Consider an NPN bipolar junction transistor common emitter amplifier:
In this circuit, recall, v th
  and
R
might represent the Thevenin equivalent th circuit of a preceding amplifier stage, a signal generator or a sensor of some sort that is connected to the input terminals of the amplifier. Note that, for now, no load is connected across the output of the amplifier so that it operates under open circuit conditions.
We now set ourselves to the task of finding the Thevenin equivalent circuits of this amplifier for signals at its input and output terminals. First, we replace the transistor by its linear equivalent circuit:

2
To analyze this circuit, recall, we divide the independent sources into two groups, signal and bias, and apply the superposition theorem to permit calculation of their effects separately. The signal sources are all independent sources that generate what we consider to be signals. In our case, v th
  is the only signal source. The bias sources are all other independent sources. In our case, the bias sources are
*
V be
,
I c
*
and
V
. In applying the superposition theorem, all dependent sources cc must, of course, be left on all the time.
The superposition theorem tells us that a part of each voltage or current in the circuit results from the signal sources and another part from the bias sources. To apply superposition, we first set the bias sources to zero to obtain the signal equivalent circuit (sometimes called the a-c equivalent circuit). Note that, for example, setting
V cc

0
connects
R b 1
and
R b 2
in parallel and connects
R c
to ground:

where
R b

R || b 1
R b 2
. If we replace the signal equivalent circuit shown here for the bipolar junction transistor with the equivalent standard hybrid  version discussed earlier, we get:
3
We now analyze the midband performance of this amplifier. By midband, we mean the range of frequencies high enough so that
C b
and
C c
behave like short circuits, but low enough so that the dotted capacitors in the high frequency model of the transistor, and other stray capacitances, can be neglected. In midband, in other words, we replace those capacitors that we can see in the signal equivalent circuit by short circuits and those that we cannot see (the strays) with open circuits. The signal equivalent circuit therefore becomes: b
From this equivalent circuit, we want to derive Thevenin equivalent circuits for signals at both the input and output terminals of the amplifier. Specifically, we seek an equivalent circuit in the following form:

We will find the Thevenin equivalent circuit for the output terminals by finding, first, the output voltage under open circuit conditions and then by finding the output current under short circuit conditions. To find the open circuit output voltage, we write the following node equations: b ' : e : v b '
 v in r bb '
 v b '
 v e r

0 v e
 v b ' r
  v e
 v c
 g ce
 g m
 v b '
 v e
  v e
R e

0 c :
 v c
 v e
 g ce
 g m
 v b '
 v e
  v c
R c

0
We can write these equations in matrix form as follows:













1 r bb ' r
1
 g m
 r
1 g m


 
  r
1
 r
1
 g ce
 g m

1
R e


  g ce
 g m


0 g ce

 g ce

1
R c









 v b v v e c


 v in r bb '



 
0

 
 0 
 
For simplicity (and with little loss of accuracy), we set g

0
. (Recall that ce g ce

G
so that this assumption is equivalent to assuming that the collectoremitter characteristic curves of the BJT are perfectly horizontal – that they have zero slope – a pretty good approximation.) We could keep terms involving gce, but they add much more complexity than accuracy to the analysis.
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We begin finding the solution to this set of equations (with g

0
) by calculating ce
Cramer's delta for the coefficient matrix in the set of equations above:
 
1
 
 
1
R c
 r bb '
 r
1

 r
1
 g m

1
R e



1
 r  r
1
 g m

 
 
1

1


1
R c r bb '
 r
 g m

1
R e



1
 r  r
1
 g m



1 1

1
 r R r r e
1
 g m




 
1

1


1
R c r bb '
 r
 g m

1
R e



1 1 r R e


The Thevenin voltage for the output terminals in these equations is simply the open-circuit output voltage, v
. We find it using Cramer's rule: c v c

1
 r v in bb '



 r
1
 g m

 g m
 g m

 r
1
 g m

1
R e

  v c

1
R c
 r
1 bb '

 r
1 v r in bb '



 g m
 g m
R e


1
R e



1 1 r R e

 v c

1

R e
 g R v m c in

 r
1
 g m


 r bb ' r
If we define the open circuit voltage gain of the amplifier as
A voc
 v v in then we can write the Thevenin voltage at the output terminals as v
  
A v voc in
 
5

and, by recognizing that v
 v c
, see that
A voc
 v c
 v in
1
 r
 r bb '
 g R m c
R e

 r
1
 g m


We can easily rewrite this result in terms of  and h ie
:
A voc

1
 r bb ' r
 r


R c r
R e

1
  
 h ie



1


R c
 
R e
We now calculate the output current under short circuit conditions so that we can calculate the Thevenin resistance at the output terminals,
R
. Consider the out signal equivalent circuit under short circuit conditions.
6
Note that, as before, we have set g

