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.

Answer: C

The input resistance rises in the order common base (2), common emitter (1), emitter follower (3), and emitter follower using a Darlington pair (4): the common base stage has a very low input resistance, the common emitter stage a moderate one of about \( h_{ie} \), the emitter follower a high one of about \( \beta R_E \), and the Darlington pair increases it further because its current gain is the product of two stages. That order is 2, 1, 3, 4, option C. Options A and B place the Darlington pair below the simple follower, and option D puts the common emitter stage below the common base stage.

Answer: B

The matching is (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii), which is option (B). RC coupling gives the least non-linear distortion. Inductive (choke) coupling has little dc drop across the choke, so a low collector supply voltage can be used. Transformer coupling gives higher voltage gain together with impedance matching, and direct coupling can amplify dc and very low frequency signals because there is no coupling capacitor. Options (A), (C) and (D) pair direct coupling or transformer coupling with the wrong property.

Answer: A

The input impedance rises in the order CB, CE, CC and then the op-amp, so the sequence is 4, 1, 2, 3, the first option. The common base stage has a very low input impedance, the common emitter stage a moderate one, the common collector stage a high one because of the \((1+\beta)R_E\) term, and an op-amp has the highest.

The other options put the CC stage or the CB stage in the wrong place, for example the second makes the CE stage lower than the CB stage.