -100.0\n",
+ "(b)percentage ripple is -0.0\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x75605c0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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rPw5KhUG87KkZ3h3zLnx7+yJ4czBu/XZLdByDI/wVzbGKvIQOCMWWyC2I3B6J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\nWORERDLHIicikjkWORGRzP0vdswUY8DlrCUAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x7533ba8>"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C2_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C2_1.ipynb new file mode 100644 index 00000000..c91c71a1 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C2_1.ipynb @@ -0,0 +1,212 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:First Experiences with Op Amp"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 2.1 Page No 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vplus=15\n",
+ "Vminus=-15\n",
+ "Vsatp=13\n",
+ "Vsatm=-13 #All in Volts\n",
+ "Aol=200000.0 #gain\n",
+ "#Example 2-1(a)\n",
+ "Vam=-10*(10**-6) #voltage at minus input\n",
+ "Vap=-15*(10**-6) #voltage at plus input\n",
+ "Ed1=Vap-Vam #Differential Input Voltage\n",
+ "Vout1=Ed1*Aol #Output Voltage\n",
+ "if Vout1>15:\n",
+ " print\"Value of o/p voltage1 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout1<-15):\n",
+ " print\"Value of o/p voltage1 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage1 = \",Vout1*Vsatp,\"V\"\n",
+ "\n",
+ "#Example 2-1(b)\n",
+ "Vbm=-10*(10**-6) #voltage at minus input\n",
+ "Vbp=+15*(10**-6) #voltage at plus input\n",
+ "Ed2=Vbp-Vbm #Differential Input Voltage\n",
+ "Vout2=Ed2*Aol #Output Voltage\n",
+ "if Vout2>15:\n",
+ " print\"Value of o/p voltage2 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout2<-15):\n",
+ " print\"Value of o/p voltage2 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage2 = \",Vout2,\"V\"\n",
+ "\n",
+ "#Example 2-1(c)\n",
+ "Vcm=-10*(10**-6) #voltage at minus input\n",
+ "Vcp=-5*(10**-6) #voltage at plus input\n",
+ "Ed3=Vcp-Vcm #Differential Input Voltage\n",
+ "Vout3=Ed3*Aol #Output Voltage\n",
+ "if(Vout3>15):\n",
+ " print\"Value of o/p voltage3 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout3<-15):\n",
+ " print\"Value of o/p voltage3 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage3 = \",Vout3,\"V\"\n",
+ "\n",
+ "\n",
+ "#Example 2-1(d)\n",
+ "Vdm=+1.000001 #voltage at minus input\n",
+ "Vdp=+1.000000 #voltage at plus input\n",
+ "Ed4=Vdp-Vdm #Differential Input Voltage\n",
+ "Vout4=Ed4*Aol #Output Voltage\n",
+ "if(Vout4>15):\n",
+ " print\"Value of o/p voltage4 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout4<-15):\n",
+ " print\"Value of o/p voltage4 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage4 = \",round(Vout4,2),\"V\"\n",
+ "\n",
+ "\n",
+ "#Example 2-1(e)\n",
+ "Vem=+5*(10**-3) #voltage at minus input\n",
+ "Vep=0 #voltage at plus input\n",
+ "Ed5=Vep-Vem #Differential Input Voltage\n",
+ "Vout5=Ed5*Aol #Output Voltage\n",
+ "if(Vout5>15):\n",
+ " print\"Value of o/p voltage5 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout5<-15):\n",
+ " print\"Value of o/p voltage5 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage5 = \",Vout5\n",
+ "\n",
+ "#Example 2-1(f)\n",
+ "Vfm=0 #voltage at minus input\n",
+ "Vfp=+5*(10**-3) #voltage at plus input\n",
+ "Ed6=Vfp-Vfm #Differential Input Voltage\n",
+ "Vout6=Ed6*Aol #Output Voltage\n",
+ "if(Vout6>15):\n",
+ " print\"Value of o/p voltage6 = 13.0000V\" #positive saturation voltage\n",
+ "elif(Vout6<-15):\n",
+ " print\"Value of o/p voltage6 = -13.0000V\" #negative saturation voltage\n",
+ "else:\n",
+ " print\" Value of o/p voltage6 = V \",Vout6\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of o/p voltage1 = -13.0 V\n",
+ " Value of o/p voltage2 = 5.0 V\n",
+ " Value of o/p voltage3 = 1.0 V\n",
+ " Value of o/p voltage4 = -0.2 V\n",
+ "Value of o/p voltage5 = -13.0000V\n",
+ "Value of o/p voltage6 = 13.0000V\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page No 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50.0 #in Hz\n",
+ "Vtemp=4.0 #input signal in volts\n",
+ "Ecm=10.0 \n",
+ "\n",
+ "#Calculation\n",
+ "#Example 2-2(a)\n",
+ "T=1/f\n",
+ "Th=(Vtemp*T)/Ecm #High time in seconds\n",
+ "\n",
+ "#Example 2-2(b)\n",
+ "d=(Th/T)*100 #duty cycle in percentage\n",
+ "\n",
+ "#Result\n",
+ "print\" High Time is \",Th*1000,\"ms\"\n",
+ "print\" Duty cycle is\",d,\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " High Time is 8.0 ms\n",
+ " Duty cycle is 40.0 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page No 37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vtemp=4.0 #in volts\n",
+ "Ecm=5.0 #maximum peak voltage of a sawtooth carrier wave\n",
+ "T=0.01 #in seconds\n",
+ "\n",
+ "#calculation\n",
+ "Th=T*(1-(Vtemp/Ecm)) #High Time\n",
+ "\n",
+ "#Result\n",
+ "print\" High Time is \",Th*1000,\"ms\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " High Time is 2.0 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C3_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C3_1.ipynb new file mode 100644 index 00000000..e799a9b7 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C3_1.ipynb @@ -0,0 +1,555 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3: Inverting and Non Inverting Amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page No 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf=100*(10**3) #Feedback Resistance in ohms\n",
+ "Ri=10*(10**3) #Input Resistance in ohms\n",
+ "Ei=1.0 #Input volts\n",
+ "\n",
+ "#Calculation \n",
+ "I=Ei/Ri #Equation for current through Rf\n",
+ "Vout=-(Rf/Ri)*Ei #Equation for Output Voltage\n",
+ "Acl=-(Rf/Ri) #Closed loop gain of the amplifier \n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Current through Rf is\",I*1000,\"mA\"\n",
