{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 6 Applications of Operational Amplifier"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.1 Pg 140"
]
},
{
"cell_type": "code",
"execution_count": 116,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the input resistance is = 20.00 kohm\n",
"The resistance R2 is = 100.00 kohm\n"
]
}
],
"source": [
"# Design an inverting amplifier\n",
"Av = -5 #\n",
"#V1 = 0.1 sin wt #\n",
"V1 = 0.1 # # *sin wt #\n",
"i = 5*10**-6 #\n",
"\n",
"# the input resistance \n",
"R1 = V1/i / 1000 # kohm\n",
"print 'the input resistance is = %0.2f'%R1,'kohm'#\n",
"\n",
"# The resistance R2\n",
"#Av = -(R2/R1)#\n",
"R2 = -(Av*R1)#\n",
"print 'The resistance R2 is = %0.2f'%R2,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.2 Pg 141"
]
},
{
"cell_type": "code",
"execution_count": 117,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the input resistance is = 20.00 kohm\n",
"The resistance R2 is = 80.00 kohm\n"
]
}
],
"source": [
"# Design an non inverting amplifier\n",
"Av = 5 #\n",
"#V1 = 0.1 sin wt #\n",
"V1 = 0.1 #\n",
"i = -5*10**-6 #\n",
"\n",
"# the input resistance \n",
"R1 = -V1/i/1000 # kohm\n",
"print 'the input resistance is = %0.2f'%R1,'kohm'#\n",
"\n",
"# The resistance R2\n",
"#Av = 1+(R2/R1)#\n",
"R2 = (Av-1)*R1#\n",
"print 'The resistance R2 is = %0.2f'%R2,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.3 Pg 146"
]
},
{
"cell_type": "code",
"execution_count": 118,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the cut off frequency of phase shifter is = 723.43 Hz\n",
"The phase shift is = -15.75\n"
]
}
],
"source": [
"# To calculate phase shift between two extremes \n",
"C = 0.22*10**-6 #\n",
"R = 1*10**3 #\n",
"f = 1*10**3 #\n",
"\n",
"# the cut off frequency of phase shifter \n",
"fc = 1/(2*pi*R*C) #\n",
"print 'the cut off frequency of phase shifter is = %0.2f'%fc,'Hz'#\n",
"f\n",
"# the phase shift\n",
"f = 1 # # KHz\n",
"fc = 7.23 # # KHz \n",
"from math import atan ,degrees\n",
"PS = -2*degrees(atan(f/fc))\n",
"print 'The phase shift is = %0.2f'%PS"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.4 Pg 146"
]
},
{
"cell_type": "code",
"execution_count": 119,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of resistance is = 1.92 kohm\n"
]
}
],
"source": [
"# To design a phase shifter\n",
"f = 2*10**3 #\n",
"PS = -135 #\n",
"# the phase shift\n",
"# PS = -2*atand(2*pi*R*C)#\n",
"#RC = 192.1*10**-6 #\n",
"C = 0.1*10**-6 #\n",
"R = (192.1*10**-6)/C/1000 # kohm\n",
"print 'The value of resistance is = %0.2f'%R,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.5 Pg 153"
]
},
{
"cell_type": "code",
"execution_count": 120,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of resistance R1 is = 25.00 kohm\n",
"The value of resistance R3 is = 25.00 kohm\n",
"The value of resistance R2 is = 750.00 kohm\n",
"The value of resistance R4 is = 750.00 kohm\n"
]
}
],
"source": [
"# Design a difference amplifier\n",
"Ri = 50 # kohm\n",
"Ad = 30 \n",
"\n",
"R1 = Ri/2 #\n",
"print 'The value of resistance R1 is = %0.2f'%R1,'kohm'#\n",
"R3 = R1 #\n",
"print 'The value of resistance R3 is = %0.2f'%R3,'kohm'#\n",
"\n",
"# the differential gain\n",
"#Ad = R2/R1 #\n",
"R2 = 30*R1 #\n",
"print 'The value of resistance R2 is = %0.2f'%R2,'kohm'#\n",
"\n",
"R4 = R2 #\n",
"print 'The value of resistance R4 is = %0.2f'%R4,'kohm'# "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.6 Pg 154"
]
},
{
"cell_type": "code",
"execution_count": 121,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The common mode rejection ratio is = 26.58 dB\n"
]
}
],
"source": [
"# Calculate CMRR ratio\n",
"Ad = 10.24 #\n",
"Acm = 0.48 #\n",
"\n",
"# the common mode rejection ratio CMRR is defined as\n",
"CMRRdB = 20*log10(Ad/Acm)#\n",
"print 'The common mode rejection ratio is = %0.2f'%CMRRdB,' dB'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.7 Pg 156"
]
},
{
"cell_type": "code",
"execution_count": 122,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The feedback resistance is = 100.00 kohm\n",
" The value of resistance R1 is = 100.00 kohm\n"
]
}
],
"source": [
"# Design current to voltage converter\n",
"Vo =-10 #\n",
"Is = 100*10**-6 #\n",
"\n",
"# the output voltage of current to voltage converter is defined as\n",
"#Vo =-1s*R2 \n",
"R2 = -Vo/Is/1000 #kohm\n",
"print ' The feedback resistance is = %0.2f'%R2,'kohm'#\n",
"\n",
"R1 = R2 #\n",
"print ' The value of resistance R1 is = %0.2f'%R1,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.8 Pg 157"
]
},
{
"cell_type": "code",
"execution_count": 123,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of resistance R2 is = 44.78 kohm\n"
]
}
],
"source": [
"# Design high sensitivity current to voltage converter\n",
"R1 = 5 # kohm\n",
"Is = 1 #\n",
"KR = 0.01/10**9 # # V / nA\n",
"\n",
"# the output voltage of high sensitivity current to voltage converter\n",
"Vo =-KR*Is #\n",
"KR = 10*10**6 #\n",
"R = 1*10**6 # #we assume then\n",
"K = 10 #\n",
"#1 + (R2/R1)+(R2/R) = 10 #\n",
"# solving above equation we get\n",
"\n",
"R2 = 9*((5*10**6)/(10**3+5))/1000 # kohm\n",
"print 'The value of resistance R2 is = %0.2f'%R2,'kohm'# "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.9 Pg 160"
