A simplified small-signal equivalent circuit of a BJT-based amplifier is given below. The small-signal voltage gain \( \frac{V_o}{V_S} \) (in V/V) is _________.
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When analyzing small-signal BJT circuits, remember that the voltage gain involves the transistor’s current gain \( \beta \), and the total resistance seen by the input is the sum of \( R_S \) and \( r_\pi \), the base-emitter resistance.
Step 1: Current gain
The small-signal current gain of the transistor is given by \( \beta \), where:
\[
i_b = \frac{V_S}{R_S} \quad (\text{the current through } R_S)
\]
The current gain from the base to the collector is \( i_c = \beta i_b \).
Step 2: Voltage gain
The voltage gain is the ratio of the output voltage \( V_o \) to the input voltage \( V_S \). The output voltage is:
\[
V_o = -i_c \times R_L = -\beta i_b R_L
\]
Substituting \( i_b = \frac{V_S}{R_S} \), we get:
\[
V_o = -\beta \left( \frac{V_S}{R_S} \right) R_L
\]
Therefore, the voltage gain is:
\[
\frac{V_o}{V_S} = -\frac{\beta R_L}{R_S}
\]
Step 3: Consider the \( r_\pi \) term
The resistance \( r_\pi \) represents the resistance between the base and emitter of the transistor. When considering the effect of \( r_\pi \), the total resistance seen by the input is \( R_S + r_\pi \). Hence, the voltage gain becomes:
\[
\frac{V_o}{V_S} = -\frac{\beta R_L}{R_S + r_\pi}
\]
Thus, the correct voltage gain is \( \frac{-\beta R_L}{R_S + r_\pi} \), which corresponds to option (A).
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