# Chapter-12 Power Amplifiers¶

## Example-12.1 Page Number-676¶

In :
#from the given question:
Vcc=20                 #in volts
Rb=1                   #base resistance in Kohm
B=25                   #gain
Rc=20                  #collector resistance in ohm

#calculation of Q-point Parameters:
Ibq=(Vcc-0.7)/Rb           #base current
Icq=B*Ibq                  #collector current
Vce=Vcc-Icq*Rc              #collector -emitter voltage

#when applying ac signal:
Ib=10                      #peak base current value in mA
Ic=B*Ib                    #peak collector current in mA
Ic=Ic*(10**-3)             #converting Ic to ampere

Po=(Ic*Ic*Rc)/2            #output ac power in watt
Pi=Vcc*(Icq*(10**-3))       #input dc power in watt

n=(Po/Pi)*100              #efficiency in %

print "Input dc power Pi=",Pi,"W"
print "output ac power Po=",Po,"W"
print "Efficiency n=",round(n,1),"%"

Input dc power Pi= 9.65 W
output ac power Po= 0.625 W
Efficiency n= 6.5 %


## Example-12.2 Page Number-678¶

In :
#for the given transformer:
N1=15.0                     #no. of turns in primary coil
N2=1.0                      #no. of turns in secondary coil

#as seen looking into the primary coil of the transformer:
a=N1/N2
Rle=a*a*Rl

print "effective resistance Rl=",(Rle/1000),"Kohm"

effective resistance Rl= 1.8 Kohm


## Example-12.3 Page Number-679¶

In :
import math
#for the given transformer:
Rle=10*(10**3)               #effective load resistance in ohm

#as seen looking into the primary coil of the transformer:
x=Rle/Rl
Tr=math.sqrt(x)

print "Turn ratio =",int(Tr),": 1 "

Turn ratio = 25 : 1


## Example-12.5 Page Number-682¶

In :
#from the given Question:
Vcc=10                      #supply voltage in volts
Icq=140*(10**-3)            #collector current in ampere
Po=0.477                    #output ac power in watt

#input dc power Pi:
Pi=Vcc*Icq                  #in watt
#power dissipiated Pq:
Pq=Pi-Po                    #in watt

#efficiency n:
n=(Po/Pi)*100               #in %

print "DC input power Pi=",Pi,"W"
print "power dissipiated by the transistor Pq=",Pq,"W"
print "Efficiency n=",round(n,1),"%"

DC input power Pi= 1.4 W
power dissipiated by the transistor Pq= 0.923 W
Efficiency n= 34.1 %


## Example-12.6(a) Page Number-683¶

In :
#for the transformer-coupled class A amplifier:

Vcc=12                     #supply voltage in volts
Vce=Vcc                    #collector-emitter voltage
Vp=12                      #output voltage in volts

Vcemax=Vce+Vp              #maximum value of Vce
Vcemin=Vce-Vp              #minimum value of Vce
x=(Vcemax-Vcemin)/(Vcemax+Vcemin)
n=50*(x*x)

print "Efficiency of the amplifier n=",n,"%"

Efficiency of the amplifier n= 50 %


## Example-12.6(b) Page Number-683¶

In :
#for the transformer-coupled class A amplifier:

Vcc=12                     #supply voltage in volts
Vce=Vcc                    #collector-emitter voltage
Vp=6.0                     #output voltage in volts

Vcemax=Vce+Vp              #maximum value of Vce
Vcemin=Vce-Vp              #minimum value of Vce

x=((Vcemax-Vcemin)/(Vcemax+Vcemin))
n=50*x*x

print "Efficiency of the amplifier n=",n,"%"

Efficiency of the amplifier n= 12.5 %


## Example-12.6(c) Page Number-683¶

In :
#for the transformer-coupled class A amplifier:

Vcc=12                     #supply voltage in volts
Vce=Vcc                    #collector-emitter voltage
Vp=2.0                     #output voltage in volts

Vcemax=Vce+Vp              #maximum value of Vce
Vcemin=Vce-Vp              #minimum value of Vce

x=((Vcemax-Vcemin)/(Vcemax+Vcemin))
n=50*x*x

print "Efficiency of the amplifier n=",round(n,2),"%"

Efficiency of the amplifier n= 1.39 %


## Example-12.7 Page Number-686¶

In :
#for the transformer-coupled class B amplifier:

Vcc=30                     #supply voltage in volts
Vp=20                      #output voltage in volts

