# BJT: Is $R_E || [\frac{R_s'}{\beta}+r_e] = (R_s'+ \beta r_e)|| \beta R_E$?

I am trying to find the effect of $$\ C_E \$$ on voltage gain.

Circuit under AC conditions:

Doubt:

Is $$\ R_E || [\frac{R_s'}{\beta}+r_e] = (R_s'+ \beta r_e)|| \beta R_E\$$ ?

Reasoning:

Small signal model of BJT, in general, is given by:

To find $$\ R_E \$$ , we short-circuit the voltage signal source:

Effectively, the input portion of the circuit reduces to:

Therefore, it may be assumed that $$\ i_b \$$ flows through $$\ R_E(1+\beta) \$$.

Now, coming back to :

We have

$$\ R_e= (R_s'+ \beta r_e)|| (\beta+1) R_E \$$

$$\ R_e= (R_s'+ \beta r_e)|| (\beta) R_E \$$ (approx)

$$\ =\frac{(R_s'+ \beta r_e)* (\beta) R_E}{(R_s'+ \beta r_e)+(\beta) R_E} \$$

$$\ = \frac{ (\frac{R_s'}{\beta}+ r_e)*(\beta) R_E}{ (\frac{R_s'}{\beta}+ r_e)+ R_E} \$$

$$\ = \beta [R_E|| (\frac{R_s'}{\beta}+ r_e)] \$$

$$\ \ne [R_E|| (\frac{R_s'}{\beta}+ r_e)] \$$

Why this discrepancy?

• Well, in fact from the emitter point of view, the $R_S$ will be seen at the emitter as (beta +1) times smaller $\frac{R_S}{\beta +1}$ – G36 Jan 3 at 17:54
• Is $R_E || [\frac{R_s'}{\beta}+r_e] = (R_s'+ \beta r_e)|| \beta R_E$? No. Compute Vb then Ib then Vc=β Ib Rc – Sunnyskyguy EE75 Jan 3 at 20:31
• Have a look at slide 40 in this APEC seminar cbasso.pagesperso-orange.fr/Downloads/PPTs/… – Verbal Kint Jan 3 at 22:05
• That slide is 1st order with Vs on base, his is 2nd order with RC to base so there is an input attenuation then a transimpedance gain. – Sunnyskyguy EE75 Jan 3 at 23:16
• It seems that both coupling capacitors (input and output) are considered short circuited here otherwise it becomes a third-order circuit. The equivalent small-signal circuit presented by the OP is only first-order. – Verbal Kint Jan 4 at 9:21