Efficiency in Power Amplifiers
Consider a power amplifier in which, at maximum signal amplitude, the transistor
current exhibits a conduction angle  where  can take on values between 0
and 2  radians.
For convenience, we take the time origin at the peak of the current so that we
have conduction for  2   t   2 . For the moment, we assume for
simplicity that    . Then we can write, for the current at maximum signal,
 I  cos  t  cos 2  ;  2   t   2
iT  t    m
0;
otherwise

What we have done, essentially, is to subtract the constant cos 2 from the
cosine so that the difference goes to 0 at  t    2 . We simply define iT  t  to
be zero for larger absolute values of  t .
For smaller input signals, the output current variations are smaller by a factor  ,
0    1 , so that we can write
 I  cos  t  cos 2  ;
iT  t    m
0;

 cos  t  cos 2
otherwise
where  ranges from 0 for no signal to 1 for the largest possible signal (that is,
the signal such that if it were larger, the transistor collector to emitter voltage
would try to change sign). For conduction angles    , iT  t  is identically zero
for sufficiently small signals (sufficiently small  's ). That is, for sufficiently small
 's ,  cos  t is never greater than cos 2 . In particular, iT  t   0 if   cos 2 .
If   cos 2 , then
 I  cos  t  cos 2  ;
iT  t    m
0;

 2   t    2
otherwise
where

1

arccos  cos  / 2    cos  / 2
 /2  


0
  cos / 2

Note that  is the conduction angle for the largest possible signal, whereas  is
the (smaller) conduction angle for smaller signals. If   cos / 2 , then   0 (no
conduction). If   1 (largest signal), then    , the usual conduction angle.
We can easily generalize these results for 0    2  :
 I  cos  t  cos 2  ;
iT  t    m
0;

 2   t    2
otherwise
where

1

arccos  cos  / 2    cos  / 2
 /2 


 H    
  cos / 2

where H  is the Heaviside unit step function:
1 x  0
H ( x)  
0 x  0
These generalized results show that for sufficiently small signals   cos / 2  ,
the current is either always zero   0  if    , or never zero     if    .
Sometimes it is important to relate I m to the permissible peak current through the
transistor, I peak . Note from the results above that:
I peak  I m (1  cos / 2)
so that we can express I m as:
Im 
I peak
(1  cos / 2)
For calculating amplifier efficiency, however, we work with I m for simplicity in
notation.
With these results, we can calculate the average power drawn from the power
supply. If the power supply voltage is V , then the instantaneous power delivered
is V iT  t  so that the average power over a cycle is
Pdc  V
1
T

T / 2
iT  t  dt
T / 2
where  T  2  . Using the expression for iT  t  , we find
Pdc  V I m
1
T

  cos  t  cos / 2  dt

where, by definition,  is chosen such that    2 . Let's change variables of
integration: let    t . Then
Pdc  V I m
1
2
 / 2
  cos 
 / 2
 cos / 2  d
Because the integrand is even in  and the limits of integration are symmetric
about the origin, we can multiply the integral by 2 and integrate over only half the
range:
Pdc  V I m
1

 / 2
  cos 
 cos / 2 d
0
Perform the integration:
Pdc  V I m
1

 cos 
  cos / 2 0
 / 2
Pdc  V I m
1
 

 sin  cos 


2
2
2
Here is a plot of the normalized DC power, pdc  Pdc VI m :
Notice that for   360 (class A amplifiers), the required power is constant,
independent of the relative amplitude of the signal, while for smaller conduction
angles, the required power increases with signal amplitude. For   180 , notice
that no power is required for small signals because they fail to produce current in
the transistor.
We next calculate the power dissipated by the transistor. Suppose the voltage
across the transistor is vT  t  . The instantaneous power dissipated by the
transistor is vT  t  iT  t  . The average power dissipated over a cycle is
PT 
1 
vT  t  iT  t  dt
T  
The (ideal, undistorted) form of the voltage across the transistor that sees a
resistive load is sinusoidal with a minimum corresponding to the current
maximum:
vT  t   V * 1   cos  t 
where V * depends upon the circuit configuration. For the usual transformer
coupled class A and B amplifiers, for typical tuned circuit class A, B, and C
amplifiers and for direct-coupled split power supply configurations, V * is just the
power supply voltage, V:
V*  V
For RC-coupled class A amplifiers, typically V *  V 2 . In any case, the power
dissipated is
PT  V * I m
1
T

