{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1: Introduction to Differential Amplifier and Op-amp"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.1, Page No. 4"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# output voltage of differential amplifier\n",
"\n",
"import math\n",
"#Variable declaration\n",
"v1 = 300*10**-6 # voltage at terminal 1\n",
"v2 = 240*10**-6 # voltage at terminal 2\n",
"Aid = 5000.0 # differential gain of amplifier \n",
"cmmr1 = 100 # CMMR for case 1\n",
"cmmr2 = 10**5 # CMMR for case 2\n",
"\n",
"\n",
"#Calculations\n",
"# case 1\n",
"Vid = v1-v2\n",
"Vcm = (v1+v2)/2\n",
"Acm = Aid/cmmr1\n",
"Vout = Aid*Vid+Acm*Vcm\n",
"Vout = Vout*1000 # mV\n",
"\n",
"# case 2\n",
"Acm2 = Aid/cmmr2\n",
"Vout2 = Aid*Vid+Acm2*Vcm\n",
"Vout2 = Vout2*1000 # mV\n",
"\n",
"Vout_ideal = Aid*Vid*1000 # mV\n",
"#Result\n",
"print(\"(i) CMMR = %d:\\nVout = %.1f mV\\n\\n(ii) CMMR = %d:\\nVout = %.4f mV\\n\\nIdeal Vout = %.0f mV\"%(cmmr1,Vout,cmmr2,Vout2,Vout_ideal))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) CMMR = 100:\n",
"Vout = 313.5 mV\n",
"\n",
"(ii) CMMR = 100000:\n",
"Vout = 300.0135 mV\n",
"\n",
"Ideal Vout = 300 mV\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.2, Page No.9"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Operating point values\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Rc = 4.7*10**3 # collector resistor\n",
"Re = 3.3*10**3 # emitter resistor\n",
"Vcc = 12 # + supply voltage\n",
"Vee = -12 # - supply voltage\n",
"Vbe = 0.7 # base-emitter junction voltage\n",
"\n",
"\n",
"#Calculations\n",
"Ie = (Vcc-Vbe)/(2*Re)\n",
"Ic = Ie\n",
"Vce = Vcc+Vbe-(Ic*Rc) \n",
"\n",
"#Result\n",
"print(\"Q point is (%.3f mA, %.3f V)\"%(Ic*1000,Vce))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Q point is (1.712 mA, 4.653 V)\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.3, Page No. 14"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# differential amplifier parameters(refer fig. 1.14)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"hie = 2.8*10**3 # input impedance\n",
"Rc = 4.7*10**3 # collector resistance\n",
"Vcc = 15 # supply voltage\n",
"hfe = 100 # transistor current gain\n",
"Re = 6.8*10**3 # emitter resistance\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"Rin = 100 # input resistance\n",
"v1 = 70*10**-3 # input voltage 1\n",
"v2 = 40*10**-3 # input voltage 2\n",
"\n",
"# Calculations\n",
"# Operating point values\n",
"Ie = Ic = (Vcc-Vbe)/( (2*Re) + (Rin/hfe))\n",
"Vce = Vcc+Vbe-Ic*Rc\n",
"# Differential gain\n",
"Aid = (hfe*Rc)/(Rin+hie)\n",
"Aid = math.floor(Aid*1000)/1000\n",
"#Common mode gain\n",
"Acm = hfe*Rc/(2*Re*(1+hfe)+Rin+hie)\n",
"#CMMR\n",
"CMMR = 20*math.log10(Aid/Acm)\n",
"CMMR = math.floor(CMMR*1000)/1000\n",
"#output voltage\n",
"Vid = v1-v2\n",
"Vcm = (v1+v2)/2\n",
"Vout = Aid*Vid+Acm*Vcm\n",
"\n",
"#Result\n",
"print(\"(i) Operating point values:\\nIcq = %.3f mA\\nVcq = %.3f V\\n\\n(ii) Differential gain:\\nAid = %.3f\"%(Ic*1000,Vce,Aid))\n",
"print(\"\\n(iii) Common mode gain:\\nAcm = %.4f\"%(math.floor(Acm*10000)/10000))\n",
"print(\"\\n(iv) CMMR:\\nCMMR in dB = %.3f dB\\n\\n(v) Output Voltage:\\nVout = %.2f V\"%(CMMR,Vout))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) Operating point values:\n",