+ "print\"(b) Value of output voltage is \",Vout,\"V\"\n",
+ "print\"(c) Value of closed loop gain is\",Acl"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Current through Rf is 0.1 mA\n",
+ "(b) Value of output voltage is -10.0 V\n",
+ "(c) Value of closed loop gain is -10\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page No 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vout=10.0 #output voltage\n",
+ "I=0.1*(10**-3) #current through Rf in amperes\n",
+ "Rl=25*(10**3) #Load resistance in ohms\n",
+ "\n",
+ "#Calculation\n",
+ "Il=Vout/Rl\n",
+ "Iout=I+Il\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Value of load current is\",Il*1000,\"m\"\n",
+ "print\"(b) Total current into the output pin of the opamp is \",Iout*1000,\"mA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Value of load current is 0.4 m\n",
+ "(b) Total current into the output pin of the opamp is 0.5 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page No 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf=250*(10**3)#Feedback Resistance in Ohms\n",
+ "Ri=10*(10**3)#Input Resistance in Ohms\n",
+ "Ei=0.5#Input voltage\n",
+ "\n",
+ "#Calculation\n",
+ "I=Ei/(Ri)\n",
+ "VRf=I*Rf\n",
+ "Ei1=-0.5\n",
+ "Vout=-(Rf/Ri)*Ei1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Value of current through Rf is\",I,\"A\"\n",
+ "print\"(b) Voltage through Rf is \",VRf,\"V\"\n",
+ "print\"(c) Output Voltage is\",Vout,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Value of current through Rf is 5e-05 A\n",
+ "(b) Voltage through Rf is 12.5 V\n",
+ "(c) Output Voltage is 12.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page No 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vout=12.5 #in volts\n",
+ "Il=2*10**-3 #in amperes\n",
+ "\n",
+ "#Calculation\n",
+ "Rl=Vout/Il #Load resistance in ohm\n",
+ "I=0.05*10**-3\n",
+ "Iout=I+Il\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Value of Load Resistance is\",Rl/1000,\"kohm\"\n",
+ "print\"(b) Value of output current is \",Iout*1000,\"mA\"\n",
+ "print\"(c)The circuit input resistance is Rl=10Kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Value of Load Resistance is 6.25 kohm\n",
+ "(b) Value of output current is 2.05 mA\n",
+ "(c)The circuit input resistance is Rl=10Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page No 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf=20*10**3\n",
+ "Ri=10*10**3\n",
+ "\n",
+ "#calculation\n",
+ "Acl=-(Rf/Ri) #Voltage Gain\n",
+ "\n",
+ "#Result\n",
+ "print\" Value of Voltage Gain is\",Acl"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of Voltage Gain is -2\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page No 51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei=-5 #input voltage\n",
+ "Acl=-2 #Voltage Gain\n",
+ "\n",
+ "#Calculation\n",
+ "Vout=Ei*Acl #output voltage\n",
+ "\n",
+ "#Result\n",
+ "print\" Value of Output Voltage is\",Vout,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of Output Voltage is 10 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.7 Page No :51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ri=10*10**3 #input resistance in ohm\n",
+ "Gain= 25\n",
+ "\n",
+ "#Calculation\n",
+ "Rf=Gain*Ri #feedback resistance in ohm\n",
+ "\n",
+ "#result\n",
+ "print\" Value of Rf is\",Rf/1000,\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of Rf is 250 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.8 Page No 52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E1=2\n",
+ "E2=3\n",
+ "E3=1 #input voltage\n",
+ "R=10*10**3 #in ohm\n",
+ "\n",
+ "#calculation\n",
+ "Vout=-(E1+E2+E3)\n",
+ "\n",
+ "#result\n",
+ "print\" Value of o/p voltage is \",Vout,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of o/p voltage is -6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page No 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E1=2.0\n",
+ "E2=3.0\n",
+ "E3=-1.0 #input voltage\n",
+ "R=10*10**3 #in ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Vout=-(E1+E2+E3)\n",
+ "\n",
+ "#result\n",
+ "print\" Value of o/p voltage is\",Vout"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of o/p voltage is -4.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.11 Page No 56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Channel 1\n",
+ "Ri= 10*10**3 #Choosing Input resistance\n",
+ "Acl = -10.0 \n",
+ "\n",
+ "#Calculation\n",
+ "Rf1 = - (Acl * Ri)\n",
+ "#Channel 2\n",
+ "Acl1 = -5\n",
+ "Rf2 = - (Acl1 * Ri)\n",
+ "#channel 3\n",
+ "Acl2 = -2\n",
+ "Rf3 = - (Acl2 * Ri)\n",
+ "\n",
+ "#result\n",
+ "print\" Value of Rf1 = \",Rf1/1000,\"kohm\"\n",
+ "print\" Value of Rf2 = \",Rf2/1000,\"kohm\"\n",
+ "print\" Value of Rf3 = \",Rf3/1000,\"kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of Rf1 = 100.0 kohm\n",
+ " Value of Rf2 = 50 kohm\n",
+ " Value of Rf3 = 20 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.12 Page No 57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=100000.0\n",
+ "R2=100000.0\n",
+ "R3=100000.0\n",
+ "R=100000.0 #in ohm\n",
+ "Rf=33*10**3 #in ohm\n",
+ "E1=5\n",
+ "E2=5\n",
+ "E3=-1 #in volts\n",
+ "n=3.0 #number of inputs\n",
+ "\n",
+ "#Calculation\n",
+ "Vout=-(E1+E2+E3)/n #output voltage\n",
+ "\n",
+ "#Result\n",
+ "print\" Value of output voltage is\",Vout,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of output voltage is -3.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.13 Page No 59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf = 40*10**3\n",
+ "R1 = 10*10**3\n",
+ "\n",
+ "#calculation\n",
+ "Acl = (Rf + R1)/R1\n",
+ "print\"(a) Value of Voltage Gain = \",Acl\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Value of Voltage Gain = 5\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.14 Page No 59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "Acl = 10.0 # Gain is positive, so choose noninvering amplifier\n",
