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{
"metadata": {
"name": "Chapter_5"
},
"nbformat": 2,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"source": [
"<h1>Chapter 5: Transistor Bias Circuits<h1>"
]
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.1, Page Number: 146<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"%matplotlib inline"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"",
"Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].",
"For more information, type 'help(pylab)'."
]
}
],
"prompt_number": 1
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaration",
"V_BB=10.0; #voltage in volt",
"V_CC=20.0; #voltage in volt",
"B_DC=200.0; #B_DC value",
"R_B=47.0*10**3; #resistance in ohm",
"R_C=330.0; #resistance in ohm",
"V_BE=0.7; #voltage in volt",
"",
"#current",
"I_B=(V_BB-V_BE)/R_B; #base current",
"I_C=B_DC*I_B; #Q POINT",
"V_CE=V_CC-I_C*R_C; #Q POINT",
"I_C_sat=V_CC/R_C; #saturation current",
"I_c_peak=I_C_sat-I_C; #peak current ",
"I_b_peak=I_c_peak/B_DC; #peak current in ampere",
"",
"#result",
"print \"Q point of I_C = %.3f amperes\" %I_C",
"print \"Q point of V_CE = %.2f volts\" %V_CE",
"print \"peak base current = %.4f amperes\" %I_b_peak"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Q point of I_C = 0.040 amperes",
"Q point of V_CE = 6.94 volts",
"peak base current = 0.0001 amperes"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.2, Page Number: 149<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaration",
"B_DC=125.0; #DC value",
"R_E=10.0**3; #resistance in ohm",
"",
"#calculation",
"R_IN_base=B_DC*R_E; #base resistance",
"",
"#Result",
"print \"DC input resistance, looking at base of transistor = %d ohm\" %R_IN_base"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"DC input resistance, looking at base of transistor = 125000 ohm"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.3, Page Number: 151<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaration",
"B_DC=100; #DC value",
"R1=10*10**3; #resistance in ohm",
"R2=5.6*10**3; #resistance in ohm",
"R_C=1*10**3; #resistance in ohm",
"R_E=560; #resistance in ohm",
"V_CC=10; #voltage in volt",
"V_BE=0.7 #voltage in volt",
"",
"#calculation",
"R_IN_base=B_DC*R_E; #calculate base resistance",
"#We can neglect R_IN_base as it is equal to 10*R2",
"print \"input resistance seen from base = %d ohm\" %R_IN_base",
"print \"which can be neglected as it is 10 times R2\"",
"",
"V_B=(R2/(R1+R2))*V_CC; #base voltage",
"V_E=V_B-V_BE; #emitter voltage",
"I_E=V_E/R_E; #emitter current",
"I_C=I_E; #currents are equal",
"V_CE=V_CC-I_C*(R_C+R_E); #voltage in volt",
"",
"#result",
"print \"V_CE = %.2f volts\" %V_CE",
"print \"I_C = %.3f amperes\" %I_C",
"print \"Since V_CE>0V, transistor is not in saturation\""
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"input resistance seen from base = 56000 ohm",
"which can be neglected as it is 10 times R2",
"V_CE = 1.95 volts",
"I_C = 0.005 amperes",
"Since V_CE>0V, transistor is not in saturation"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.4, Page Number: 154<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaration",
"V_EE=10.0; #voltage in volt",
"V_BE=0.7; #voltage in volt",
"B_DC=150.0; #DC value ",
"R1=22.0*10**3; #resistance in ohm",
"R2=10.0*10**3; #resistance in ohm",
"R_C=2.2*10**3; #resistance in ohm",
"R_E=1.0*10**3; #resistance in ohm",
"",
"#calculation",
"R_IN_base=B_DC*R_E; #R_IN_base>10*R2,so it can be neglected",
"print \"input resistance as seen from base = %d ohm\" %R_IN_base",
"print \"it can be neglected as it is greater than 10 times R2\"",
"V_B=(R1/(R1+R2))*V_EE; #base voltage",
"V_E=V_B+V_BE; #emitter voltage",
"I_E=(V_EE-V_E)/R_E; #emitter current",
"I_C=I_E; #currents are equal",
"V_C=I_C*R_C; #collector voltage",
"V_EC=V_E-V_C; #emitter-collector voltage",
"",
"#result",
"print \"I_C collector current = %.4f amperes\" %I_C",
"print \"V_EC emitter-collector voltage = %.2f Volts\" %V_EC"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"input resistance as seen from base = 150000 ohm",
"it can be neglected as it is greater than 10 times R2",
"I_C collector current = 0.0024 amperes",
"V_EC emitter-collector voltage = 2.24 Volts"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.5, PAge Number: 154<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaration",
"R1=68.0*10**3; #resistance in ohm",
"R2=47.0*10**3; #resistance in ohm",
