# Chapter 5 : Power Amplifiers¶

## Example 5.1, Page No. 167¶

In [3]:
# efficiency

import math
#Variable declaration
Pac=0.1             #in W
Vcc=20.0            #in V
Ic=20.0             #in mA

#Calculations
Pdc=Vcc*Ic*10**-3   #in W
eta=(Pac/Pdc)*100   #efficiency

#Result
print("efficiency is ,(%%)= %.f"%eta)

efficiency is ,(%)= 25


## Example 5.2, Page No. 167¶

In [5]:
# collector current

import math
#Variable declaration
Pac=2.0                 #in W
Vcc=12.0                #in V

#Calculations
Ic=(Pac*math.sqrt(2)*math.sqrt(2))/Vcc

#Result
print("maximum collector current is ,(A)= %.2f" %Ic)

maximum collector current is ,(A)= 0.33


## Example 5.3, Page No. 167¶

In [11]:
# collector efficiency and power rating

import math
#Variable declaration
Pac=3.0                 #in W
Pdc=10.0                #in W

#Calculations
eta=(Pac/Pdc)*100       #percentage efficieny

#Result
print("collector efficiency is ,      (%%) = %.f"%eta)
print("power rating of transistor is ,(W) = %.f" %Pdc)

collector efficiency is ,      (%) = 30
power rating of transistor is ,(W) = 10


## Example 5.4, Page No. 168¶

In [2]:
# power

import math
#Variable declaration
dIc=100.0                             #in mA
Rl=6.0                                #in ohm
Rl=6.0                                #in ohm
dVc=10.0                              #in V

#Calculations
mv=dIc*Rl*10**-3                      #in V
pd=mv*dIc                             #in mW
oi=(dVc/dIc)*10**3                    #in ohm
n=math.sqrt(oi/Rl)                    #turn ratio of transformer
tsv=dVc/n                             #om V
Il=tsv/Rl                             #in A
ptr= Il**2*Rl*10**3                   #in mW

#Result
print("(i)  power developed in loudspeaker is ,  (mW)= %.f"%pd)
print("(ii) power transferred to loudspeaker is ,(mw)= %.f"%ptr)
#in textbook in second case there is one point deviation in the answer.

(i)  power developed in loudspeaker is ,  (mW)= 60
(ii) power transferred to loudspeaker is ,(mw)= 1000


## Example 5.5, Page No. 168¶

In [20]:
# power

import math
#Variable declaration
n=10.0                          #turn ratio
Rl=10.0                         #ohm
Ic=100.0                        #in mA

#Result
Rld=n**2*Rl                     #in ohm
Irms=Ic/(math.sqrt(2))          #in mA
P=Irms**2*Rld                    #in W

#Result
print("maximum power output is ,(W)= %.f"%(P*10**-6))

maximum power output is ,(W)= 5


## Example 5.6, Page No. 169¶

In [6]:
# harmonic distortions and change in power

import math
#Variable declaration
#ie=15*sin 400*t+1.5*sin 800*t + 1.2*sin 1200*t + 0.5*sin 1600*t  given equation
I2=1.5                               #in A
I1=15.0                              #in A
I3=1.2                               #in A
I4=0.5                               #in A
D2=(I2/I1)*100                       #Second percentage harmonic distortion
D3=(I3/I1)*100                       #Third percentage harmonic distortion
#in book I2 is mentioned wrongly in place of I1
D4=(I4/I1)*100                       #Fourth percentage harmonic distortion
D=math.sqrt(D2**2+D3**2+D4**2)/100   #Distortion Factor
P1=1.0                               #assume
P=(1+D**2)*P1                        #in W
peri=((P-P1)/P1)*100                 #percentage increase in power due to distortion

#Result
print("part (i)")
print("Second percentage harmonic distortion (D2) is ,(%%) = %.f"%D2)
print("Third percentage harmonic distortion  (D3) is ,(%%) = %.f"%D3)
print("Fourth percentage harmonic distortion (D4) is ,(%%) = %.2f"%D4)
print("\npart (ii)")
print("percentage increase in power due to distortion is ,(%%) = %.2f"%peri)
# answer for % increase is slightly different than book

part (i)
Second percentage harmonic distortion (D2) is ,(%) = 10
Third percentage harmonic distortion  (D3) is ,(%) = 8
Fourth percentage harmonic distortion (D4) is ,(%) = 3.33

part (ii)
percentage increase in power due to distortion is ,(%) = 1.75


## Example 5.7, Page No. 169¶

In [28]:
# power dissipated

import math
#Variable declaration
Vcc=15.0                        #in V
Vpeak=24.0/2                    #in V
Rl=100.0                        #in ohm

#Calculations
Ipeak= Vpeak/Rl                 #in A
Pdc=Vcc*(2/(math.pi))*Ipeak     #in W

power dissipated by each transistor is,(mW)= 213