"Icq = 1.051 mA\n",
"Vcq = 10.758 V\n",
"\n",
"(ii) Differential gain:\n",
"Aid = 162.068\n",
"\n",
"(iii) Common mode gain:\n",
"Acm = 0.3414\n",
"\n",
"(iv) CMMR:\n",
"CMMR in dB = 53.527 dB\n",
"\n",
"(v) Output Voltage:\n",
"Vout = 4.88 V\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.4, Page No. 16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# operating point , voltage gain and input/output impedance(refer fig 1.15)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Rc = 3.3*10**3 # collector resistor\n",
"Rin = 150 # inputr resistance\n",
"Re = 8.2*10**3 # emitter resistor\n",
"Vcc = 12 # supply voltage\n",
"beta = 100 # beta value\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"\n",
"#Calculations\n",
"# opearating point\n",
"Ie = Ic = (Vcc-Vbe)/( (2*Re) + (Rin/beta))\n",
"Ie = Ic = math.floor(Ic*10**6)/10**6\n",
"Vce = Vcc+Vbe-Ic*Rc\n",
"# Voltage gain\n",
"re = (26*10**-3)/Ie\n",
"Aid = Rc/re\n",
"# input and output impedance\n",
"Ri = 2*re*beta\n",
"Ro = Rc\n",
"\n",
"#Result\n",
"print(\"(i) Operating point values:\\nIcq = %.3f mA\\nVcq = %.4f V\\n\\n(ii) Voltage gain:\\nAid = %.2f\"%(Ic*1000,Vce,Aid))\n",
"print(\"\\n(iii)input and output impedances:\\nRi = %.3f k-ohm\\nRo = %.1f k-ohm\"%(Ri/1000,Ro/1000)) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) Operating point values:\n",
"Icq = 0.688 mA\n",
"Vcq = 10.4296 V\n",
"\n",
"(ii) Voltage gain:\n",
"Aid = 87.32\n",
"\n",
"(iii)input and output impedances:\n",
"Ri = 7.558 k-ohm\n",
"Ro = 3.3 k-ohm\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.5, Page No. 19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# constant current in the circuit(refer fig 1.19(a)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vee = 20.0 # supply voltage\n",
"R1 = 4.7*10**3 # resistor R1\n",
"R2 = 4.7*10**3 # resistor R2\n",
"R3 = 2.2*10**3 # resistor R3\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"\n",
"#Calcualtions\n",
"Vb = (-Vee*R1)/(R1+R2)\n",
"Ve = Vb - Vbe\n",
"Ie = (Ve-(-Vee))/(R3) \n",
"I = Ie = math.floor(Ie*10**5)/10**5\n",
"\n",
"#Result\n",
"print(\"I = %.2f mA\"%(I*1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I = 4.22 mA\n"
]
}
],
"prompt_number": 47
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.6, Page No.23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Widlar current source(refer fig. 1.24)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Ic2 = 20*10**-6 # required collector current of 2nd Trasistor\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"Vcc = 30 # supply voltage\n",
"R = 33.3*10**3 # Resistor R\n",
"Vt = 26*10**-3 # thermal voltage \n",
"\n",
"#Calculations\n",
"Ic1 = (Vcc-Vbe)/R\n",
"Re = Vt*math.log(Ic1/Ic2)/Ic2\n",
"\n",
"#Result\n",
"print(\"Re = %.2f k-ohm\"%(Re/1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Re = 4.92 k-ohm\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.7, Page No. 24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Widlar current source: value of Ic2\n",
"\n",
"import math\n",
"# Variable declaration\n",
"Iref = 1*10**-3 # reference current\n",
"Re = 5*10**3 # emitter resistor\n",
