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| author | hardythe1 | 2015-06-03 15:27:17 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-06-03 15:27:17 +0530 |
| commit | 47d7279a724246ef7aa0f5359cf417992ed04449 (patch) | |
| tree | c613e5e4813d846d24d67f46507a6a69d1a42d87 /Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter5.ipynb | |
| parent | 435840cef00c596d9e608f9eb2d96f522ea8505a (diff) | |
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diff --git a/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter5.ipynb b/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter5.ipynb new file mode 100755 index 00000000..982187ad --- /dev/null +++ b/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter5.ipynb @@ -0,0 +1,477 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 05 : Transistor Characteristic"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1a, Page No 133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "B=100.0 #Beta\n",
+ "Ico=20.0 #in nA \n",
+ "Rc=3.0\n",
+ "Rb=200.0\n",
+ "Vbb=5.0 #in V\n",
+ "Vcc=10 #in V\n",
+ "Vbe=0.7 #in Active region\n",
+ "\n",
+ "#Applying KVL to base circuit\n",
+ "\n",
+ "#Vbb+Rb*Ib+Vbe=0\n",
+ "\n",
+ "#Calculations\n",
+ "Ib=(Vbb-Vbe)/Rb #in mA\n",
+ "#Ico<<Ib\n",
+ "Ic=B*Ib #in mA\n",
+ "#To verify the Active region Assumption\n",
+ "#Vcc+Rc*Ic+Vcb+Vbe=0\n",
+ "\n",
+ "Vcb=(-Rc*Ic)+Vcc-Vbe #in V\n",
+ "\n",
+ "#Results\n",
+ "print(\"Vcb = %.2f V \" %Vcb)\n",
+ "if Vcb>0 :\n",
+ " print('Positive value of Vcb represents reversed biased collector junction and Transistor in active region')\n",
+ "\n",
+ "\n",
+ "print(\"Current in transistor(Ic) is %.2f mA \" %Ic)\n",
+ "print(\"Current in transistor(Ib) is %.2f mA \" %Ib)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Vcb = 2.85 V \n",
+ "Positive value of Vcb represents reversed biased collector junction and Transistor in active region\n",
+ "Current in transistor(Ic) is 2.15 mA \n",
+ "Current in transistor(Ib) is 0.02 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1b, Page No 133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "B=100.0 #Beta\n",
+ "Ico=20.0 #in nA \n",
+ "Rc=3.0\n",
+ "Ico=20 #in nA\n",
+ "Rb=200.0\n",
+ "Re=2.0\n",
+ "Vbb=5.0 #in V\n",
+ "Vcc=10.0 #in V\n",
+ "Vbe=0.7 #in Active region\n",
+ "\n",
+ "#Ico<<Ib Assuming\n",
+ "#Itot=Ib+Ic=Ib+B*Ib=(B+1)*Ib\n",
+ "#Applying KVL to base circuit\n",
+ "#Vbb+Rb*Ib+Vbe+Re*Itot=0\n",
+ "\n",
+ "#Calculations\n",
+ "Ib=(Vbb-Vbe)/(Rb+(Re*(B+1))) #in mA\n",
+ "\n",
+ "Ic=B*Ib #in mA\n",
+ "\n",
+ "#Hence Ico<<Ib\n",
+ "#To verify the Active region Assumption\n",
+ "#Vcc+Rc*Ic+Vcb+Vbe=0\n",
+ "\n",
+ "Vcb=(-Rc*Ic)+Vcc-Vbe-(Re*(B+1)*Ib) #in V\n",
+ "\n",
+ "print(\"Vcb = %.2f V \" %Vcb)\n",
+ "\n",
+ "if Vcb>0 :\n",
+ " print('Positive value of Vcb represents reversed biased collector junction and Transistor in active region')\n",
+ "\n",
+ "#Results\n",
+ "print(\"Current in transistor(Ic) is %.2f mA \" %Ic)\n",
+ "print(\"Current in transistor(Ib) is %.2f mA \" %Ib)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Vcb = 3.93 V \n",
+ "Positive value of Vcb represents reversed biased collector junction and Transistor in active region\n",
+ "Current in transistor(Ic) is 1.07 mA \n",
+ "Current in transistor(Ib) is 0.01 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 Page No 134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Rc=3.0\n",
+ "Rb=50.0 \n",
+ "Vbb=5.0 #in V\n",
+ "Vcc=10.0 #in V\n",
+ "Vce=0.2 #in V\n",
+ "Vbe=0.8 #in Active region\n",
+ "hFE=100.0 \n",
+ "\n",
+ "#Assuming transistor in saturated region\n",
+ "#Applying KVL to base circuit\n",
