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While trying to calculate the high frequency pole frequency corner of the common emitter amplifier I came across the following small signal model (found here in page 332)

Small signal model CE

The small signal equivalent simplified by Miller effect usage shown seems to apply Miller's theorem to \$C_\pi\$ and \$r_\pi\$ using only the equivalent inputs impedance formula:

$$ C_\pi' = (1 - v_e/v_b)\cdot C_\pi\quad r_\pi' = (1 - v_e/v_b)\cdot r_\pi $$

Where, curiously \$r_\pi' = r_\pi + (b+1)\cdot R_4\$. I do not understand how this is valid, or if the theorem is even being applied here. I am aware that the current source may be connected to the ground, since it does not alter the value fed to the output resistance. However, that leaves us with \$C_\pi\$ parallel to \$r_\pi\$ in series with \$R_4\$:

Equivalent impedances

How can both impedances be equivalent and how do I reach the given coefficient \$1 - v_e/v_b\$ (where \$v_e/v_b = \frac{g_m + g_\pi}{g_m + g_\pi + G_4}\$, which is the gain calculated for the CC amplifier).

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There are many ways to see this, but your small signal hand drawing is missing the collector current flowing through it (the CCCS). The controlled current source is NOT connected to ground when that R4 resistor is connecting the emitter to ground (via the cap at high frequencies).

This is the reason why there's a beta+1 scaling R4. From the base, you see not only the rpi resistance but also R4 amplified by the current amplification factor of the BJT.

From loop gain analysis stand point, it simply means that when we have series-series feedback with R4, and its impedance R4 is seen amplified by the loop gain of the circuit.

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  • \$\begingroup\$ Could you expand on the last paragraph? Thank you. \$\endgroup\$
    – 13A
    May 19, 2023 at 13:30

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