"Vt = 26*10**-3 # thermal voltage \n",
"\n",
"# Calculations\n",
"Ic2_1 = 12*10**-6 # assumed value 1\n",
"lhs1 = (Vt*math.log(Iref/Ic2_1))-(Ic2_1*Re)\n",
"Ic2_2 = 22*10**-6 # assumed value 2\n",
"lhs2 = (Vt*math.log(Iref/Ic2_2))-(Ic2_2*Re)\n",
"lhs2 = math.ceil(lhs2*10**5)/10**5\n",
"Ic2_3 = 20*10**-6 # assumed value 3\n",
"lhs3 = (Vt*math.log(Iref/Ic2_3))-(Ic2_3*Re)\n",
"\n",
"#Result\n",
"print(\"for Ic2 = %.0f micro-A, L.H.S = %.5f\"%(Ic2_1*10**6,lhs1))\n",
"print(\"for Ic2 = %.0f micro-A, L.H.S = %.5f\"%(Ic2_2*10**6,lhs2))\n",
"print(\"\\nhence actual value is less than but closer to %.0f micro-A\"%(Ic2_2*10**6))\n",
"print(\"\\nfor Ic2 = %f micro-A, L.H.S = %.4f\"%(Ic2_3*10**6,lhs3))\n",
"print(\"LHS is almost zero, so the value of Ic2 = %.0f micro-A\"%(Ic2_3*10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"for Ic2 = 12 micro-A, L.H.S = 0.05499\n",
"for Ic2 = 22 micro-A, L.H.S = -0.01076\n",
"\n",
"hence actual value is less than but closer to 22 micro-A\n",
"\n",
"for Ic2 = 20.000000 micro-A, L.H.S = 0.0017\n",
"LHS is almost zero, so the value of Ic2 = 20 micro-A\n"
]
}
],
"prompt_number": 61
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.8, Page No.34"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# level shifting network (refer fig. 1.42)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vi = 6.84 # input dc level\n",
"R2 = 270 # resistor R2\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"\n",
"# Calculations\n",
"R1 = ((Vi-Vbe)*R2)-R2\n",
"R1 = R1/1000\n",
"#Result\n",
"print(\"Required value of R to get 0V output level is %.3f k-ohm\"%(math.floor(R1*10**3)/1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Required value of R to get 0V output level is 1.387 k-ohm\n"
]
}
],
"prompt_number": 69
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.9, Page No. 36"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# differential and common mode output \n",
"\n",
"import math\n",
"#Variable declaration\n",
"Aid = 80.0 # differential gain\n",
"CMMR = 95.0 # common mode rejection ratio\n",
"V1 = 2*10**-6 # input voltage 1\n",
"V2 = 1.6*10**-6 # input voltage 2\n",
"\n",
"#Calculations\n",
"Aid = 10**(Aid/20)\n",
"CMMR = 10.0**(CMMR/20)\n",
"Vid = Aid*(V1-V2)\n",
"Acm = Aid/CMMR\n",
"Vcm = Acm*(V1+V2)/2\n",
"\n",
"#Result\n",
"print(\"Vid = %.0f mV\\nVcm = %.2f micro-V\"%(Vid*10**3,Vcm*10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Vid = 4 mV\n",
"Vcm = 0.32 micro-V\n"
]
}
],
"prompt_number": 75
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.10, Page No. 37"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Common mode output\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Aid = 125.0 # differential gain \n",
"CMMR = 60.0 # common mode rejection ratio\n",
"Vp = 4.0 # peak voltage \n",
"#Calculations\n",
"x = 10**(CMMR/20) # Aid/Acm\n",
"Acm = Aid/x\n",
"\n",
"#Result\n",
"print(\"Commode mode output is given by Acm*Vcm = %.1f*sin(200*pi*t)\"%(Vp*Acm))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Commode mode output is given by Acm*Vcm = 0.5*sin(200*pi*t)\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.11, Page No.37"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Collector voltage and and differential voltage gain(refer fig 1.48)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"hfe = 100.0 # current gain\n",