+ "R1 = 10*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "Rf = (Acl*R1)-R1\n",
+ "\n",
+ "#Result\n",
+ "print\" Value of Rf is\",Rf/1000,\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.15 Page No 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei=4.0 #in volts\n",
+ "Rl=10*10**3 #in ohm\n",
+ "I=0 #in ampere\n",
+ "\n",
+ "#Calculation\n",
+ "Vout=Ei #output voltage\n",
+ "Il=Vout/Rl #load current\n",
+ "I0=I+Il #output current \n",
+ "\n",
+ "#Result\n",
+ "print\" Value of o/p voltage is\",Vout,\"V\"\n",
+ "print\" Value of load current is \",Il*1000,\"mA\"\n",
+ "print\" Value of output current is \",I0*1000,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of o/p voltage is 4.0 V\n",
+ " Value of load current is 0.4 mA\n",
+ " Value of output current is 0.4 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C4_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C4_1.ipynb new file mode 100644 index 00000000..7fb262ec --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C4_1.ipynb @@ -0,0 +1,236 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4: Comparators and Controls"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.1 Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vsat = 14.0 #Saturation Voltage\n",
+ "R1 = 1000.0 \n",
+ "R2 = 100.0 #Load resistances\n",
+ "\n",
+ "#Calculation\n",
+ "Vut = (R2/(R1*R2))*Vsat\n",
+ "\n",
+ "#Result\n",
+ "print\" Value of Upper Threshold Voltage = \",Vut*1000,\"mV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of Upper Threshold Voltage = 14.0 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vsat = -13.0 #Saturation Voltage\n",
+ "R1 = 1000.0\n",
+ "R2 = 100.0 #Load resistances\n",
+ "\n",
+ "#Calculation\n",
+ "Vlt = (R2/(R1*R2))*Vsat\n",
+ "\n",
+ "#Result\n",
+ "print\" Value of Lower Threshold Voltage = \",Vlt*1000,\"mV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of Lower Threshold Voltage = -13.0 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page No 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vut = 12.0\n",
+ "Vlt = 8.0 #Upper and Lower Threshold Voltages \n",
+ "Vsatp = 15.0\n",
+ "Vsatm = -15.0 # Saturation Voltages\n",
+ "R = 10*10**3 #Choosing R\n",
+ "\n",
+ "#Calculation\n",
+ "Vh = Vut - Vlt # Hysteresis Voltage \n",
+ "Vctr = (Vut + Vlt)/2.0 #Center Voltage\n",
+ "n = (Vsatp - Vsatm)/Vh # Resistor Factor\n",
+ "Vref = Vctr / (1 + (1/n)) #Reference Voltage\n",
+ "Resistance = n * R\n",
+ "\n",
+ "#Result\n",
+ "print\" Center Voltage = \", Vctr ,\"V\"\n",
+ "print\" Resistor Factor = \" , n \n",
+ "print\" Reference Voltage = \", round(Vref,2)\n",
+ "print\" Feedback resistor is \", Resistance"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Center Voltage = 10.0 V\n",
+ " Resistor Factor = 7.5\n",
+ " Reference Voltage = 8.82\n",
+ " Feedback resistor is 75000.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page No 95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vsatp = 15.0\n",
+ "Vsatm = -15.0 # Saturation Voltages\n",
+ "Vh = 4.0 #Hysteresis Voltage\n",
+ "Vctr = 10.0 #Center voltage\n",
+ "\n",
+ "#Calculation\n",
+ "n = ((Vsatp - Vsatm)/Vh)-1 \n",
+ "R = 10*10**3 #Input Resistance\n",
+ "Vref = ((n + 1)*(Vctr))/n # Reference Voltage\n",
+ "Resistance = n * R # Feedback Resistance\n",
+ "\n",
+ "#Result\n",
+ "print\" Resistance Factor = \", n\n",
+ "print\" Reference Voltage = \", round(Vref,2),\"V\"\n",
+ "print\" Feedback Resistance = \", Resistance/1000,\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Resistance Factor = 6.5\n",
+ " Reference Voltage = 11.54 V\n",
+ " Feedback Resistance = 65.0 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page No 98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vut = 13.5\n",
+ "Vlt = 10.5 #Upper and Lower Threshold Voltages\n",
+ "Vref = -15 # Reference Voltage\n",
+ "Vsatp = 13.0\n",
+ "Vsatm = -13.0 #Saturation Voltages \n",
+ "R = 10*10**3 # Input Resistance\n",
+ "\n",
+ "#Calculation\n",
+ "Vctr = (Vut + Vlt)/2.0\n",
+ "Vh = Vut - Vlt \n",
+ "m = -(Vref / Vctr)\n",
+ "Resistance = m * R \n",
+ "n = (Vsatp-Vsatm)/Vh \n",
+ "Resistance1 = n * R \n",
+ "\n",
+ "#Result\n",
+ "print\" Hysteresis Voltage = \", Vh \n",
+ "print\" Center Voltage = \", Vctr \n",
+ "print\" Resistor mR = \", Resistance/1000,\"kohm\"\n",
+ "print\" Resistor nR = \", round(Resistance1/1000,2),\"kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Hysteresis Voltage = 3.0\n",
+ " Center Voltage = 12.0\n",
+ " Resistor mR = 12.5 kohm\n",
+ " Resistor nR = 86.67 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C5_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C5_1.ipynb new file mode 100644 index 00000000..e04d327d --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C5_1.ipynb @@ -0,0 +1,733 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5 :Selected applications of Op Amps"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.1 Page No 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei = 0.5 #Input voltage\n",
+ "Ri = 1*10**3 #Input resistance in ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Im = Ei / Ri #Meter Current\n",
+ "\n",
+ "#Result\n",
+ "print\"Meter Current = \", Im*1000,\"mA\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Meter Current = 0.5 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.2 Page No 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Efs = 5.0 #Full scale Voltage\n",
+ "Ifs = 50*10**-6 #Full scale Meter Current\n",
+ "#Calculation\n",
+ "Ri = Efs / Ifs # Input Resistance\n",
+ "\n",
+ "#Result\n",
+ "print\" Input Resistance =\", Ri/1000,\"Kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Input Resistance = 100.0 Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.3 Page No 122"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Edc = 5.0\n",
