Chapter8-Analysis of transistor Amplifier using Hybrid Equivalent Circuit

Ex1-pg262

In [1]:
##Ex8_1
import math
print("(a)")
Vce=0.
Ic=2.*10**-3
Ib=30.*10**-6
Vbe=50.*10**-3
print'%s %.f %s'%("Vce = ",(Vce),"V")##collector-emitter voltage
print'%s %.2e %s'%("Ic = ",(Ic),"A")##collector current
print'%s %.f %s'%("Ib = ",(Ib),"A")##base current
print'%s %.2f %s'%("Vbe = ",(Vbe),"V")##base-emitter voltage
hfe=Ic/Ib
print'%s %.2f %s'%("hfe = Ic/Ib = ",(hfe),"")##current gain in CE amplifier
hie=Vbe/Ib
print'%s %.2f %s'%("hie = Vbe/Ib = ",(hie),"ohm")##input impedance in CE amplifier
print("(b)")
Ib=0.
Vce=1.
Vbe=0.3*10**-3
Ic=0.1*10**-3
print'%s %.f %s'%("Vce = ",(Vce),"V")##collector-emitter voltage
print'%s %.2e %s'%("Ic = ",(Ic),"A")##collector current
print'%s %.f %s'%("Ib = ",(Ib),"A")##base current
print'%s %.2e %s'%("Vbe = ",(Vbe),"V")##base-emitter voltage
hoe=Ic/Vce
print'%s %.2e %s'%("hoe = Ic/Vce = ",(hoe),"mho")##output conductance in CE amplifier
hre=Vbe/Vce
print'%s %.2e %s'%("hre = Vbe/Vce = ",(hre),"")##voltage gain in CE amplifier

## note: textbook answers has printing mistake, regaeding hre.

(a)
Vce =  0 V
Ic =  2.00e-03 A
Ib =  0 A
Vbe =  0.05 V
hfe = Ic/Ib =  66.67 
hie = Vbe/Ib =  1666.67 ohm
(b)
Vce =  1 V
Ic =  1.00e-04 A
Ib =  0 A
Vbe =  3.00e-04 V
hoe = Ic/Vce =  1.00e-04 mho
hre = Vbe/Vce =  3.00e-04

Ex2-pg263

In [2]:
##Ex8_2
import math
RL = 8.*10**3
hie=1.0*10**3
hre=2.5*10**-4
hfe=50.
hoe=25.*10**-6
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
##h-parameters for CE transistor amplifier are as follows:
print'%s %.2f %s'%("hie = ",(hie),"ohm")##input resistance of CE transistor
print'%s %.2e %s'%("hre = ",(hre),"")##voltage gain of CE transistor
print'%s %.2f %s'%("hfe = ",(hfe),"")##current gain of CE transistor
print'%s %.2e %s'%("hoe = ",(hoe),"mho")##output conductance of CE transistor
##calculation for current gain:
Ai=-hfe/(1.+(hoe*RL))
print'%s %.2f %s'%("Ai = -hfe/(1+(hoe*RL)) = ",(Ai),"")
print'%s %.2f %s'%("Ai = ",(abs(Ai)),"")
##calculation for input resistance:
Ri = hie+(hre*Ai*RL)
print'%s %.2f %s'%("Ri = hie+(hre*Ai*RL) = ",(Ri),"ohm")

##note : answer in the textbook regarding above problem is not accuratly calculated.

RL =  8000.00 ohm
hie =  1000.00 ohm
hre =  2.50e-04 
hfe =  50.00 
hoe =  2.50e-05 mho
Ai = -hfe/(1+(hoe*RL)) =  -41.67 
Ai =  41.67 
Ri = hie+(hre*Ai*RL) =  916.67 ohm