]
},
{
"cell_type": "code",
"execution_count": 124,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The load current iL is = 5.00 mA\n",
"The current i3 is = 0.50 mA\n",
"The current iA is = 5.50 mA\n",
"The output voltage is = 6.00 V\n",
"The current i1 is = 49.40 A\n",
"The current i2 is = 49.40 A\n"
]
}
],
"source": [
"# Determine a load current in a V to I converter\n",
"R1 = 10 # kohm\n",
"R2 = 10 # koohm\n",
"R3 = 1 # kohm\n",
"R4 = 1 # kohm\n",
"VI = -5 #\n",
"\n",
"# The Load Current\n",
"iL = -VI/R3 #\n",
"print 'The load current iL is = %0.2f'%iL,'mA'#\n",
"\n",
"VL = 0.5 #\n",
"# The Current i3 and iA\n",
"i3 = VL/R3 #\n",
"print 'The current i3 is = %0.2f'%i3,'mA'#\n",
"\n",
"iA = i3+iL #\n",
"print 'The current iA is = %0.2f'%iA,'mA'#\n",
"\n",
"# the output voltage \n",
"Vo = (iA*R3)+VL #\n",
"print 'The output voltage is = %0.2f'%Vo,' V'#\n",
"\n",
"ZL =100 #\n",
"# The current i1 and i2 \n",
"#i1 = (VI-iL*ZL)/R1 #\n",
"i1 = (iL*ZL-Vo)/R2 #\n",
"print 'The current i1 is = %0.2f'%i1,' A'#\n",
"\n",
"i2 = i1 #\n",
"print 'The current i2 is = %0.2f'%i2,' A'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.10 Pg 163"
]
},
{
"cell_type": "code",
"execution_count": 125,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of resistance R1f is = 0.0606 K ohm \n",
"The value of resistance R2 is = 75.5 K ohm \n"
]
}
],
"source": [
"# Design an instrumentation amplifier\n",
"#A = 5 to 500 # adjustable gain\n",
"VR = 100*10**3 #\n",
"\n",
"# the maximum differential gain of instrumentation amplifier is 500 \n",
"#Amax = (R4/R3)*(1+(2R2/R1))#\n",
"#by solving above equation we get following equation\n",
"# 2R2 -249R1f = 0 equation 1\n",
"\n",
"# the minimum differential gain of instrumentation amplifier is 5\n",
"# Amin = (R4/R3)*(1+(2R2/R1)) #\n",
"#by solving above equation we get following equation\n",
"# 2R2 -1.5R1f = 150*10**3 equation 2\n",
"\n",
"#by solving equation 1 and 2 we get\n",
"print 'The value of resistance R1f is = 0.0606 K ohm '#\n",
"\n",
"print 'The value of resistance R2 is = 75.5 K ohm '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.11 Pg 164"
]
},
{
"cell_type": "code",
"execution_count": 127,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of resistane R1 is = 9.09 kohm\n"
]
}
],
"source": [
" # To find the value of resistance R1 for instrumentation amplifier\n",
"A =100 #\n",
"R2 = 450*10**3 #\n",
"R3 = 1*10**3 #\n",
"R4 = 1*10**3 #\n",
"\n",
"# The gain of differential amplifier \n",
"# A = (R4/R3)*(1+(2R2/R1)) #\n",
"#but R3 = R4 then\n",
"# A = 1+(2R2/R1) #\n",
"R1 = 2*R2/(A-1)/1000 # kohm\n",
"print 'The value of resistane R1 is = %0.2f'%R1,'kohm'# "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.12 Pg 167"
]
},
{
"cell_type": "code",
"execution_count": 130,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" At t = 1 msec the time constant RC is = 0.10 m sec\n",
" if C = 0.01 uF then R of RC time constant is = 10 K ohm \n",
" if C = 0.001 uF then R of RC time constant is = 100 K ohm \n"
]
}
],
"source": [
"# determine the time constant of an integrator\n",
"Vo = 10 # # at t= 1 m sec\n",
"t = 1 # # m sec\n",
"\n",
"# the output of integrator \n",
"#Vo = t/RC # when t is from 0 to 1\n",
"RC = t/Vo #\n",
"print ' At t = 1 msec the time constant RC is = %0.2f'%RC,' m sec'#\n",
"\n",
"print ' if C = 0.01 uF then R of RC time constant is = 10 K ohm '#\n",
"\n",
"print ' if C = 0.001 uF then R of RC time constant is = 100 K ohm '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.13 Pg 168"
]
},
{
"cell_type": "code",
"execution_count": 131,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The feedback resistance Rf is = 100.00 kohm\n",
" The frequency fa is = 2.00 kHz\n",
" The value of capacitor C is = 0.8 nF \n"
]
}
],
"source": [
"# Design an integrator circuit\n",
"A = 10 #\n",
"f =20*10**3 #\n",
"R = 10*10**3 # # we assume \n",
"Rf =10*R #\n",
"\n",
"print ' The feedback resistance Rf is = %0.2f'%(Rf/1000),'kohm'#\n",
"\n",
"# for proper integration f>= 10fa \n",
"fa = f/10/1000 #\n",
"print ' The frequency fa is = %0.2f'%fa,'kHz'#\n",
"\n",
"# in practical integrator\n",
"#fa = 1/(2*pi*Rf*C)#\n",
"\n",
"C = 1/(2*pi*Rf*fa)*1e6# nF\n",
"print ' The value of capacitor C is = %0.1f'%C,'nF '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.1 Pg 185"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance R1 is = 0.199 Mohm\n",
"the value of resistance R2 is = 1 Mohm\n"
]
}
],
"source": [
"from __future__ import division\n",
"# design an inverting amplifier with a closed loop voltage gain of Av = -5\n",
"Av = -5 #\n",
"Is = 5*10**-6 # # A\n",
"Rs = 1*10**3 # # ohm\n",
"# input voltage source Vs = sinwt volts\n",
"\n",
"# in an inverting amplifier frequency effect is neglected then i/p volt Vin = 1 V and total resistance equal to Rs+R1\n",
"\n",
"# the input current can be written as Iin=Is\n",
"# Is = (Vin/Rs+R1)#\n",
"Iin = Is#\n",
"Vin = 1 # # V\n",
"R1 = (1-(Iin*Rs))/Iin #\n",
"print 'the value of resistance R1 is = %0.3f'%(R1/1e6),'Mohm'#\n",
"\n",
"# closed loop voltage gain of an inverting amplifier\n",
"#Av = -(R2/Rs+R1)\n",
"R2 = -(Av*(Rs+R1))#\n",