"hie = 3.9*10**3 # inpput impedance\n",
"Rin = 1*10**3 # Source resistance\n",
"Re = 27*10**3 # Emitter resistance\n",
"Rc = 6.8*10**3 # collector resistance\n",
"Vcc = 15 # supply voltage\n",
"\n",
"\n",
"#Calculations\n",
"Ie = Ic = (Vcc-Vbe)/((Rin/hfe)+2*Re)\n",
"Vout1 = Vout2 = Vcc-Ic*Rc\n",
"Aid = abs(hfe*Rc/(Rin+hie))\n",
"\n",
"#Result\n",
"print(\"Vc1 = Vc2 = %.2f V\\nAid = %f \"%(Vout1,Aid))\n",
"#Answer for Aid is wrong in the book"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Vc1 = Vc2 = 13.20 V\n",
"Aid = 138.775510 \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.12, Page No. 38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Value of Resistance Re\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vcc = 10.0 # Supply voltage\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"Rc = 47*10**3 # Collector resistance\n",
"hfe = 100.0 # current gain\n",
"Vceq = 8.6 # quiescent collector emitter voltage\n",
"\n",
"#Calculations\n",
"Ic = (Vcc+Vbe-Vceq)/Rc\n",
"Re =(Vcc-Vbe)/(2*((1+hfe)/hfe))\n",
"Re = math.floor(Re*10000)/10000\n",
"Re = Re/(Ic*1000) # k-ohm\n",
"\n",
"#Result\n",
"print(\"Re = %.2f k-ohm\"%(math.floor(Re*100)/100))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Re = 103.03 k-ohm\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.13, Page No. 38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Differential amplifier parameter\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vbe = 0.712 # emitter-base junction voltage drop\n",
"hfe = 500.0 # current gain\n",
"hie = 18 *10**3 # inpput impedance\n",
"Rin = 50 # Source resistance\n",
"Re = 6.8*10**3 # Emitter resistance\n",
"Rc = 4.7*10**3 # collector resistance\n",
"Vcc = 10 # supply voltage\n",
"\n",
"#Calcualtions\n",
"#(i)\n",
"Ie = (Vcc -Vbe)/(2*Re)\n",
"Ic = (hfe/(1+hfe))*Ie\n",
"Vce = Vcc+Vbe-Ic*Rc\n",
"Vce = math.floor(Vce*1000)/1000\n",
"Ic = math.floor(Ic*10**7)/10**7\n",
"Ic = Ic *1000 # mA\n",
"#(ii)\n",
"Ad = hfe*Rc/(2*(Rin+hie))\n",
"Ad = math.floor(Ad*100)/100\n",
"#(iii)\n",
"Ri = 2*(Rin+hie)\n",
"Ro = Rc\n",
"\n",
"#Result\n",
"print(\"(i)\\nIcq =%.4f mA\\t\\tVce = %.3f V\\n\\n(ii)\\nAd = %.2f\\n\\n(iii)\\nRi = %.1f k-ohm\\t\\tRo = %.1f k-ohm\"%(Ic,Vce,Ad,Ri/1000.0,Ro/1000))\n",
"# answer for Ri is wrong in the book"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)\n",
"Icq =0.6815 mA\t\tVce = 7.508 V\n",
"\n",
"(ii)\n",
"Ad = 65.09\n",
"\n",
"(iii)\n",
"Ri = 36.1 k-ohm\t\tRo = 4.7 k-ohm\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.14, Page No. 39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Current through the transistors(refer fig 1.50)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Q2 = 0.5 # Area of Q2 /Area of Q1\n",
"Q3 = 0.25 # Area of Q3 /Area of Q1\n",
"Q4 = 0.125 # Area of Q4 /Area of Q1\n",
"Vcc = 15 # Supply voltage\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"R = 20*10**3 # Resistor R\n",
"\n",
"#Calculations\n",
"I1 = (Vcc-Vbe)/R\n",
"I2 = I1*Q2\n",
"I3 = I1*Q3\n",
"I4 = I1*Q4\n",
"\n",
"\n",
"#Result\n",
