Small Signal Amplifiers Questions and Answers

Answer: D

The overall gain is 30 dB.

The voltage gains of cascaded stages multiply, and since decibels are logarithmic, multiplying gains means adding their decibel values:

\[ G = 5 + 10 + 15 = 30\ \text{dB} \]

750 dB is the mistake of multiplying the decibel figures (5 × 10 × 15). 15 dB takes only the largest stage, and 25 dB adds only two stages.

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.

Answer: A

Johnson noise is \(\bar{v}_{\text{rms}} = (4KTR\Delta f)^{1/2}\).

Thermal (Johnson-Nyquist) noise arises from the random thermal motion of charge carriers in any resistor. Nyquist showed that its mean square voltage over a bandwidth \(\Delta f\) is \(4kTR\Delta f\), independent of frequency (white noise) and present even with no current flowing.

For example, a 1 k\(\Omega\) resistor at 300 K over 10 kHz gives about 0.4 \(\mu\)V rms. \((2qI\Delta f)^{1/2}\) is shot noise current, which depends on a DC current and multiplied by R gives a shot noise voltage. The forms \(A/f\) and \(A/f^{2}\) describe flicker (1/f) and other frequency-dependent noise.

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.

Answer: A

A dc multistage amplifier needs direct coupling (the paper prints 'output state' for output stage).

A dc amplifier must amplify signals down to zero frequency. Direct coupling joins the output of one stage to the input of the next with a wire or resistor, so there is no component that blocks dc, and the low-frequency response extends down to 0 Hz.

RC (capacitive) coupling places a series capacitor between stages, which blocks dc and gives a lower cut-off frequency. Transformer coupling also blocks dc, since a transformer passes only changing currents. So all three would remove the very signals a dc amplifier exists to amplify.

Answer: A

The small-signal model is valid when the transistor operates in the active region.

Small-signal models such as the hybrid-pi and h-parameter models replace the BJT's non-linear characteristics with a linear approximation around a fixed operating point (Q-point). This is accurate only for small variations about a Q-point in the forward-active region, where the collector current is controlled by the base-emitter voltage.

In cutoff and saturation the transistor behaves as an open or closed switch and the linear relations fail. Breakdown is a non-linear, destructive region that no small-signal model describes.