Electronic – How to decide bypass emitter resistance value

The value of \$r_{o}\$ depends on the transistor's characteristics and its biasing:

$$r_{o} = \frac{V_{\text{A}} + V_{\text{CE}}}{I_{\text{C}}} \approx \frac{V_{\text{A}}}{I_{\text{C}}}$$

where \$V_{\text{A}}\$ is the transistor's Early voltage and can vary widely between transistors, but is usually on the order of \$10 - 100\text{ V}\$. Assuming a typical bias of \$I_{\text{C}} \approx 1\text{ mA}\$, \$r_{o}\$ should be on the order of \$10 - 100\text{ k}\Omega\$. If \$R_{\text{L}} = 1\text{ k}\Omega\$ and \$r_{o}\$ is only \$10\text{ k}\Omega\$ then you could have a significant error if you assume \$r_{o} \parallel R_{\text{L}} \approx R_{\text{L}}\$. But if \$V_{\text{A}}\$ is higher or \$I_{\text{C}}\$ is lower then it may be safe to ignore \$r_{o}\$, depending on how accurate you are looking to be.

Also, it's worth noting that for calculating \$R_{\text{in}}\$ the resistance \$R_{\text{E}} = r_{o}||R_{\text{L}}\$ at the emitter is magnified by \$\beta + 1\$ so an error in \$R_{\text{E}}\$ due to ignoring \$r_{o}\$ could be magnified significantly. On the other hand, for calculating \$R_{\text{out}}\$ the emitter resistance \$R_{\text{E}}\$ is divided by \$\beta + 1\$ so the error would not be as significant.

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.