Electronic – Calculating gain of Amplifiers

One option not mentioned is to use packaged difference amplifiers or instrumentation amplifiers, where all the internal resistors are very effectively trimmed to high accuracy. This effectively minimized the risk of mismatched resistors, which can really kill common mode rejection (which is often a bigger problem than non-exact gains).

If high precision gain is a real must, an instrumentation amplifier sets the gain using only one resistor, and that can be as high precision as you're willing to pay for. Alternatively, a lower resistor and a trimmer pot can be used for adjustment

For my opinion, it is somewhat confusing to use this term "re" (and your question can serve as a "good" example). This parameter re (called "intrinsic emitter resistance") is nothing else than the inverse of the transconductance re=1/gm. I think, it is not correct to call it "resistance" because it relates the output current and the input voltage (gm=dIc/dVBE). In contrast, the classical resistor definition relates the voltage across and the current through the two-pole element.

This transconductance appears in each gain formula, for example in common emitter configuration:

Gain A=-gmRc/(1+gmRe)=-Rc/(Re+1/gm)=-Rc/(Re+re) .

The value of gm=1/re can be found by differentiating the transfer function Ic=f(Vbe) at the selected quiescent current Ic and can be found as gm=Ic/Vt (Vt=temperature voltage). Hence, it is a parameter which is computed at the DC bias point.

Does this answer your question?

EDIT

From the gain formula (with emitter resistor RE) we can derive that the gain is reduced by the factor 1/(1+gmRE). From this we can again derive that the term (gmRE) is the loop gain of the feedback system. From feedback theory we know that the disrortions are reduced also by the same factor. Rewriting this factor we get

1/(1+gm*RE)=(1/gm)/(RE+1/gm)=re/(RE+re).

This prooves the result you have shown.