Lecture Notes for Analog Electronics - The Electronic Universe ...

The ratio of the differential to common-mode gain (ignoring the sign) then gives the
CMRR:
CMRR = RE +2REE
2RE
≈ REE
RE
where for a typical design REE ≫ RE.
Building on what we learned, we can easily improve our differential amplifier design by
adding an emitter-follower stage to the output and a replacing the resistor REE with a
current source. This is discussed briefly in the next two sections.
5.7.2 Adding a Follower
The output impedance of the common-emitter configuration, as used in the differential amplifer,
Zout ≈ RC, is not always as small as one would like. This can be easily improved
by adding an emitter follower to the output. Hence, the input of the follower would be
connected to the output of the differential amplifier. As discussed before, the follower then
produces an output impedance which is ≈ β + 1 times smaller than the preceding stage.
Hence, in this case, we would have Zout ≈ RC/β. The follower’s emitter resistor (call it
R ′ E), of course, has to be consistent with our impedance non-loading criteria, in this case
βR ′ E ≫ RC.
5.7.3 Adding a Current Source
In our expression for CMRR above, we see that a large REE improves performance. However,
this can also significantly load the voltage source of VEE, producing non-ideal behavior.
Hence, REE is limited in practice. A solution to this which is commonly used is to replace
REE by the output of a current source. In other words, point “A” in Fig. 25 would be
connected to the collector of a current source such as that of Fig. 20. This can be justified
by noting that the dynamic impedance provided by REE is given by vA/iEE. By limiting
variations in iEE, as provided by a current source, one effectively achieves a large dynamic
impedance.
To implement this one has to decide what quiescent current is required for the current
source and what the quiescent voltage of the collector should be. The latter is given by the
quiescent voltage at the inputs of the differential amplifier. For example, if the inputs are
DC ground, then point “A” will be at approximately −0.6 V, depending upon any voltage
drop across RE.
5.7.4 No Emitter Resistor
One variation of the above is to remove the emitter resistor. In this case one replaces RE in
the expressions above with the intrinsic emitter resistance discussed in Section 5.8.1 below:
RE → re = 25mV/IC
To be exact, one should replace RE in our equations with the series resistance of RE and re:
RE → RE + re. However, in most practical situations RE ≫ re.
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