Ex3-pg263

In [3]:
##Ex8_3
import math
RL = 8.*10**3
Rs= 500.
hie=1.0*10**3
hre=2.5*10**-4
hfe=50.
hoe=25.*10**-6
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
print'%s %.2f %s'%("Rs = ",(Rs),"ohm")##source resistance
##h-parameters for CE transistor amplifier are as follows:
print'%s %.2f %s'%("hie = ",(hie),"ohm")##input resistance of CE transistor
print'%s %.2e %s'%("hre = ",(hre),"")##voltage gain of CE transistor
print'%s %.2f %s'%("hfe = ",(hfe),"")##current gain of CE transistor
print'%s %.2e %s'%("hoe = ",(hoe),"mho")##output conductance of CE transistor

Ai=-hfe/(1.+(hoe*RL))
print'%s %.2f %s'%("Ai = -hfe/(1+(hoe*RL)) = ",(Ai),"")##calculation for current gain

Ri = hie+(hre*Ai*RL)
print'%s %.2f %s'%("Ri = hie+(hre*Ai*RL) = ",(Ri),"ohm")##calculation for input resistance

Ais=(Ai*Rs)/(Ri+Rs)
print'%s %.2f %s'%("Ais = (Ai*Rs)/(Ri+Rs)= ",(Ais),"")##current gain with source resistance

Avs = Ai*RL/Ri
print'%s %.2f %s'%("Avs = Ai*RL/Ri = ",(Avs),"")##voltage gain with source resistance

##note : in the textbook above problem has given two values for hie BUT no value for hfe ... 
##       thus assuming hie=50 as hfe =50, as given in the previous example 8_2

##note : answer in the textbook is not accuratly calculated.

RL =  8000.00 ohm
Rs =  500.00 ohm
hie =  1000.00 ohm
hre =  2.50e-04 
hfe =  50.00 
hoe =  2.50e-05 mho
Ai = -hfe/(1+(hoe*RL)) =  -41.67 
Ri = hie+(hre*Ai*RL) =  916.67 ohm
Ais = (Ai*Rs)/(Ri+Rs)=  -14.71 
Avs = Ai*RL/Ri =  -363.64

Ex4-pg264

In [4]:
##Ex8_4
import math
RL=5.*10**3
Rs=1.2*10**3
hre=2.5*10**-4
hie=1.1*10**3
hfe=100.
hoe=25.*10**-6
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
print'%s %.2f %s'%("Rs = ",(Rs),"ohm")##source resistance
##h-parameters for CE transistor amplifier are as follows:
print'%s %.2f %s'%("hie = ",(hie),"ohm")##input resistance of CE transistor
print'%s %.2e %s'%("hre = ",(hre),"")##voltage gain of CE transistor
print'%s %.2f %s'%("hfe = ",(hfe),"")##current gain of CE transistor
print'%s %.2e %s'%("hoe = ",(hoe),"mho")##output conductance of CE transistor
##calculation for current gain:
Ai=-hfe/(1.+(hoe*RL))
print'%s %.2f %s'%("Ai = -hfe/(1+(hoe*RL)) = ",(abs(Ai)),"")
##calculation for input resistance:
Ri = hie+(hre*Ai*RL)
print'%s %.2f %s'%("Ri = hie+(hre*Ai*RL) = ",(Ri),"ohm")
##calculation for voltage gain:
Av = Ai*RL/Ri
print'%s %.2f %s'%("Av = Ai*RL/Ri = ",(Av),"")
##calculation for output resistance:
Go=hoe-((hre*hfe)/(hie+Rs))
Ro = 1./Go
print("Ro = 1/Go")
print'%s %.2e %s'%("Go = hoe-((hre*hfe)/(hie+Rs)) = ",(Go),"mho")
print'%s %.2f %s'%("Ro = ",(Ro),"ohm")

##note : in the textbook, above problem has given two values for "hfe" and no value for "hre"... 
##        thus assuming value for "hre = 2.5*10^-4" as taken in previous example 8_2
##        and "hfe=100" 

##note : in text LOAD RESISTANCE is noted as Rc in question, but RL in solution.
##       I have work with Load Resistance with notification RL.

