{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2 : Operational Amplifier"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1 Page No.44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given Data\n",
"\n",
"R1 = 5*10**3\n",
"Rf = 20*10**3\n",
"vi = 1 \n",
"RL = 5*10**3\n",
"\n",
"# calculating the values \n",
"vo = float((1+(Rf/R1))*vi) \n",
"ACL = int(vo/vi)\n",
"iL = int((vo/RL)*10**3)\n",
"i1 = float(((vo - vi)/Rf))*(10**3)\n",
"io = iL+i1\n",
" \n",
" \n",
"# printing the values\n",
"print \"Output voltage vo = \",vo,\"V\"\n",
"print \"Gain ACL = \",ACL\n",
"print \"Load current iL = \",iL,\"mA\"\n",
"print \"The value of i1 = \",i1,\"mA\"\n",
"print \"Output current io = \", io,\"mA\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Output voltage vo = 5.0 V\n",
"Gain ACL = 5\n",
"Load current iL = 1 mA\n",
"The value of i1 = 0.2 mA\n",
"Output current io = 1.2 mA\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2 Page No.44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"\n",
"R1 = 10*10**3 #R1 input resistance \n",
"Rf = 100*10**3 # Rf feedback resistance\n",
"vi = float(1) #input voltage \n",
"RL = 25*10**3\n",
"#calculating the values \n",
"\n",
"i1 = float((vi/R1)*10**3) # input resistace id the ratio of input voltage to the input resitance \n",
"vo = float(-(Rf/R1)*vi) # finding the output voltage \n",
"iL = float((abs(vo)/RL)*10**3) # calculating the load current \n",
"io = float((i1+iL)) # calculating the output current which is equal to the sum of input current and load current\n",
"\n",
"#printing the values \n",
"\n",
"print \"The input current i1 =\",i1,\"mA\"\n",
"print \"The output voltage vo =\",vo,\"V\"\n",
"print \"The load current iL =\",iL,\"mA\"\n",
"print \"The output current io =\",io,\"mA\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The input current i1 = 0.1 mA\n",
"The output voltage vo = -10.0 V\n",
"The load current iL = 0.4 mA\n",
"The output current io = 0.5 mA\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3 Page No.49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"\n",
"ACL = 5 # Gain of the amplifier\n",
"R1 = 10*10**3 # input resisitance in ohms \n",
"\n",
"# calculations\n",
"\n",
"Rf = (5-1) * R1 # calculating the resistance of feedback resistor \n",
"\n",
"# printing the values \n",
"\n",
"print \"The value of feedback resistor = \", (Rf/10**3),\"KiloOhms\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of feedback resistor = 40 KiloOhms\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4 Page No.49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Given Data\n",
"\n",
"R1 = 5*10**3\n",
"Rf = 20*10**3\n",
"vi = 1 \n",
"RL = 5*10**3\n",
"\n",
"# calculating the values \n",
"vo = float((1+(Rf/R1))*vi) \n",
"ACL = int(vo/vi)\n",
"iL = int((vo/RL)*10**3)\n",
"i1 = float(((vo - vi)/Rf))*(10**3)\n",
"io = iL+i1\n",
" \n",
" \n",
"# printing the values\n",
"print \"Output voltage vo = \",vo,\"V\"\n",
"print \"Gain ACL = \",ACL\n",
"print \"Load current iL = \",iL,\"mA\"\n",
"print \"The value of i1 = \",i1,\"mA\"\n",
"print \"Output current io = \", io,\"mA\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Output voltage vo = 5.0 V\n",
"Gain ACL = 5\n",
"Load current iL = 1 mA\n",
