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{
"metadata": {
"name": "",
"signature": "sha256:3942a48c51f6e66ddeb010a1a5acaeebd4c9f56b964a269d92588d67ccc10453"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2 : Introduction to operational amplifier"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex 2.1 Page no. 88"
]
},
{
"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": [
"Ex 2.2 Page no.89"
]
},
{
"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),\"kohms\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of feedback resistor = 40 kohms\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex 2.4 Page no.92"
]
},
{
"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": [
"Ex 2.5 Page No.94"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given Data\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)),\"kohm\" #converting the answer into kohm"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of gm = 4 mMho\n",
" The value of Rc = 25 kohm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex 2.6 Page no. 95"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# 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)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ACM is = -0.05\n"
]
}
],
"prompt_number": 10
},
{
"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": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex 2.8 Page no.97"
]
},
{
"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
}
],
"metadata": {}
}
]
}
|
