# Chapter 6 - Power Amplifiers [Large Signal Amplifiers]¶

## Example 6.1 Page No. 6-64¶

In [2]:
from math import sqrt
from __future__ import division
ibq=(20-0.7)/1.5
print "(i) I_BQ = V_CC-V_BE / R_B =%0.2f mA"%ibq
icq=50*12.87
print "I_CQ = beta * I_BQ =%0.2f mA"%icq
print "(ii) V_CC = I_CQ*R_L + V_CEQ"
vceq=20-(643.5*16*10**-3)
print "Therefore,  V_CEQ(in V) = V_CC - I_CQ*R_L =%0.2f V"%vceq
pdc=20*643.5*10**-3
print "(iii) P_DC = V_CC * I_CQ =%0.2f W"%pdc
print "(iv) P_ac Peak current i_b = 9 mA"
ic=50*9
print "i_c = beta * i_b =%0.2f mA"%ic
icm=450/sqrt(2)
print "Therefore,  i_c(rms) = I_rms(in mA) = i_c(peak) / sqrt(2) =%0.2f"%icm
pac=318.19**2*16*10**-6
print "Therefore,  P_ac = (I_rms)**2 * R_L =%0.2f W"%pac
n=(1.619*100)/12.87
print "(v) Efficiency eta(in percentage) = P_ac/P_DC * 100 =%0.2f"%n

(i) I_BQ = V_CC-V_BE / R_B =12.87 mA
I_CQ = beta * I_BQ =643.50 mA
(ii) V_CC = I_CQ*R_L + V_CEQ
Therefore,  V_CEQ(in V) = V_CC - I_CQ*R_L =9.70 V
(iii) P_DC = V_CC * I_CQ =12.87 W
(iv) P_ac Peak current i_b = 9 mA
i_c = beta * i_b =450.00 mA
Therefore,  i_c(rms) = I_rms(in mA) = i_c(peak) / sqrt(2) =318.20
Therefore,  P_ac = (I_rms)**2 * R_L =1.62 W
(v) Efficiency eta(in percentage) = P_ac/P_DC * 100 =12.58


## Example 6.2 Page No. 6-64¶

In [3]:
from __future__ import division
print "R_L = 4 ohm,  N1 = 200,  N2 = 20"
n=20/200
print "Therefore,  n = N2 / N1 =%0.2f"%n
rl=4/(0.1**2)
print "Therefore,  R''_L = R1 / n**2 =%0.2f ohm"%rl
print "As N2 < N1, the transformer is step down and hence R''_L > R_L, as the primary winding is high voltage winding."

R_L = 4 ohm,  N1 = 200,  N2 = 20
Therefore,  n = N2 / N1 =0.10
Therefore,  R''_L = R1 / n**2 =400.00 ohm
As N2 < N1, the transformer is step down and hence R''_L > R_L, as the primary winding is high voltage winding.


## Example 6.3 Page No. 6-65¶

In [4]:
from __future__ import division
print "R_L = 8 ohm, R''_L = 648 ohm"
print "Now  R''_L = R_L / n**2"
n=8/648
print "Therefore,  n**2 = R_L / R''_L =%0.2f"%n
print "Therefore,  n = 0.1111 = Turn ratio"
print "But,  n = N2 / N1 = 0.1111"
print "Therefore,  N1/N2 = 9"
print "Generally the turns ratio is specified as Ni/N2 : 1 i.e. for this transformer it is 9:1"

R_L = 8 ohm, R''_L = 648 ohm
Now  R''_L = R_L / n**2
Therefore,  n**2 = R_L / R''_L =0.01
Therefore,  n = 0.1111 = Turn ratio
But,  n = N2 / N1 = 0.1111
Therefore,  N1/N2 = 9
Generally the turns ratio is specified as Ni/N2 : 1 i.e. for this transformer it is 9:1


## Example6.4 Page No. 6-66¶

In [6]:
from math import sqrt
from __future__ import division
print "R_L=8ohm, I_CQ=140 mA, V_CC=10V"
print "P_ac= 0.48 W"
print "The turns ratio are specified as N1/N2:1 i.e 3:1"
print "Therefore N1/N2=3"
n=1/3
print "n=N2/N1=1/3=%0.2f"%n
r=8/(0.333)**2
print "Therefore R''_L=R_L/n**2=%0.2f"%r
print "1. As the transformer is ideal, whatever is the power delivered to the load,same is the power developed across primary."
print "Therefore P_ac(across primary)=0.48W"
print "2. Using equation (9),"
print "we get, P_ac=(V_1rms**2)/(R''_L)"
print "Therefore 0.48=(V_1rms**2)/72"
v=sqrt(34.56)
print "Therefore V_1rms(in V)=%0.2f"%v,
print "But rms value of the load voltage is V_2rms"
print "So (V1_rms)/(V2_rms)=N1/N2=3/1"
v=5.8787/3
print "Therefore (V2_rms)(in V)=(V1_rms)/3=%0.2f"%v
print "This is the rms value of the load voltage."
print "3. The rms value of the primary voltage is (V1_rms) as calculated above."
print "Therefore (V1_rms)=5.8787 V"
print "4. The power delivered to the load = (I_2rms**2)*R_L       ..Refer equation 13."
print "0.48=(I_2rms**2)*8"
i=sqrt(0.06)
print "(I_2rms)[in A]=%0.2f"%i
print "This is the rms value of the load current as the resistance value used is R_L and not R''_L"
print "5. The rms values of primary and secondary are related through the transformation ratio."
print "Therefore (I_1rms)/(I_2rms)=N2/N1=n=0.333"
i=0.2449*0.333
print "Thererfore (I_1rms)[in A]=0.2449*0.333= %0.2f"%i
print "6. The dc power input is,"
p=140*10**-2
print "P_DC =(V_CC)*(I_CQ)=%0.2f W"%p
n=(0.48*100)/1.4
print "7. %%eta=(P_ac *100)/(P_dc)=%0.2f"%n
d=1.4-0.48
print "P_d = %0.2f W"%d
print "This is the power dissipation."

