Discussion
The voltage gain for the following amplifier circuit
In the circuit, a transistor with \(\beta\) = 200 is supplied from +20 V; its base is biased by a 50K resistor from +20 V and a 50K resistor to ground and receives the signal through a capacitor, a 10K resistor connects its emitter to ground, and \(V_{out}\) is taken from the emitter through a capacitor across a 500 Ω load.
is nearly equal to :
Answer: B
The voltage gain is nearly 1.0.
The output is taken from the emitter, so this is an emitter follower (common collector) stage. Its voltage gain is:
\[ \begin{aligned} A_{v} &= \frac{(1 + \beta)R'_{E}}{r_{\pi} + (1 + \beta)R'_{E}} \\ &\approx 1 \end{aligned} \]
Here \(R'_{E}\) is 10 kΩ in parallel with the 500 Ω load, about 476 Ω, and \((1 + \beta)R'_{E} \approx 96\) kΩ, far larger than \(r_{\pi}\), so the gain is just under 1. The emitter follows the base: it buffers rather than amplifies voltage, so gains of 1.5 or 2 are impossible and 0.5 is far too low.
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