This document discusses power amplifiers and provides information on several key topics: - Power amplifiers are classified based on the percentage of time the collector current is non-zero, with classes including A, B, and C. - Class A amplifiers have the lowest theoretical maximum efficiency of 25% since the collector current is always on. - The document provides equations to calculate input power, output power, and efficiency for different amplifier classes and configurations. - A transformer-coupled class A amplifier is described that uses a transformer to couple the output signal from the amplifier to the load.

1. Kinjal Koradiya(13EC019)
Priyal Mehta (13EC020)
Dolly Suchak (13EC032)

2.  Induction of Power Amplifier
 Power and Efficiency
 Amplifier Classification
 Basic Class A Amplifier
 Transformer Coupled Class A Amplifier

3.  Power amplifiers are used to deliver a relatively high amount
of power, usually to a low resistance load.
 Typical load values range from 300W (for transmission
antennas) to 8W (for audio speaker).
 Although these load values do not cover every possibility,
they do illustrate the fact that power amplifiers usually drive
low-resistance loads.
 Typical output power rating of a power amplifier will be 1W
or higher.
 Ideal power amplifier will deliver 100% of the power it draws
from the supply to load. In practice, this can never occur.
 The reason for this is the fact that the components in the
amplifier will all dissipate some of the power that is being
drawn form the supply.

4. P1 = I1
2R1
P2 = I2
2R2
ICQ
RC
RE
R1
R2
VCC
I1
I2
ICC
PC = ICQ
2
RC
PT = ITQ
2
RT
PE = IEQ
2
RE
IEQ
The total amount of power
being dissipated by the
amplifier, Ptot , is
Ptot = P1 + P2 + PC + PT + PE
The difference between this
total value and the total power
being drawn from the supply is
the power that actually goes to
the load – i.e. output power.
 Amplifier Efficiency h

5.  A figure of merit for the power amplifier is its efficiency, h .
 Efficiency ( h ) of an amplifier is defined as the ratio of ac
output power (power delivered to load) to dc input power .
 By formula :
 As we will see, certain amplifier configurations have much
higher efficiency ratings than others.
 This is primary consideration when deciding which type of
power amplifier to use for a specific application.
  Amplifier Classifications
%
100
)
(
)
(
%
100 



dc
P
ac
P
power
input
dc
power
output
ac
i
o
h

6.  Power amplifiers are classified according to the
percent of time that collector current is nonzero.
 The amount the output signal varies over one cycle
of operation for a full cycle of input signal.
vin vout
Av Class-A
vin vout
Av Class-B
vin vout
Av Class-C

7.  The maximum theoretical efficiency ratings
of class-A, B, and C amplifiers are:
Amplifier Maximum Theoretical
Efficiency, hmax
Class A 25%
Class B 78.5%
Class C 99%

8.  output waveform  same shape  input waveform +
 phase shift.
 The collector current is nonzero 100% of the time.
 inefficient, since even with zero input signal, ICQ
is nonzero
(i.e. transistor dissipates power in the rest, or
quiescent, condition)
vin vout
Av

9. Common-emitter (voltage-divider) configuration (RC-coupled amplifier)
RC
+VCC
RE
R1
R2
RL
vin
ICQ
I1
ICC

11.  Previous figure shows an example of a
sinusoidal input and the resulting collector
current at the output.
 The current, ICQ , is usually set to be in the
center of the ac load line. Why?
(DC and AC analyses  discussed in previous
sessions)

12. RC
+VCC
RE
R1
R2
RL
vin
ICQ
I1
ICC
The total dc power, Pi(dc) , that an
amplifier draws from the power
supply :
CC
CC
i
I
V
dc
P 
)
(
1
I
I
I CQ
CC


CQ
CC
I
I  )
( 1
I
ICQ

CQ
CC
i
I
V
dc
P 
)
(
Note that this equation is valid for most amplifier power analyses.
We can rewrite for the above equation for the ideal amplifier as
CQ
CEQ
i
I
V
dc
P 2
)
( 

13. R1//R2
vce
vin
vo
ic
RC
//RL
rC
AC output (or load) power, Po(ac)
Above equations can be used to
calculate the maximum possible
value of ac load power. HOW??
L
rms
o
rms
o
rms
c
o
R
v
v
i
ac
P
2
)
(
)
(
)
(
)
( 

Disadvantage of using class-A amplifiers is the fact that their
efficiency ratings are so low, hmax  25% .
Why?? A majority of the power that is drawn from the supply by a
class-A amplifier is used up by the amplifier itself.
 Class-B Amplifier

14. IC
(mA)
VCE
VCE(off) = VCC
IC(sat) = VCC/(RC+RE)
DC Load Line
IC
VCE
IC(sat) = ICQ + (VCEQ/rC)
VCE(off) = VCEQ + ICQrC
ac load line
IC
VCE
Q - point
ac load line
dc load line
L
PP
CQ
CEQ
CQ
CEQ
o
R
V
I
V
I
V
ac
P
8
2
1
2
2
)
(
2















%
25
%
100
2
2
1
%
100
)
(
)
(





CQ
CEQ
CQ
CEQ
dc
i
ac
o
I
V
I
V
P
P
h

16. Calculate the input power [Pi(dc)], output power [Po(ac)],
and efficiency [h] of the amplifier circuit for an input
voltage that results in a base current of 10mA peak.
RC
RB
+VCC
= 20V
IC
Vi
25



20

k
1
Vo
 )
%
5
.
6
%
100
6
.
9
)
48
.
0
)(
20
(
625
.
0
)
20
(
2
10
250
2
250
)
10
(
25
20
1
1000
20
20
4
.
10
)
20
)(
48
.
0
(
20
48
.
0
5
.
482
)
3
.
19
(
25
3
.
19
1
7
.
0
20
)
(
)
(
)
(
2
3
2
)
(
)
(
)
(
)
(
)
(
)
(






































dc
i
ac
o
CQ
CC
dc
i
C
peak
C
ac
o
C
CC
sat
c
B
P
P
W
A
V
I
V
P
W
A
R
I
P
peak
mA
peak
mA
I
I
V
V
V
A
mA
V
R
V
I
V
A
V
R
I
V
V
A
mA
mA
I
I
mA
k
V
V
R
V
V
I
peak
b
peak
C
CC
cutoff
CE
C
C
CC
CEQ
CQ
B
BE
CC
BQ
h



17. Input
+VCC
RE
R1
R2
RL
N1
:N2
Z2 = RL
Z1
A transformer-coupled class-A amplifier
uses a transformer to couple the output
signal from the amplifier to the load.
The relationship between the primary
and secondary values of voltage, current
and impedance are summarized as:
L
R
Z
Z
Z
N
N
I
I
V
V
N
N
1
2
1
2
2
1
1
2
2
1
2
1












N1, N2 = the number of turns in the primary and secondary
V1, V2 = the primary and secondary voltages
I1, I2 = the primary and secondary currents
Z1, Z2 = the primary and seconadary impedance ( Z2 = RL )

18.  An important characteristic of the transformer is
the ability to produce a counter emf, or kick emf.
 When an inductor experiences a rapid change in
supply voltage, it will produce a voltage with a
polarity that is opposite to the original voltage
polarity.
 The counter emf is caused by the
electromagnetic field that surrounds the
inductor.