R_L=8ohm, I_CQ=140 mA, V_CC=10V
P_ac= 0.48 W
The turns ratio are specified as N1/N2:1 i.e 3:1
Therefore N1/N2=3
n=N2/N1=1/3=0.33
Therefore R''_L=R_L/n**2=72.14
1. As the transformer is ideal, whatever is the power delivered to the load,same is the power developed across primary.
Therefore P_ac(across primary)=0.48W
2. Using equation (9),
we get, P_ac=(V_1rms**2)/(R''_L)
Therefore 0.48=(V_1rms**2)/72
Therefore V_1rms(in V)=5.88 But rms value of the load voltage is V_2rms
So (V1_rms)/(V2_rms)=N1/N2=3/1
Therefore (V2_rms)(in V)=(V1_rms)/3=1.96
This is the rms value of the load voltage.
3. The rms value of the primary voltage is (V1_rms) as calculated above.
Therefore (V1_rms)=5.8787 V
4. The power delivered to the load = (I_2rms**2)*R_L       ..Refer equation 13.
0.48=(I_2rms**2)*8
(I_2rms)[in A]=0.24
This is the rms value of the load current as the resistance value used is R_L and not R''_L
5. The rms values of primary and secondary are related through the transformation ratio.
Therefore (I_1rms)/(I_2rms)=N2/N1=n=0.333
Thererfore (I_1rms)[in A]=0.2449*0.333= 0.08
6. The dc power input is,
P_DC =(V_CC)*(I_CQ)=1.40 W
7. %eta=(P_ac *100)/(P_dc)=34.29
P_d = 0.92 W
This is the power dissipation.


## Example 6.6 Page No. 6-67¶

In [7]:
from __future__ import division
print "R_L = 4 k-ohm,  (P_ac)_D = 0.85 W"
print "The current without signal is I_CQ = 31 mA"
print "The current with signal is I_CQ + B0 = 34 mA"
print "The increase is due to harmonic content in the signal"
print "Therefore,  B0 = 34 - 31 = 3 mA"
print "But,  B2 = B0 = 3 mA"
print "Now  (P_ac)_D = P_ac * [1+D2**2]      ... Assuming only second harmonic"
print "Therefore,  (P_ac)_D = 1/2*B1**2*R_L * [1 + B2**2/B1**2]"
print "Therefore,  (P_ac)_D = 1/2*B1**2*R_L + 1/2*B2**2*R_L"
print "0.85 = 1/2*B1**2*(4*10**3) + 1/2*(9*10**-6)*(4*10**3)"
print "Therefore,  B1 = 20.396 mA"
d2=300/20.396
print "Therefore,  D2(in percentage) = |B2|/|B1| * 100 =%0.2f"%d2

R_L = 4 k-ohm,  (P_ac)_D = 0.85 W
The current without signal is I_CQ = 31 mA
The current with signal is I_CQ + B0 = 34 mA
The increase is due to harmonic content in the signal
Therefore,  B0 = 34 - 31 = 3 mA
But,  B2 = B0 = 3 mA
Now  (P_ac)_D = P_ac * [1+D2**2]      ... Assuming only second harmonic
Therefore,  (P_ac)_D = 1/2*B1**2*R_L * [1 + B2**2/B1**2]
Therefore,  (P_ac)_D = 1/2*B1**2*R_L + 1/2*B2**2*R_L
0.85 = 1/2*B1**2*(4*10**3) + 1/2*(9*10**-6)*(4*10**3)
Therefore,  B1 = 20.396 mA
Therefore,  D2(in percentage) = |B2|/|B1| * 100 =14.71


## Example 6.8 Page No. 6-68¶

In [9]:
from math import pi,sqrt
from __future__ import division
print "R_L = 12 ohm,  n = N2/N1 = 1/3 = 0.333, eta_trans = 78.5%"
rl=12/(0.333**2)
print "Therefore,  R''_L = R_L / n**2 =%0.2f"%rl
pac=(0.5*20**2)/108
print "(i) For P_max, V_m = V_CC"
print "Threfore,  (P_ac)_max = 1/2 * V_CC**2/R''_L =%0.2f W"%pac
print "But  eta_trans = 78.5%"
pl=0.785*1.8518
print "Therefore,  P_L = eta_trans * (P_ac)_max =%0.2f W"%pl
vm=(2*20)/pi
print "(ii) Condition for (P_d)_max is V_m(in V) = 2*V_CC/pi =%0.2f"%vm
pd=(2*20**2)/(108*pi**2)
print "Therefore,  (P_d)_max(in W) = 2*V_CC**2 / pi**2*R''_L =%0.2f W"%pd
pdm=0.7505/2
print "Therefore,  (P_d)_max per transistor =%0.2f W"%pdm
print "(iii) (P_ac)_max = V_rms * I_rms = V_m/sqrt(2) * I_m/sqrt(2) = V_m*I_m / 2  and  V_m = V_CC"
print "Therefore,  1.8518 = 20*I_m / 2"
im=(2*1.8518)/20
print "Therefore,  I_m = (I_c)_max =%0.2f A"%im
ibm=(0.1851/25)*10**3
print "and  (i_b)_max = (i_c))max / h_fe =%0.2f mA"%ibm