#calculation:
Ilp=Vp/Rl                #peak load current in ampere
Idc=(2*Ilp)/3.14         #dc value of current drawn from power supply im ampere
Pi=Vcc*Idc               #input dc power in watt
Po=(Vp*Vp)/(2*Rl)        #output ac power in watt

n=(Po/Pi)*100            #efficiency in %

print "Input dc power Pi=",round(Pi,1),"W"
print "output ac power Po=",Po,"W"
print "Efficiency n=",round(n,1),"%"

Input dc power Pi= 23.9 W
output ac power Po= 12.5 W
Efficiency n= 52.3 %


## Example-12.8 Page Number-687¶

In :
#for the transformer-coupled class B amplifier:

Vcc=30                     #supply voltage in volts

#calculation:
Po=(Vcc*Vcc)/(2*Rl)                      #output ac power in watt
Pi=(2*Vcc*Vcc)/(Rl*3.14)                 #input dc power in watt
n=(Po/Pi)*100                            #efficiency in %
Pmax=(0.5*2*Vcc*Vcc)/(Rl*3.14*3.14)      #maximum power dissipiated

print "DC input power Pi=",round(Pi,2),"W"
print "AC output power Po=",round(Po,3),"W"
print "maximum power dissipiated by each transistor Pmax=",round(Pmax,1),"W"

DC input power Pi= 35.83 W
AC output power Po= 28.125 W
maximum power dissipiated by each transistor Pmax= 5.7 W


## Example-12.9(a) Page Number-687¶

In :
#for the transformer-coupled class B amplifier:

Vcc=24.0                     #supply voltage in volts
Vp=22.0                      #output voltage in volts

#calculation:
n=78.54*(Vp/Vcc)            #efficiency in %

print "Efficiency n=",round(n,1),"%"

Efficiency n= 72.0 %


## Example-12.9(b) Page Number-687¶

In :
#for the transformer-coupled class B amplifier:

Vcc=24.0                     #supply voltage in volts
Vp=06.0                      #output voltage in volts

#calculation:
n=78.54*(Vp/Vcc)            #efficiency in %

print "Efficiency n=",round(n,1),"%"

Efficiency n= 19.6 %


## Example-12.10 Page Number-692¶

In :
import math
#for the given circuit:
Vrms=12                     #supply volts in rms voltage
Vcc=25                      #in volts

#Calculation:
Vi=math.sqrt(2)*Vrms               #peak input voltage in volts
Vl=Vi                       #voltage across load as gain=1
Po=(Vl*Vl)/(2*Rl)           #Output power across load in watt

Il=Vl/Rl                    #peak load current in ampere
Idc=(2*Il)/3.14             #dc current from supplies

Pi=Vcc*Idc                  #power supplied to circuit in watt

Pq=(Pi-Po)/2                #power dissipiated

n=(Po/Pi)*100               #efficiency in %

print "DC input power Pi=",round(Pi,2),"W"
print "AC output power Po=",round(Po,2),"W"
print "maximum power dissipiated by each transistor Pmax=",round(Pq,1),"W"
print "Efficiency n=",round(n,1),"%"

DC input power Pi= 67.56 W
AC output power Po= 36.0 W
maximum power dissipiated by each transistor Pmax= 15.8 W
Efficiency n= 53.3 %


## Example-12.11 Page Number-693¶

In :
#for the given circuit:
Vrms=12                     #supply volts in rms voltage
Vcc=25                      #in volts

#Calculation:
Pi=(2*Vcc*Vcc)/(Rl*3.142)    #Input power
Po=(Vcc*Vcc)/(2*Rl)         #Output power in watt
n=(Po/Pi)*100               #efficiency in %

Pq=(Pi-Po)                  #power dissipiated
Vl=Vp                       #condition to achieve maximum power operation

print "DC input power Pi=",round(Pi,2),"W"
print "AC output power Po=",round(Po,3),"W"
print "maximum power dissipiated by each transistor Pmax=",round(Pq,1),"W"
print "Efficiency n=",round(n,1),"%"

DC input power Pi= 99.46 W
AC output power Po= 78.125 W
maximum power dissipiated by each transistor Pmax= 21.3 W
Efficiency n= 78.5 %


## Example-12.12 Page Number-693¶

In :
#for the given circuit:
Vrms=12                     #supply volts in rms voltage
Vcc=25                      #in volts

#Calculation:
Pmax=(2*Vcc*Vcc)/(3.142*3.142*Rl)        #maximum power dissipiated in watt
Vl=0.636*Vcc                             #input voltage for maximum power dissipiation in volts

print "maximum power dissipiated=",round(Pmax,2),"W"
print "Input voltage for maximum power dissipiated=",Vl,"V"

maximum power dissipiated= 31.65 W
Input voltage for maximum power dissipiated= 15.9 V


