Chapter 5:Large Signals Amplifiers

Example 5.1,Page number 280

In [1]:
import math

#Variable declaration
Rb=1*10**3                    #base resistance(ohms)
Vcc=20                        #supply voltage(V)
Rc=20                         #collector resistance(ohms) 
beeta=25                      #current gain 
Vbe=0.7                       #base to emitter voltage(V) 
ib=10*10**-3                  #base current(ohms)

#Calculations
Ibq=(Vcc-Vbe)/Rb               #current(A)
Icq=beeta*Ibq                  #current(A)
Vceq=Vcc-(Icq*Rc)              #collector voltage(V)
ic=beeta*ib                    #collector current(A)
Po=((ic/(math.sqrt(2)))**2)*Rc #output voltage(V)
Pi=Vcc*Icq                     #input power(W)
eta=(Po/Pi)*100                #efficiency     
Pd=Pi-((Icq**2)*Rc)-Po         #power dissipated(W)   

#Results
print"input power is Pi",Pi,"W"
print"output power is Po",Po,"W"
print"power dissipated is",round(Pd,1),"W"

input power is Pi 9.65 W
output power is Po 0.625 W
power dissipated is 4.4 W

Example 5.2,Page number 283

In [1]:
import math

#Variable declaration 
Rl=500                           #load resistance(ohms)
Vceq=50                          #queinscent collector voltage(V)
beetamin=30                      #current gain minimum(at Q)
Icq=0.4                          #queinscent collector current(A)
Ibq=8                            #queinscent base current(mA)

#Calculations
Rac=Vceq/Icq                       #ac resistance(ohms)
beeta=(Icq*10**-3)/Ibq             #current gain
Re=5/Icq                           #emitter resistance(ohms)
Rc=(512.5*Rac)/(512.5-Rac)         #as Re+Rl=500+12.5=512.5
Vcc=5+Vceq+(Icq*Rc)                #supply voltage(V)   
Rb=(beetamin*Re)/10                #base resistance(ohms)
R1=39.5                            #solving 125=Rc||(Rl+Re) and Vbb=Vcc*(R1/(R1+R2))
R2=750
Pi=120*Icq                         #Vcc chosen as 120
r=(Rc*Rl)/(Rc+Rl)
Poac=(100/(2*math.sqrt(2)))**2/r  #output power(W)
etamax=Poac/Pi                    #efficiency
Poac1=(100/(2*math.sqrt(2)))**2/Rl #ac power absorbed by load(W)
eta=Poac1/Pi   
Pc=(Icq**2)*Rc                  #power lost in Rc(W)
Pe=(Icq**2)*Re                  #power lost in Re(W)
Pd=Pi-Pc-Pe-Poac                  #power consumed(W)

#Results
print"input power is Pi",Pi,"W"
print"output power is Po",round(Poac,2),"W"
print"dissipated power is",round(Pd,2),"W"
print"values of R1,R2,Re and Rc are",R1,"ohms,",R2,"ohms,",Re,"ohms and",round(Rc),"ohms resp. (Calculated value of Rc is wrong in the book)"

input power is Pi 48.0 W
output power is Po 10.06 W
dissipated power is 9.49 W
values of R1,R2,Re and Rc are 39.5 ohms, 750 ohms, 12.5 ohms and 165.0 ohms resp. (Calculated value of Rc is wrong in the book)

Example 5.3,Page number 285

In [2]:
import math

#Variable declaration
Pmax=10                         #power maximum(W)
Ic=1                            #collector current(A)
Vcemax=100                      #max collector to emitter current(V)
Vcemin=2                        #min collector to emitter current(V)

#Calculations
#Part a
Vceq=46                       #Vce at Q point             
Icq=0.21                      #Ic  at Q point             
Vcc=92                        #supply voltage(V)
ic=0.42                       #collector current(A) 

#Part b
Rl=Vceq/Icq                       #load resistance(ohms)

#Part c
Pi=Vcc*Icq                        #input power(W)
Po=((ic/(2*math.sqrt(2)))**2)*Rl  #output power(W)
eta=(Po/Pi)*100                   #efficiency

#Results
print"Rl for maximum power input is",round(Rl),"ohms"
print"input power is is",Pi,"W"
print"Po is",Po
print"eta is",eta,"%"

Rl for maximum power input is 219.0 ohms
input power is is 19.32 W
Po is 4.83
eta is 25.0 %

Example 5.4,Page number 286

In [24]:
import math

#Variable declaration
Vcc=15                   #supply voltage(V)
beeta=40.                #current gain
Icq=5.                   #Ic at Q(mA)
Vceq=7.5                 #Vce at Q(V)      
icswing=10               #swing in ic(mA) 

#Calculations
#Part a
Rl=Vceq/Icq*10**-3       #load resistance(ohms)    

#Part b
Ibq=Icq/beeta            #base current at Q(uA)

