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|
{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"CHAPTER 17 DIFFERENTIAL AMPLIFIERS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-1, Page 625"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.1.py\n",
"#What are the ideal currents & voltages in figure 17-6a?\n",
"\n",
"#Variable declaration\n",
"VCC=15 #supply voltage(V)\n",
"RE=7.5 #Emitter resistance(KOhm)\n",
"\n",
"#Calculation\n",
"IT=VCC/RE #tail current (mA)\n",
"IE=IT/2 #emitter current (mA)\n",
"\n",
"#Result\n",
"print 'tail current = ',IT,'mA'\n",
"print 'emitter current = ',IE,'mA'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"tail current = 2.0 mA\n",
"emitter current = 1.0 mA\n"
]
}
],
"prompt_number": 94
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-2, Page 626"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.2.py\n",
"#Recalculate the currents and voltages for figure 17-6a using second approximation.\n",
"\n",
"#Variable declaration\n",
"VCC=15 #supply voltage(V)\n",
"RE=7.5 #Emitter resistance(KOhm)\n",
"VBE=0.7 #base-emitter voltage(V)\n",
"RC=5 #collector resistance(KOhm)\n",
"\n",
"#Calculation\n",
"IT=(VCC-VBE)/RE #tail current (mA)\n",
"IE=IT/2 #emitter current (mA)\n",
"VC=VCC-(IE*RC) #collector quiescent voltage(V) \n",
"\n",
"#Result\n",
"print 'tail current = ',round(IT,2),'mA'\n",
"print 'emitter current = ',round(IE,2),'mA'\n",
"print 'collector quiescent voltage VC = ',round(VC,2),'V'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"tail current = 1.91 mA\n",
"emitter current = 0.95 mA\n",
"collector quiescent voltage VC = 10.23 V\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-3, Page 626"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.3.py\n",
"#What are the ideal currents & voltages in figure 17-7a?\n",
"\n",
"#Variable declaration\n",
"VCC=12.0 #supply voltage(V)\n",
"RE=5.0 #Emitter resistance(KOhm)\n",
"RC=3.0 #collector resistance(KOhm)\n",
"\n",
"#Calculation\n",
"IT=VCC/RE #tail current (mA)\n",
"IE=IT/2 #emitter current (mA)\n",
"VC=VCC-(IE*RC) #collector quiescent voltage(V) \n",
"#with second approximation,\n",
"IT1=(VCC-VBE)/RE #tail current (mA)\n",
"IE1=IT1/2 #emitter current (mA)\n",
"VC1=VCC-(IE1*RC) #collector quiescent voltage(V) \n",
"\n",
"#Result\n",
"print 'tail current = ',IT,'mA'\n",
"print 'emitter current = ',IE,'mA'\n",
"print 'collector quiescent voltage VC = ',VC,'V'\n",
"print 'with second approximation,'\n",
"print 'tail current = ',IT1,'mA'\n",
"print 'emitter current = ',IE1,'mA'\n",
"print 'collector quiescent voltage VC = ',VC1,'V'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"tail current = 2.4 mA\n",
"emitter current = 1.2 mA\n",
"collector quiescent voltage VC = 8.4 V\n",
"with second approximation,\n",
"tail current = 2.26 mA\n",
"emitter current = 1.13 mA\n",
"collector quiescent voltage VC = 8.61 V\n"
]
}
],
"prompt_number": 92
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-4, Page 631"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.4.py\n",
"#In figure 17-11, what is the ac output voltage? If B = 300, what is input impedance?\n",
"\n",
"#Variable declaration\n",
"VCC=12.0 #supply voltage(V)\n",
"RE=7.5 #Emitter resistance(KOhm)\n",
"RC=5.0 #collector resistance(KOhm)\n",
"IE=1 #emitter current as per preceding example(mA)\n",
"Vin=1 #input ac voltage(mV)\n",
"B=300 #current gain\n",
"\n",
"#Calculation\n",