"print 'the value of resistance R2 is = %0.f'%(R2/1e6),'Mohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.2 Pg 186"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance R1 is = 8 kohm\n",
"the value of resistance R2 is = 72 kohm\n"
]
}
],
"source": [
" # design an inverting amplifier with a closed loop voltage gain of Av = 10\n",
"Av = 10 #\n",
"Vin = 0.8 # #V\n",
"Iin = 100*10**-6 # # A\n",
"# in an non- inverting amplifier the input voltage Vin=V1=V2 because of vortual short effect then the i/p current In = Vin/R1\n",
"R1 = Vin/Iin/1e3\n",
"print 'the value of resistance R1 is = %0.f'%R1,'kohm'#\n",
"\n",
"# closed loop voltage gain of an non-inverting amplifier\n",
"#Av = Vo/Vin = (1+R2/R1)\n",
"R2 = (Av-1)*R1 # kohm\n",
"print 'the value of resistance R2 is = %0.f'%R2,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.3 Pg 187"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance R1 is = 20.00 kohm\n",
"the value of resistance R2 is = 80.00 kohm\n",
"the output current I2 is = 50 uA\n"
]
}
],
"source": [
"# design an non-inverting amplifier with colsed loop gain of 5 limited voltage of -5 V <= Vo <= 5 V and maximum i/p c/n 50 uA\n",
"R1 = 8*10**3 # # ohm\n",
"R2 = 72*10**3 # # ohm\n",
"Iin = 50*10**-6 # # A\n",
"Vo = 5 # # V \n",
"\n",
"# closed loop gain\n",
"#Av = Vo/Vin = (1+R2/R1)\n",
"Av = 1+(R2/R1)#\n",
"# but \n",
"Av = 5 #\n",
"# then\n",
"# (R2/R1) = 4 #\n",
"\n",
"# the output voltage of the amplifier is Vo = 5 V \n",
"#i.e\n",
"Vin = 1 # # V\n",
"# Iin = Vin/R1 #\n",
"R1 = Vin/Iin/1e3\n",
"print 'the value of resistance R1 is = %0.2f'%R1,'kohm'#\n",
"\n",
"R2 = 4*R1 #\n",
"print 'the value of resistance R2 is = %0.2f'%R2,'kohm'#\n",
"\n",
"# the output current I2 is given as\n",
"I2 = (Vo-Vin)/R2*1e3 # uA\n",
"print 'the output current I2 is = %0.f'%I2,'uA'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.4 Pg 188"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance Ria is = 20.00 kohm\n",
"the value of resistance Rib is = 15.00 kohm\n",
"the value of resistance Ric is = 30.00 kohm\n"
]
}
],
"source": [
"# Design a op-amp circuit to provide the output voltage Vo = -2(3 V1 +4 V2 +2 V3)\n",
"# Vo = -2(3 V1 + 4 V2+ 2 V3)# equation 1\n",
"# the output of the summer circuit is given as\n",
"# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric)) equation 2\n",
"\n",
"# compare equation 1 and 2 of Vo we get \n",
"\n",
"# (R2/Ria)= 6 #\n",
"# (R2/Rbi=8 #\n",
"# (R2/Ric)=4 #\n",
"\n",
"R2 = 120*10**3/1e3 # # we choose then \n",
"\n",
"Ria = R2/6 #\n",
"print 'the value of resistance Ria is = %0.2f'%Ria,'kohm'#\n",
"\n",
"Rib = R2/8 #\n",
"print 'the value of resistance Rib is = %0.2f'%Rib,'kohm'#\n",
"\n",
"Ric = R2/4 #\n",
"print 'the value of resistance Ric is = %0.2f'%Ric,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.5 Pg 188"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance Ria is = 90.00 kohm\n",
"the value of resistance Rib is = 42.00 kohm\n",
"the value of resistance Ric is = 63.00 kohm\n",
"the value of resistance Rid is = 210.00 kohm\n"
]
}
],
"source": [
" # Design a summing amplifier circuit to provide the output voltage Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)\n",
"R2 = 630# kohm # Assume feedback resistance\n",
"# Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)# equation 1\n",
"# the output of the summer circuit is given as\n",
"# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric)+(Vid/Rid)) equation 2\n",
"\n",
"# compare equation 1 and 2 of Vo we get \n",
"\n",
"# (R2/Ria)= 7 #\n",
"# (R2/Rbi= 15 #\n",
"# (R2/Ric)= 10 #\n",
"# (R2/Rid)= 3 #\n",
"\n",
"Ria = R2/7 #\n",
"print 'the value of resistance Ria is = %0.2f'%Ria,' kohm'#\n",
"\n",
"Rib = R2/15 #\n",
"print 'the value of resistance Rib is = %0.2f'%Rib,' kohm'#\n",
"\n",
"Ric = R2/10 #\n",
"print 'the value of resistance Ric is = %0.2f'%Ric,' kohm'#\n",
"\n",
"Rid = R2/3 #\n",
"print 'the value of resistance Rid is = %0.2f'%Rid,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.6 Pg 190"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance R2 is = 300.00 kohm\n",
"the value of resistance R4 is = 33333.33 kohm\n"
]
}
],
"source": [
"# Design a op-amp circuit to provide the output voltage Vo = V2 - 3 V1 with Ri1 =Ri2 = 100*10**3\n",
"Ri1 = 100 # # kohm\n",
"Ri2 = 100 # # kohm\n",
"# the i/p resistance \n",
"R1 = Ri1 #\n",
"R3 = Ri2 #\n",
"\n",
"# Vo = V2 - 3 V1# equation 1\n",
"# the output of the summer circuit is given as\n",
"# Vo = [(R4/(R3+R4)*(1+(R2/R1))*Vi2-(R2/R1)*Vi1] equation 2\n",
"\n",
"# compare equation 1 and 2 of Vo we get \n",
"# (R4/(R3+R4)*(1+(R2/R1)) = 1 # equation 3\n",
"# R2/R1 = 3 # equation 4\n",
"\n",
"# by subsituting the value of R1 and R3 in equation 3 and 4\n",
"\n",
"# from equation 4\n",
"R2 = 3*R1 #\n",
"print 'the value of resistance R2 is = %0.2f'%R2,'kohm'#\n",
"\n",
"# from equation 3\n",
"R4 = (100*10**3)/3 #\n",
"print 'the value of resistance R4 is = %0.2f'%R4,'kohm'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.7 Pg 191"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The load current iL is = 5 mA\n",
"The voltage across load VL is = 1.00 V\n",
