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diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14_1.ipynb new file mode 100644 index 00000000..b66475f7 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14_1.ipynb @@ -0,0 +1,643 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14 Special Function ICs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1 Pg 415" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the adjustable voltage regulator is = 22.25 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "\n", + "# to determine the regulated voltage \n", + "R1 = 250 # #ohm \n", + "R2 = 2500 # # ohm \n", + "Vref = 2 # #V #reference voltage\n", + "Iadj = 100*10**-6# # A # adjacent current\n", + "\n", + "#the output voltage of the adjustable voltage regulator is defined by\n", + "Vo = (Vref*((R2/R1)+1)+(Iadj*R2)) #\n", + "print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2 Pg 416" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the total power dissipation of the IC is = 25.00 mA \n" + ] + } + ], + "source": [ + "# to determine the current drawn from the dual power supply \n", + "V = 10 # # V\n", + "P = 500 # # mW\n", + "\n", + "# we assume that each power supply provides half power supply to IC\n", + "P1 = (P/2)#\n", + "\n", + "# the total power dissipation of the IC\n", + "# P1 = V*I #\n", + "I = P1/V #\n", + "print 'the total power dissipation of the IC is = %0.2f'%I,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3 Pg 416" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the adjustable voltage regulator is = 7.50 V \n" + ] + } + ], + "source": [ + "# to determine the output voltage \n", + "R1 = 100*10**3 # #ohm \n", + "R2 = 500*10**3 # # ohm \n", + "Vref = 1.25 # #V #reference voltage\n", + "\n", + "#the output voltage of the adjustable voltage regulator is defined by\n", + "Vo = Vref*(R1+R2)/R1#\n", + "print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.4 Pg 417" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 3.50 V \n" + ] + } + ], + "source": [ + "# determine the output voltage of the switching regulator circuit\n", + "d = 0.7 # # duty cycle\n", + "Vin = 5 # # V # input voltage\n", + "\n", + "# The output voltage of switching regulator circuit is given by\n", + "Vo = d*Vin #\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.5 Pg 417" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.96 \n" + ] + } + ], + "source": [ + "# determine the duty cycle of the switching regulator circuit\n", + "Vo = 4.8 # # V # output voltage\n", + "Vin = 5 # # V # input voltage\n", + "\n", + "# The output voltage of switching regulator circuit is given by\n", + "# Vo = d*Vin #\n", + "\n", + "# Duty cycle is given as\n", + "d =Vo/Vin #\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6 Pg 418" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.50 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the duty cycle of the switching regulator circuit\n", + "T =120 # #msec # total pulse time\n", + "# T = ton + toff #\n", + "ton = T/2 #\n", + "\n", + "# The duty cycle of switching regulator circuit is given by\n", + "d = ton/T#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7 Pg 418" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.67 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the duty cycle of the switching regulator circuit\n", + "ton = 12 # #msec # on time of pulse\n", + "# ton = 2*toff # given\n", + "# T = ton + toff #\n", + "toff = ton/2 #\n", + "T = ton+toff # # total time\n", + "\n", + "# The duty cycle of switching regulator circuit is given by\n", + "d = ton/T#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.8 Pg 419" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter bias voltage is = 3.80 V \n", + "The output voltage of the IC LM380 is = 7.90 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the output voltage of the audio power amplifier IC LM380\n", + "Vcc = 12 # # V\n", + "Ic3 = 12*10**-6 # # A # collector current of the transistor Q3\n", + "Ic4 = 12*10**-6 # # A # collector current of the transistor Q4\n", + "R11 = 25*10**3 # # ohm\n", + "R12 = 25*10**3 # # ohm\n", + "\n", + "# the collector current of Q3 is defined as\n", + " # Ic3 = (Vcc-3*Veb)/(R11+R12)#\n", + "Veb = (Vcc-(R11+R12)*Ic3)/3 #\n", + "print 'The emitter bias voltage is = %0.2f'%Veb,' V '\n", + "\n", + "# the output voltage of