 1   cos  t  cos  t  cos / 2  dt

Change the variable of integration to    t .
PT  V * I m
 2
1
2
 1   cos   cos 
 2
 cos / 2  d
As before, the integrand is even and the limits of the integral are symmetric so
we can simplify the limits of integration:
PT  V * I m
PT  V * I m
1

 2
 1   cos   cos   cos / 2  d
0
1

 / 2
0
PT  V * I m
1

 / 2
0
 cos    2 cos 2   cos  / 2   cos  cos  / 2  d
 1  cos  / 2  cos   cos  / 2   2 cos 2   d
V *Im  


1

PT 
 1  cos  sin    cos   2   sin  cos   

  
2
2
2
0
PT 
 / 2
V *Im  
  

1 

 
 1  cos  sin  cos   2   sin cos  

  
2
2 2
2
2 2
2
2 
This expression gives the power dissipated in the transistor in terms of the
conduction angle,  , and the signal size (normalized to the maximum signal),  .
For class A amplifiers   360 , notice that the greatest dissipation occurs for
the smallest signals. For class B amplifiers   180 , close examination of the
plot shows that the transistors work harder for intermediate sized signals than for
large signals.
Let's specialize to the cases for which V *  V . Such a restriction excludes mainly
RC coupled amplifiers. In these amplifiers, a resistor in series with the transistor
drops the voltage across the transistor to a value significantly less than the power
supply voltage and in the process drops the efficiency by dissipating power in
addition to that necessary in other configurations. As a consequence, RC
coupled amplifiers are not very useful for power amplifiers, anyway. For the
common case V *  V , we can calculate the power in the signal as
Psig  Pdc  PT
because there is no resistor in the circuit to dissipate any power. Thus
Psig 
V Im 
 
  
  


  
2 1 
 sin  cos    1  cos  sin  cos     sin cos   
 
2 2
2  
2
2 2
2
2 2
2
2  
Psig 
V Im 
 

  

1 

 

 sin  cos   1  cos  sin  cos   2   sin cos  

 
2 2
2
2
2 2
2
2 2
2
2 

Psig 
V Im 
 



 

1 

 
 sin  cos   sin   cos sin  cos   2   sin cos  

 
2 2
2
2
2
2 2
2
2 2
2
2 
Psig 
V Im 


1 

 
  cos sin   2   sin cos  

 
2
2
2 2
2
2 
Psig  V I m
  




    sin cos   cos sin 
 2 2
2
2
2
2
1
Here is a plot of the normalized signal power, psig  Psig VI m :
Not surprisingly, the signal power increases with signal amplitude. For   180 ,
note again that the signal amplitude must exceed some threshold before any
signal power is produced. The decrease of the signal power with decreasing
conduction angle may seem to be inconsistent with the use of conduction angles
of   180 (class B and C) in most high power amplifiers. For high power
amplifiers, however, the efficiency   Psig Pdc is a primary concern.
  




    sin cos   cos sin 
2
2
2
2
2 2
  ,  

 

Pdc
 sin  cos
Psig
2
Here is a plot of the efficiency:
2
2
This plot clearly shows the high efficiencies available at small conduction angles
and large amplitudes. Operating too close to the maximum, however, can mean
that a small change in signal amplitude can result in no power output at all. A
typical compromise is to choose large amplitude operation   1 with 
120 .
To examine the efficiency at large amplitudes in more detail, we specialize to the
case   1 . In that case,    and  becomes:
max 
Psig
Pdc

1 




  sin sin   cos sin
2 2
2
2
2
2
sin

2


2
cos

2
max



 sin sin 

1 2
2
2




2 sin  cos
2 2
2
max



  2sin sin 

1 
2
2

4 sin    cos 
2 2
2
If we employ the trigonometric identity
sin 2 x  2sin x cos x
we can write
max 
1   sin 
4 sin    cos 
2 2
2
It is easy to see that for class A amplifiers   2  , max  0.5 . (With RC
coupling, it is lower by the factor V * V  0.5 .) For class B amplifiers     ,
max   4  78% . For class C amplifiers (  small), it is not so easily shown (with
L’Hospital’s rule) that max approaches 1 .
Here is a plot of maximum efficiency vs. conduction angle:
Power Amplifier Efficiency
1.2
maximum efficiency, etamax
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
conduction angle, theta
5
6
7