0
. Also note that ce
R
has been removed c because it appears in parallel with the short circuit at the output terminals. We solve for the short circuit current by writing mesh equations for mesh currents i b and i
and then solving for sc i
: sc base loop: v in
 b
 bb '
 r
  i b
 i sc

R e collector loop: i sc
  g v m '
  g m
 i r
'


We can write these two equations in matrix form:

 r bb '
 r

R e g r

1
R e i b   
 v
0 in
 
We solve these equations for isc by using Cramer's rule: i sc
 r bb '
 g r v m ' in
 r

R e

R g r e m ' or i sc

1
 r
 r bb '
 g v m in
R e

 r
1
 g m


We find the Thevenin resistance for the output terminals from the open circuit output voltage and the short circuit output current:
R out
 v i sc

1

R e


 g R v in r
1
 g m


 r bb ' r
1
 r r bb '

R e

 r
 g v m in
1
 g m


After cancellation of terms, we find
R out

R c
In summary, we have derived the following Thevenin equivalent circuit for the amplifier at its output terminals:
7

Notice that we have eliminated the negative sign on the open circuit voltage gain by reversing the sign on the Thevenin voltage source.
We now derive the Thevenin equivalent circuit for the amplifier at its input terminals. The relevant signal equivalent circuit is:
8
Notice that there are no independent sources in this circuit unless one is connected to the input terminals. It follows that the open circuit voltage, and hence the Thevenin voltage, at the input terminals is zero: i in
 i b
 v in

R b v in
 v b ' r bb '
 v in
R b
To find
R
, therefore, we must apply an input voltage, call it in v
, and calculate the in resulting input current. From the signal equivalent circuit, we see that i in
 i b
 v in

R b v in
 v b ' r bb '
 v in
R b
The quotient v i in in
then gives
R in
. Note that to find i in
, we need to solve for v b '
.
This signal equivalent circuit is the same as the one we considered for the analysis of the output terminals with two changes. First, g ce
is not shown since it has been set to zero for simplicity, as explained before. Second, a load resistor is connected across the output because the i-v characteristics at the input terminals could well be affected by what load is connected to the amplifier output since the

input and output circuits share
R
, and hence are coupled. To the extent that e g ce can be neglected, however, the collector current is driven by an ideal current source and hence is independent of both
R
and c
R
L
. For convenience, we let
R
  . In that case, the node equations that we wrote before still apply,
L provided that we apply a signal voltage v in
to the input of the amplifier:













1 r bb ' r
1
 g m
 r
1 g m


 
  r
1
 r
1
 g m

1
R e


 g m
0
0






1
R c

 v b v v e c


 r v in bb '



 
0

 
 0 
 
We can solve these equations for v b '
by recalling Cramer's delta and using
Cramer's rule. v b '
 r v in bb '

 r
1
1

1


1
R c r bb '
 r
 g m
 g m


1
R e
1



1
R R e c
 r
1 1
R e


Thus, v in
 v b '

1

1 r r bb '


1 1 r R e g m
 v in
1
R e



1 1 r R e
The base current is therefore i b
 v in
 v b ' r bb '

1

1 r r bb '
1 1 1 r r R bb ' ' e v in
 g m

1
R e



1 1 r R e
9

10 or i b
 v in
R e
 r
' g R e
 r
 r bb ' i b

R e


1
 r v in r
 

 h ie i b
 h ie
 v in

1
  
R e
Thus, i in
 i b
 v in

R h b ie
 v in

1
  
 v in
R R e b and i in



 h ie
 
1
1
  

1
R R e b


 v in
The Thevenin resistance at the input terminals is therefore
R in
 v in
 i in
1 h ie
 
1
1
  

1
R R e b or
R in
 h ie
 
1
  
R e

R b
The Thevenin equivalent circuits for the input and output terminals thus become:

11
Notice that from this equivalent circuit, we can easily calculate the signal output of the amplifier for any load connected across the amplifier terminals.
If we follow the biasing procedure outlined earlier, we'll see later that h ie
   
R e and
   
R R e b so that the equivalent circuit simplifies to:
This approximate equivalent circuit for the signal behavior of the amplifier, simple and valuable for design, shows that the gain, output resistance and input resistance of the amplifier can be set by the designer, largely independent of the transistor characteristics, simply by choosing the ratio
R R c e
,
R c
and
R b
, respectively.
R e
and
R b
depend mainly, recall, on the value of the collector current at the operating point,
I cq
.
R
depends on c
I cq
and the power supply voltage,
V cc
. The main design compromise is the choice of
R c
: a designer often would like to choose
R c
large to realize a large open circuit voltage gain but choose it small to realize a low output resistance or accommodate a low power supply voltage. If no suitable combination of output resistance and gain can be

found with a single stage of amplification, one possibility is to cascade two stages, the first with higher
R
to achieve higher gain and the second with lower c
R
to obtain a lower output resistance. An alternative second stage could be a c voltage follower, a circuit that we'll see later has an exceedingly low output resistance but a voltage gain slightly less than one.
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