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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 3 Current Voltage Sources and Differential Amplifiers"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.1 Pg 53"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The collector current of difference amplifier Ic1 = Ic2 = 0.50 mA \n",
"The collector voltages of transistors Q1 and Q2 are Vc1 = Vc2 = 5.00 volt \n",
"For Ve = -0.7 Volt the collector - emitter voltage Vce1 = 5.70 Volt\n",
"For Ve = 4.3 Volt the collector - emitter voltage Vce1 = 0.70 Volt\n",
"For Ve = -5.7 Volt the collector - emitter voltage Vce1 = 10.70 Volt\n"
]
}
],
"source": [
"from __future__ import division\n",
"# Determine the collector current Ic1 and collector-emitter voltage Vce1 for the difference amplifier circuit\n",
"\n",
"V1 = 0 # # volt\n",
"V2 = -5 # #volt\n",
"Vcm = 5 # #volt\n",
"Vcc = 10# #volt\n",
"Vee = -10 # #volt\n",
"Ie = 1 # #mA\n",
"Rc = 10 # #kilo ohm\n",
"\n",
"# Transistor parameters\n",
"# base current are negligible\n",
"Vbe = 0.7 # # volt\n",
"\n",
"# The collector current of difference amplifier is\n",
"Ic1 = Ie/2 # \n",
"print 'The collector current of difference amplifier Ic1 = Ic2 = %0.2f'%Ic1,' mA '\n",
"\n",
"# The collector voltages of transistors Q1 and Q2 are expressed as\n",
"\n",
"Vc1 = Vcc-Ic1*Rc #\n",
"print 'The collector voltages of transistors Q1 and Q2 are Vc1 = Vc2 = %0.2f'%Vc1,' volt '\n",
"\n",
"# We know common mode voltage (Vcm) , from this the emitter voltage can be identified as follows\n",
"# For the common mode voltage Vcm = 0 V , the emitter voltage is Ve = -0.7 V\n",
"# For the common mode voltage Vcm = 5 V , the emitter voltage is Ve = 4.3 V\n",
"# For the common mode voltage Vcm = -5 V , the emitter voltage is Ve = -5.7 V\n",
"\n",
"# For the different emitter voltages the collector-emitter voltage can be calculated as\n",
"\n",
"Ve = -0.7 # # volt\n",
"Vce1 = Vc1-Ve#\n",
"print 'For Ve = -0.7 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'\n",
"\n",
"Ve = 4.3 # # volt\n",
"Vce1 = Vc1-Ve#\n",
"print 'For Ve = 4.3 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'\n",
"\n",
"Ve = -5.7 # # volt\n",
"Vce1 = Vc1-Ve#\n",
"print 'For Ve = -5.7 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.2 Pg 54"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The differential mode gain Ad = 184.6\n",
" The common mode gain Acm = -0.237\n"
]
}
],
"source": [
"# To determine the difference-mode and common-mode gain of the difference amplifier\n",
"\n",
"Vcc = 10 # # volt\n",
"Vee = -10 # #volt\n",
"Iq = 0.8 # #mA\n",
"Ie = 0.8 # #mA\n",
"Rc = 12 # #kilo-Ohm\n",
"Vt = 0.026 # # volt\n",
"\n",
"# Transistor parameter\n",
"beta = 100 #\n",
"Rs = 0 # #Ohm\n",
"Ro = 25 # #kilo-Ohm \n",
"# The differential mode gain Ad\n",
"gm = (Ie/ 2*Vt) #\n",
"# Ad = (gm*r*Rc/r+Rc) # # where r is r-pi\n",
"# For Rb=0 , the differential mode gain is\n",
"\n",
"Ad = (Ie/(2*Vt))*Rc#\n",
"#But\n",
"print ' The differential mode gain Ad = %0.1f'%Ad\n",
"\n",
"#The common mode gain Acm\n",
"# Acm = - (gm*Rc/1+2*gm*Re+2*Re/r)\n",
"Acm =-(Ad/(1+(((1+beta)*Ie*Ro)/(beta*Vt))))\n",
"print ' The common mode gain Acm = %0.3f'%Acm"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.3 Pg 56"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The output of a difference amplifier is Vo = -47.40 sinwt uV \n"
]
}
],
"source": [
"# To find the output of a difference amplifier when only common mode signal is applied\n",