the IC LM380\n", + "Vo = (1/2)*Vcc+(1/2)*Veb#\n", + "print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.9 Pg 420" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter bias voltage is = 3.33 V \n", + "The output voltage of the IC LM380 is = 6.67 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the output voltage of the audio power amplifier IC LM380\n", + "Vcc = 10 # # V\n", + "Ic3 = 0.01*10**-6 # # A # collector current of the transistor Q3\n", + "Ic4 = 0.01*10**-6 # # A # collector current of the transistor Q4\n", + "R11 = 25*10**3 # # ohm\n", + "R12 = 25*10**3 # # ohm\n", + "\n", + "# the collector current of Q3 is defined as\n", + " # Ic3 = (Vcc-3*Veb)/(R11+R12)#\n", + "Veb = (Vcc-(R11+R12)*Ic3)/3 #\n", + "print 'The emitter bias voltage is = %0.2f'%Veb,' V '\n", + "\n", + "# the output voltage of the IC LM380\n", + "Vo = (1/2)*Vcc+(1/2)*Veb#\n", + "print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.10 Pg 421" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter resistor of Q3 is = 52.00 ohm ( at temperature 25 degree celsius) \n", + "The trans conductance of transistor is = 38.5 mA/V \n", + "The base emitter resistor rbe is = 1.30 K ohm \n", + "The emitter capacitor Ce = 7.65 pF \n", + "The value of resistance RL is = 264.55 ohm \n", + "The pole frequency fa is = 601.91 M Hz \n", + "The pole frequency fb is = 1073.74 M Hz \n", + "The pole frequency fc is = 3060.67 M Hz \n", + "Hence fa is a dominant pole frequency \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from numpy import inf\n", + "from math import sqrt, pi\n", + "# Design a video amplifier of IC 1550 circuit\n", + "Vcc = 12 # # V\n", + "Av = -10 #\n", + "Vagc = 0 # # at bandwidth of 20 MHz\n", + "hfe = 50 # # forward emitter parameter\n", + "rbb = 25 # # ohm # base resistor\n", + "Cs = 1*10**-12 # # F # source capacitor\n", + "Cl = 1*10**-12 # # F # load capacitor\n", + "Ie1 = 1*10**-3 # # A # emitter current of Q1\n", + "f = 1000*10**6 # # Hz\n", + "fT = 800*10**6 # # Hz\n", + "Vt = 52*10**-3 #\n", + "Vt1 = 0.026 #\n", + "\n", + "# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3\n", + "# i.e Ic1=Ie1=Ie3\n", + "Ie3 = 1*10**-3 # # A # emitter current of Q3\n", + "Ic1 = 1*10**-3 # # A # collector current of the transistor Q1\n", + "\n", + "# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite\n", + "\n", + "re2 = inf #\n", + "\n", + "# emitter resistor of Q3 \n", + "re3 = (Vt/Ie1)#\n", + "print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm ( at temperature 25 degree celsius) '\n", + "\n", + "# the trans conductance of transistor is\n", + "gm = (Ie1/Vt1)#\n", + "print 'The trans conductance of transistor is = %0.1f'%(gm*1000),' mA/V ' # Round Off Error\n", + "\n", + "# the base emitter resistor rbe\n", + "rbe = (hfe/gm)#\n", + "print 'The base emitter resistor rbe is = %0.2f'%(rbe/1000),' K ohm ' # Round Off Error\n", + "\n", + "# the emitter capacitor Ce \n", + "\n", + "Ce = (gm/(2*pi*fT))#\n", + "print 'The emitter capacitor Ce = %0.2f'%(Ce*1e12),' pF ' # Round Off Error\n", + "\n", + "# the voltage gain of video amplifier is\n", + "# Av = (Vo/Vin) #\n", + "# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl)))) \n", + " # At Avgc = 0 i.e s=0 in the above Av equation\n", + "alpha3 = 1 #\n", + "s = 0 #\n", + "# Rl = -((alpha3*gm)/(rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av)))# \n", + "\n", + "# After solving above equation for Rl We get Rl Equation as\n", + "Rl = 10/(37.8*10**-3)#\n", + "print 'The value of resistance RL is = %0.2f'%Rl,' ohm '\n", + "\n", + "# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency \n", + "Rl = 675 #\n", + "# fa = 1/(2*pi*Rl*(Cs+Cl))#\n", + "# after putting value of Rl ,Cs and Cl we get\n", + "fa = 1/(2*3.14*264.55*1*10**-12)#\n", + "print 'The pole frequency fa is = %0.2f'%(fa*10**-3/1000),' M Hz '# Round Off Error\n", + "\n", + "\n", + "#fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))#\n", + "# after putting value of Ce rbb and rbe we get\n", + "fb = 1/(2*pi*6.05*10**-12*24.5)#\n", + "print 'The pole frequency fb is = %0.2f'%(fb*10**-3/1000),' M Hz '\n", + "\n", + "fc = 1/(2*pi*Cs*re3)#\n", + "print 'The pole frequency fc is = %0.2f'%(fc*10**-3/1000),' M Hz '\n", + "\n", + "print 'Hence fa is a dominant pole frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.11 Pg 423" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter resistor of Q3 is = 52.00 ohm \n", + "The trans conductance of transistor is = 38.5 mA/V \n", + "The base emitter resistor rbe is = 1.3 kohm \n", + "The emitter capacitor is = 6.12 pF \n", + "The value of resistance RL is = 265.00 ohm \n", + "The pole frequency fa is = 600.58 MHz \n", + "The pole frequency fb is = 1060.00 MHz \n", + "The pole frequency fc is = 3060.67 MHz \n", + "Hence fa is a dominant pole frequency \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from numpy import inf,pi\n", + "# Design a video amplifier of IC 1550 circuit\n", + "Vcc = 12 # # V\n", + "Av = -10 #\n", + "Vagc = 0 # # at bandwidth of 20 MHz\n", + "hfe = 50 # # forward emitter parameter\n", + "rbb = 25 # # ohm # base resistor\n", + "Cs = 1*10**-12 # # F # source capacitor\n", + "Cl = 1*10**-12 # # F # load capacitor\n", + "Ie1 = 1*10**-3 # # A # emitter current of Q1\n", + "f = 1000*10**6 # # Hz\n", + "Vt = 52*10**-3 #\n", + "Vt1 = 0.026 #\n", + "\n", + "# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3\n", + "# i.e Ic1=Ie1=Ie3\n", + "Ie3 = 1*10**-3 # # A # emitter current of Q3\n", + "Ic1 = 1*10**-3 # # A # collector current of the transistor Q1\n", + "\n", + "# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite\n", + "re2 = inf #\n", + "\n", + "# emitter resistor of Q3 \n", + "re3 = (Vt/Ie1)#\n", + "print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm '\n", + "\n", + "# the trans conductance of transistor is\n", + "gm = (Ie1/Vt1)#\n", + "print 'The trans conductance of transistor is = %0.1f'%(gm*1e3),' mA/V '\n", + "\n", + "# the base emitter resistor rbe\n", + "rbe = (hfe/gm)#\n", + "print 'The base emitter resistor rbe is = %0.1f'%(rbe/1e3),' kohm '\n", + "\n", + "# the emitter capacitor Ce \n", + "Ce = (gm/(2*pi*f))#\n", + "print 'The emitter capacitor is = %0.2f'%(Ce*1e12),' pF '\n", + "\n", + "# the voltage gain of video amplifier is\n", + "# Av = (Vo/Vin) #\n", + "# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl)))) \n", + " # At Avgc = 0 i.e s=0 in the above Av equation\n", + "alpha3 = 1 #\n", + "s = 0 #\n", + "Av =-10 #\n", + "Rl = -((alpha3*gm)/((rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av))))# \n", + "Rl = (1/Rl)#\n", + "print 'The value of resistance RL is = %0.2f'%Rl,' ohm '\n", + "\n", + "# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency \n", + "Rl = 265\n", + "fa = 1/(2*pi*Rl*(Cs))/1e6#\n", + "print 'The pole frequency fa is = %0.2f'%fa,'MHz '\n", + "\n", + "\n", + "fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))/1e6\n", + "print 'The pole frequency fb is = %0.2f'%fb,'MHz '\n", + "\n", + "fc = 1/(2*pi*Cs*re3)/1e6\n", + "print 'The pole frequency fc is = %0.2f'%fc,'MHz '\n", + "\n", + "print 'Hence fa is a dominant pole frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.12 Pg 425" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input current is = 0.50 mA \n", + "The output of an op-amp is = 27.50 V \n" + ] + } + ], + "source": [ + "# Determine the output voltage of an isolation amplifier IC ISO100\n", + "Vin = 5.0 # # V\n", + "Rin = 10*10**3 # \n", + "Rf = 55*10**3 # # ohm # feedback resistance\n", + "\n", + "# the input voltage of an amplifier 1\n", + "# Vin = Rin*Iin\n", + "Iin = Vin/Rin # \n", + "print 'The input current is = %0.2f'%(Iin*1e3),'mA '\n", + "\n", + "# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction\n", + "# Iin = -Iout\n", + "# the output of an op-amp\n", + "# Vo = -Rf*Iout\n", + "Vo = Rf*Iin#\n", + "print 'The output of an op-amp is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.13 Pg 426" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input current is = 12 mA \n", + "The output of an op-amp is = 204 V \n" + ] + } + ], + "source": [ + "# Determine the output voltage of an isolation amplifier IC ISO100\n", + "Vin = 12.0 # # V\n", + "Rin = 1*10**3 # \n", + "Rf = 17*10**3 # # ohm # feedback resistance\n", + "\n", + "# the input voltage of an amplifier 1\n", + "# Vin = Rin*Iin\n", + "Iin = Vin/Rin # \n", + "print 'The input current is = %0.f'%(Iin*1e3),'mA '\n", + "\n", + "# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction\n", + "# Iin = -Iout\n", + "# the output of an op-amp\n", + "# Vo = -Rf*Iout\n", + "Vo = Rf*Iin#\n", + "print 'The output of an op-amp is = %0.f'%Vo,' V '" + ] + } + ], + "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 +} |