"R_C=1.8*10**3; #resistance in ohm",
"R_E=2.2*10**3; #resistance in ohm",
"V_CC=-6.0; #voltage in volt",
"V_BE=0.7; #voltage in volt",
"B_DC=75.0; #DC value",
"",
"#calculation",
"R_IN_base=B_DC*R_E;",
"print \"input resistance as seen from base\"",
"print \"is not greater than 10 times R2 so it should be taken into account\"",
"#R_IN_base in parallel with R2",
"V_B=((R2*R_IN_base)/(R2+R_IN_base)/(R1+(R2*R_IN_base)/(R2+R_IN_base)))*V_CC;",
"V_E=V_B+V_BE; #emitter voltage",
"I_E=V_E/R_E; #emitter current",
"I_C=I_E; #currents are equal",
"V_C=V_CC-I_C*R_C; #collector voltage",
"V_CE=V_C-V_E; #collector-emitter voltage",
"",
"#result",
"print \"collector current = %.4f amperes\" %I_C",
"print \"collector emitter voltage = %.2f volts\" %V_CE"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"input resistance as seen from base",
"is not greater than 10 times R2 so it should be taken into account",
"collector current = -0.0006 amperes",
"collector emitter voltage = -3.46 volts"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.6, Page Number: 156<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaration",
"V_CC=12.0; #voltage in volt",
"R_B=100.0*10**3; #resistance in ohm",
"R_C=560.0; #resistance in ohm",
"#FOR B_DC=85 AND V_BE=0.7V",
"B_DC=85.0; #DC value",
"V_BE=0.7; #base-emitter voltage",
"",
"#calculation",
"I_C_1=B_DC*(V_CC-V_BE)/R_B; #collector current",
"V_CE_1=V_CC-I_C_1*R_C; #collector-emittor voltage",
"#FOR B_DC=100 AND V_BE=0.6V",
"B_DC=100.0; #DC value ",
"V_BE=0.6; #base emitter voltage",
"I_C_2=B_DC*(V_CC-V_BE)/R_B; #collector current",
"V_CE_2=V_CC-I_C_2*R_C; #voltage in volt",
"p_del_I_C=((I_C_2-I_C_1)/I_C_1)*100; #percent change in collector current ",
"p_del_V_CE=((V_CE_2-V_CE_1)/V_CE_1)*100; #percent change in C-E voltage",
"",
"#result",
"print \"percent change in collector current = %.2f\" %p_del_I_C",
"print \"percent change in collector emitter voltage = %.2f\" %p_del_V_CE"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"percent change in collector current = 18.69",
"percent change in collector emitter voltage = -15.18"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.7, Page Number: 159<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaration",
"V_CC=20.0; #voltage in volt",
"R_C=4.7*10**3; #resistance in ohm",
"R_E=10.0*10**3; #resistance in ohm",
"V_EE=-20.0; #voltage in volt",
"R_B=100*10**3; #resistance in ohm",
"#FOR B_DC=85 AND V_BE=0.7V",
"B_DC=85; #DC value",
"V_BE=0.7; #base-emitter voltage",
"I_C_1=(-V_EE-V_BE)/(R_E+(R_B/B_DC));",
"V_C=V_CC-I_C_1*R_C; #colector voltage",
"I_E=I_C_1; #emittor current",
"V_E=V_EE+I_E*R_E; #emittor voltage",
"V_CE_1=V_C-V_E; #CE voltage",
"print \"I_C_1 = %.3f\" %I_C_1",
"print \"V_CE_1 = %.2f\" %V_CE_1",
"#FOR B_DC=100 AND V_BE=0.6V",
"B_DC=100; #DC value ",
"V_BE=0.6; #base-emitter voltage",
"I_C_2=(-V_EE-V_BE)/(R_E+(R_B/B_DC));",
"V_C=V_CC-I_C_2*R_C;#colector voltage",
"I_E=I_C_2; #emittor current",
"V_E=V_EE+I_E*R_E; #emittor voltage",
"V_CE_2=V_C-V_E; #CE voltage",
"print \"I_C_2 = %.3f\" %I_C_2",
"print \"V_CE_2 = %.2f\" %V_CE_2",
"",
"p_del_I_C=((I_C_2-I_C_1)/I_C_1)*100;",
"p_del_V_CE=((V_CE_2-V_CE_1)/V_CE_1)*100;",
"print \"percent change in collector currrent = %.2f\" %p_del_I_C",
"print \"percent change in collector emitter voltage = %.2f\" %p_del_V_CE",
"print \"answers in book are approximated\""
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I_C_1 = 0.002",
"V_CE_1 = 14.61",
"I_C_2 = 0.002",
"V_CE_2 = 14.07",
"percent change in collector currrent = 2.13",
"percent change in collector emitter voltage = -3.69",
"answers in book are approximated"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"source": [
"<h3>Example 5.8, Page Number: 161<h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"",
"# variable declaratio",
"V_CC=10.0; #voltage in volt",
"B_DC=100.0; #Dc value",
"R_C=10.0*10**3; #resistance in ohm",
"R_B=100.0*10**3; #resistance in ohm",
"V_BE=0.7; #base-emittor voltage",
"",
"#calculation",
"I_C=(V_CC-V_BE)/(R_C+(R_B/B_DC)); #collector current",
"V_CE=V_CC-I_C*R_C; #CE voltage",
"",
"#result",
"print \"Q point of collector current %.4f amperes\" %I_C",
"print \"Q point of collector-emitter voltage %.3f volts\" %V_CE"
],
"language": "python",
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Q point of collector current 0.0008 amperes",
"Q point of collector-emitter voltage 1.545 volts"
]
}
],
"prompt_number": 9
}
]
}
]
}
|