"re=25/IE #ac emitter resistance(Ohm)\n",
"Av=1000*RC/re #voltage gain\n",
"Vout=Av*Vin #Output voltage(mV)\n",
"Zinb=2*B*re #input impedance of diff amp(Ohm)\n",
"\n",
"#Result\n",
"print 'voltage gain Av = ',Av\n",
"print 'Output voltage Vout = ',Vout,'mV'\n",
"print 'input impedance of diff amplifier, Zin(base) = ',Zinb/1000,'KOhm'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"voltage gain Av = 200.0\n",
"Output voltage Vout = 200.0 mV\n",
"input impedance of diff amplifier, Zin(base) = 15 KOhm\n"
]
}
],
"prompt_number": 91
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-5, Page 632"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.5.py\n",
"#Repeat preceding example using second approximation to calculate ICQ.\n",
"\n",
"#Variable declaration\n",
"VCC=12.0 #supply voltage(V)\n",
"RE=7.5 #Emitter resistance(KOhm)\n",
"RC=5.0 #collector resistance(KOhm)\n",
"IE=0.955 #emitter current as per previous example(mA)\n",
"Vin=1 #input ac voltage(mV)\n",
"B=300 #current gain\n",
"\n",
"#Calculation\n",
"re=25/IE #ac emitter resistance(Ohm)\n",
"Av=1000*RC/re #voltage gain\n",
"Vout=Av*Vin #Output voltage(mV)\n",
"Zinb=2*B*re #input impedance of diff amp(Ohm)\n",
"\n",
"#Result\n",
"print 'voltage gain Av = ',Av\n",
"print 'Output voltage Vout = ',Vout,'mV'\n",
"print 'input impedance of diff amplifier, Zin(base) = ',round((Zinb/1000),2),'KOhm'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"voltage gain Av = 191.0\n",
"Output voltage Vout = 191.0 mV\n",
"input impedance of diff amplifier, Zin(base) = 15.71 KOhm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-6, Page 633"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.6.py\n",
"#Repeat example 17-4 for v2=1mV & v1=0.\n",
"\n",
"#Variable declaration\n",
"VCC=12.0 #supply voltage(V)\n",
"RE=7.5 #Emitter resistance(KOhm)\n",
"RC=5.0 #collector resistance(KOhm)\n",
"IE=1 #emitter current as per preceding example(mA)\n",
"V2=1 #input ac voltage(mV)\n",
"B=300 #current gain\n",
"\n",
"#Calculation\n",
"re=25/IE #ac emitter resistance(Ohm)\n",
"Av=1000*RC/re #voltage gain\n",
"Vout=Av*V2 #Output voltage(mV)\n",
"Zinb=2*B*re #input impedance of diff amp(Ohm)\n",
"\n",
"#Result\n",
"print 'V2 input is at inverting input, So, ideally same as previous case.'\n",
"print 'Voltage gain Av = ',Av\n",
"print 'Output voltage Vout = ',Vout,'mV'\n",
"print 'input impedance of diff amplifier, Zin(base) = ',Zinb/1000,'KOhm'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"V2 input is at inverting input, So, ideally same as previous case.\n",
"Voltage gain Av = 200.0\n",
"Output voltage Vout = 200.0 mV\n",
"input impedance of diff amplifier, Zin(base) = 15 KOhm\n"
]
}
],
"prompt_number": 96
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-7, Page 633"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.7.py\n",
"#What is ac output voltage in figure 17-12? if B=300, what is input impedance?\n",
"\n",
"#Variable declaration\n",
"VCC=15.0 #supply voltage(V)\n",
"RC=1.0**10**6 #collector resistance(Ohm)\n",
"RE=1.0**10**6 #emitter resistance(Ohm)\n",
"IE=1 #emitter current as per preceding example(mA)\n",
"Vin=7 #input ac voltage(mV)\n",
"B=300 #current gain\n",
"\n",
"#Calculation\n",
"IT=VCC/RE #tail current (uA)\n",
"IE=IT/2 #emitter current (uA)\n",
"re=25/IE/1000 #ac emitter resistance(Ohm)\n",
"Av=RC/(2*re) #voltage gain\n",
"Vout=Av*Vin #Output voltage(V)\n",