+ "Erms = 5.0\n",
+ "Epeak = 5.0\n",
+ "Eptop = 5.0 # Voltages of meters \n",
+ "Ifs = 50*10**-6 # Full scale Meter Current\n",
+ "\n",
+ "#Calculation\n",
+ "Ri1 = Edc / Ifs # DC Voltmeter\n",
+ "Ri2 = 0.90 * (Erms / Ifs ) # Rms ac voltmeter ( math.sine wave only )\n",
+ "Ri3 = 0.636 * (Epeak / Ifs ) #Peak Reading Voltmeter ( math.sine wave only )\n",
+ "Ri4 = 0.318 * (Eptop / Ifs ) #Peak-to-Peak ac Voltmeter(math.sine wave only)\n",
+ "\n",
+ "#Result\n",
+ "print\" Ri1 \", Ri1/1000,\"kohm\"\n",
+ "print\" Ri2 =\",Ri2/1000 ,\"kohm\"\n",
+ "print\" Ri3 = \", Ri3/1000,\"kohm\"\n",
+ "print\" Ri4 = \", Ri4/1000,\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Ri1 100.0 kohm\n",
+ " Ri2 = 90.0 kohm\n",
+ " Ri3 = 63.6 kohm\n",
+ " Ri4 = 31.8 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.4 Page No 123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vo = 10.3 #Voltage across the load resistor\n",
+ "Ei = 5.0 #Input voltage\n",
+ "Ri = 1*10**3 #Input Resistance\n",
+ "\n",
+ "#calculation\n",
+ "I = Ei / Ri #Zener Current\n",
+ "Vt = Vo - Ei #Zener Voltage\n",
+ "\n",
+ "#Result\n",
+ "print\" Zener Current = \", I,\"A\"\n",
+ "print\" Zener Voltage = \", Vt ,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Zener Current = 0.005 A\n",
+ " Zener Voltage = 5.3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.5 Page No 124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei = 1.0 #Reference voltage\n",
+ "Ri = 1*10**3 #Input Resistance\n",
+ "Vo = 0.6 #Outpur Voltage\n",
+ "\n",
+ "#Calculation\n",
+ "I = Ei / Ri #Diode Current\n",
+ "Vdiode = Vo \n",
+ "\n",
+ "#Result\n",
+ "print\" Diode Current = \", I*1000 ,\"mA\"\n",
+ "print\" Voltage drop across the diode = \", Vdiode,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Diode Current = 1.0 mA\n",
+ " Voltage drop across the diode = 0.6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.6 Page No 127"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R = 10*10**3 #Resistance\n",
+ "E2 = 0 #Source across negative terminal\n",
+ "Rl = 5*10**3 # Load Resistance\n",
+ "E1 = 5.0 # source across positive terminal\n",
+ "\n",
+ "#calculation\n",
+ "Il = (E1 - E2)/R #Load Current\n",
+ "Vl = Il * Rl # Voltage across Rl \n",
+ "Vo = (2*Vl)-E2 #Output voltage\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Load current across Rl = \" , Il*1000 ,\"mA\"\n",
+ "print\"(b) Voltage across load resistance = \" , Vl,\"V\"\n",
+ "print\"(c) Output Voltage = \", Vo ,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Load current across Rl = 0.5 mA\n",
+ "(b) Voltage across load resistance = 2.5 V\n",
+ "(c) Output Voltage = 5.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.7 Page No 127"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R = 10*10**3 #Resistance\n",
+ "E2 = 5.0 #Source across negative terminal\n",
+ "Rl = 5*10**3 # Load Resistance\n",
+ "E1 = 0 # source across positive terminal\n",
+ "\n",
+ "#Calculation\n",
+ "Il = (E1 - E2)/R #Load Current\n",
+ "Vl = Il * Rl # Voltage across Rl \n",
+ "Vo = (2*Vl)-E2 #Output voltage\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Load current across Rl = \" , Il*1000,\"mA\"\n",
+ "print\"(b) Voltage across load resistance = \" , Vl,\"V\"\n",
+ "print \"(c) Output Voltage = \", Vo ,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Load current across Rl = -0.5 mA\n",
+ "(b) Voltage across load resistance = -2.5 V\n",
+ "(c) Output Voltage = -10.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.8 Page No 132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vo = 5.0 #Output Voltage\n",
+ "Rf = 100*10**3 #Feedback Resistance\n",
+ "\n",
+ "#calculation\n",
+ "Isc = Vo / Rf #Short Circuit Current\n",
+ "\n",
+ "#Result\n",
+ "print\" Short Circuit Current = \" , Isc*10**6,\"microA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Short Circuit Current = 50.0 microA\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.9 Page No 132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "Rf = 10*10**3 #Feedback Resistance\n",
+ "I = 10*10**-6 #Current through Photo Detector\n",
+ "\n",
+ "#calculation\n",
+ "Vo = Rf * I #Vo for Dark Condition\n",
+ "I1 = 1*10**-3 #Current in presence of sunlight\n",
+ "Vo1 = Rf * I1 #output voltage in light condition\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)Output Voltage for dark Condition = \", Vo,\"V\"\n",
+ "print\"(b) Output voltage in light condition = \", Vo1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Output Voltage for dark Condition = 0.1 V\n",
+ "(b) Output voltage in light condition = 10.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.10 Page No 133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf = 100*10**3 #Feedback Resistance\n",
+ "Il1 = 1*10**-6 #Load current 1\n",
+ "\n",
+ "#calculation\n",
+ "Vo1 = Rf * Il1 #Output voltage in photo detector\n",
+ "Il2 = 50*10**-6 # Load current 2\n",
+ "Vo2 = Rf * Il2 #Output Voltage in photo detector\n",
+ "\n",
+ "#Result\n",
+ "print\" (a)Output Voltage in photo detector for Il1 = \",Vo1 ,\"V\"\n",
+ "print \" (b)Output Voltage in photo detector for Il2 = \",Vo2 ,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a)Output Voltage in photo detector for Il1 = 0.1 V\n",
+ " (b)Output Voltage in photo detector for Il2 = 5.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.11 Page No 134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R = 1*10**3\n",
+ "R1 = 99*10**3 \n",
+ "\n",
+ "#calculation\n",
+ "m = R1 / R #multiplier\n",
+ "Isc = 10*10**-6 #Current on short-circuit condition\n",
+ "Il = (1 + m)*Isc \n",
+ "\n",
+ "#result\n",
+ "print\" Load current = \", Il*1000,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Load current = 1.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.12 Page No 136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Im = 100*10**-6 #Meter current \n",