R_L = 12 ohm,  n = N2/N1 = 1/3 = 0.333, eta_trans = 78.5%
Therefore,  R''_L = R_L / n**2 =108.22
(i) For P_max, V_m = V_CC
Threfore,  (P_ac)_max = 1/2 * V_CC**2/R''_L =1.85 W
But  eta_trans = 78.5%
Therefore,  P_L = eta_trans * (P_ac)_max =1.45 W
(ii) Condition for (P_d)_max is V_m(in V) = 2*V_CC/pi =12.73
Therefore,  (P_d)_max(in W) = 2*V_CC**2 / pi**2*R''_L =0.75 W
Therefore,  (P_d)_max per transistor =0.38 W
(iii) (P_ac)_max = V_rms * I_rms = V_m/sqrt(2) * I_m/sqrt(2) = V_m*I_m / 2  and  V_m = V_CC
Therefore,  1.8518 = 20*I_m / 2
Therefore,  I_m = (I_c)_max =0.19 A
and  (i_b)_max = (i_c))max / h_fe =7.40 mA


## Example6.9 Page No. 6-69¶

In [10]:
from math import sqrt,pi
from __future__ import division
print "R_L=16 ohm, V_CC=25 V"
print "Now 2N1=200, N2=50"
n=200/2
print "Therefore N1=%0.2f"%n
n=50/100
print "Therefore n=N2/N1=%0.2f"%n
r=16/(0.5**2)
print "Therefore R''_L =(R_L)/(n**2)=%0.2f"%r
print "For maximum power output, V_m=V_CC"
p=(25**2)/(2*64)
print "i) (P_ac)_max  =(V_CC**2)/(2*R_L)=%0.2f W"%p
print "ii) (P_dc)=(2*V_CC*I_m)/pi"
print "Now  (V_m)/(I_m)=(R''_L)"
print "and V_m=V_CC"
i=25/64
print "Therefore (I_m)=(V_CC)/(R''_L)=%0.2f"%i
p=(2*25*0.3906)/(pi)
print "Therefore (P_DC) = %0.2f W"%p
n=(4.8848*100)/6.2169
print "iii) %%eta=(P_ac*100)/(P_DC)=%0.2f"%n
p=(2*4.8828)/(pi**2)
print "iv) (P_d)_max[in W]=(2*(P_ac)_max)/(pi**2)=%0.2f W"%p

R_L=16 ohm, V_CC=25 V
Now 2N1=200, N2=50
Therefore N1=100.00
Therefore n=N2/N1=0.50
Therefore R''_L =(R_L)/(n**2)=64.00
For maximum power output, V_m=V_CC
i) (P_ac)_max  =(V_CC**2)/(2*R_L)=4.88 W
ii) (P_dc)=(2*V_CC*I_m)/pi
Now  (V_m)/(I_m)=(R''_L)
and V_m=V_CC
Therefore (I_m)=(V_CC)/(R''_L)=0.39
Therefore (P_DC) = 6.22 W
iii) %eta=(P_ac*100)/(P_DC)=78.57
iv) (P_d)_max[in W]=(2*(P_ac)_max)/(pi**2)=0.99 W


## Example 6.12 Page No. 6-70¶

In [11]:
from math import sqrt,pi
from __future__ import division
print "R_L = 8 ohm, V_CC = +-12 V hence dual supply version"
pac=0.5*(12**2/8)
print "(1)  (P_ac)_max = 1/2 * V_CC**2/R_L =%0.2f W"%pac
print "(2)  P_DC = V_CC*I_DC  but I_DC = 2*I_m / pi"
print "          = V_CC * (2*I_m/pi)"
print "Now  R_L = V_m/I_m i.e. I_m = V_m/R_L and V_m = V_CC"
pdc=(12**2*2)/(8*pi)
print "Therefore,  P_DC = V_CC * 2 * V_CC/R_L * 1/pi =%0.2f W"%pdc
pdt=11.4591-9
print "Therefore,  Total P_D = P_DC - P_ac =%0.2f W"%pdt
pd=2.4591/2
print "Therefore,  P_D per transistor =%0.2f W"%pd
n=900/11.4591
print "(3)  %%eta(in percentage) = P_ac/P_DC * 100 =%0.2f"%n

R_L = 8 ohm, V_CC = +-12 V hence dual supply version
(1)  (P_ac)_max = 1/2 * V_CC**2/R_L =9.00 W
(2)  P_DC = V_CC*I_DC  but I_DC = 2*I_m / pi
= V_CC * (2*I_m/pi)
Now  R_L = V_m/I_m i.e. I_m = V_m/R_L and V_m = V_CC
Therefore,  P_DC = V_CC * 2 * V_CC/R_L * 1/pi =11.46 W
Therefore,  Total P_D = P_DC - P_ac =2.46 W
Therefore,  P_D per transistor =1.23 W
(3)  %eta(in percentage) = P_ac/P_DC * 100 =78.54