"Zinb=2*B*re #input impedance of diff amp(MOhm)\n",
"\n",
"#Result\n",
"print 'Voltage gain Av = ',Av\n",
"print 'Output voltage Vout = ',Vout/1000,'V'\n",
"print 'input impedance of diff amplifier, Zin(base) = ',Zinb,'MOhm'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltage gain Av = 150.0\n",
"Output voltage Vout = 1.05 V\n",
"input impedance of diff amplifier, Zin(base) = 2.0 MOhm\n"
]
}
],
"prompt_number": 97
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-8, Page 639"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.8.py\n",
"#The diff amp of figure 17-17 has Av=200, Iin(bias)=3uA, Iin(off) =0.5 uA. and Vin(off)= 1mV. What is output error voltage?\n",
"#what is output error voltage?\n",
"\n",
"#Variable declaration\n",
"VCC=15.0 #supply voltage(V)\n",
"RC=5.0*10**3 #collector resistance(Ohm)\n",
"RE=7.5*10**3 #emitter resistance(Ohm)\n",
"RB=1*10**3 #base resistance(Ohm)\n",
"Vin=10 #input ac voltage(mV)\n",
"Av=200 #voltage gain\n",
"Iinb=3*10**-6 #bias current(A)\n",
"Iino =0.5*10**-6 #Iin(off) (A) \n",
"Vino=1.0 #Vin(off) (mV)\n",
"\n",
"#Calculation\n",
"V1err=RB*Iinb*1000 #dc error input1 (mV)\n",
"V2err=RB*(Iino/2)*1000 #dc error input2 (mV)\n",
"V3err=Vino #dc error input3 (mV)\n",
"Verror=Av*(V1err+V2err+V3err) #output error voltage(mV)\n",
"#if matching base resistor is used,\n",
"V1err1=0 #dc error input1 (mV)\n",
"V2err1=RB*Iino*1000 #dc error input2 (mV)\n",
"V3err1=Vino #dc error input3 (mV)\n",
"Verror1=Av*(V1err1+V2err1+V3err1) #output error voltage(mV)\n",
"\n",
"#Result\n",
"print 'output error voltage Verror = ',Verror,'mV'\n",
"print 'output error voltage Verror = ',Verror1,'mV'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"output error voltage Verror = 850.0 mV\n",
"output error voltage Verror = 300.0 mV\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-9, Page 640"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.9.py\n",
"#The diff amp of figure 17-18 has Av=300, Iin(bias)=80nA, Iin(off) =20 nA. and Vin(off)= 5mV. What is output error voltage?\n",
"\n",
"#Variable declaration\n",
"VCC=15.0 #supply voltage(V)\n",
"RC=1.0*10**6 #collector resistance(Ohm)\n",
"RE=1.0*10**6 #emitter resistance(Ohm)\n",
"RB=10*10**3 #base resistance(Ohm)\n",
"Vin=10 #input ac voltage(mV)\n",
"Av=300 #voltage gain\n",
"Iinb=80*10**-9 #bias current(A)\n",
"Iino=20*10**-9 #Iin(off) (A) \n",
"Vino=5.0 #Vin(off) (mV)\n",
"\n",
"#Calculation\n",
"V1err=0 #dc error input1 (mV)\n",
"V2err=RB*(Iino/2)*1000 #dc error input2 (mV)\n",
"V3err=Vino #dc error input3 (mV)\n",
"Verror=Av*(V1err+V2err+V3err) #output error voltage(mV)\n",
"\n",
"#Result\n",
"print 'output error voltage Verror = ',Verror/1000,'V'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"output error voltage Verror = 1.53 V\n"
]
}
],
"prompt_number": 98
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-10, Page 643"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.10.py\n",
"#In figure 17-2, what is common-node voltage gain & output voltage?\n",
"\n",
"#Variable declaration\n",
"VCC=15.0 #supply voltage(V)\n",
"RC=1.0*10**6 #collector resistance(Ohm)\n",
"RE=1.0*10**6 #emitter resistance(Ohm)\n",
"Vin=1 #input ac voltage(mV)\n",
"\n",
"#Calculation\n",
"Av_CM=RC/(2*RE) #common mode voltage gain\n",
"Vout=Av_CM*Vin #output voltage(mV) \n",
"\n",
"#Result \n",
"print 'common mode voltage gain : Av(CM) = ',Av_CM\n",