+ "Isc = 0.5 # Current in short-circuit condition\n",
+ "Rf = 20 # Feedback resistance\n",
+ "Rm = 0.8*10**3 #Meter resistance\n",
+ "\n",
+ "#calculation\n",
+ "d = Isc / Im #Current divider\n",
+ "R1 = d * Rf \n",
+ "Rscale = R1 - Rm \n",
+ "\n",
+ "#Result\n",
+ "print\" Resistance dRf = \", R1/1000 ,\"kohm\"\n",
+ "print\" Rscale = \", Rscale/1000,\"kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Resistance dRf = 100.0 kohm\n",
+ " Rscale = 99.2 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.13 Page No 138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f = 10**3 #Frequency of Ei in Hz\n",
+ "Ci = 0.01*10**-6 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m = math.tan(math.pi/4)\n",
+ "Ri = m / (2*math.pi*f*Ci)\n",
+ "\n",
+ "#result\n",
+ "print\" Value of Ri = \",round(Ri/1000,1),\"kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of Ri = 15.9 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.14 Page No 138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f = 10**3 \n",
+ "Ri = 100*10**3 \n",
+ "Ci = 0.01*10**-6\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "phaseangle = 2*math.atan(2*math.pi*f*Ri*Ci)\n",
+ "\n",
+ "#result\n",
+ "print\" Phase angle =\", round(phaseangle*180/3.14,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Phase angle = 162.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.15 Page no 143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From fig 5-14(a) and b\n",
+ "Ei=2 #V, voltage\n",
+ "Vf=4 #Vf force\n",
+ "\n",
+ "#Calculation\n",
+ "Vo=-2*Ei #Vo forces\n",
+ "Vcap=3*Ei\n",
+ "\n",
+ "#Result\n",
+ "print\"Vo is\",Vo,\"V\"\n",
+ "print\"Vcap is\",Vcap,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Vo is -4 V\n",
+ "Vcap is 6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.16 Page no 144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei=4 #V\n",
+ "Vo=8 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Vr=-Vo\n",
+ "Vcap=Ei+Vo\n",
+ "\n",
+ "#Result\n",
+ "print\"Vcap is\",Vcap,\"V\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Vcap is 12 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.17 Page no 144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ri=1*10**5\n",
+ "C=10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "T=3*Ri*C\n",
+ "Et=5*T\n",
+ "\n",
+ "#Result\n",
+ "print\"Equilibrium Time is\",Et,\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium Time is 1.5 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 5.18 Page no 146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Ci=0.1*10**-6 #capacitance, farady\n",
+ "Ci_=0.1 #microF\n",
+ "f=1000 #frequency\n",
+ "#Calculation\n",
+ "import math\n",
+ "Ri=1/(2*math.pi*f*Ci)\n",
+ "Rf=20/(2*math.pi*f*Ci)\n",
+ "Cf=1/(2*math.pi*f*Rf)\n",
+ "#a=0.4*math.sin(2*math.pi*1000*t)\n",
+ "#Vo=-Cf*Ci*a\n",
+ "\n",
+ "from sympy import *\n",
+ "import numpy as np\n",
+ "t = Symbol('t')\n",
+ "y = -Rf*Ci_/10**6*0.4*sin(2*math.pi*1000*t)\n",
+ "y_ = y.diff(t)\n",
+ "\n",
+ "#Reslt\n",
+ "print Ci_\n",
+ "print\"a)Rf is \",round(Rf/1000,1),\"kohm\"\n",
+ "print\"Cf is \",round(Cf*10**6,9),\"microF\"\n",
+ "print\"(b)Output voltage is \",y_,\"V\"\n",
+ "\n",
+ "#NOTE: Answer is same .In the book it is written 2*math.pi*1000 ."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.1\n",
+ "a)Rf is 31.8 kohm\n",
+ "Cf is 0.005 microF\n",
+ "(b)Output voltage is -8.0*cos(6283.18530717959*t) V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C6_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C6_1.ipynb new file mode 100644 index 00000000..7bc13612 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C6_1.ipynb @@ -0,0 +1,490 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6: Signal Generators"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.1 Page No 155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1 = 100*10**3 \n",
+ "R2 = 86*10**3 \n",
+ "Vsatp = 15.0\n",
+ "Vsatm = -15.0 #Saturation voltages\n",
+ "\n",
+ "#calculation\n",
+ "Vut = (R2 * Vsatp)/(R1 + R2)\n",
+ "Vlt = (R2 * Vsatm)/(R1 + R2)\n",
+ "\n",
+ "#Result\n",
+ "print\" Upper Threshold Voltage = \",round(Vut,0),\"V\"\n",
+ "print\" Lower Threshold Voltage = \",round(Vlt,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Upper Threshold Voltage = 7.0 V\n",
+ " Lower Threshold Voltage = -7.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.2 Page No 155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf = 100*10**3 #Feedback Resistance\n",
+ "C = 0.1*10**-6 \n",
+ "\n",
+ "#calculation\n",
+ "T = 2 * Rf * C\n",
+ "\n",
+ "#result\n",
+ "print\" Period \", T,\"s or \",T*1000,\"ms\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Period 0.02 s or 20.0 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.3 Page No 155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T = 20*10**-3 #Period \n",
+ "\n",
+ "#calculation\n",
+ "f = 1 / T\n",
+ "\n",
+ "#Result\n",
+ "print\" Frequency = \", f,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Frequency = 50.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.5 Page No 159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf = 100*10**3 #Feedback Resistance\n",
+ "C = 0.1*10**-6 \n",
+ "\n",
+ "#Calculation\n",
+ "t = (Rf * C)/ 5.0 \n",
+ "\n",
+ "#Result\n",
+ "print\" Duration of output pulse of one-shot = \", t ,\"sec or\",t*1000,\"ms\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Duration of output pulse of one-shot = 0.002 sec or 2.0 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.6 Page No 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vsatm = -13.8 \n",
+ "Vut = 5.0 #Upper Threshold Voltage\n",
+ "R = 10*10**3 \n",
+ "f = 1000.0 #Frequency\n",
