Consider the transistor circuit shown in the figure below:

Assume that the biasing details are left out and the transistors are working in active region with $T_{1}$ having transconductance $g_{m 1}$ and input base-emitter small signal resistance $r_{π 1}$ and for transistor $T_{2}$ the transconductance $g_{m 2}$ and input small signal base-emitter resistance $r_{π 2}$ . Then the voltage gain $| A_{v} |$ is given as

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Solution

The small signal equivalent of the above circuit can be given as

From the above circuit it is clear that,
$V_{o u t} = g_{m 1} r_{π 2} v_{π 1}$
thus, $V_{i n} = v_{π 1}$
thus, $V_{o u t} = g_{m 1} r_{π 2} v_{i n}$
$∴ Voltage gain | A_{v} | = ∣ ∣ ∣ \frac{V_{0}}{V_{i n}} ∣ ∣ ∣ = g_{m 1} r_{π 2}$
now, $g_{m} r_{π} = β$
Thus, $r_{π 2} = \frac{β_{2}}{g_{m 2}}$
$∴ | A_{v} | = β_{2} (\frac{g_{m 1}}{g_{m 2}})$

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