# Chapter 11 - Power Amplifier¶

## Example E1 - Pg 536¶

In [6]:
#Ex11_1 Pg-536
Vcc=15. #supply voltage in V
R1=2.*10.**(3.) #resistor R1 in ohm
R2=470. #resistor R2 in ohm
Rc=680. #collector resistor in ohm
Re=220. #emitter resistor
Idc=Vcc/(Rc+Re) #saturation current
print '%s %.1f' %("(1) Idc_sat = mA",Idc*1e3)
Vb=R2/(R1+R2)*Vcc #base voltage
Icq=(Vb-0.7)/Re #collector current
print '%s %.1f' %("(4) Icq = mA",Icq*1e3)
#answer in the book is wrong
Vc=Vcc-Icq*Rc #collector emitter voltage
Vceq=Vc-Icq*Re
print '%s %.1f' %("(5) Vceq = V",Vceq)
#answer in the book is wrong
Pac=Vcc**2./(8.*Rl) #ac power
Pdc=Vcc*Idc #dc power
n=Pac/Pdc*100. #efficiency
print '%s %.0f' %("Efiiciency =",n)

(1) Idc_sat = mA 16.7
(2) DC load = ohm 680
(3) AC load = ohm 543
(4) Icq = mA 9.8
(5) Vceq = V 6.2
Efiiciency = 4


## Example E2 - Pg 551¶

In [2]:
#Ex11_2 Pg-551
#calculate Effective power to be transfered,Effective load,Maximum transistor voltage,Maximum transistor curren,Transformer specification
Po=4. #power in watts
n=80./100. #transformer efficiency in percentage
Vcc=30. #supply voltage
Pout=Po/n #effective power
print '%s %.0f' %("Effective power to be transfered = W",Pout)
print '%s' %("\nImpedance seen when ""looking into"" the whole winding of centertapped transformer ")
Vp=Vcc #peak voltage
print '%s' %("\nTransformer specification Po=4W,RL=16ohm,RL""=360ohm")
Vce=2*Vcc #Maximum transistor voltage
print '%s %.0f' %("\nMaximum transistor voltage = V",Vce)
Ip=2*Pout/Vp #Maximum transistor current
Ic=Ip
print '%s %.0f' %("\nMaximum transistor current = mA",Ip*1e3)
# answer in the book is different due to approximate value
print '%s %.0f' %("\nTransformer specification Vce=60V,Ic = mA",Ic*1e3)
# answer in the book is different  due to approximate value

Effective power to be transfered = W 5

Impedance seen when looking into the whole winding of centertapped transformer

Transformer specification Po=4W,RL=16ohm,RL=360ohm

Maximum transistor voltage = V 60

Maximum transistor current = mA 333

Transformer specification Vce=60V,Ic = mA 333


## Example E3 - Pg 564¶

In [3]:
#Ex11_3 Pg-564
#calculate Resonant frequency,Bandwidth
import math
L=2.*10.**(-6.) #inductance in H
C=220.*10.**(-12.) #capacitance in F
f0=1./(2.*math.pi*math.sqrt(L*C)) #resonant frequency (textbook answer is wrong)
print '%s %.1f' %("Resonant frequency = MHz",f0*1e-6)
Q=125. #quality factor
BW=f0/Q #Bandwidth (textbook answer is wrong)
print '%s %.0f' %("Bandwidth = kHz",BW*1e-3)

Resonant frequency = MHz 7.6
Bandwidth = kHz 61


## Example E4 - Pg 564¶

In [4]:
#Ex11_4 Pg-564
#calculate Bias voltag,Emitter current,DC collector voltage,DC collector to emitter voltage,Power dissipation
Vcc=10. #supply volage in V
Rc=3600. #collector resistor in ohm
Re=680. #emitter resistor in ohm
Ri=10000. #input resistor in ohm
R2=2.2 #resistor R2 in ohm
R1=10. #resistor R1 in ohm
Vb=R2/(R1+R2)*Vcc #bias voltage
print '%s %.1f' %("(1) Bias voltage = V",Vb)
Ie=(Vb-0.7)/Re #emitter current
print '%s %.2f' %(" Emitter current = mA",Ie*1e3)
Vc=Vcc-Rc*Ie #Dc collector voltage
print '%s %.2f' %(" DC collector voltage = V",Vc)
Vceq=Vc-Ie*Re #DC collector to emitter voltage
print '%s %.2f' %(" DC collector to emitter voltage = V",Vceq)
Pd=Vceq*Ie #power dissipation
print '%s %.2f' %(" Power dissipation = mW",Pd*1e3)
print '%s' %("\n(2)If collector resistance Rc is replaced by tank circuit there is no voltage drop across it")
Vc=Vcc
print '%s %.0f' %(" DC collector voltage = V",Vc)
Vceq=Vc-Ie*Re #DC collector to emitter voltage
print '%s %.2f' %(" DC collector to emitter voltage = V",Vceq)
Pd=Vceq*Ie #power dissipation
print '%s %.2f' %(" Power dissipation = mW",Pd*1e3)

(1) Bias voltage = V 1.8
Emitter current = mA 1.62
DC collector voltage = V 4.16
DC collector to emitter voltage = V 3.06
Power dissipation = mW 4.96

(2)If collector resistance Rc is replaced by tank circuit there is no voltage drop across it
DC collector voltage = V 10
DC collector to emitter voltage = V 8.90
Power dissipation = mW 14.43


## Example E5 - Pg 565¶

In [5]:
#Ex11_5 Pg-565
#calculate The peak value,peak-to-peak value of input voltage,Base to ground voltage
import math
Vin=5. #input voltage
Vp=Vin*math.sqrt(2.) #peak voltage
print '%s' %("The peak value(maximum amplication) of input signal")
print '%s %.2f' %("= V",Vp)
Vin_pp=2*Vp #peak-to-peak value of input voltage
print '%s' %("\nPeak-to-peak value of input voltage")
print '%s %.2f' %("= V",Vin_pp)
Vbg=-1*(Vp-0.7) #base to ground voltage 0.7 is the voltage drop
print '%s %.2f' %("\nBase to ground voltage = V",Vbg)

The peak value(maximum amplication) of input signal
= V 7.07

Peak-to-peak value of input voltage
= V 14.14

Base to ground voltage = V -6.37