"The non-inverting current across i3 is = 1.00 mA\n",
"The non-inverting current across i4 is = 6.00 mA\n",
"The output voltage of given voltage to current converter is = 6.00 V\n"
]
}
],
"source": [
" # determine the load current and output voltage\n",
"Vin = -5 # # V\n",
"ZL = 200 # # ohm\n",
"R1 = 10*10**3 # # ohm\n",
"R2 = 10*10**3 # # ohm\n",
"R3 = 1*10**3 # # ohm\n",
"R4 = 1*10**3 # # ohm\n",
"\n",
"# the load c/n of the given voltage to c/n converter circuit is given by\n",
"iL =-Vin/(R1*R4)*R2*1e3 # m\n",
"print 'The load current iL is = %0.f'%iL,'mA'#\n",
"\n",
"# the voltage across the load \n",
"VL = iL/1e3*ZL#\n",
"print 'The voltage across load VL is = %0.2f'%VL,' V'#\n",
"\n",
"# the non-inverting current across i3 and i4 are\n",
"i3 = VL/R3*1000 #mA\n",
"print 'The non-inverting current across i3 is = %0.2f'%i3,'mA'#\n",
"\n",
"i4 = iL+i3 # mA\n",
"print 'The non-inverting current across i4 is = %0.2f'%i4,'mA'#\n",
"\n",
"# the output voltage of given voltage to current converter is given by\n",
"Vo = (iL/1e3*R3)+VL #\n",
"print 'The output voltage of given voltage to current converter is = %0.2f'%Vo,' V'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.8 Pg 192"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The common mode rejection ratio CMRR is = 120.50 \n",
"The common mode rejection ratio CMRR in decibel is = 41.62 dB \n"
]
}
],
"source": [
"from math import log10\n",
"# determine the common mode rejection ratio CMRR\n",
"# R2/R1 = 10 #\n",
"# R4/R3 = 11 #\n",
"\n",
"# the output of the difference amplifier is given by\n",
"# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
"\n",
"# putting R1 R2 R3 R4 value in above equation we get Vo as\n",
"\n",
"# Vo =(121/12)*VI2-10VI1 # equation 1\n",
"\n",
"# the differential mode input of difference amplifier is given by\n",
"# Vd = VI2-VI1 # eqution 2\n",
"\n",
"# the common mode input of difference amplifier is given by\n",
"# VCM = (VI2+VI1)/2 # equation 3\n",
"\n",
"# from equation 2 and 3 \n",
"\n",
"# VI1 = VCM-Vd/2 # equation 4\n",
"\n",
"# VI2 = VCM+Vd/2 # equation 5\n",
"\n",
"# substitute equation 4 and 5 in 1 we get \n",
"# Vo = (VCM/12)+(241Vd/24)# equation6\n",
"\n",
"# Vd = Ad*Vd+ACM*VCM # equation 7\n",
"\n",
"#equation from equation 6 and 7 we get\n",
"\n",
"Ad = 241/24 #\n",
"ACM = 1/12 #\n",
"\n",
"# the common mode rejection ratio CMRR is \n",
"CMRR = abs(Ad/ACM)#\n",
"print 'The common mode rejection ratio CMRR is = %0.2f'%CMRR,' '#\n",
"\n",
"# in decibal it can be expressed as\n",
"\n",
"CMRR = 20*log10(CMRR)#\n",
"print 'The common mode rejection ratio CMRR in decibel is = %0.2f'%CMRR,' dB '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.9 Pg 194"
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The output of the difference amplifier is = -8.12 V \n",
"The output of the difference amplifier is = 0.12 V \n",
"the common mode input of difference amplifier is = 2.00 \n",
"the common mode gain ACM of difference amplifier is = 0.06 \n",
"the differential gain of the difference amplifier is = 2.00 \n",
"The common mode rejection ratio CMRR is = 32.00 \n",
"The common mode rejection ratio CMRR in decibel is = 30.10 dB \n"
]
}
],
"source": [
"# determine Vo when 1) VI1 = 2 V VI2 = -2 V and 2) VI1 = 2 V VI2 = 2 V\n",
"# and common mode rejection ratio CMRR\n",
"R1 = 10*10**3 # # ohm\n",
"R2 = 20*10**3 # # ohm\n",
"R3 = 10*10**3 # # ohm\n",
"R4 = 22*10**3 # # ohm\n",
"\n",
"\n",
"# the output of the difference amplifier is given by\n",
"# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
"\n",
"# Case 1 when VI1 = 2 V VI2 = -2 V\n",
"VI1 = 2 #\n",
"VI2 = -2 #\n",
"\n",
"Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
"print 'The output of the difference amplifier is = %0.2f'%Vo,' V '#\n",
"\n",
"# case 2 when VI1 = 2 V VI2 = 2 V\n",
"VI1 = 2 #\n",
"VI2 = 2 #\n",
"\n",
"Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n",
"print 'The output of the difference amplifier is = %0.2f'%Vo,' V '#\n",
"\n",
"# the common mode input of difference amplifier is given by\n",
"VCM = (VI2+VI1)/2 #\n",
"print 'the common mode input of difference amplifier is = %0.2f'%VCM,' '#\n",
"\n",
"# the common mode gain ACM of difference amplifier is given by\n",
"ACM = Vo/VCM\n",
"print 'the common mode gain ACM of difference amplifier is = %0.2f'%ACM,' '#\n",
"\n",
"# the differential gain of the difference amplifier is given \n",
"Ad = R2/R1 # \n",
"print 'the differential gain of the difference amplifier is = %0.2f'%Ad,' '#\n",
"\n",
"# the common mode rejection ratio CMRR is \n",
"CMRR = abs(Ad/ACM)#\n",
"print 'The common mode rejection ratio CMRR is = %0.2f'%CMRR,' '#\n",
"\n",
"# in decibal it can be expressed as\n",
"CMRR = 20*log10(CMRR)#\n",
"print 'The common mode rejection ratio CMRR in decibel is = %0.2f'%CMRR,' dB '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.10 Pg 195"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the maximum differential voltage gain of the instrumentation amplifier is = 101.00 \n",
"the minimum differential voltage gain of the instrumentation amplifier is = 5.00 \n",
" the range of the differential voltage gain of the instrumentation amplifier is \n",