"print(\"I2 = %.3f mA\\nI3 = %.4f mA\\nI4 = %.3f mA\"%(math.floor(I2*10**6)/1000,math.floor(I3*10**7)/10000,I4*1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I2 = 0.357 mA\n",
"I3 = 0.1787 mA\n",
"I4 = 0.089 mA\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
" example 1.15, Page No. 39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Current calculation in the circuit\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vz = 6.2 # zener voltage\n",
"R1 = 4.7*10**3 # resistor R1\n",
"R3 = 2.2*10**3 # resistor R3\n",
"Vee = -20 # supply voltage\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"\n",
"#Calcualtions\n",
"Vb = Vee + Vz\n",
"Ve = Vee+Vz-Vbe\n",
"I = Ie = (Ve-Vee)/R3\n",
"\n",
"#Result\n",
"print(\"I = %.1f mA\"%(I*1000))\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I = 2.5 mA\n"
]
}
],
"prompt_number": 53
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.16, Page No.39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Differential amplifier\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Ie = 10**-3 # Emitter current\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"Rc = 4.7*10**3 # colloector Resistance\n",
"Vcc = 12 # Supply voltage\n",
"\n",
"\n",
"#Calculations\n",
"Re = (Vcc-Vbe)/(Ie)\n",
"Ic = Ic1 = Ic2 = Ie/2\n",
"Vout = Vcc-Ic*Rc\n",
"\n",
"#Result\n",
"print(\"(i) Re = %.1f k-ohm\\n\\n(ii) Vout = %.2f V\"%(Re/1000,Vout))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) Re = 11.3 k-ohm\n",
"\n",
"(ii) Vout = 9.65 V\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.17, Page No. 40"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Ie3, Vce1 and differential mode gain(refer fig. 1.54)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"hfe = 250.0 # current gain\n",
"hie = 560 # inpput impedance for Q1 and Q2\n",
"Rin = 1.0*10**3 # Source resistance\n",
"Rb4 = 1.5*10**3 # Emitter resistance\n",
"Rc = 1.0*10**3 # collector resistance\n",
"Vcc = 12 # supply voltage\n",
"\n",
"# Calculations\n",
"I = (Vcc-Vbe)/Rb4\n",
"Ie3 = I\n",
"Ie1 = Ie2 = Ie3/2\n",
"Ic1 = Ie1\n",
"Vce1 = Vcc+Vbe-Ic1*Rc\n",
"Aid = hfe*Rc/(Rc+hie)\n",
"\n",
"#Result\n",
"print(\"Ie3 = %.3f mA\\nVce1 = %.3f V\\nAid = %.2f\"%(Ie3*10**3,Vce1,math.floor(Aid*100)/100))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ie3 = 7.533 mA\n",
"Vce1 = 8.933 V\n",
"Aid = 160.25\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.18, Page No. 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"print(\"Theoretical example\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Theoretical example\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.19, Page No. 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# DC shift and curretn in the circuit (refer fig. 1.55)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vcc = 15.0 # supply voltage\n",
"R1 = 10.0*10**3 # Resistor R1\n",
"R2 = 5.0*10**3 # Resistor R2\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"\n",
"#Calculations\n",
"shift = (Vcc*R2/R1)+Vbe*(1-(R2/R1))\n",
"I = (Vcc-Vbe)/R1\n",
"\n",
"#Result\n",
"print(\"(Vin-Vout) = %.2f V\\n\\t I = %.2f mA\"%(shift, I*10**3))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(Vin-Vout) = 7.85 V\n",