#Part c
ibswing=icswing/beeta                   #swing in ib(mA)
Pac=Rl*(icswing/(2*math.sqrt(2)))**2    #ac power(W)
Pdc=Vcc*(Icq*10**-3)                    #dc power(W)
eta=(Pac/Pdc)*100                       #efficiency

#Results
print"a)value of Rl is",round(Rl/1E-6),"ohms"
print"b)Ibq is",round(Ibq/1E-3),"uA"
print"c)ac power output is",round((Pac/1E-3),2),"mW"
print"efficiency is",eta,"%"
print"corresponding swing in ib is",ibswing,"mA"

a)value of Rl is 1500.0 ohms
b)Ibq is 125.0 uA
c)ac power output is 18.75 mW
efficiency is 25.0 %
corresponding swing in ib is 0.25 mA

Example 5.5,Page number 288

In [5]:
import math

#Variable declaration
Vcc=Vce=10                  #supply voltage(V)  
Icq=140*10**-3              #Ic at Q point(A)
Rl=8                        #load resistance(ohms)
vce=16                      #instantaneous collector to emitter voltage(V)
ic=235*10**-3               #instantaneous collector current(A)

#Calculations
RL=Vcc/Icq
r=math.sqrt(RL/Rl)                             #load resistance for max ac swing(ohms)
Po=(vce*ic)/(2*math.sqrt(2)*2*math.sqrt(2))    #output power(W)
Pi=Vcc*Icq                                     #input power(W)   
eta=Po/Pi                                      #efficiency
Pd=Pi-Po                                       #dissipated power(W)  

#Results
print"a)transformation ratio is",round(r)
print"c)power output is",Po,"W"
print"efficiency is",round(eta*100,2),"%"

a)transformation ratio is 3.0
c)power output is 0.47 W
efficiency is 33.57 %

Example 5.6,Page number 290

In [36]:
import math

#Variable declaration
Rl=4.5                    #load resistance(ohms)
Vceq=50                   #Vc at point Q(V)
Icq=400*10**-3            #Ic at Q(A)
Re=12.5                   #emitter resistance(ohms)
Vcemax=90                 #from figure 
Vcemin=10                 #from figure
Icmax=730                 #max Ic(mA)
Icmin=30                  #min Ic(mA)   

#Calculations
#Part a
Rac=Vceq/Icq                #ac resistance(ohms)
n=math.sqrt(Rac/Rl)         #as n=N1/N2 and Rac=(N1/N2)^2*Rl

#Part b
Vcc=Vceq+(Icq*Re)           #supply voltage(V)   

#Part c
vce=Vcemax-Vcemin                                    #instantaneous collector to emitter voltage(V)
ic=Icmax-Icmin                                       #instantaneous collector current(mA)
Po=(vce*ic)/((2*math.sqrt(2))*(2*math.sqrt(2)))      #output voltage(V)
Pi=Vcc*Icq                                           #input voltage(V)  
eta=(Po/Pi)*100                                      #efficiency
Pd=Pi-(Icq**2*Re)-Po*10**-3                          #dissipated power(W)

#Results
print"a)transformation ratio is",round(n,2)
print"b)Vcc is",Vcc,"V"
print"c)power efficiency for the load is",round((eta/1E+3),1),"%"
print"power dissipated is",Pd,"W"

a)transformation ratio is 5.27
b)Vcc is 55.0 V
c)power efficiency for the load is 31.8 %
power dissipated is 13.0 W

Example 5.7,Page number 295

In [5]:
#Finding input power,output power,Pd,efficiency,

import math

#Variable declaration
Vcc=30                        #supply voltage(V)
Rl=16                         #load resistance(ohms)    
n=2                           #transformation ratio
Im=1                          #peak value of current(A)
etamax=78.54                  #max efficiency(%)

#Calculations
#Part a
Rl1=Rl*(n/2)**2                  #load resistance(ohms)
Pi=(2*Vcc*Im)/math.pi            #input power(W)
Pimax=(2*Vcc**2)/((math.pi)*Rl1) #input power max(W)

#Part b
Po=((Im**2)*Rl1)/2               #output power(W)
Pomax=(Vcc**2)/(2*Rl1)           #output power max(W)

#Part c
eta=Po/Pi                        #efficiency
                

#Part d
P=((2*Vcc*Im)/math.pi)-((Im**2*Rl1)/2) #Power dissipated by transistors(W)
Pd=P/2                                 #power dissipated by each transistors
Pmax=(2*Vcc**2)/((math.pi)**2*Rl1)     #max power dissipated by transistors
Pdmax=Pmax/2                           #max power dissipated by each transistor