## Example6.13 Page No. 6-71¶

In [2]:
from math import pi
from __future__ import division
print "V_CC=10V ,R_L=5 ohm"
p=100/10
print "i) (P_ac)_max = (V_CC**2)/(2*R_L)=(10**2)/(2*5)=%0.2f W"%p
print "ii) To decide Power rating of transistors means to find (P_D)_max"
v=(2*10)/(pi)
print "V_m = %0.2f V"%v
print "Now, R_L=(V_m)/(I_m)"
i=6.3662/5
print "Therefore (I_m) = %0.2f A"%i
print "Therefore (P_DC)=(V_CC)*(I_DC)=(V_CC)*(2*I_m)/pi    (I_DC)=(2*I_m)/pi"
p=(10*2*1.2732)/(pi)
print " =(10*2*1.2732)/(pi) =%0.2f"%p
p=(6.3662*1.2732)/2
print "and  (P_ac)[in W]=(V_m*I_m)/2=%0.2f W"%p
p=8.1056-4.5027
print "(P_D)_max = (P_DC)-(P_ac)=%0.2f W"%p
p=4.0528/2
print "Therefore P_D rating fora each transistor =(P_D)_max/2=%0.2f"%p
print "iii) For (P_ac)_max, V_m=V_CC=10 V"
i=10/5
print "I_m = (V_m)/R_L=%0.2f A"%i
p=(10*2*2)/pi
print "P_DC = %0.2f W"%p

V_CC=10V ,R_L=5 ohm
i) (P_ac)_max = (V_CC**2)/(2*R_L)=(10**2)/(2*5)=10.00 W
ii) To decide Power rating of transistors means to find (P_D)_max
V_m = 6.37 V
Now, R_L=(V_m)/(I_m)
Therefore (I_m) = 1.27 A
Therefore (P_DC)=(V_CC)*(I_DC)=(V_CC)*(2*I_m)/pi    (I_DC)=(2*I_m)/pi
=(10*2*1.2732)/(pi) =8.11
and  (P_ac)[in W]=(V_m*I_m)/2=4.05 W
(P_D)_max = (P_DC)-(P_ac)=3.60 W
Therefore P_D rating fora each transistor =(P_D)_max/2=2.03
iii) For (P_ac)_max, V_m=V_CC=10 V
I_m = (V_m)/R_L=2.00 A
P_DC = 12.73 W


## Example 6.15 Page No. 6-72¶

In [3]:
from math import pi
from __future__ import division
print "V_CC = 20 V,  R_L = 4 ohm"
vm=(2*20)/pi
print "For (P_d)max,  V_m(in V) = 2/pi * V_CC = %0.2f"%vm
print "R_L = V_m / I_m"
im=12.7324/4
print "Therefore,  I_m = V_m / R_L =%0.2f A"%im
idc=(2*3.183)/pi
print "Therefore,  I_dc = 2*I_m / pi =%0.2f A"%idc
pac=(0.5*12.7324**2)/4
print "Therefore,  P_ac = 1/2 * V_m**2/R_L =%0.2f W"%pac
pdc=20*2.0254
print "and  P_dc = V_CC * I_DC =%0.2f W"%pdc
pdm=40.508-20.2542
print "Therefore,  Total (P_d)max = P_dc - P_ac =%0.2f W"%pdm
pdma=20.2538/2
print "Therefore,  (P_d)max per transistor =%0.2f W"%pdma

V_CC = 20 V,  R_L = 4 ohm
For (P_d)max,  V_m(in V) = 2/pi * V_CC = 12.73
R_L = V_m / I_m
Therefore,  I_m = V_m / R_L =3.18 A
Therefore,  I_dc = 2*I_m / pi =2.03 A
Therefore,  P_ac = 1/2 * V_m**2/R_L =20.26 W
and  P_dc = V_CC * I_DC =40.51 W
Therefore,  Total (P_d)max = P_dc - P_ac =20.25 W
Therefore,  (P_d)max per transistor =10.13 W


## Example6.16 Page No. 6-73¶

In [4]:
from math import pi,sqrt
from __future__ import division
print "For a given transistor,"
print "Maximum collector current =I_cm =1A"
print "Maximum power dissipation =P_d=10W"
print "Maximum V_CEO =40V"
print "For maximum output power,"
print "I_cm=2*I_CQ"
i=1/2
print "I_CQ=1/2=%0.2f"%i
print "and V_CEO=2*V_CC"
v=40/2
print "V_CC = V_CEO/2= %0.2f V"%v
print "and  V_cc=V_m=20V for (P_ac)_max"
print "(P_ac)_max=(V_cc**2)/(2*R_L)"
print "R''_L=(V_m)/I_m and I_m=I_CQ=0.5 A"
r=20/0.5
print "R''_L =%0.2f ohm"%r
p=(20**2)/80
print "(P_ac)_max = (20**2)/(2*40)=%0.2f W"%p
print "Now, R''_L=R_L/n**2"
n=sqrt(0.0625)
print "n=N2/N1=%0.2f"%n
n=1/0.25
print "Therefore N1/N2=1/n=%0.2f"%n
print "Hence the turns ratio of output transformer is 4:1"

