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diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb deleted file mode 100755 index d64bea33..00000000 --- a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb +++ /dev/null @@ -1,636 +0,0 @@ -{ - "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": 5, - "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": 6, - "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": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The output voltage of switching regulator circuit is = 0.50 \n" - ] - } - ], - "source": [ - "# 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": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The output voltage of switching regulator circuit is = 0.67 \n" - ] - } - ], - "source": [ - "# 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": 9, - "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": [ - " # 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": 10, - "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": [ - "# 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": 6, - "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 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": 13, - "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 = 5.00 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": [ - "# 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": 18, - "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": 24, - "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 -} |