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

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Circuit under AC conditions:

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


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

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

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Effectively, the input portion of the circuit reduces to:

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

Now, coming back to :

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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?

  • \$\begingroup\$ 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}\$ \$\endgroup\$ – G36 Jan 3 at 17:54
  • \$\begingroup\$ 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 \$\endgroup\$ – Sunnyskyguy EE75 Jan 3 at 20:31
  • \$\begingroup\$ Have a look at slide 40 in this APEC seminar cbasso.pagesperso-orange.fr/Downloads/PPTs/… \$\endgroup\$ – Verbal Kint Jan 3 at 22:05
  • \$\begingroup\$ 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. \$\endgroup\$ – Sunnyskyguy EE75 Jan 3 at 23:16
  • \$\begingroup\$ 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. \$\endgroup\$ – Verbal Kint Jan 4 at 9:21

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