Equivalent resistance in BJT small-signal : Rpi in parallel with Beta*Ib

Question

I'm currently reading Vorperian's book on Fast Analytical Circuits techniques. On page 129 he states :

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I'm not sure I understand correctly the statement I highlighted in yellow. Is it some kind of use of the Norton<->Thevenin conversion?

If anybody could make me understand what he means mathematically and/or schematically.

I know the resistance reflection rule in a BJT, but I'm not sure I understand it in terms of "parallel combination between \$i_b \beta\$ and \$r_{\pi}\$".

It doesn't seem the resulting equivalent circuit has \$r_e\$ in series with \$R_E\$ given eq. 4.59.

Any help will be appreciated. Thanks!

Answer 1

TThe dynamic resistance \$r_e\$ is nothing else than the resistance of the parallel combination of \$r_{\pi}\$ and the current source (nomenclature as in the figure):

Test voltage: \$v_T=i_T r_e\$ with \$i_T=i_b + \beta i_b\$ and with \$i_b=\frac{v_T}{r_{\pi}}\$.

From this you can solve for \$r_e=\frac{r_{\pi}}{\beta+1}\$

By the way (physical interpretation): the resistance \$r_e\$ is nothing else than the inverse transconductance \$g_e\$ for the transistor: \$r_e=\frac{1}{g_e}=\frac{dV_{be}}{dI_e}\$.

From the diagram you can derive that \$r_e\$ is in parallel with the sum of \$R_E\$ and \$R_s||R_L\$