"The value of i1 = 0.2 mA\n",
"Output current io = 1.2 mA\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.6 Page No.61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Given Data\n",
"from decimal import Decimal, ROUND_HALF_UP\n",
"import math\n",
"Beta = 200\n",
"ICQ = 100*10**-6\n",
"ADM = 100\n",
"CMRR = 80\n",
"\n",
"# finding the solution \n",
"# for VT =25 milli volt \n",
"VT = 25*10**-3\n",
"gm = float(ICQ/VT)\n",
"Rc = (ADM/gm) \n",
"CMRR = 10**(80/20) # log inverse is equal to powers of 10\n",
"RE = float((CMRR-1)/gm)\n",
"x = Decimal((RE/10**6))\n",
"\n",
"# printing the values \n",
"\n",
"print \" The value of gm =\",int(math.ceil((gm*10**3))),\"mMho\" #converting the answer into milli Mho\n",
"print \" The value of Rc =\",int((Rc/10**3)),\"kiloOhm\" #converting the answer into Kilo Ohm\n",
"print \" The value of RE =\", x.quantize(Decimal('2.5')),\"MegaOhm\" #converting the answer into MegaOhm"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of gm = 4 mMho\n",
" The value of Rc = 25 kiloOhm\n",
" The value of RE = 2.5 MegaOhm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.7 Page No.61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"# Given Data\n",
"RC = 2*10**3\n",
"RE = 4.3*10**3\n",
"VCC = VEE = 5\n",
"beta0 = 200\n",
"VBE = 0.7\n",
"VT = 25*10**-3 # VT = 25 mV \n",
"\n",
"#calculations \n",
"\n",
"Ibq = (((VEE - VBE)/(2*(1+beta0)*RE)))*10**6 # converting the into mA\n",
"IBQ = int(Ibq*10)/10.0 # rounding the answer 2.48 to 2.4\n",
"ICQ = beta0 * IBQ*10**-3 # finding ICQ and convert it into milli Ambere\n",
"V01 = V02 = VCC - (RC * ICQ*10**-3)\n",
"VCEQ = V01 + VBE\n",
"gm = ICQ / VT # Finding the gm value in milli Mho\n",
"rpi = float(beta0/gm) # finding rpi interms of KiloOhms \n",
"ADM = -(gm*RC)/10**3 # Calculating the gain \n",
"ACM = float((-beta0*RC)/(rpi*10**3 + (2*beta0) * RE)) # Calulating the value of ACM and coverting it into a float \n",
"CMRR = float(round(ADM,2)/round(ACM,2)) # rounding ACM and ADM to 2 digits after decimals and taking the\n",
" #ratio\n",
"CMRRdb = float(20*math.log10(round(CMRR,1))) # Find the CMRR in dB using equation 20*log(CMRR)\n",
"\n",
"# Printing all values \n",
"\n",
"print \"The IBQ =\",IBQ,\"uA\"\n",
"print \"The ICQ =\",ICQ,\"mA\"\n",
"print \"The V01 =\",V01,\"mV\"\n",
"print \"The VCEQ =\",VCEQ,\"V\"\n",
"print \"The gm =\",gm,\"m Mho\"\n",
"print \"The rpi =\",round(rpi,1),\"kilo Ohm\"\n",
"print \"The ADM =\",round(ADM,2)\n",
"print \"The ACM =\",round(ACM,2)\n",
"print \"The CMRR =\",round(CMRR,1)\n",
"print \"The CMRR in dB =\",int(CMRRdb*10)/10.0,\"dB\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The IBQ = 2.4 uA\n",
"The ICQ = 0.48 mA\n",
"The V01 = 4.04 mV\n",
"The VCEQ = 4.74 V\n",
"The gm = 19.2 m Mho\n",
"The rpi = 10.4 kilo Ohm\n",
"The ADM = -38.4\n",
"The ACM = -0.23\n",
"The CMRR = 167.0\n",
"The CMRR in dB = 44.4 dB\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.8 Page No.62"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# v1 = 15*sin*2*pi(60)t + 5*sin*2*pi(1000)t mV\n",
"# v1 = 15*sin*2*pi(60)t - 5*sin*2*pi(1000)t mV\n",
"\n",
"# Given data \n",
"\n",
"gm = 4*10**-3\n",
"RC = 125*10*3\n",
"RE = 1.25*10**3\n",
"beta0 = 200\n",
"\n",
"# calculating the values\n",
"\n",