For a given transistor,
Maximum collector current =I_cm =1A
Maximum power dissipation =P_d=10W
Maximum V_CEO =40V
For maximum output power,
I_cm=2*I_CQ
I_CQ=1/2=0.50
and V_CEO=2*V_CC
V_CC = V_CEO/2= 20.00 V
and  V_cc=V_m=20V for (P_ac)_max
(P_ac)_max=(V_cc**2)/(2*R_L)
R''_L=(V_m)/I_m and I_m=I_CQ=0.5 A
R''_L =40.00 ohm
(P_ac)_max = (20**2)/(2*40)=5.00 W
Now, R''_L=R_L/n**2
n=N2/N1=0.25
Therefore N1/N2=1/n=4.00
Hence the turns ratio of output transformer is 4:1


## Example 6.19 Page No. 6-76¶

In [5]:
from math import pi,sqrt
from __future__ import division
print "Using equation (2) from section 6.7, we can determine I_BQ."
i=(18-0.7)/(1.2*10**3)
print "I_BQ = %0.2f A"%i
i=40*14.4167
print "Now (I_CQ) =(beta*I_BQ)=%0.2f mA"%i
v=18-(576.67*16*10**-3)
print "And (V_CEQ) = (V_CC)-(I_CQ*R_L)=%.2f V"%v
p=18*576.67
print "So  P_dc =(V_CC)*(I_CQ)=%0.2f W"%p
print "This is the input power."
print "Now input a.c. voltage causes a base current of 5mA rms"
print "Therefore (I_b)_rms=5 mA"
i=40*5
print "Therefore i_c_rms = 40*5= %0.2f mA"%i
print "This is nothing but the output collector current,rms value I_rms"
print "Therefore I_rms = 200mA"
print "Using equation (13) from section 6.8, we can write,"
p=16*(200*10**-3)**2
print "P_ac = (I_rms**2)**R_L=%0.2f W"%p
print "This is the power delivered to the load."
print "Hence the efficiency of the amplifier is,"
n=(64000*10**-3)/10.38
print "%%eta=(P_ac*100)/P_dc= %0.2f"%n

Using equation (2) from section 6.7, we can determine I_BQ.
I_BQ = 0.01 A
Now (I_CQ) =(beta*I_BQ)=576.67 mA
And (V_CEQ) = (V_CC)-(I_CQ*R_L)=8.77 V
So  P_dc =(V_CC)*(I_CQ)=10380.06 W
This is the input power.
Now input a.c. voltage causes a base current of 5mA rms
Therefore (I_b)_rms=5 mA
Therefore i_c_rms = 40*5= 200.00 mA
This is nothing but the output collector current,rms value I_rms
Therefore I_rms = 200mA
Using equation (13) from section 6.8, we can write,
P_ac = (I_rms**2)**R_L=0.64 W
This is the power delivered to the load.
Hence the efficiency of the amplifier is,
%eta=(P_ac*100)/P_dc= 6.17


## Example 6.20 Page No. 6-77¶

In [6]:
from math import pi,sqrt
from __future__ import division
print "V_CC = 20 V,  R_L = 20 ohm, turns ratio 1.58:1"
n=1/1.58
print "  n = 1/1.58 = %0.2f"%n
rl=20/0.6329**2
print "Therefore,  R''_L = R_L / n**2 =%0.2f ohm"%rl
print "(i) For maximum possible peak to peak output voltage, the power output is also maximum possible. For this condition the slope of the a.c. load line can be expressed as"
print "R''_L = V_m/I_m = V_CC/I_CQ"
icq=20/49.928
print "Therefore,  I_CQ =%0.2f A"%icq
ibq=0.4/40
print "Therefore,  I_BQ = I_CQ/beta =%0.2f A"%ibq
print "This is the required value of the base current"
print "(ii)  P_ac = I_Irms**2 * R''_L"
print "But for maximum power output condition,"
irms=0.4/sqrt(2)
print "I_Irms = I_Im/sqrt(2) = I_CQ/sqrt(2) =%0.2f A"%irms
pac=49.928*0.2828**2
print "Therefore,  P_ac = %0.2f W"%pac
print "(iii)  %eta = P_ac/P_DC * 100"
pdc=20*0.4
print "Now  P_DC = V_CC * I_CQ =%0.2f W"%pdc
eta=400/8
print "%%eta(in percentage) =%0.2f"%eta

V_CC = 20 V,  R_L = 20 ohm, turns ratio 1.58:1
n = 1/1.58 = 0.63
Therefore,  R''_L = R_L / n**2 =49.93 ohm
(i) For maximum possible peak to peak output voltage, the power output is also maximum possible. For this condition the slope of the a.c. load line can be expressed as
R''_L = V_m/I_m = V_CC/I_CQ
Therefore,  I_CQ =0.40 A
Therefore,  I_BQ = I_CQ/beta =0.01 A
This is the required value of the base current
(ii)  P_ac = I_Irms**2 * R''_L
But for maximum power output condition,
I_Irms = I_Im/sqrt(2) = I_CQ/sqrt(2) =0.28 A
Therefore,  P_ac = 3.99 W
(iii)  %eta = P_ac/P_DC * 100
Now  P_DC = V_CC * I_CQ =8.00 W
%eta(in percentage) =50.00