RL =  5000.00 ohm
Rs =  1200.00 ohm
hie =  1100.00 ohm
hre =  2.50e-04 
hfe =  100.00 
hoe =  2.50e-05 mho
Ai = -hfe/(1+(hoe*RL)) =  88.89 
Ri = hie+(hre*Ai*RL) =  988.89 ohm
Av = Ai*RL/Ri =  -449.44 
Ro = 1/Go
Go = hoe-((hre*hfe)/(hie+Rs)) =  1.41e-05 mho
Ro =  70769.23 ohm

Ex5-pg264

In [5]:
##Ex8_5
import math
RL = 22.*10**3
hfb=-0.98
hob=7.6*10**-7
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
print'%s %.2f %s'%("hfb = ",(hfb),"")##forward current gain in CB amplifier
print'%s %.2e %s'%("hob = ",(hob),"mho")##output conductance in CB amplifier
Ai = -hfb/(1.+(hob*RL))
print'%s %.2f %s'%("Ai = -hfb/(1+(hob*RL)) = ",(Ai),"")##current gain

RL =  22000.00 ohm
hfb =  -0.98 
hob =  7.60e-07 mho
Ai = -hfb/(1+(hob*RL)) =  0.96

Ex6-pg265

In [6]:
##Ex8_6
import math
hfb = -0.999
hib = 50.
hob = 0.82*10**-6
hrb = 4.*10**-6
RL = 22.*10**3
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load impedence
##h-parameters for CB transistor amplifier are as follows:
print'%s %.2f %s'%("hib = ",(hib),"ohm")##input resistance of CB transistor
print'%s %.2e %s'%("hrb = ",(hrb),"")##voltage gain of CB transistor
print'%s %.2f %s'%("hfb = ",(hfb),"")##current gain of CB transistor
print'%s %.2e %s'%("hob = ",(hob),"mho")##output conductance of CB transistor
Av = -(hfb*RL)/((RL*(hib*hob-hfb*hrb))+hib)
print'%s %.2f %s'%("Av = -(hfb*RL)/((RL*(hib*hob-hfb*hrb))+hib) = ",(Av),"")##voltage gain


##note : answer provided in the textbook is not precised.

RL =  22000.00 ohm
hib =  50.00 ohm
hrb =  4.00e-06 
hfb =  -1.00 
hob =  8.20e-07 mho
Av = -(hfb*RL)/((RL*(hib*hob-hfb*hrb))+hib) =  431.03

Ex7-pg265

In [7]:
##Ex8_7
import math
RL = 1.2*10**3
##assuming Rs = RL as given in problem
Rs = RL
##assuming values for h-parameters
hie = 1.0*10**3
hre=2.5*10**-4
hfe = 50.
hoe = 25.*10**-6
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
print'%s %.2f %s'%("Rs = RL = ",(RL),"ohm")##source resistance
##h-parameters for CE transistor amplifier are as follows:
print'%s %.2f %s'%("hie = ",(hie),"ohm")##input resistance of CE transistor
print'%s %.2e %s'%("hre = ",(hre),"")##voltage gain of CE transistor
print'%s %.2f %s'%("hfe = ",(hfe),"")##current gain of CE transistor
print'%s %.2e %s'%("hoe = ",(hoe),"mho")##output conductance of CE transistor
##calculation for current gain:
Ai=-hfe/(1+(hoe*RL))
print'%s %.2f %s'%("Ai = -hfe/(1+(hoe*RL)) = ",(Ai),"")
##calculation for input impedence:
Ri = hie+(hre*Ai*RL)
print'%s %.2f %s'%("Ri = hie+(hre*Ai*RL) = ",(Ri),"ohm")
##calculation for voltage gain:
print("Av = Ai*RL/Ri")
Av = Ai*RL/Ri
print'%s %.2f %s'%("   = ",(Av),"")
##calculation for output impedence:
Ro = 1/((hoe - (hfe*hre)/(hie+Rs)))
print("Ro = 1/((hoe - (hfe*hre)/(hie+Rs)))")
print'%s %.2f %s'%("     = ",(Ro),"ohm")
##current gain with source impedence:
Ais=(Ai*Rs)/(Ri+Rs)
print'%s %.2f %s'%("Ais = (Ai*Rs)/(Ri+Rs)= ",(Ais),"")
##voltage gain with source impedence:
Avs = Av*Ri/(Ri+Rs)
print'%s %.2f %s'%("Avs = Av*Ri/(Ri+Rs) = ",(Avs),"")