"rpi = beta0/gm # value is in ohms \n",
"ADM =-500 # Given Value\n",
"ACM = -((200*RC)/(402*RE)+rpi)*10**-6\n",
"print \"ACM is =\",round(ACM,2)\n",
"\n",
"# derivation part is as follows \n",
"# VDM = (v1 - v2)/2 = 5*sin*2*pi(1000)*t\n",
"# VCM = (v1 - v2)/2 = 15*sin*2*pi(60)*t\n",
"# VO1 = ADM*VDM + ACM*VCM\n",
"# VO2 = ADM*VDM - ACM*VCM\n",
"# V01 = -2500*sin*2*pi(1000)t - 0.75*sin*2*pi(60)t mV\n",
"# V01 = 2500*sin*2*pi(1000)t - 0.75*sin*2*pi(60)t mV"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ACM is = -0.05\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.9 Page No.63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from fractions import Fraction \n",
"# Given data\n",
"\n",
"beta0 = 100\n",
"IQ = 5*10**-4\n",
"RC = 10*10**3\n",
"RE = 150\n",
"VT = 25*10**-3 \n",
"\n",
"# calculations \n",
"\n",
"ICQ = float(IQ/2)\n",
"gm = float(ICQ / VT)\n",
"rpi = beta0/gm\n",
"# calculaing the gain in Differential mode\n",
"ADM = ((0.5)*(beta0*RC))/(rpi+((1+beta0)*RE))\n",
"# To get the differentila mode gain multiply the value by 2\n",
"ADM2 = (ADM*2)\n",
"\n",
"# print the values \n",
"\n",
"print \"ICQ value is =\",ICQ*10**3,\"mA\"\n",
"print \"gm value is =\",Fraction(gm).limit_denominator(100),\"Mho\" \n",
"print \"rpi value is =\",int(rpi/10**3),\"kilo Ohm\"\n",
"print \"THe gain is =\",int(math.ceil(ADM2)),\"V/V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ICQ value is = 0.25 mA\n",
"gm value is = 1/100 Mho\n",
"rpi value is = 10 kilo Ohm\n",
"THe gain is = 40 V/V\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.10 Page No.67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given Data \n",
"VCC = 10.0 # initializing as floats \n",
"beta = 125.0\n",
"VBE = 0.7\n",
"\n",
"# calculations \n",
"# Define a function for calculating the value of R1 according to the equation given in 2.67\n",
"def R1(I):\n",
" R1 = (beta/(2+beta))*((VCC-VBE)/I) \n",
" return (R1/1000)\n",
"# Cosisdering Ic = 1mA\n",
"print \"The value of R1 when I is 1 mA = \",round(R1(10**-3),2),\"Kilo Ohm\" # calling the function R1 with arguiment as current value 1 mA\n",
"print \"The value of R1 when I is 10 uA = \",int(round(R1(10*10**-6),0)),\"Kilo Ohm\" # calling the function R1 with arguiment as current value 10 uA"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of R1 when I is 1 mA = 9.15 Kilo Ohm\n",
"The value of R1 when I is 10 uA = 915 Kilo Ohm\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.11 Page No.69"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"import math\n",
"I0 = 10*10**-6\n",
"VCC =10\n",
"VBE = 0.7\n",
"beta = 125\n",
"VT = 25*10**-3\n",
"\n",
"# Solution of the problem is \n",
"Iref = 10**-3 # Assumption\n",
"\n",
"R1 = (VCC - VBE)/Iref\n",
"# Finding the value RE from the equation 2.74\n",
"RE = (VT/(1+(1/beta)*I0))*math.log(Iref/I0)\n",
"\n",
"# printing the values \n",
"\n",
"print \"The value of R1 =\",R1/10**3,\"Kilo Ohms\"\n",
"print \"The value of RE =\",round(RE*100,1),\"Kilo Ohms\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of R1 = 9.3 Kilo Ohms\n",
"The value of RE = 11.5 Kilo Ohms\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.13 Page No.74"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data \n",
"\n",
"VCC = 10.0\n",
