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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": 1,
+ "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": 2,
+ "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": 3,
+ "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": 4,
+ "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": 5,
+ "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": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " The differential mode gain Ad is = 321\n"
+ ]
+ }
+ ],
+ "source": [
+ "from __future__ import division\n",
+ "# 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"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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