" 5 <= Av <= 101 \n"
]
}
],
"source": [
"# To determine the range of the differential voltage gain\n",
"#R1 = 1 K ohm to 25 K ohm #\n",
"R2 = 50 # # K ohm\n",
"R3 = 10 # # K ohm\n",
"R4 = 10 # # K ohm\n",
"\n",
"# the output of instrumentation amplifier is given by\n",
"#Vo = (R4/R3)*(1+(2*R2/R1))*(VI@-VI1)#\n",
"\n",
"# the differential voltage gain of the instrumentation amplifier can be written as\n",
"#Av = (Vo/(VI2-VI1)) = (R4/R3)*(1+(2R2/R1))#\n",
"\n",
"# For R1 = 1 K ohm the maximum differential voltage gain of the instrumentation amplifier is\n",
"R1 = 1 # # K ohm\n",
"Av = (R4/R3)*(1+(2*R2/R1))#\n",
"print 'the maximum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n",
"\n",
"# For R1 = 25 K ohm the mminimum differential voltage gain of the instrumentation amplifier is\n",
"R1 = 25 # # K ohm\n",
"Av = (R4/R3)*(1+(2*R2/R1))#\n",
"print 'the minimum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n",
"\n",
"print ' the range of the differential voltage gain of the instrumentation amplifier is '#\n",
"print ' 5 <= Av <= 101 '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.11 Pg 196"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The variable resistance R1 varies is 401 ohm <= R1 <= 100.2 K-ohm \n"
]
}
],
"source": [
"# To design an instrumentation amplifier\n",
"# 4 <= Av <= 1000 # gain\n",
"Ad = 2 #\n",
"Res = 100 # # K ohm\n",
"\n",
"# we cosider the variable resistance is R1 , the maximum and the minimum range of variable resistance \n",
"# R1min = R1 # \n",
"# R1max = R1+100 #\n",
"\n",
"# the gain of difference amplifier \n",
"#A3 = Ad = Vo/(Vo2-Vo1) = (R4/R3)\n",
"\n",
"# the maximum range of differential voltage gain Avmax = 1000 when R1min = R1\n",
"#Avmax = R4/R3*(1+(2*R2/R1min))#\n",
"\n",
"# by solvin we get following equation\n",
"# 499*R1-2*R2=0 equation 1\n",
"\n",
"# the maximum range of differential voltage gain Avmin =4 when R1max = R1+100 K ohm\n",
"# Avmin = (R4/R3)*(1+(2R2/R1max))#\n",
"\n",
"# by solving above equation we get\n",
"# R1 -2 R2 = -200 K ohm equation 2\n",
"\n",
"#by solving equation 1 and 2 we get\n",
"R1 = 401 # # ohm\n",
"R2 = 100.2 # # ohm\n",
"print 'The variable resistance R1 varies is 401 ohm <= R1 <= 100.2 K-ohm ' #"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.12 Pg 198"
]
},
{
"cell_type": "code",
"execution_count": 41,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The time constant of the given filter is RC = 0.20 msec \n"
]
}
],
"source": [
" # Determine the time constant of the integrator\n",
"Vo = 10 #\n",
"t = 2*10**-3 #\n",
"VI = -1 # # at t =0 #\n",
"\n",
"# The output voltage of an integrator is define as\n",
"RC = t/10*1e3 # ms\n",
"print ' The time constant of the given filter is RC = %0.2f'%RC,'msec '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.13 Pg 198"
]
},
{
"cell_type": "code",
"execution_count": 48,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The time constant of the given filter is RC = 0.1 msec \n",
"The capacitor value is = 0.1 F\n"
]
}
],
"source": [
"from __future__ import division\n",
"# Determine the time constant of the integrator\n",
"Vo = 20 #\n",
"t = 1*10**-3 #\n",
"VI = -1 # # at t =0 #\n",
"\n",
"# The output voltage of an integrator is define as\n",
"RC = t/10 #\n",
"print ' The time constant of the given filter is RC = %0.1f'%(RC*1000),'msec '#\n",
"\n",
"R = 1*10**3 # # we assume \n",
"C = RC/R*10**6 #\n",
"print 'The capacitor value is = %0.1f'%C,'F'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.14 Pg 199"
]
},
{
"cell_type": "code",
"execution_count": 49,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of resistance RIa is = 90.00 K ohm \n",
"The value of resistance RIb is = 22.50 K ohm \n",
"The value of resistance RIc is = 18.00 K ohm \n",
"The value of resistance RId is = 15.00 K ohm \n"
]
}
],
"source": [
" # to design a summing amplifier\n",
"\n",
"# the output of the summing amplifier is given by\n",
"#Vo = -R2*((VIa/RIa)+(VIb/RIb)+(VIc/RIc)+(VId/RId))# equation 1\n",
"\n",
"# the equation given is\n",
"#Vo = -(3*VIa+12*VIb+15*VIc+18*VId)# equation 2\n",
"\n",
"# comparing equation 1 and 2\n",
"#R2/RIa = 3 #\n",
"#R2/RIb = 12 #\n",
"#R2/RIc = 15 #\n",
"#R2/RId = 18 # \n",
"\n",
"# the feedback resistance R2= 270 K ohm \n",
"R2 = 270 # # K ohm\n",
"RIa = R2/3 #\n",
"print 'The value of resistance RIa is = %0.2f'%RIa,' K ohm '#\n",
"\n",
"RIb = R2/12 #\n",
"print 'The value of resistance RIb is = %0.2f'%RIb,' K ohm '#\n",
"\n",
"RIc = R2/15 #\n",
"print 'The value of resistance RIc is = %0.2f'%RIc,' K ohm '#\n",
"\n",
"RId = R2/18 #\n",
"print 'The value of resistance RId is = %0.2f'%RId,' K ohm '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.15 Pg 200"
]
},
{
"cell_type": "code",
"execution_count": 50,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The output of first op-amp A1 is = -275*sin wt mV \n",
"The output of second op-amp A2 is = 275*sin wt mV \n",
"The output of third op-amp A3 is = 825*sin wt mV \n",
"current through the resistor R1 and R2 is = 5 sin wt uA \n",
"current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA \n",
"current through the inverting terminal resistor R3 and R4 = 22 sin wt uA \n"