"beta = 100\n",
"VBE = 0.7\n",
"RE = 10*10**3\n",
"\n",
"# solution\n",
"# fom the KVL loop1\n",
"Iref = ((VCC - VBE)/RE) # Finding the Iref \n",
"# Assuming the transistor are identical \n",
"# Iref = 2IE\n",
"# IE = IC + IB\n",
"IC = (beta*Iref/(2*(1+beta)))\n",
"I0 = IC\n",
"#Displaying the values \n",
"\n",
"print \"The value of Iref =\",Iref*10**3,\"mA\" # Converting the value into mA\n",
"print \"The value of IC =\",round(IC*10**3,2),\"mA\" # Converting the value into mA and rounding the answer\n",
"print \"The value of I0 =\",round(I0*10**3,2),\"mA\" # Converting the value into mA and rounding the answer"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Iref = 0.93 mA\n",
"The value of IC = 0.46 mA\n",
"The value of I0 = 0.46 mA\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.14 Page No.75"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data \n",
"\n",
"VCC = 12.0\n",
"Rref = 15*10**3 \n",
"R1 = 2.8*10**3\n",
"beta = 200\n",
"VBE = 0.7\n",
"V0 = 6.0\n",
"\n",
"#Calculations \n",
"#Calculations for finding the IC1 and IC2\n",
"Iref = (VCC - VBE)/Rref\n",
"I1 = VBE / R1\n",
"IC1 = (Iref - I1)/(1+(2/beta))\n",
"IC2 = IC1 # Due to mirror effect \n",
"RC = (VCC - V0)/IC1\n",
"\n",
"# Displaying the values \n",
"\n",
"print \"The value of Iref =\",round(Iref*10**3,2),\"mA\" # convert into milli amps and rounding the output\n",
"print \"The value of I1 =\",round(I1*10**3,2),\"mA\" # convert into milli amps and rounding the output\n",
"print \"The value of IC1 =\",round(IC1*10**3,1),\"mA\" # convert into milli amps and rounding the output\n",
"print \"The value of Rc =\",int(round(RC*10**-3,0)),\"Kilo Ohm\" # convert into kilo Ohms and rounding the output"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Iref = 0.75 mA\n",
"The value of I1 = 0.25 mA\n",
"The value of IC1 = 0.5 mA\n",
"The value of Rc = 12 Kilo Ohm\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.15 Page No.75"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data \n",
"\n",
"VCC = 10.0\n",
"VBE = 0.75\n",
"R1 = 4.7*10**3\n",
"\n",
"# solution \n",
"# I is approximately equals to Ic(2 + (1/beta)), since 1/beta is a small value its considerd as 2Ics \n",
"I = (VCC - VBE)/R1\n",
"# IE3 is approxiamtly equal to IC3 = I/2\n",
"#finding the value of IC \n",
"IC = I /2\n",
"print \" The value of I =\",round(I*10**3,2),\"mA\"\n",
"print \" The value of IC =\",round(IC*10**3,2),\"mA\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of I = 1.97 mA\n",
" The value of IC = 0.98 mA\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.16 Page No.80"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"VCC = 15.0\n",
"VBE = 0.7\n",
"R = 10*10**3\n",
"RC2 = 5*10**3\n",
"# consider beta is a large value \n",
"I = (VCC - VBE)/R\n",
"IC1 = IC2 = I\n",
"#V12 = V1 - V2\n",
"V12 = VBE + IC2 * RC2 \n",
"\n",
"# displaying the values\n",
"print \"The value of I =\",round(I*10**3,2),\"mA\"\n",
"print \"The value of V1 - V2 =\",V12,\"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of I = 1.43 mA\n",
"The value of V1 - V2 = 7.85 V\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.17 Page No.83"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Given data\n",
"\n",
"VEE = 6.0\n",
"VCC = 6.0\n",