+ "C = 0.05*10**-6 \n",
+ "\n",
+ "#calculation\n",
+ "p = -Vsatm / Vut \n",
+ "pR = round(p,1) * R \n",
+ "Ri = p / (4*f*C) \n",
+ "\n",
+ "#Result\n",
+ "print\" Value of p = \", p \n",
+ "print\" Value of pR =\", pR/1000,\"kohm\"\n",
+ "print\" Value of Ri = \", round(Ri/1000,0),\"kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Value of p = 2.76\n",
+ " Value of pR = 28.0 kohm\n",
+ " Value of Ri = 14.0 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.7 Page No 163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p = 2.8 \n",
+ "Vsatm = -13.8 \n",
+ "Ri = 28*10**3 \n",
+ "C = 0.05*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "Vut = - ((Vsatm+0.6)/p)\n",
+ "f = p / (2*Ri*C)\n",
+ "\n",
+ "#Result\n",
+ "print\" Peak Voltage = \", round(Vut,2),\"V\"\n",
+ "print\" frequency = \", f ,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Peak Voltage = 4.71 V\n",
+ " frequency = 1000.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.8 Page No 167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Checking Frequency value\n",
+ "Ri = 10*10**3 \n",
+ "Ci = 0.1*10**-6 \n",
+ "Ei = 1.0 \n",
+ "Vref = 10.0 \n",
+ "\n",
+ "#Calculation\n",
+ "f = Ei / (Ri*Ci*Vref) \n",
+ "\n",
+ "#Result\n",
+ "print\" Frequency is \", f,\"Hz\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Frequency is 100.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.9 Page No 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ri = 10*10**3 #Resistance\n",
+ "Ci = 0.1*10**-6 \n",
+ "Ei = 2.0 \n",
+ "Vref = 2.0 \n",
+ "\n",
+ "#Calculation\n",
+ "f = Ei / (Ri*Ci*Vref) \n",
+ "\n",
+ "#Result\n",
+ "print \" New Frequency of oscillation is \", f/1000 ,\"KHz\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Frequency is 200.0 Hz. So It means Frequency is doubled.\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.10 Page no 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vref1 = 10.0 \n",
+ "Vref2 = 5.0 \n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "f1 =1000/Vref1\n",
+ "f2 = 1000/Vref2 \n",
+ "\n",
+ "#Result\n",
+ "print\"For Vref=10V, frequency is\",f1,\"Hz and For Vref=5V, frequency is\",f2,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For Vref=10V, frequency is 100.0 Hz and For Vref=5V, frequency is 200.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.11 Page no 172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=5 #Peak voltage\n",
+ "f=1 #Khz frequency\n",
+ "Vref=5 #V\n",
+ "C=0.01*10**-6 #Capacitance\n",
+ "\n",
+ "#Calculation\n",
+ "Ri=1/(4*f*1000*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of Ri is\",Ri/1000,\"Kohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Ri is 25.0 Kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.12 Page No 175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ei = 1.0 \n",
+ "#Angle given in degree\n",
+ "t1 = 45.0 \n",
+ "t2 = 90.0\n",
+ "t3 = 225.0\n",
+ "t4 = 405.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vang1 = (20*10**-3)*t1 \n",
+ "Vang2 = (20*10**-3)*t2 \n",
+ "Vang3 = (20*10**-3)*t3 \n",
+ "Vang4 = (20*10**-3)*t4\n",
+ "Vo1 = Ei*math.sin(t1*3.14/180.0)\n",
+ "Vo2 = Ei*math.sin(t2*3.14/180.0)\n",
+ "Vo3 = Ei*math.sin(t3*3.14/180.0)\n",
+ "Vo4 = Ei*math.sin(t4*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\" Input Voltages are \", Vang1,\"V, \",Vang2,\"V ,\",Vang3,\"V,\",Vang4 , \"V\"\n",
+ "print\" Output Voltages are \", round(Vo1,3),\"V,\",round(Vo2,3),\"V,\",round(Vo3,3),\"V,\",round(Vo4,3),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Input Voltages are 0.9 V, 1.8 V , 4.5 V, 8.1 V\n",
+ " Output Voltages are 0.707 V, 1.0 V, -0.706 V, 0.705 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.13 Page No 178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1 = 10*10**3 \n",
+ "R2 = 100*10**3 \n",
+ "C = 0.025*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "f1 = 1 / (4*R1*C) \n",
+ "f2 = 1 / (4*R2*C) \n",
+ "\n",
+ "#Result\n",
+ "print\" Frequency when Ri is 10KiloOhm is \" , f1/1000.0,\"KHz\"\n",
+ "print\" Frequency when Ri is 100KiloOhm is \" , f2,\"Hz\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Frequency when Ri is 10KiloOhm is 1.0 KHz\n",
+ " Frequency when Ri is 100KiloOhm is 100.0 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C7_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C7_1.ipynb new file mode 100644 index 00000000..5a018ea4 --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C7_1.ipynb @@ -0,0 +1,63 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Op Amps with diodes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 7.1 Page no 205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V = 15.0 #V,Voltage\n",
+ "mR = 30*10**3 #, ohm, Resistance \n",
+ "R = 10*10**3 #ohm, 2nd resistor\n",
+ "Ei = -10 #V, Voltage supply\n",
+ "\n",
+ "#calculation\n",
+ "Vref = V / 3.0 \n",
+ "Voa = -Ei-Vref \n",
+ "Vob = Ei / 2.0 \n",
+ "\n",
+ "#Result\n",
+ "print\"Reference voltage is\",Vref,\"V\"\n",
+ "print\"Values of Voa =\",Voa,\"V\",\"and Vob \", Vob,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reference voltage is 5.0 V\n",
+ "Values of Voa = 5.0 V and Vob -5.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C8_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C8_1.ipynb new file mode 100644 index 00000000..c5da6d5b --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C8_1.ipynb @@ -0,0 +1,544 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eb16b5620fb956842995f8d06be59d7e1e1d7123054fdad5a84f667167801f94"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8: Differential, Instrumentation and Bridge amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.1 Page No 218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m = 100.0 #Differential Gain\n",
+ "E1 = 10*10**-3\n",