## Example-12.13 Page Number-694¶

In :
#for the given output signal:
A1=2.5                         #fundamental amplitude in volts
A2=0.25                        #second harmonic amplitude in volts
A3=0.1                         #Third harmonic amplitude in volts
A4=0.05                        #Fourth harmonic amplitude in volts

#calculating Harmonic Distortions:
D2=(A2/A1)*100
D3=(A3/A1)*100
D4=(A4/A1)*100

print " Second harmonic distortion D2=",D2,"%"
print " Third harmonic distortion D3=",D3,"%"
print " Fourth harmonic distortion D4=",D4,"%"

 Second harmonic distortion D2= 10.0 %
Third harmonic distortion D3= 4.0 %
Fourth harmonic distortion D4= 2.0 %


## Example-12.14 Page Number-694¶

In :
import math
#for the given output signal:
A1=2.5                         #fundamental amplitude in volts
A2=0.25                        #second harmonic amplitude in volts
A3=0.1                         #Third harmonic amplitude in volts
A4=0.05                        #Fourth harmonic amplitude in volts

#calculating Harmonic Distortions:
D2=(A2/A1)
D3=(A3/A1)
D4=(A4/A1)

THD=math.sqrt((D2*D2)+(D3*D3)+(D4*D4))*100

print "Total harmonic Distortion THD=",round(THD,2),"%"

Total harmonic Distortion THD= 10.95 %


## Example-12.15(a) Page Number-696¶

In :
Vcemin=1.0                             #maximum value of collector emitter voltage in volts
Vcemax=22.0                            #minimum value of collector emitter voltage in volts
Vceq=12.0                              #collector emitter voltage in volts at Q-point

x=(Vcemax+Vcemin)/2.0                          #temporary variable

D2=(abs(x-Vceq)/abs(Vcemax-Vcemin))*100        #in %

print "second harmonic distortion D2=",round(D2,2),"%"

second harmonic distortion D2= 2.38 %


## Example-12.15(b) Page Number-696¶

In :
Vcemin=4.0                             #maximum value of collector emitter voltage in volts
Vcemax=20.0                            #minimum value of collector emitter voltage in volts
Vceq=12.0                              #collector emitter voltage in volts at Q-point

x=(Vcemax+Vcemin)/2.0                       #temporary variable

D2=(abs(x-Vceq)/abs(Vcemax-Vcemin))*100     #in %

print "second harmonic distortion D2=",round(D2,2),"%"

second harmonic distortion D2= 0.0 %


## Example-12.16 Page Number-696¶

In :
#given the distortion reading:
D2=0.1               #second harmonic distortion
D3=0.02              #third harmonic distortion
D4=0.01              #fourth harmonic distortion

I1=4                 #in ampere
Rc=8                 # load resistance in ohm

#Calculation:
THD=math.sqrt((D2*D2)+(D3*D3)+(D4*D4))        #Total harmonic distortion
P1=(I1*I1*Rc)/2                               #Fundamental power in watt
P=(1+THD*THD)*P1                              #Total power in watt

print "Total harmonic distortion=",round(THD,2),"%"
print "Fundamental power P=",P1,"W"
print "Total power P=",round(P,2),"W"

Total harmonic distortion= 0.1 %
Fundamental power P= 64 W
Total power P= 64.67 W


## Example-12.17 Page Number-698¶

In :
#for the given silicon transistor:
T1=125                        #temperature in degree celsius
T2=25                         #temperature in degree celsius
Df=0.5                        #derating factor in W/degree C
Pd=80                         #powerdissipiation at 25 degree celsius

PdT1=Pd-(T1-T2)*Df             #power dissipiation at T1=125 degree celsius

print "maximum dissipiation at 125 degreeC=",PdT1,"W"

 maximum dissipiation at 125 degreeC= 30.0 W


## Example-12.18 Page Number-700¶

In :
#for the given silicon power transistor:
Tsa=1.5             #in degreeC/W (heat sink thermal resistance)
Tjc=0.5             #in degreeC/W (transistor thermal resistance)
Tcs=0.6             #in degreeC/W (insulator thermal resistance)

Tj=200              #maximum junction temperature in celsius
Ta=40               #ambient temperature in celsius

Pd=(Tj-Ta)/(Tjc+Tcs+Tsa)
print " maximum power dissipiation Pd=",round(Pd,2),"W"

 maximum power dissipiation Pd= 61.54 W