## NOTE : calculation in the textbook for the above problem is wrong.
##        while calculating Ri author has use "hie = 1.2*10^3" instead of ASSUMED9 value i.e., "hie = 1.0*10^3"   

RL =  1200.00 ohm
Rs = RL =  1200.00 ohm
hie =  1000.00 ohm
hre =  2.50e-04 
hfe =  50.00 
hoe =  2.50e-05 mho
Ai = -hfe/(1+(hoe*RL)) =  -48.54 
Ri = hie+(hre*Ai*RL) =  985.44 ohm
Av = Ai*RL/Ri
   =  -59.11 
Ro = 1/((hoe - (hfe*hre)/(hie+Rs)))
     =  51764.71 ohm
Ais = (Ai*Rs)/(Ri+Rs)=  -26.65 
Avs = Av*Ri/(Ri+Rs) =  -26.65

Ex8-pg267

In [8]:
##Ex8_8
import math
Ai = -60.
hfe = 100.
hoe = 10.*10**-6
print'%s %.2f %s'%("hfe = ",(hfe),"")##forward current gain
print'%s %.2e %s'%("hoe = ",(hoe),"A/V")##output conductance
print'%s %.2f %s'%("Ai = ",(Ai),"")##current gain


RL = -(1./hoe)*(1.+(hfe/Ai))
RL = -(1./hoe)*(1.+(hfe/Ai))
print'%s %.2f %s'% (" ",(RL),"ohm") ##load resistance

hfe =  100.00 
hoe =  1.00e-05 A/V
Ai =  -60.00 
  66666.67 ohm

Ex9-pg267

In [9]:
##Ex8_9
import math
Ai = -60.
Ri = 2.0*10**3
RL = 15.*10**3
print'%s %.2f %s'%("Ai = ",(Ai),"")##current gain
print'%s %.2f %s'%("Ri = ",(Ri),"ohm")##input resistance
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
Av = Ai*RL/Ri
print'%s %.2f %s'%("Av = Ai*RL/Ri = ",(Av),"")##voltage gain

##note : in textbook,
##       author notify LOAD RESISTANCE as 'Rc' in question BUT 'RL' in solution.
##       I have work with "load resistance notified as RL".

Ai =  -60.00 
Ri =  2000.00 ohm
RL =  15000.00 ohm
Av = Ai*RL/Ri =  -450.00

Ex10-pg267

In [10]:
##Ex8_10
import math
Av = -200.
Ri = 10.*10**3
RL = 3.*10**3
Ai = Av*Ri/RL
print'%s %.2f %s'%("Av = ",(Av),"")##voltage gain
print'%s %.2f %s'%("Ri = ",(Ri),"ohm")##input resistance
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
print'%s %.2f %s'%("Ai = Av*Ri/RL = ",(Ai),"")##current gain

## note : there are mis-printring in the textbook for the above problem regading formula and notations.
##        answer in the textbook for above problem is wrong.

Av =  -200.00 
Ri =  10000.00 ohm
RL =  3000.00 ohm
Ai = Av*Ri/RL =  -666.67

Ex11-pg267

In [11]:
##Ex8_11
import math
Av = -250.
Ai = -50.
RL = 12.*10**3
print'%s %.2f %s'%("Av = ",(Av),"")##voltage gain
print'%s %.2f %s'%("Ai = ",(Ai),"")##current gain
print'%s %.2f %s'%("RL = ",(RL),"ohm")##load resistance
Ri = Ai*RL/Av
print'%s %.2f %s'%("Ri = Ai*RL/Av = ",(Ri),"ohm")##input resistance

Av =  -250.00 
Ai =  -50.00 
RL =  12000.00 ohm
Ri = Ai*RL/Av =  2400.00 ohm