]
}
],
"source": [
"# for the instrumentation amplifier find Vo1 , Vo2 , Vo \n",
"# Vi1 = -25 sin wt # # mV\n",
"# Vi2 = 25 sin wt # # mV\n",
"R1 = 10*10**3 #\n",
"R2 = 20*10**3 #\n",
"R3 = 20*10**3 #\n",
"R4 = 10*10**3 #\n",
"\n",
"# the output of first op-amp A1 is given by\n",
"# Vo1 = (1+(R2/R1))*Vi1-(R2/R1)*Vi2 #\n",
"#by solving above equation we get\n",
"print 'The output of first op-amp A1 is = -275*sin wt mV '#\n",
"\n",
"# the output of second op-amp A2 is given by\n",
"# Vo2 = (1+(R2/R1))*Vi2-(R2/R1)*Vi1 #\n",
"#by solving above equation we get\n",
"print 'The output of second op-amp A2 is = 275*sin wt mV '#\n",
"\n",
"# the output of third op-amp A3 is given by\n",
"# Vo = (R4/R3)-(1+(2R2/R1)*(Vi2-Vi1) #\n",
"#by solving above equation we get\n",
"print 'The output of third op-amp A3 is = 825*sin wt mV '#\n",
"\n",
"# current through the resistor R1 and R2 is\n",
"#i = (Vi1-Vi2)/R1 #\n",
"print 'current through the resistor R1 and R2 is = 5 sin wt uA '#\n",
"\n",
"# current through the non-inverting terminal resistor R3 and R4 \n",
"#i3 = Vo2/(R3+R4)#\n",
"print 'current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA '#\n",
"\n",
"# current through the inverting terminal resistor R3 and R4 \n",
"#i2 = Vo1-(R3/(R3+R4))*Vo2/R3 #\n",
"print 'current through the inverting terminal resistor R3 and R4 = 22 sin wt uA '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.16 Pg 202"
]
},
{
"cell_type": "code",
"execution_count": 52,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The input resistance Rin is = 0.00999 ohm \n",
"The value of Resistance Rs is = 1.0990 K ohm \n"
]
}
],
"source": [
" # for the a current to voltage converter show a) Rin = (Rf/1+Aop) b) Rf = 10 K ohm Aop = 1000 \n",
"\n",
"#a) The input resistance given as\n",
"#Rin = (Rf)/(1+Aop) #\n",
"\n",
"# The input resistance of the circuit can be written as\n",
"#Rin = (V1/i!)#\n",
"\n",
"# the feedback current of the given circuit is defined as\n",
"#i1 =(V1-Vo)/RF #\n",
"\n",
"# the feedback resistance RF is \n",
"#RF =(V1-Vo)/i1 #\n",
"\n",
"# The output voltage Vo is\n",
"#Vo = -Aop*V1 #\n",
"\n",
"#by using this output feedback currenty i1 can be reformed as\n",
"#i1 = (V1-(-Aop*V1))/RF #\n",
"\n",
"#i1 = V1*(1+Aop)/RF #\n",
"\n",
"# Then Rin Becomes \n",
"#Rin =Rf/(1+Aop)#\n",
"\n",
"Rf =10*10**3 #\n",
"Aop = 1000 #\n",
"\n",
"# the input current and output voltage of the circuit are defined as\n",
"#i1 =(Rs)/(Rs+Rin) #\n",
"# Vo = -(Aop*(RF/1+Aop))*i1 #\n",
"\n",
"#the input resistance Rin is \n",
"Rin =(Rf/(1+Aop)) #\n",
"\n",
"# subsituting the value of RF Aop Rin and Vo we get \n",
"RF = 10 #\n",
"Rin = RF/(1+Aop)\n",
"print 'The input resistance Rin is = %0.5f'%Rin,' ohm '#\n",
"\n",
"Aop = 1000 #\n",
"#(1000/1001)*(Rs/(Rs*0.00999))> 0.99 #\n",
"# by solving above equation we get \n",
"Rs = 1.099 # # K ohm \n",
"print 'The value of Resistance Rs is = %0.4f'%Rs,' K ohm '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.17 Pg 204"
]
},
{
"cell_type": "code",
"execution_count": 54,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"for Aop = 10**4 closed loop gain is = 0.9999 \n",
"for Aop = 10**3 closed loop gain is = 0.9990 \n",
"for Aop = 10**2 closed loop gain is = 0.9901 \n",
"for Aop = 10**1 closed loop gain is = 0.9091 \n"
]
}
],
"source": [
" # determine the closed loop gain\n",
"\n",
"# the output of the voltage follower is given as\n",
"#Vo = Aop(V1-Vo)#\n",
"\n",
"# the closed loop gain of the voltage follower \n",
"#A = 1/(1+(1/Aop))#\n",
" \n",
"# for Aop = 10**4 closed loop gain\n",
"Aop = 10**4 #\n",
"A = 1/(1+(1/Aop))#\n",
"print 'for Aop = 10**4 closed loop gain is = %0.4f'%A,' '#\n",
"\n",
"# for Aop = 10**3 closed loop gain\n",
"Aop = 10**3 #\n",
"A = 1/(1+(1/Aop))#\n",
"print 'for Aop = 10**3 closed loop gain is = %0.4f'%A,' '#\n",
"\n",
"# for Aop = 10**2 closed loop gain\n",
"Aop = 10**2 #\n",
"A = 1/(1+(1/Aop))#\n",
"print 'for Aop = 10**2 closed loop gain is = %0.4f'%A,' '#\n",
"\n",
"# for Aop = 10**1 closed loop gain\n",
"Aop = 10**1 #\n",
"A = 1/(1+(1/Aop))#\n",
"print 'for Aop = 10**1 closed loop gain is = %0.4f'%A,' '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.18 Pg 205"
]
},
{
"cell_type": "code",
"execution_count": 56,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"for the frequency f = 10 Hz the output is = 106.10 V \n",
"for the frequency f = 1000 Hz the output is = 1.06 V \n",
"for the frequency f = 10000 Hz the output is = 0.106 V \n"
]
}
],
"source": [
"from math import pi\n",
"# To determine the output voltage of integrator\n",
"Vin = 1 #\n",
"R = 150*10**3 ## ohm\n",
"C = 1*10**-9 # # F\n",
"\n",
"# the output voltage of an integrator is given as\n",
"#Vo = (fc/f)*Vin #\n",
"\n",
"#fc = 1/(2*pi*R*C)#\n",
"\n",
"#Vo = (1/(2*pi*R*C*f))*Vin#\n",
"\n",
"#for the frequency f = 10 Hz the output is\n",
"f = 10 # # Hz\n",
"Vo = (1/(2*pi*R*C*f))*Vin#\n",
"print 'for the frequency f = 10 Hz the output is = %0.2f'%Vo,' V '#\n",
"\n",
"#for the frequency f = 1000 Hz the output is\n",
"f = 1000 # # Hz\n",
"Vo = (1/(2*pi*R*C*f))*Vin#\n",
"print 'for the frequency f = 1000 Hz the output is = %0.2f'%Vo,' V '#\n",
