Discussion
The AC Gain of the following transistor amplifier
is nearly equal to :
Answer: C
The AC gain is nearly 400.
Find the DC emitter current first. The Thevenin equivalent of the base divider is \(V_{TH} = 30 \times \frac{15.7}{30} = 15.7\) V and \(R_{TH} = \frac{14.3 \times 15.7}{30} \approx 7.5\) K:
\[ \begin{aligned} I_{E} &= \frac{15.7 - 0.7}{3 + 7.5/201} \approx 4.9\ \text{mA} \\ r_{e} &= \frac{26\ \text{mV}}{4.9\ \text{mA}} \approx 5.3\ \Omega \\ A_{v} &= \frac{R_{C}}{r_{e}} = \frac{2000}{5.3} \approx 380 \end{aligned} \]
The emitter resistor is bypassed by the capacitor, so only \(r_{e}\) appears in the emitter for AC. The gain of about 380 is nearest 400. Leaving the 3 K unbypassed would give a gain below 1, and 200 is \(\beta\), the current gain, not the voltage gain.
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