## Example 6.21 Page No. 6-78¶

In [7]:
from math import sqrt,pi
from __future__ import division
print "R_L = 8 ohm,  P_ac(max) = 40 W"
print "2*N1 = 160,  N2 = 40"
print "N1 = 80"
n=40/80
print "n = N2/N1 =%0.2f"%n
rl=8/0.5**2
print "Therefore,  R''_L = R_L / n**2 =%0.2f ohm"%rl
print "Under maximum condition, V_CC = V_m"
print "Therefore,  P_ac(max) = 1/2 * V_CC**2/R''_L"
vcc=sqrt(40*2*32)
print "Therefore,  V_CC = %0.2f V"%vcc
print "This is the required value of V_CC"

R_L = 8 ohm,  P_ac(max) = 40 W
2*N1 = 160,  N2 = 40
N1 = 80
n = N2/N1 =0.50
Therefore,  R''_L = R_L / n**2 =32.00 ohm
Under maximum condition, V_CC = V_m
Therefore,  P_ac(max) = 1/2 * V_CC**2/R''_L
Therefore,  V_CC = 50.60 V
This is the required value of V_CC


## Example 6.22 Page No. 6-79¶

In [8]:
from math import sqrt,pi
from __future__ import division
print "For a common collector configuration the voltage gain is 1"
print "Therefore,  V_in(peak) = V_out(peak) = 20 V"
print "i.e.  V_m = 20 V"
print "Now  V_m/I_m = R_L"
im=20/16
print "Therefore,  I_m = V_m/R_L =%0.2f A"%im
print "while  V_CC = 25 V"
pdc=(2*25*1.25)/pi
print "Now  P_DC = 2*V_CC*I_m / pi =%0.2f W"%pdc
pac=(20*1.25)/2
print "P_ac = V_m*I_m / 2 =%0.2f W"%pac
eta=1250/19.8943
print "Therefore,  %%eta(in percentage) = P_ac*100 / P_DC =%0.2f"%eta

For a common collector configuration the voltage gain is 1
Therefore,  V_in(peak) = V_out(peak) = 20 V
i.e.  V_m = 20 V
Now  V_m/I_m = R_L
Therefore,  I_m = V_m/R_L =1.25 A
while  V_CC = 25 V
Now  P_DC = 2*V_CC*I_m / pi =19.89 W
P_ac = V_m*I_m / 2 =12.50 W
Therefore,  %eta(in percentage) = P_ac*100 / P_DC =62.83


## Example 6.26 Page No. 6-81¶

In [9]:
from math import sqrt,pi
from __future__ import division
print "(i) As single transistor is used, even harmonic components will not get eliminated"
d2=1000/120
d3=400/120
d4=200/120
d5=100/120
print "  D2 = |B2| / |B1| =%0.2f %%"%d2
print "  D3 = |B3| / |B1| =%0.2f %%"%d3
print "  D4 = |B4| / |B1| =%0.2f %%"%d4
print "  D5 = |B5| / |B1| =%0.2f %%"%d5
print "The total harmonic distortion is,"
print "%D = sqrt(D2**2 + D3**2 + D4**2 + D5**2) * 100"
d=sqrt((0.0833**2)+(0.0333**2)+(0.01667**2)+(0.00833**2))*100
print "Therefore,  %%D =%0.2f %%"%d
print "(ii) When identical second transistor is used, then all even harmonics get eliminated. So only D3 and D5 will present"
dp=sqrt((0.033**2)+(0.00833**2))*100
print "Therefore,  %%D = sqrt(D3**2 + D5**2)*100 =%0.2f %%"%dp

(i) As single transistor is used, even harmonic components will not get eliminated
D2 = |B2| / |B1| =8.33 %
D3 = |B3| / |B1| =3.33 %
D4 = |B4| / |B1| =1.67 %
D5 = |B5| / |B1| =0.83 %
The total harmonic distortion is,
%D = sqrt(D2**2 + D3**2 + D4**2 + D5**2) * 100
Therefore,  %D =9.16 %
(ii) When identical second transistor is used, then all even harmonics get eliminated. So only D3 and D5 will present
Therefore,  %D = sqrt(D3**2 + D5**2)*100 =3.40 %


## Example 6.27 Page No. 6-82¶

In [10]:
from math import sqrt,pi
from __future__ import division
print "From the fig 6.50 we can write,"
print "V_CC=20V and R_L=12 ohm"
print "i) The maximum ac power that can be delivered to the load is,"
p=(20**2)/24
print "(P_ac)_max = %0.2f W"%p
print "Let new power delivered to load be (P_ac)''."
print "The corresponding new supply voltage be (V''_cc)"
print "(P_ac)''[in W]=1.36(P_ac)_max     ..36% more"
p=1.36*16.67
print "= 1.36*16.67=%0.2f"%p
print "And (P_ac)''=(V''_cc**2)/R_L"
print "Therefore 22.67=(V''_cc**2)/(2*12)"
v=sqrt(544.1088)
print "V''_cc = %0.2f V"%v
print "Hence the percentage increase in supply voltage is,"
p=(23.326-20)/0.2
print "= ((V''_cc-V_cc)*100)/V_cc= %0.2f"%p
print "The mimimum breakdown voltage per transistor this condition is,"
v=2*23.326
print "=2*V''_cc=2*23.326=%0.2f"%v