"VD = VD1 = VD2 = VD3 = VD4 = 0.7\n",
"R5 = 3.2*10**3\n",
"R4 = 1.5*10**3\n",
"VBE1 = VBE2 =VBE3 = VBE4 = VBE5 = VBE6 = VBE7 = VBE8 = 0.7\n",
"R1 = 2.2*10**3\n",
"R2 = 7.75*10**3\n",
"R3 = 5.2*10**3\n",
"R6 = 1.5*10**3\n",
"R7 = 3*10**3\n",
"R8 = 3.4*10**3\n",
"R9 = 6*10**3\n",
"R10 =30*10**3\n",
"hfe = 100\n",
"VT = 26*10**-3\n",
"\n",
"# Solution for part A\n",
"\n",
"VBN1 = round((-VEE+VD+VD)*R5/(R4+R5),2)\n",
"I1 = (VEE+VBN1-VBE1)/R1\n",
"IQ = I1 # If the base current of Q1 is neglected\n",
"IC2 = IC3 = IQ2 = round(IQ/2,6)\n",
"VC2 = VC3 = round((VCC - R2*IC2),2)\n",
"VE4 = VC2 - VBE4 \n",
"I6 = round(VE4/R6,6)\n",
"IC4 = IC5 = round((I6/2),6)\n",
"VC5 = round(VCC - IC5*R7,2)\n",
"VE6 = round(VC5 - VBE6,2)\n",
"IC7 = I8 = round((VEE - VD3)/R8,5)\n",
"VB8 = VBE8 + VD4 - VEE\n",
"I9 = round((VE6 - VB8)/R9,5)\n",
"I10 = IC7 - I9\n",
"Vo = (I10 * R10) + VB8\n",
"V0 = int(Vo) # Vo value is assumed to be very small\n",
"\n",
"# Displaying the values \n",
"print \"PART A\"\n",
"print \"==================\"\n",
"print \"the value of VBN1 =\",VBN1,\"V\"\n",
"print \"The value of I1 =\",I1*10**3,\"mA\"\n",
"print \"The value of IQ =\",IQ*10**3,\"mA\"\n",
"print \"The value of IC2 =\",IC2*10**3,\"mA\"\n",
"print \"The value of VC3 =\",VC3,\"V\"\n",
"print \"The value of VE4 =\",VE4,\"V\"\n",
"print \"The value of IC4 =\",IC4*10**3,\"mA\"\n",
"print \"The value of VC5 =\",VC5,\"V\"\n",
"print \"The value of VE6 =\",VE6,\"V\"\n",
"print \"The value of IC7 =\",IC7*10**3,\"mA\"\n",
"print \"The value of VB8 =\",VB8,\"V\"\n",
"print \"The value of I9 =\",I9*10**3,\"mA\"\n",
"print \"The value of I10 =\",I10*10**3,\"mA\"\n",
"print \"The value of V0 =\",V0,\"V\"\n",
"\n",
"# Solution for part B\n",
"VT = 26*10**-3 #the volt equivalent of temperature \n",
"IC2 = IC3 = IC4 = IC5 = IC = 0.5*10**-3\n",
"hie = (hfe*VT)/IC\n",
"RL2 = RL3 = (R2*R3)/(R2+R3)\n",
"AV1 = round((hfe*RL2)/hie,0)\n",
"AV2 = round(-(hfe*R7)/(2*hie),2)\n",
"AV3 = 1 # the thrid stage emitter follower differential gain \n",
"AV4 = int(-R10/R9)\n",
"AV = AV1*round(AV2,1)*AV4 # again rounding AV2 to get the desired value\n",
"\n",
"# Displaying the values\n",
"print \"PART B\"\n",
"print \"==================\"\n",
"print \"The value of hie =\",int(hie)\n",
"print \"The value of RL2 =\",RL2*10**-3,\"Kilo Ohms\"\n",
"print \"The value of AV1 =\",int(AV1)\n",
"print \"The value of AV2 =\",round(AV2,1)\n",
"print \"The value of AV3 =\",AV3\n",
"print \"The value of AV4 =\",AV4\n",
"print \"The value of AV =\",int(AV)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"PART A\n",
"==================\n",
"the value of VBN1 = -3.13 V\n",
"The value of I1 = 0.986363636364 mA\n",
"The value of IQ = 0.986363636364 mA\n",
"The value of IC2 = 0.493 mA\n",
"The value of VC3 = 2.18 V\n",
"The value of VE4 = 1.48 V\n",
"The value of IC4 = 0.494 mA\n",
"The value of VC5 = 4.52 V\n",
"The value of VE6 = 3.82 V\n",
"The value of IC7 = 1.56 mA\n",
"The value of VB8 = -4.6 V\n",
"The value of I9 = 1.4 mA\n",
"The value of I10 = 0.16 mA\n",
"The value of V0 = 0 V\n",
"PART B\n",
"==================\n",
"The value of hie = 5200\n",
"The value of RL2 = 3.11196911197 Kilo Ohms\n",
"The value of AV1 = 60\n",
"The value of AV2 = -28.9\n",
"The value of AV3 = 1\n",