"\n",
"#for the frequency f = 10000 Hz the output is\n",
"f = 10000 # # Hz\n",
"Vo = (1/(2*pi*R*C*f))*Vin#\n",
"print 'for the frequency f = 10000 Hz the output is = %0.3f'%Vo,' V '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.19 Pg 206"
]
},
{
"cell_type": "code",
"execution_count": 63,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The cutoff frequency of the integrator is = 13.263 kHz\n",
"The gain of the integrator is = 3.18\n"
]
}
],
"source": [
"from math import sqrt\n",
"# To determine the magnitude gain of the integrator\n",
"Vin = 1 #\n",
"f = 50*10**3 #\n",
"Rf = 120*10**3 #\n",
"R = 10*10**3 #\n",
"C = 0.1*10**-9 #\n",
"\n",
"# the magnitude gain of the integrator is given by\n",
"#A = (Rf/R)/(sqrt(1+(f/fc)**2))#\n",
"\n",
"# the cutoff frequency of the integrator \n",
"fc = 1/(2*pi*Rf*C)/1e3\n",
"print 'The cutoff frequency of the integrator is = %0.3f'%fc,'kHz'#\n",
"\n",
"\n",
"A = (Rf/R)/(sqrt(1+(f/fc)**2))*1e3#\n",
"print 'The gain of the integrator is = %0.2f'%A"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.20 Pg 207"
]
},
{
"cell_type": "code",
"execution_count": 67,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the break frequency fa is = 31.83 kHz \n",
"the break frequency fb is = 21.22 kHz \n",
"The gain of the differentiator is = 0.6667 \n"
]
}
],
"source": [
" # To determine the magnitude gain of the differentiator\n",
"Vin = 1 #\n",
"f = 50*10**3 #\n",
"R = 75*10**3 #\n",
"R1 = 50*10**3 #\n",
"C = 0.1*10**-9 #\n",
"\n",
"# the magnitude gain of the differentiator is given by\n",
"#A = (f/fa)/(sqrt(1+(f/fb)**2))#\n",
"\n",
"# the break frequency fa is defined as\n",
"fa = 1/(2*pi*R1*C) / 1e3\n",
"print 'the break frequency fa is = %0.2f'%fa,'kHz '#\n",
"\n",
"# the break frequency fb is defined as\n",
"fb = 1/(2*pi*R*C) /1e3\n",
"print 'the break frequency fb is = %0.2f'%fb,'kHz '#\n",
"\n",
"\n",
"A = (f/fa)/(sqrt(1+(f/fb)**2))#\n",
"print 'The gain of the differentiator is = %0.4f'%A,' '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.21 Pg 209"
]
},
{
"cell_type": "code",
"execution_count": 70,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The input voltage of an op-amp is = -40 mV\n"
]
}
],
"source": [
" # to determine the input voltage of an op-amp\n",
"Vo = 2 # # V\n",
"R1 = 20*10**3 # # ohm\n",
"R2 = 1*10**6 # # ohm\n",
"\n",
"# the input voltage of an op-amp\n",
"Vin = -(R1/R2)*Vo *1000 # mV\n",
"print 'The input voltage of an op-amp is = %0.f'%Vin,'mV'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.22 Pg 209"
]
},
{
"cell_type": "code",
"execution_count": 71,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the output voltage of follower Vo1 is = 2.00 V\n",
"The output voltage of an inverting amplifier is = -20.00 V \n"
]
}
],
"source": [
"# To determine the output voltage\n",
"Vin = 2 #\n",
"R2 = 20*10**3 #\n",
"R1 = 2*10**3 #\n",
"\n",
"# the output voltage of follower Vo1 is\n",
"Vo1 = Vin #\n",
"print 'the output voltage of follower Vo1 is = %0.2f'%Vo1,' V'#\n",
"# the output voltage of an inverting amplifier\n",
"Vo = -(R2/R1)*Vo1 #\n",
"print 'The output voltage of an inverting amplifier is = %0.2f'%Vo,' V '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.23 Pg 210"
]
},
{
"cell_type": "code",
"execution_count": 72,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The output of the inverting amplifier is = -15.00 V\n",
"The output of the non-inverting amplifier is = 20.00 V\n"
]
}
],
"source": [
"# to determine the output voltage of an op-amp\n",
"Vin = 5 # # V\n",
"R1 = 25*10**3 # # ohm\n",
"R2 = 75*10**3 # # ohm\n",
"\n",
"# in this problem op-amp A1 perform the voltage follower and op-amp A2 perform inverting amplifier and op-amp A3 perform non-inverting amplifier\n",
"\n",
"# the output voltage of follower op-amp A1\n",
"Vo1 = Vin #\n",
"\n",
"# the output of the inverting amplifier A2\n",
"Vo2 = -((R2/R1)*Vo1) #\n",
"print 'The output of the inverting amplifier is = %0.2f'%Vo2,' V'#\n",
"\n",
"# the output of the non-inverting amplifier A3\n",
"Vo =(1+(R2/R1))*Vo1 #\n",
"print 'The output of the non-inverting amplifier is = %0.2f'%Vo,' V'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.24 Pg 211"
]
},
{
"cell_type": "code",
"execution_count": 73,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The output voltage of an inverting amplifier is = 27.50 V \n",
"the output voltage of follower Vo1 is = 2.50 V\n",
"the output of the inverting summing amplifier is = 101.25 V \n"
]
}
],
"source": [
" # To determine the output voltage\n",
"Vin = 2.5 #\n",
"Rf = 100*10**3 #\n",
"R1 = 10*10**3 #\n",
"RI1 = 25*10**3 #\n",
"RI2 = 10*10**3 #\n",
"R2 = 100*10**3 #\n",
"\n",
"# the output voltage of an inverting amplifier\n",
"Vo1 = (1+(R2/R1))*Vin # #\n",
"print 'The output voltage of an inverting amplifier is = %0.2f'%Vo1,' V '#\n",
"\n",
"# the output voltage of follower Vo2 is\n",
"Vo2 = Vin #\n",
"print 'the output voltage of follower Vo1 is = %0.2f'%Vo2,' V'#\n",
"\n",
"# the output of the inverting summing amplifier\n",
"R2 = 75*10**3 #\n",
"Vo = R2*((Vo1/RI1)+(Vo2/RI2))#\n",
"print 'the output of the inverting summing amplifier is = %0.2f'%Vo,' V '#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.25 Pg 212"