+ "#Vbb+Rb*Ib+Vbe=0\n",
+ "\n",
+ "#Calculations\n",
+ "Ib=(Vbb-Vbe)/Rb #in mA\n",
+ "\n",
+ "#Applying KVL to Collector circuit\n",
+ "#Vcc+Rc*Ic+Vce=0\n",
+ "\n",
+ "Ic=(Vcc-Vce)/Rc #in mA\n",
+ "\n",
+ "Ib_min=Ic/hFE\n",
+ "\n",
+ "print(\"Minimum Ib = %.2f mA \" %Ib_min)\n",
+ "\n",
+ "if Ib>Ib_min :\n",
+ " print('Transistor in saturated Region')\n",
+ "\n",
+ "#Results\n",
+ "print(\"Current in transistor(Ic) is %.2f mA \" %Ic)\n",
+ "print(\"Current in transistor(Ib) is %.2f mA \" %Ib)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum Ib = 0.03 mA \n",
+ "Transistor in saturated Region\n",
+ "Current in transistor(Ic) is 3.27 mA \n",
+ "Current in transistor(Ib) is 0.08 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3, Page No 134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ib=0.01 #mA\n",
+ "Ic=100.0*Ib\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "Vcb=5-Ic-0.7-(101*Ib)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of Vcb is= %.2f V \" %(Vcb))\n",
+ "print(\"Since Vcb is positive, the transistor is in the active region \")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vcb is= 2.29 V \n",
+ "Since Vcb is positive, the transistor is in the active region \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4, Page No 135"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ve= 4.0 #V\n",
+ "Ie=2 #mA\n",
+ "Vc=12-(2*2)\n",
+ "beta=19.0\n",
+ "\n",
+ "#Calculations\n",
+ "Rb=2*(1.0+beta)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of Vcb is= %.2f V \" %(Rb))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vcb is= 40.00 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5, Page No 135"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ve= 4.0 #V\n",
+ "Ie=2 #mA\n",
+ "Vc=12-(2*2)\n",
+ "beta=100.0\n",
+ "\n",
+ "#Calculations\n",
+ "Ib=(2.7-0.7)/beta\n",
+ "Ic=beta*Ib\n",
+ "Vc=(10.0-Ic)/2\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of Vcb is= %.2f V \" %(Vc))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vcb is= 4.00 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6, Page No 135"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ie1=1.0 #mA\n",
+ "Vc1=10.7*Ie1-10\n",
+ "Vbe2=0.7\n",
+ "\n",
+ "#Calculations\n",
+ "Ie2=(10+Vc1-Vbe2)/10.0\n",
+ "Vc2=10-Ie1\n",
+ "Vcb2=Vc2-Vbe2\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of Vcb is= %.2f V \" %(Vcb2))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vcb is= 8.30 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7a, Page No 142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "hfe=100\n",
+ "Ib=4.2/50 #mA\n",
+ "Ic=9.8/3\n",
+ "Ib=Ic/hfe\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of Ib is= %.3f V \" %(Ib))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Ib is= 0.033 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9a, Page No 144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ie1=1.0 #mA\n",
+ "Ic1=0.99 #mA\n",
+ "Vcb1=12-(5*Ic1)-5\n",
+ "Ve=5-0.7\n",
+ "Vbe2=2.5-Ve\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of Vbe2 is= %.2f V \" %(Vbe2))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vbe2 is= -1.80 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9b, Page No 144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ie2=1.0 #mA\n",
+ "Ic2=0.99 #mA\n",
+ "Vcb2=12-(5*Ic2)-2.5\n",
+ "\n",
+ "#Calculations\n",
+ "Ve2=2.5-0.7\n",
+ "Vbe1=0-Ve\n",
+ "V0=12-(5*Ic2)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of V0 is= %.2f V \" %(V0))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V0 is= 7.05 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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