"The value of AV4 = -5\n",
"The value of AV = 8670\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.18 Page No.87"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data \n",
"VEE = 15.0\n",
"VCC = 15.0\n",
"VBE1 = VBE2 = VBE3 = VBE4 = VBE5 = VBE= 0.7\n",
"R = 28.6*10**3\n",
"RC1 = 20*10**3\n",
"RC6 = 3*10**3\n",
"RC8 = 2.3*10**3\n",
"RA = 15.7*10**3\n",
"hfe = 100\n",
"VT = 25*10**-3\n",
"\n",
"# Solution for part 1 DC - ANALYSIS\n",
"\n",
"I = (VEE - VBE3)/R\n",
"ICQ4 = I\n",
"ICQ1 = ICQ2 = ICQ4/2\n",
"# Q1 and Q2 are biased at 0.25mA so their Collector voltages are\n",
"IC1 = IC = ICQ1\n",
"VCQ1 = VCQ2 = VCC - (IC1*RC1)\n",
"VEQ5 = VEQ6 = VCQ1 - VBE\n",
"IQ7 = 4*I\n",
"# The collector current of Q5 and Q6 are \n",
"ICQ5 = ICQ6 = IQ7/2\n",
"VCQ6 = VCC - (ICQ6 * RC6)\n",
"VEQ8 = VCQ6 + VBE\n",
"IEQ8 = (VCC - VEQ8)/RC8\n",
"# The voltage VA at the collector of Q8 is \n",
"VA = -VCC + (IEQ8 * RA)\n",
"IEQ9 = (0-(-VCC))/RC6\n",
"\n",
"# Displaying all values of Part 1\n",
"\n",
"print \"The value of I =\",int(I*10**3),\"mA\"\n",
"print \"The value of ICQ1 =\",int(ICQ1*10**3),\"mA\"\n",
"print \"The value of VCQ1 =\",int(VCQ1),\"V\"\n",
"print \"The value of VEQ5 =\",int(VEQ5),\"V\"\n",
"print \"The value of IQ7 =\",int(IQ7*10**3),\"mA\"\n",
"print \"The value of ICQ5 =\",int(ICQ5*10**3),\"mA\"\n",
"print \"The value of VCQ6 =\",int(VCQ6),\"V\"\n",
"print \"The value of VEQ8 =\",int(VEQ8),\"V\"\n",
"print \"The value of IEQ8 =\",int(IEQ8*10**3),\"mA\"\n",
"print \"The value of VA =\",VA,\"V\"\n",
"print \"The value of IEQ9 =\",int(IEQ9*10**3),\"mA\"\n",
"\n",
"# Part 2 AC -ANALYSIS\n",
"\n",
"hieQ12 = (hfe * VT)/IC\n",
"# The ac emitter resistance of transister Q5 - Q6\n",
"hieQ56 = (hfe * VT)/ICQ5\n",
"RL1 = RL2 = (RC1*hieQ56)/(RC1 + hieQ56) # parallel combination of resistor RC1 and hieQ56\n",
"ADM1 = (hfe * RL2)/hieQ12\n",
"ADM2 = -(hfe * RC6)/(2*hieQ56)\n",
"\n",
"\n",
"# Displaying the values \n",
"\n",
"print \"The value of ac emiiter resistace of the transistor Q1-Q2 = \",int(round(hieQ12*10**-3,0)),\"Kilo Ohm\"\n",
"print \"The value of ac emiiter resistace of the transistor Q5-Q6 = \",round(hieQ56*10**-3,1),\"Kilo Ohm\"\n",
"print \"The value of effective load of Q1-Q2 = \",round(RL1*10**-3,1),\"Kilo Ohm\"\n",
"print \"The value of voltage gain of first differential pair = \",int(round(ADM1,0)),\"Kilo Ohm\"\n",
"print \"The value of voltage gain of second differential pair = \",int(ADM2),\"Kilo Ohm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of I = 0 mA\n",
"The value of ICQ1 = 0 mA\n",
"The value of VCQ1 = 10 V\n",
"The value of VEQ5 = 9 V\n",
"The value of IQ7 = 2 mA\n",
"The value of ICQ5 = 1 mA\n",
"The value of VCQ6 = 12 V\n",
"The value of VEQ8 = 12 V\n",
"The value of IEQ8 = 1 mA\n",
"The value of VA = 0.7 V\n",
"The value of IEQ9 = 5 mA\n",
"The value of ac emiiter resistace of the transistor Q1-Q2 = 10 Kilo Ohm\n",
"The value of ac emiiter resistace of the transistor Q5-Q6 = 2.5 Kilo Ohm\n",
"The value of effective load of Q1-Q2 = 2.2 Kilo Ohm\n",
"The value of voltage gain of first differential pair = 22 Kilo Ohm\n",
"The value of voltage gain of second differential pair = -60 Kilo Ohm\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}