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For the sake of understanding, let's assume we take the small signal output
from the collector of Q2, and input the small signal voltage to the
base of Q1. This is shown in Fig. 5.3.
Figure 5.3:
Differential Amplifier Single-Ended Differential Mode Gain Configuration
|
The small signal voltage gain will then be defined as the change in the
collector voltage at Q2 divided by the change in the base voltage of Q1,
or .Taking the small signal change of (5.5),
we find the small signal output voltage Vo.
| |
(106) |
To find the small signal change
in the input, we start with the large signal KVL equation
|
Vb1 -Vb2= Vbe1-Vbe2
|
(107) |
Now, if we ground Vb2, and make a small signal change in Vb1
we obtain
| |
(108) |
Recall from previous labs that a first order Taylor
series gives .We then find that
| |
(109) |
Since
and, under DC conditions IC1=IC2, then
Now,
recall that the total current in the diff-amp is determined by the DC bias current
IEE which is fixed. Thus, if we make a small increase in Ic1 then there
will be a corresponding small decrease in Ic2.
We express this mathematically by
taking the small signal change of
equation (5.2), which gives
| |
(111) |
which leads to
| |
(112) |
Substituting the previous expressions into equation (5.9) leads to
| |
(113) |
Taking the ratio of equations (5.7) and (5.14) leads to the
following expression for
voltage gain Av12
of a differential amplifier when the output is take from the opposite
transistor as the input.
| |
(114) |
If the output were to be taken from the collector of Q1 instead of Q2, we could
follow virtually the same analysis and would find that the voltage gain Av11
would differ only by a minus sign.
| |
(115) |
If we apply a signal to both inputs, and take the output from the collector
of Q2, then we can define the differential gain to be
| |
(116) |
Next: Input and Output Resistance
Up: Differential Amplifiers
Previous: Differential Pair DC Bias
Neil Goldsman
10/23/1998