"\t I = 1.43 mA\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.20, Page No. 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# maximum possible closed loop gain\n",
"\n",
"import math\n",
"#Variable declaration\n",
"sr = 3*10**6 # slew rate in V/sec\n",
"dV = 0.4 # input voltage variation\n",
"dt = 12*10**-6 # time in which voltage varies\n",
"\n",
"#Calculations\n",
"A = sr/(dV/dt)\n",
"\n",
"#Result\n",
"print(\"Maximum closed loop gain = %.0f\"%A)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum closed loop gain = 90\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.21, Page No. 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# input bias and input offset current\n",
"\n",
"import math\n",
"#Variable Declaration\n",
"Vbe = 0.7 # emitter-base junction voltage drop\n",
"beta1 = 100.0 # current gain for Q1\n",
"beta2 = 125.0 # current gain for Q1\n",
"Re = 39*10**3 # Emitter resistance\n",
"Rc = 4.7*10**3 # collector resistance\n",
"Vcc = 12 # supply voltage\n",
"\n",
"# Calculations\n",
"Ie = (Vcc-Vbe)/Re\n",
"Ie1 = Ie2 = Ie/2\n",
"Ib1 = Ie1/beta1\n",
"Ib2 = Ie2/beta2\n",
"Ib = (Ib1+Ib2)/2\n",
"Iios = abs(Ib1-Ib2)\n",
"\n",
"#Result\n",
"print(\"Ib = %.4f micro-A\\nIios = %.5f micro-A\"%(Ib*10**6,Iios*10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ib = 1.3038 micro-A\n",
"Iios = 0.28974 micro-A\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.22, Page No. 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Maximum possible amplitude of output\n",
"\n",
"import math\n",
"#Variable declaration\n",
"f = 7000.0 # frequency of input sine wave\n",
"Icq = 8*10**-6 # collector current\n",
"Cc = 27*10**-12 # Capacitance\n",
"\n",
"#Calculations\n",
"S = Icq/Cc\n",
"Vm = S/(2*math.pi*f)\n",
"\n",
"#Result\n",
"print(\"Maximum amplitude of output = %.3f V\"%(math.floor(Vm*1000)/1000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum amplitude of output = 6.736 V\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.23, Page No. 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Slew rate\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vm = 3.8 # output amplitude\n",
"tr = 4.5*10**-6 # required rise time\n",
"S_741 = 0.5*10**-6 # slew rate of IC741 op-amp\n",
"\n",
"#Calculations\n",
"del_V = (0.9-0.1)*Vm\n",
"S = del_V/tr\n",
"\n",
"#Result\n",
"print(\"Required slew rate = %.3f V/micro-s\"%(math.floor(S/10**3)/1000))\n",
"print(\"Slew rate of IC 741 is 0.5V/usec which is too low as compared to the required value. Hence Ic 741 cannot be used.\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Required slew rate = 0.675 V/micro-s\n",
"Slew rate of IC 741 is 0.5V/usec which is too low as compared to the required value. Hence Ic 741 cannot be used.\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.24, Page No.43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Maximum gain using op-amp\n",
"\n",
"import math\n",
"#Variable declaration\n",
"S = 0.4*10**6 # op-amp slew rate in V/sec\n",
"Vm = 0.04 # maximum input \n",
"w = 1.13*10**5 # freq in rad/sec\n",
"\n",
"#Calculations\n",
"fm = w/(2*math.pi)\n",
"V = S/(2*math.pi*fm)\n",
"G = V/Vm\n",
"\n",
"#Result\n",