"print 'Output voltage Vout = ',Vout,'mV'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"common mode voltage gain : Av(CM) = 0.5\n",
"Output voltage Vout = 0.5 mV\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-11, Page 643"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.11.py\n",
"#In figure 17-22 has Av=150, Av(CM)=0.5, Vin =1 mV. \n",
"#If base leads are picking up a common-mode signal of 1mV, what is the output voltage?\n",
"\n",
"#Variable declaration\n",
"Vin=1 #input ac voltage(mV)\n",
"Av=150 #voltage gain\n",
"Av_CM=0.5 #common mode voltage gain\n",
"\n",
"#Calculation\n",
"Vout1=Av*Vin #output voltage1(mV)\n",
"Vout2=Av_CM*Vin #output voltage2(mV)\n",
"Vout=Vout1+Vout2 #output volatge(mV)\n",
"\n",
"#Result\n",
"print 'output voltage Vout1 = ',Vout1,'mV'\n",
"print 'output voltage Vout2 = ',Vout2,'mV'\n",
"print 'output voltage Vout = ',Vout,'mV'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"output voltage Vout1 = 150 mV\n",
"output voltage Vout2 = 0.5 mV\n",
"output voltage Vout = 150.5 mV\n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-12, Page 644"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.12.py\n",
"#OP-AMP with Av=2,00,000 & CMRR(dB)=90dB. what is the common mode voltage gain? \n",
"#If desired & common-mode signal is of 1uV, what is the output voltage?\n",
"\n",
"#Variable declaration\n",
"CMRR_dB=90 #CMRR(dB)\n",
"Av=200000 #voltage gain\n",
"Vin=1*10**-6 #input voltage(V)\n",
"\n",
"#Calculation\n",
"CMRR=10**(CMRR_dB/20.0) #CMRR \n",
"Av_CM=Av/CMRR #common mode voltage gain \n",
"Vout1=Av*Vin #desired output: voltage1(V)\n",
"Vout2=Av_CM*Vin #common mode output: voltage2(V)\n",
"\n",
"#Result\n",
"print 'output voltage Vout1 = ',Vout1,'V'\n",
"print 'output voltage Vout2 = ',round((Vout2*10**6),2),'uV'\n",
"print 'see, desired output is much larger than common mode output.'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"output voltage Vout1 = 0.2 V\n",
"output voltage Vout2 = 6.32 uV\n",
"see, desired output is much larger than common mode output.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-13, Page 651"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.13.py\n",
"#What is load voltage in figure 17-32a when RL=15KOhm?\n",
"\n",
"#Variable declaration\n",
"VCC=15.0 #supply voltage(V)\n",
"RC=7.5*10**3 #collector resistance(Ohm)\n",
"RE=7.5*10**3 #emitter resistance(Ohm)\n",
"V1=10 #input ac voltage(mV)\n",
"re=25 #as per example 17-4 (Ohm) \n",
"RL=15*10**3 #load resistance(KOhm)\n",
"\n",
"#Calculation\n",
"Av=RC/re #voltage gain\n",
"Vout=Av*V1/1000 #output voltage(V) \n",
"RTH=2*RC #Thevenin resistance(Ohm)\n",
"VL=(RL/(RL+RTH))*Vout #load voltage(V)\n",
"\n",
"#Result \n",
"print 'Load voltage VL = ',VL,'V'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Load voltage VL = 1.5 V\n"
]
}
],
"prompt_number": 67
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 17-14, Page 652"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 17.14.py\n",
"#An ammeter is used for load resistance. find current through the ammeter?\n",
"\n",
"#Variable declaration\n",
"RL=15.0 #load resistance as per previous example(Ohm)\n",
"VL=3 #load voltage as per previous example(V)\n",
"\n",
"#Calculation\n",
"iL=VL/RL #load current(mA)\n",
"\n",
"#Result\n",
"print 'Load current iL = ',iL,'mA'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Load current iL = 0.2 mA\n"
]
}
],
"prompt_number": 69
}
],
"metadata": {}
}
]
}
|