]
},
{
"cell_type": "code",
"execution_count": 74,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the total gain of the circuit is = 36.00 \n",
"The output voltage of the op-amp is = 90.00 V\n"
]
}
],
"source": [
"# To calculate the output voltage\n",
"Vin = 2.5 # \n",
"R1 = 10*10**3 #\n",
"R2 = 10*10**3 #\n",
"R3 = 10*10**3 #\n",
"Rf = 30*10**3 #\n",
"\n",
"# the total gain of the circuit \n",
"#Av =A1v*A2v*A3v #\n",
"Av = (1+(Rf/R1))*(-Rf/R2)*(-Rf/R3)#\n",
"print 'the total gain of the circuit is = %0.2f'%Av,' '#\n",
"\n",
"# The output voltage of the op-amp \n",
"Vo = Av*Vin #\n",
"print 'The output voltage of the op-amp is = %0.2f'%Vo,' V'#"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.26 Pg 212"
]
},
{
"cell_type": "code",
"execution_count": 76,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The output of op-amp A1 is = -10.00 V1\n",
"The output of op-amp A2 is Vo = 40V1 - 2V2 \n",
"The output is equal to the difference between 40V1 and 2V2 \n"
]
}
],
"source": [
"# to calculate the output voltage of op-amp circuit\n",
"Rf = 100*10**3 # # ohm\n",
"R1 = 10*10**3 # # ohm\n",
"R2 = 25*10**3 # # ohm\n",
"R3 = 50*10**3 # # ohm\n",
"\n",
"# the output of op-amp A1 is\n",
"# VA1 = (-Rf/R1)*V1 #\n",
"VA1 = (-Rf/R1)#\n",
"print 'The output of op-amp A1 is = %0.2f'%VA1,'V1' # # *V1 because the output is come from 1 op-amp\n",
"\n",
"# the output of op-amp A2 is\n",
"# Vo = -Rf*((VA1/R2)+(V2/R3))#\n",
"#Vo = -100*(-0.4*V1+0.02V2)#\n",
"print 'The output of op-amp A2 is Vo = 40V1 - 2V2 '# \n",
"\n",
"print 'The output is equal to the difference between 40V1 and 2V2 '# "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.27 Pg 213"
]
},
{
"cell_type": "code",
"execution_count": 77,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the upper crossover voltage of schmitt trigger is = 1.00 V\n",
"the lower crossover voltage of schmitt trigger is = -1.00 V\n",
"the hysteresis width HW of schmitt trigger is = 2.00 V\n"
]
}
],
"source": [
" # to determine the hysteresis width of a schmitt trigger\n",
"R1 = 25*10**3 # # ohm\n",
"R2 = 75*10**3 # # ohm\n",
"VTH = 4 # # V\n",
"VTL = -4 # # V\n",
"\n",
"# the upper crossover voltage of schmitt trigger is defined as\n",
"VU = (R1/(R1+R2))*VTH#\n",
"print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n",
"\n",
"# the lower crossover voltage of schmitt trigger is defined as\n",
"VL = (R1/(R1+R2))*VTL#\n",
"print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n",
"\n",
"# the hysteresis width of schmitt trigger is\n",
"HW = VU-VL #\n",
"print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.28 Pg 214"
]
},
{
"cell_type": "code",
"execution_count": 78,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the upper crossover voltage of schmitt trigger is = 1.43 V\n",
"the lower crossover voltage of schmitt trigger is = -1.43 V\n",
"the hysteresis width HW of schmitt trigger is = 2.86 V\n"
]
}
],
"source": [
" # to determine the hysteresis width of a schmitt trigger\n",
"R1 = 15*10**3 # # ohm\n",
"R2 = 90*10**3 # # ohm\n",
"VTH = 10 # # V\n",
"VTL = -10 # # V\n",
"\n",
"# the upper crossover voltage of schmitt trigger is defined as\n",
"VU = (R1/(R1+R2))*VTH#\n",
"print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n",
"\n",
"# the lower crossover voltage of schmitt trigger is defined as\n",
"VL = (R1/(R1+R2))*VTL#\n",
"print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n",
"\n",
"# the hysteresis width of schmitt trigger is\n",
"HW = VU-VL #`\n",
"print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.29 Pg 214"
]
},
{
"cell_type": "code",
"execution_count": 80,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance R1 is = 5.00 kohm\n"
]
}
],
"source": [
"# to determine the resistance R1 when low and high saturated output states are given\n",
"R2 = 20*10**3 # # ohm\n",
"VH = 2 # # V crossover voltage\n",
"VL = -2 # # V crossover voltage\n",
"VOH = 10 # # V saturated output states\n",
"VOL = -10 # # V saturated output states\n",
"\n",
"# the upper crossover voltage of schmitt trigger is defined as\n",
"# V = (R1/(R1+R2))*VOH#\n",
"# solving above equation we get \n",
"# 2R1+2R2 = 10R1 #\n",
"R1 = (2*R2)/8/1000 # kohm\n",
"print 'the value of resistance R1 is = %0.2f'%R1,'kohm'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6.30 Pg 215"
]
},
{
"cell_type": "code",
"execution_count": 82,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the value of resistance R2 is = 30.00 kohm\n"
]
}
],
"source": [
"# to determine the value of resistance R1 and R2 when low and high saturated output states are given\n",
"VH = 3 # # V crossover voltage\n",
"VL = -3 # # V crossover voltage\n",
"VOH = 12 # # V saturated output states\n",
"VOL = -12 # # V saturated output states\n",
"\n",
"# the upper crossover voltage of schmitt trigger is defined as\n",
"# V = (R1/(R1+R2))*VOH#\n",
"# solving above equation we get \n",
"# 3R1+3R2 = 12R1 #\n",
"\n",
"# 3*R1 = R2 #\n",
"R1 = 10*10**3 # # ohm we assume\n",
"R2 = 3*R1 / 1e3#\n",
"print 'the value of resistance R2 is = %0.2f'%R2,'kohm'"
]
}
],
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