diff options
Diffstat (limited to 'Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb')
-rwxr-xr-x | Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb | 344 |
1 files changed, 0 insertions, 344 deletions
diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb deleted file mode 100755 index 7a4856f8..00000000 --- a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb +++ /dev/null @@ -1,344 +0,0 @@ -{ - "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" - ] - } - ], - "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 -} |