+ "E2 = 10*10**-3 #input voltages\n",
+ "E3 = 0*10**-3 \n",
+ "E4 = -20*10**-3 #input voltages\n",
+ "\n",
+ "#Calculation\n",
+ "Vout1 = (m*E1)-(m*E2) \n",
+ "Vout2 = (m*E1)-(m*E3)\n",
+ "Vout3 = (m*E1)-(m*E4)\n",
+ "\n",
+ "#Result\n",
+ "print\" Output Voltages are \", Vout1,\"V, \",Vout2,\"V ,\",Vout3,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Output Voltages are 0.0 V, 1.0 V , 3.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.2 Page no. 220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=5 #V\n",
+ "d=2 #mV\n",
+ "E1=V+d*10**-3\n",
+ "E2=V-d*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "Eds=E1-E2\n",
+ "Ecm=(E1+E2)/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Input signal voltage is\",Eds*1000,\"mV\"\n",
+ "print\"Common mode voltage is\",Ecm,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Input signal voltage is 4.0 mV\n",
+ "Common mode voltage is 5.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.2 Page No 225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a = 2/9.0 #Differential Gain\n",
+ "E1 = 10*10**-3\n",
+ "E2 = 5*10**-3 #Input Voltages \n",
+ "\n",
+ "#Calculation\n",
+ "Vout = (E1 - E2)*(1+(2/a))\n",
+ "\n",
+ "#Result\n",
+ "print\" output voltage = \", Vout*1000,\"mV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " output voltage = 50.0 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.3 Page No 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R = 25*10**3 \n",
+ "aR = 50.0 \n",
+ "\n",
+ "#Calculation\n",
+ "a = aR / R \n",
+ "Gain = 1 + (2/a) \n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage Gain = \" , Gain "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage Gain = 1001.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.4 Page No 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "import numpy\n",
+ "a=numpy.inf\n",
+ "\n",
+ "#Calculation\n",
+ "Gain = 1+(2/a) \n",
+ "\n",
+ "#Result\n",
+ "print\" Voltage Gain = \", Gain"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Voltage Gain = 1.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page No 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m = 1001.0 #Gain\n",
+ "E1 = 5.001\n",
+ "E2 = 5.002\n",
+ "E3 = 5.001 \n",
+ "E4 = 5.000 \n",
+ "E5 = -1.001 \n",
+ "E6 = -1.002 \n",
+ "\n",
+ "#Calculation\n",
+ "Vout1 = m*(E1-E2)\n",
+ "Vout2 = m*(E3-E4) \n",
+ "Vout3 = m*(E5-E6)\n",
+ "\n",
+ "#result\n",
+ "print\" Output Voltages are \", Vout1,\"V \",Vout2,\"V \",Vout3,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Output Voltages are -1.001 V 1.001 V 1.001 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.6 Page No 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vout = 5.0 #Output Voltage\n",
+ "\n",
+ "#Calculation\n",
+ "Vce = Vout \n",
+ "\n",
+ "#Result\n",
+ "print\" Vce=\",Vce,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Vce= 5.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.7 Page No 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vre = 1.2 \n",
+ "Re = 1*10**3 \n",
+ "Vce = 5\n",
+ "Vcc = 15 #Voltages in the circuit\n",
+ "\n",
+ "#Calculation\n",
+ "Ie = Vre / Re \n",
+ "Vcol = Vce + Vre \n",
+ "Vrl = Vcc - Vcol \n",
+ "\n",
+ "#Result\n",
+ "print\" Collector Voltage = \", Vcol,\"V\"\n",
+ "print\" Voltage across Rl =\", Vrl,\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Collector Voltage = 6.2 V\n",
+ " Voltage across Rl = 8.8 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.8 Page No 232"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rs = 1*10**3\n",
+ "E1 = 100*10**-3 \n",
+ "E2 = 0 \n",
+ "Rl = 5*10**3 \n",
+ "Gain = 10 \n",
+ "\n",
+ "#Calculation\n",
+ "Il = 10*((E1-E2)/Rs)\n",
+ "Vr = Il * Rs \n",
+ "Vref = Il * Rl \n",
+ "V9 = Vref + Gain*(E1 - E2)\n",
+ "\n",
+ "#Result\n",
+ "print\" Current across Load Resistor = \", Il*1000 ,\"mA\"\n",
+ "print\" Voltage across R = \", Vr,\"V\"\n",
+ "print\" Reference Voltage = \", Vref,\"V\" \n",
+ "print\" Voltage at terminal 9 =\", V9,\"V\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Current across Load Resistor = 1.0 mA\n",
+ " Voltage across R = 1.0 V\n",
+ " Reference Voltage = 5.0 V\n",
+ " Voltage at terminal 9 = 6.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.9 Page No 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "GF = 2.0 #Gage factor\n",
+ "DR = 0.001 \n",
+ "R = 120.0 \n",
+ "\n",
+ "#calculation\n",
+ "L = DR /(R*GF) \n",
+ "\n",
+ "#result\n",
+ "print\" Change in length is \",round(L*10**6,1),\"microinches per inch\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Change in length is 4.2 microinches per inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.10 Page No 236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E = 1.0 \n",
+ "DR = 0.001 \n",
+ "R = 120.0 #Resistance in Ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Vout = (E*DR)/(4*R) \n",
+ "\n",
+ "#Result\n",
+ "print\" Output of the Bridge = \", round(Vout*10**6,1),\"micro voltage\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Output of the Bridge = 2.1 micro voltage\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.11 Page No 241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vout = 100*10**-3 \n",
+ "E = 5.0\n",
+ "R = 120.0 \n",
+ "Gain = 1000.0 \n",
+ "E1 = 30 *10**6 \n",
+ "Gf = 2.0 #Gage factor\n",
+ "\n",
+ "#calculation\n",
+ "Vbridge = Vout / Gain \n",
+ "DL = 20*10**-6 \n",
+ "DR = (R * Vbridge)/E \n",
+ "Rratio = DR / R #Change in Resistance\n",
+ "Strain = DL / Gf #Change in Length \n",
+ "Stress = E1 * Strain \n",
+ "\n",
+ "#Result\n",
+ "print \" Change in Resistance = \", DR*1000 ,\"mohm\"\n",
+ "print \" Ratio of Resistance = \", Rratio,\"micro ohm/ohm\"\n",
+ "print \" Strain value = \", Strain ,\"micro inch/inch\"\n",
+ "print \" Stress value = \", Stress ,\"psi\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Change in Resistance = 2.4 mohm\n",