"print(\"Gain = %.1f \"%G)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Gain = 88.5 \n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.25, Page No. 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# smallest and largest possible input bias current and input offset current\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Ie = 400*10**-6 # total emitter bias current\n",
"beta_min = 80 # minimum current gain\n",
"beta_max = 200 # maximum current gain\n",
"\n",
"#Calculation\n",
"Ie1 = Ie2 = Ie/2\n",
"Ib1_1 = Ib2_1 = Ie1/(1+beta_min) # for beta =80\n",
"Ib1_2 = Ib2_2 = Ie1/(1+beta_max)\n",
"\n",
"Ib_max = (Ib1_1+Ib2_1)/2\n",
"Ib_min = (Ib1_2+Ib2_2)/2\n",
"Iios = Ib_max-Ib_min\n",
"\n",
"#Result\n",
"print(\"Largest input bias current = %.3f micro-A\\nSmallest input bias current = %.3f micro-A\"%(Ib_max*10**6,Ib_min*10**6))\n",
"print(\"Largest input offset current = %.3f micro-A\"%(Iios*10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Largest input bias current = 2.469 micro-A\n",
"Smallest input bias current = 0.995 micro-A\n",
"Largest input offset current = 1.474 micro-A\n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.26, Page No.43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Collector current for each transistor(refer fig. 1.59)\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vbe = 0.715 # emitter-base junction voltage drop\n",
"beta = 100.0 # Current gain\n",
"Vcc = 10 # input voltage\n",
"Rc1 = 5.6*10**3 # Collector resistance\n",
"Rc2 = 1.0*10**3 # Collector resistance\n",
"\n",
"#Calculations\n",
"I = (Vcc-Vbe)/Rc1\n",
"Ic2 = (beta/(beta+4))*I\n",
"\n",
"#Result\n",
"print(\"I = %.3f mA\\nIc2 = Ic3 = Ic4 = %.4f mA\"%(I*1000,math.floor(Ic2*10**7)/10000))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I = 1.658 mA\n",
"Ic2 = Ic3 = Ic4 = 1.5942 mA\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 1.27, Page No. 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Differential amplifier parameter\n",
"\n",
"import math\n",
"#Variable declaration\n",
"beta = 100.0 # current gain\n",
"Vbe = 0.715 # emitter-base junction voltage drop\n",
"Vz = 6.2 # Zener Voltage\n",
"Iz = 41*10**-3 # Zener Current \n",
"Rc = 4.7*10**3 # Collector Current\n",
"Rin = 500 # input resistance\n",
"Rb = 68*10**3 # base voltage \n",
"Re = 2.7*10**3 # emitter resistance\n",
"Vcc = 10 # Supply Voltage\n",
"hie = 1000.0 # input resistance\n",
"\n",
"#Calculations\n",
"I = (Vcc-Vbe-Vz)/Rb\n",
"I = math.floor(I*10**8)/10**8\n",
"Ie3 = (Vcc-Vbe-Rb*I)/Re\n",
"Ie3 = math.ceil(Ie3*10**7)/10**7\n",
"Ic3 = (beta/(1+beta))*Ie3\n",
"Ic3 = math.floor(Ic3*10**7)/10**7\n",
"Ie = Ic3/2\n",
"Ie = math.floor(Ie*10**7)/10**7\n",
"Ic = beta*Ie/(1+beta)\n",
"Ic = math.ceil(Ic*10**7)/10**7\n",
"Vce = Vcc+Vbe-Ic*Rc\n",
"Vce = math.floor(Vce*10**4)/10**4\n",
"Aid = beta*Rc/(Rin+hie)\n",
"Ri = Rin+2*hie\n",
"\n",
"#Result\n",
"print(\"(i) Aid = %.3f\\n\\n(ii) input Resistance = %.1f k-ohm\\n\\n(iii) Icq = %.4f mA\\t\\tVceq = %.4f V\"%(Aid,Ri/1000,Ic*1000,Vce))\n",
"#Value for input resistance is wrong in the book"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) Aid = 313.333\n",
"\n",
"(ii) input Resistance = 2.5 k-ohm\n",
"\n",
"(iii) Icq = 1.1256 mA\t\tVceq = 5.4246 V\n"
]
}
],
"prompt_number": 77
}
],
"metadata": {}
}
]
}