"\n",
"# V1 = V2 = Vcm = 200*sin(wt) # # micro volt (uV)\n",
"Acm = -0.237 #\n",
"\n",
"# When the common mode input signal is applied to the difference amplifier , the difference mode gain is zero\n",
"Vcm = 200 #\n",
"Vo = Acm*Vcm #\n",
"print 'The output of a difference amplifier is Vo = %0.2f'%Vo,'sinwt uV ' # multiply by sinwt because it is in Vcm"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.4 Pg 56"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The differential mode gain Ad = 184.6\n",
"The common mode gain Acm = -0.237\n",
"The CMRR of difference amplifier is = 389\n",
"In decibel CMRR is = 51.80\n"
]
}
],
"source": [
"from math import log10\n",
"#Determine the common mode rejection ratio(CMRR) of the difference amplifier\n",
"\n",
"Vcc = 10 # # volt\n",
"Vee = -10 # #volt\n",
"Iq = 0.8 # #mA\n",
"Ie = 0.8 # #mA\n",
"Rc = 12 # #kilo-Ohm\n",
"Vt = 0.026 # # volt\n",
"\n",
"# Transistor parameter\n",
"beta = 100 #\n",
"Rs = 0 # #Ohm\n",
"Ro = 25 # #kilo-Ohm\n",
" \n",
"# The differential mode gain Ad\n",
"gm = (Ie/ 2*Vt) #\n",
"# Ad = (gm*r*Rc/r+Rc) # # where r is r-pi\n",
"# For Rb=0 , the differential mode gain is\n",
"\n",
"Ad = (Ie/(2*Vt))*Rc#\n",
"#But\n",
"print 'The differential mode gain Ad = %0.1f'%Ad\n",
"\n",
"#The common mode gain Acm\n",
"# Acm = - (gm*Rc/1+2*gm*Re+2*Re/r)\n",
"Acm =-(Ad/(1+(((1+beta)*Ie*Ro)/(beta*Vt))))\n",
"print 'The common mode gain Acm = %0.3f'%Acm\n",
"\n",
"# The CMRR of difference amplifier is given as\n",
"Ad = Ad/2 #\n",
"CMRR = abs(Ad/Acm)\n",
"print 'The CMRR of difference amplifier is = %0.f'%CMRR\n",
"\n",
"# In decibel it can be expressed as\n",
"CMRRdb = 20*log10(CMRR)\n",
"print 'In decibel CMRR is = %0.2f'%CMRRdb"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.5 Pg 58"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The CMRR of difference amplifier is = 3.16e+04\n",
" The value of resistance RE is = 2.04 Mohm \n"
]
}
],
"source": [
"# To determine emitter resistance of the difference amplifier\n",
"\n",
"Vcc = 10 # # volt\n",
"Vee = -10 # #volt\n",
"Iq = 0.8 # #mA\n",
"Ie = 0.8 # #mA\n",
"CMRRdb = 90 # #dB\n",
"Vt = 0.026 #\n",
"\n",
"# Transistor parameter\n",
"beta = 100 #\n",
"\n",
"# CMRR = abs(Ad/Acm)\n",
"# the CMRR of the difference amplifier is defined as\n",
"#CMRR = ((1/2)*(1+((1+beta)*Ie*Re)/beta*Vt))\n",
"\n",
"# CMRRdb = 20*log10(CMRR)\n",
"CMRR = 10**(CMRRdb/20)\n",
"print ' The CMRR of difference amplifier is = %0.2e'%CMRR\n",
"\n",
"# The resistance RE is calculated as\n",
"\n",
"RE = (((2*CMRR)-1)/((1+beta)*Ie))*(beta*Vt)/1e3\n",
"print ' The value of resistance RE is = %0.2f'%RE,' Mohm '"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.6 Pg 59"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The differential mode gain Ad is = 321\n"
]
}
],
"source": [
"# determine the differential mode gain when load resistance RL = 100 k ohm\n",
"\n",
"RL = 100*10**3 # # k ohm # load resistance\n",
"IE = 0.20*10**-3 # # mA # biasing current\n",
"VA = 100 # # V # early voltage\n",
"VT = 0.026 # # threshold volt\n",
"\n",
"# the differential gain of differential amplifier with an active load circuit\n",
"#Ad = Vo/Vd = gm(ro2 || ro4 || RL )\n",
"ro2 = (2*VA)/IE#\n",
"ro4 = ro2 #\n",
"gm = IE/(2*VT) #\n",
"\n",
"Ad = gm/((1/ro2)+(1/ro4)+(1/RL))\n",
"print ' The differential mode gain Ad is = %0.f'%Ad"
]
}
],
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|