+ " Ratio of Resistance = 2e-05 micro ohm/ohm\n",
+ " Strain value = 1e-05 micro inch/inch\n",
+ " Stress value = 300.0 psi\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.12 Page No 245\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T1=25 #C Temprature\n",
+ "T2=50\n",
+ "Tref=25 #C Reference temprature\n",
+ "Rref=10000\n",
+ "Rtrans=3603 #ohm, at 50 C\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "dR=Rtrans-Rref\n",
+ "R1=10 #Kohm say\n",
+ "E=1 #V trial choice\n",
+ "I=E/(R1+Rref/1000.0)\n",
+ "Vo=-I*dR\n",
+ "\n",
+ "#result\n",
+ "print\"The value of dR for each temprature is\",dR,\"ohm\"\n",
+ "print\"The current is\",I,\"mA\"\n",
+ "print\"Vo for each value of R\",Vo,\"mV\"\n",
+ "#NOTE: In the book the value of Vo is calculated wrong"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of dR for each temprature is -6397 ohm\n",
+ "The current is 0.05 mA\n",
+ "Vo for each value of R 319.85 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/C9_1.ipynb b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C9_1.ipynb new file mode 100644 index 00000000..e0e0173e --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/C9_1.ipynb @@ -0,0 +1,297 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:DC Performance :Bias,Offsets and Drift"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Exampme 9.1 Page No. 256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ibplus=0.4 #MicroA\n",
+ "Ibminus=0.3 #MicroA\n",
+ "\n",
+ "#calculation\n",
+ "Ib=(Ibplus+Ibminus)/2.0\n",
+ "Ios=Ibplus-Ibminus\n",
+ "\n",
+ "#Result\n",
+ "print\"The average bias current is\",Ib,\"microA\"\n",
+ "print\"The offset current is\",Ios,\"microA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average bias current is 0.35 microA\n",
+ "The offset current is 0.1 microA\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Exampme 9.2 Page No. 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vo=0.4 #A\n",
+ "Rf=1*10**6 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ibminus=Vo/Rf\n",
+ "\n",
+ "#Result\n",
+ "print\"The Ibminus is\",Ibminus*10**6,\"microA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Ibminus is 0.4 microA\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Exampme 9.3 Page No.258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vo=-0.3 #A\n",
+ "Rf=1*10**6 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ibplus=-Vo/Rf\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The Ibplus is\",Ibplus*10**6,\"microA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Ibplus is 0.3 microA\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Exampme 9.4 Page No. 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rf=100.0 #kohm\n",
+ "Ri1=10.0 #kohm\n",
+ "Ri2=100.0 #kohm\n",
+ "\n",
+ "#Calculation\n",
+ "R1=Rf*Ri1/(Rf+Ri1)\n",
+ "R2=Rf*Ri2/(Rf+Ri2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The value of resistance is\",round(R1,1),\"ohm\"\n",
+ "print\"(b) The value of resistance is\",R2,\"ohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The value of resistance is 9.1 ohm\n",
+ "(b) The value of resistance is 50.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Exampme 9.5 Page No. 262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vio=1 #mV\n",
+ "Rf=1 #kohm\n",
+ "Ri=10 #ohm\n",
+ "\n",
+ "#calculation\n",
+ "Vo=(Vio+(Rf*1000/Ri)*Vio)\n",
+ "\n",
+ "#result\n",
+ "print\"The value of Vo is\",Vo,\"mV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vo is 101 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Exampme 9.6 Page No. 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T1=75.0 #Temprature ,C\n",
+ "T2=25.0 #Temprature ,C\n",
+ "Viomax=20.0 #microV, Voltage\n",
+ "Ioschange=0.1 #nm\n",
+ "Rf=10**6 #ohm\n",
+ "Ri=10*10**3 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Vio=Viomax*(T1-T2)\n",
+ "change=Vio*(1+Rf/Ri)\n",
+ "Ioschange=Ioschange*(T1-T2)\n",
+ "Vochange=Ioschange*Rf\n",
+ "\n",
+ "#Result\n",
+ "print\"The change in Vo due to change in Vio is\",change/10**3,\"mV\"\n",
+ "print\"The change in Vo due to change in Ios is\",Vochange/10**6,\"mV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in Vo due to change in Vio is 101.0 mV\n",
+ "The change in Vo due to change in Ios is 5.0 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Exampme 9.7 Page No. 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Vo=0.421 #V,\n",
+ "Vo1=0.097\n",
+ "Vo2=-0.082\n",
+ "Rf1=5000 #from fig 9-10(a)\n",
+ "Ri=50\n",
+ "Rf2=1.0 #Mohm fig 9-10 (b)\n",
+ "\n",
+ "#calculation\n",
+ "Vio=Vo/(1+Rf1/Ri)\n",
+ "Ibminus=(Vo1*1000-Vio*1000)/Rf2\n",
+ "Ibplus=-(Vo2*1000-0.0041*1000)/Rf2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of Vio is\",round(Vio*1000,1),\"mV\"\n",
+ "print\"Ibminus is\",round(Ibminus,0),\"nA\"\n",
+ "print\"Ibplus is\",round(Ibplus,0),\"nA\"\n",
+ "print\"Ios calculated as\",round(Ibplus-Ibminus,0),\"nA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vio is 4.2 mV\n",
+ "Ibminus is 93.0 nA\n",
+ "Ibplus is 86.0 nA\n",
+ "Ios calculated as -7.0 nA\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Operational_Amplifiers_and_Linear_Integrated_Circuit/README.txt b/Operational_Amplifiers_and_Linear_Integrated_Circuit/README.txt new file mode 100644 index 00000000..39ab128f --- /dev/null +++ b/Operational_Amplifiers_and_Linear_Integrated_Circuit/README.txt @@ -0,0 +1,10 @@ +Contributed By: Rahul garg +Course: btech +College/Institute/Organization: MDU,Rohtak +Department/Designation: ECE +Book Title: Operational Amplifiers and Linear Integrated Circuit +Author: R F Coughline anf F F Driscoll +Publisher: Prentice Hall, United states +Year of publication: 2000 +Isbn: 978-0130149916 +Edition: 6
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