From the fig 6.50 we can write,
V_CC=20V and R_L=12 ohm
i) The maximum ac power that can be delivered to the load is,
(P_ac)_max = 16.67 W
Let new power delivered to load be (P_ac)''.
The corresponding new supply voltage be (V''_cc)
(P_ac)''[in W]=1.36(P_ac)_max     ..36% more
= 1.36*16.67=22.67
And (P_ac)''=(V''_cc**2)/R_L
Therefore 22.67=(V''_cc**2)/(2*12)
V''_cc = 23.33 V
Hence the percentage increase in supply voltage is,
= ((V''_cc-V_cc)*100)/V_cc= 16.63
The mimimum breakdown voltage per transistor this condition is,
=2*V''_cc=2*23.326=46.65


## Example 6.30 Page No. 6-83¶

In [12]:
from math import sqrt,pi
from __future__ import division
print "The circuit used for providing proper biasing is self bias, for which the various currents can be shown in the fig 6.52"
print "Applying KVL to base emiter loop,"
print "(-V_BE)-(I_E*R_E)+(I*R2)=0"
print "Theredfore (I*100)-(1+beta)*I_B*10=V_BE"
print "100I-210(I_B)=0.5     ..(1)"
print "Applying KVL through R1 and R2,"
print "(-R1(I+I_B))-(R2*I)+V_cc=0"
print "-1000(I+I_B)-100I=-V_cc"
print "1100I+1000(I_B)=25    ..(2)"
print "Multiplying equation (1) by 11 and subtracting from equation (2) we get,"
print "3310(I_B)=19.5"
i=19.5/3310
print "Therefore I_B = %0.2f A"%i
print "Thereofore I_C=(beta*I_B)=117.82 mA=I_CQ"
print "Now n=N2/N1=1/8"
r=5*(8**2)
print "Therefore (R''_L)=(R_L)/(n**2) : "
print "i) For maximum power delivered to load,"
print "V_1m=V_CEQ"
print "Apply KVL to collector-emitter loop,"
print "(-10I_C)-(V_CEQ)-(10*I_E)+V_CC=0"
v=25-(20*(117.82*10**-3))
print "V_CEQ=V_cc-20*I_C=%0.2f     ...I_C=I_E"%v
p=(22.643**2)/640
print "(P_ac)_pri = (V_CEQ**2)/(2*R''_L)=%0.2f W"%p
p=0.9*0.8011
print "(P_ac)_max = 0.9*0.8011= %0.2f W"%p
print "This is maximum power delivered to the load."
p=25*117.82*10**-3
print "ii) Now (P_DC)[in W]=V_CC*I_CQ:"
n=(0.721*100)/2.9455
print "%%eta=(P_ac*100)/(P_dc)=%0.2f %%"%n

The circuit used for providing proper biasing is self bias, for which the various currents can be shown in the fig 6.52
Applying KVL to base emiter loop,
(-V_BE)-(I_E*R_E)+(I*R2)=0
Theredfore (I*100)-(1+beta)*I_B*10=V_BE
100I-210(I_B)=0.5     ..(1)
Applying KVL through R1 and R2,
(-R1(I+I_B))-(R2*I)+V_cc=0
-1000(I+I_B)-100I=-V_cc
1100I+1000(I_B)=25    ..(2)
Multiplying equation (1) by 11 and subtracting from equation (2) we get,
3310(I_B)=19.5
Therefore I_B = 0.01 A
Thereofore I_C=(beta*I_B)=117.82 mA=I_CQ
Now n=N2/N1=1/8
Therefore (R''_L)=(R_L)/(n**2) :
i) For maximum power delivered to load,
V_1m=V_CEQ
Apply KVL to collector-emitter loop,
(-10I_C)-(V_CEQ)-(10*I_E)+V_CC=0
V_CEQ=V_cc-20*I_C=22.64     ...I_C=I_E
(P_ac)_pri = (V_CEQ**2)/(2*R''_L)=0.80 W
(P_ac)_max = 0.9*0.8011= 0.72 W
This is maximum power delivered to the load.
ii) Now (P_DC)[in W]=V_CC*I_CQ:
%eta=(P_ac*100)/(P_dc)=24.48 %


## Example 6.32 Page No. 3-85¶

In [13]:
from __future__ import division
print "When no signal is applied, current drawn is"
print "I_CQ =200mA from V_cc= 10V"
p=10*200*10**-3
print "P_DC(in W)=V_CC*I_CQ:"
print "For maximum power output,"
print "V_1m=V_cc=10V and I_1m=I_CQ=200mA"
p=2/2
print "P_ac = (V_1rms*I_1rms)=(V_1m*I_1m)/2=(10*200*10**-3)/2=%0.2f W"%p
print "i) P_ac(max)=Maximum output power =1W"
n=100/2
print "ii) %%eta = (P_ac*100)/(P_DC)=%0.2f %%"%n
print "P_d(max)=V_cc*I_CQ= 2W"
print "The power dissipation rating of the transistor must be higher than 2W"
r=10/(200**10**-3)
print "Now  R''_L = (V_1m)/(I_1m)=%0.2f ohm"%r
n=1/5
print "Now R_L=2 ohm and n =N2/N1=1/5=%0.2f"%n
r=2/(0.2**2)
print "R''_L =(R_L)/n**2=%0.2f ohm"%r
print "As R''_L required matches with the R''_L of the circuit, impedance matching is perfect"