#Results
print"a)input power is",round(Pi,1),"W and  max input power is",round(Pimax,2),"W"
print"b)output power ",Po,"W and max output power is",round(Pomax,2),"W"
print"c)power efficiency for the load is",round((eta/1E-2),2),"% and its max value is",etamax,"%"
print"power dissipated by each transiator is",round(Pd,1),"W and  max value is",round(Pdmax,1),"W"

a)input power is 19.1 W and  max input power is 35.81 W
b)output power  8 W and max output power is 28.0 W
c)power efficiency for the load is 41.89 % and its max value is 78.54 %
power dissipated by each transiator is 5.5 W and  max value is 5.7 W

Example 5.8,Page number 296

In [14]:
import math
#Variable declaration
Pd=10

#Calculations
#Part a
Poacmax=10.                #as Pd=Po(ac)max by class A

#Part b
Pd=2*Poacmax               #power dissipated(W)
Poacmax1=146/2             #max output power by class B
f=Poacmax1/Poacmax         #factor by which power of class B is greater than class A
                
#Results
print"maximum signal output powerclass A produce is",Poacmax,"W"
print"maximum signal output powerclass  produce is",Poacmax1,"W"
print"factor by which power in class b is larger than power in class A transformer is",f

maximum signal output powerclass A produce is 10.0 W
maximum signal output powerclass  produce is 73 W
factor by which power in class b is larger than power in class A transformer is 7.3

Example 5.9,Page number 300

In [8]:
import math

#Variable declaration
Vcc=30.                          #supply voltage(V)
Im=1                             #peak value of current(A)
Rl=10.                           #load resistance(ohms)

#Calculations
#Part a
Pi=(Vcc*Im)/math.pi             #input power(W)
Pimax=(Vcc**2)/(math.pi*2*Rl)   #max input power(W)

#Part b
Po=((Im**2)*Rl)/2               #output power(W)
Pomax=(Vcc**2)/(8*Rl)           #output power max(W)

#Part c
eta=Po/Pi                     #efficiency
etamax=Pomax/Pimax            #efficiency max                

#Part d
Pd=Pi-Po                             #Power dissipated by transistors(W)                            
Pmax=(Vcc**2)/(2*(math.pi)**2*Rl)    #max power dissipated by transistors
                              
#Results
print"a)input power is ",round(Pi,2),"W and max input power is",round(Pimax,2),"W"
print"b)output power is ",Po,"W and max output power is",round(Pomax,2),"W"
print"c)power efficiency for the load is",round((eta/1E-2),2),"% and its max value is",round((etamax/1E-2),2),"%"
print"power dissipated and its max value are",round(Pd,2),"W and",round(Pmax,2),"W"

a)input power is  9.55 W and max input power is 14.32 W
b)output power is  5.0 W and max output power is 11.25 W
c)power efficiency for the load is 52.36 % and its max value is 78.54 %
power dissipated and its max value are 4.55 W and 4.56 W

Example 5.10,Page number 303

In [7]:
import math

#Variable declaration
P1=2                          #transistor power(W)
Rl=5*10**3.                   #load resistance()
Ic=35                         #collector current(mA)   

#Calculations
Bo=40-Ic                           
B1=math.sqrt((2*P1)/Rl)
B2=Bo
D2=(B2/B1)*100                #second harmonic distortion(%)

#Results
print"second harmonic distortion is",round((D2/1E+3),2),"%"

second harmonic distortion is 17.68 %

Example 5.12,Page number 314

In [9]:
import math

#Variable declaration
Vcc=15.                            #supply voltage(V)
Rl=10.                             #load resistance(ohms)

#Calculations
#Part a
Immax=Vcc/Rl                      #max peak current(A)
Irmsmax=Immax/(math.sqrt(2))      #max rms current(A)
Pomax=Irmsmax**2*Rl               #max output power(W)
Pi=(2*Vcc*Immax)/math.pi          #max input power(W)
eta=Pomax/Pi                      #efficiency

#Part b 
Im=(2*Vcc)/(math.pi*Rl)                       #peak current(A)
Pdmax=((2*Vcc*Im)/(math.pi))-((Im**2*Rl)/2)   #max power dissipated(W)
eta1=((Im**2)*Rl*math.pi)/(2*2*Vcc*Im)        #efficiency

#Results
print"a)max signal output power,collector dissipation  are",Pomax,"W,",round(Pi,2),"W and efficiency is",round((eta/1E-2),2),"%"
print"b)max dissipation of each transistor and corresponding efficiency is",round(Pdmax,2),"W and",eta1,"resp."

a)max signal output power,collector dissipation  are 11.25 W, 14.32 W and efficiency is 78.54 %
b)max dissipation of each transistor and corresponding efficiency is 4.56 W and 0.5 resp.

Example 5.13,Page number 315

In [27]:
#Calculations
eta=0.5              #As Po(ac)=Vcc^2/2*pi^2*Rl and Pi(dc)=Vcc^2/pi^2*Rl
                     #put these in eta=Po(ac)/Pi(dc) which is 1/2=0.5  
    
#Results
print"push pull amplifier efficiency is",round(eta/1E-2),"%"

push pull amplifier efficiency is 50.0 %