When no signal is applied, current drawn is
I_CQ =200mA from V_cc= 10V
P_DC(in W)=V_CC*I_CQ:
For maximum power output,
V_1m=V_cc=10V and I_1m=I_CQ=200mA
P_ac = (V_1rms*I_1rms)=(V_1m*I_1m)/2=(10*200*10**-3)/2=1.00 W
i) P_ac(max)=Maximum output power =1W
ii) %eta = (P_ac*100)/(P_DC)=50.00 %
P_d(max)=V_cc*I_CQ= 2W
The power dissipation rating of the transistor must be higher than 2W
Now  R''_L = (V_1m)/(I_1m)=9.95 ohm
Now R_L=2 ohm and n =N2/N1=1/5=0.20
R''_L =(R_L)/n**2=50.00 ohm
As R''_L required matches with the R''_L of the circuit, impedance matching is perfect


## Example 6.33 Page No. 6-86¶

In [14]:
from math import sqrt,pi
from __future__ import division
print "B1 = 5*10**-2,  B2 = 10**-4,  B3 = 3*10**-6"
print "These are the amplitudes of various frequency components"
d2=10**-4/(50*10**-2)
d3=(3*10**-6)/(50*10**-2)
d=sqrt((2*10**-4)**2+(6*10**-6)**2)*100
print "Therefore,  D2 = |B2|/|B1| =%0.2f"%d2
print "Therefore,  D2 = |B3|/|B1| =%0.2f"%d3
print "Therefore,  %%D = sqrt(D2**2 + D3**2)*100 =%0.2f %%"%d

B1 = 5*10**-2,  B2 = 10**-4,  B3 = 3*10**-6
These are the amplitudes of various frequency components
Therefore,  D2 = |B2|/|B1| =0.00
Therefore,  D2 = |B3|/|B1| =0.00
Therefore,  %D = sqrt(D2**2 + D3**2)*100 =0.02 %


## Example 6.34 Page No. 6-87¶

In [15]:
from __future__ import division
print "V_CC = 12 V,  I_PP = 100 mA,  R_L = 5 ohm"
print "Therefore,  I_m = I_PP/2 = 50 mA"
pac=((2500*10**-6)*5)/2
print "(i)  P_ac = I_m**2*R_L / 2 =%0.2f W"%pac
print "(ii)  P_ac(max) = 1/2 * V_CC**2/R_L"
print "But  P_ac = V_m*I_m/2 and V_m = V_CC for maximum power"
rl=12**2/0.6
print "Therefore,  R''_L = %0.2f ohm"%rl
print "But  R''_L = R_L/n**2  i.e. 240 = 5/n**2"
print "Therefore,  n**2 = 0.02083  i.e. n = 0.1443 = N2/N1"
print "Therefore,  N1/N2 = 6.928 : 1"

V_CC = 12 V,  I_PP = 100 mA,  R_L = 5 ohm
Therefore,  I_m = I_PP/2 = 50 mA
(i)  P_ac = I_m**2*R_L / 2 =0.01 W
(ii)  P_ac(max) = 1/2 * V_CC**2/R_L
But  P_ac = V_m*I_m/2 and V_m = V_CC for maximum power
Therefore,  R''_L = 240.00 ohm
But  R''_L = R_L/n**2  i.e. 240 = 5/n**2
Therefore,  n**2 = 0.02083  i.e. n = 0.1443 = N2/N1
Therefore,  N1/N2 = 6.928 : 1


## Example 6.35 Page No. 6-88¶

In [16]:
from math import pi,sqrt
from __future__ import division
print "I_CQ = 250 mA,  V_CEQ = 8 V"
print "V_max = 15 V,  V_min = 1 V,  I_max = 450 mA,  I_min = 40 mA"
ipp=450-40
print "Therefore,  I_pp( = I_max - I_min =%0.2f mA"%ipp
vpp=15-1
print "Therefore,  V_pp = V_max - V_min =%0.2f V"%vpp
vm=14/2
print "Therefore,  V_m = V_pp/2 =%0.2f V"%vm
im=410/2
print "Therefore,  I_m = I_pp/2 =%0.2f mA"%im
pac=(7*205*10**-3)/2
pdc=250*8*10**-3
n=71.75/2
pd=2-0.7175
print "(i)  P_ac = V_m*I_m/2 = %0.2f W               ...output power"%pac
print "(ii)  P_DC = I_CQ*V_CEQ = %0.2f W             ...inpu:"
print "   %%eta = P_ac/P_DC * 100 = %0.2f %%          ...efficien Wcy"%pd

I_CQ = 250 mA,  V_CEQ = 8 V
V_max = 15 V,  V_min = 1 V,  I_max = 450 mA,  I_min = 40 mA
Therefore,  I_pp( = I_max - I_min =410.00 mA
Therefore,  V_pp = V_max - V_min =14.00 V
Therefore,  V_m = V_pp/2 =7.00 V
Therefore,  I_m = I_pp/2 =205.00 mA
(i)  P_ac = V_m*I_m/2 = 0.72 W               ...output power
(ii)  P_DC = I_CQ*V_CEQ = %0.2f W             ...inpu:
%eta = P_ac/P_DC * 100 = 1.28 %          ...efficien Wcy