FE Electrical Electronics Practice Problems

Answer: C) Base-emitter forward biased and base-collector reverse biased

The BJT table on p. 387 of the Handbook lists the three regions by the bias of the two junctions. Active region: base emitter junction forward biased and base collector junction reverse biased. Saturation region: both junctions forward biased. Cutoff region: both junctions reverse biased.

Physically, the forward-biased base-emitter junction injects carriers from the heavily doped emitter into the thin base, and the reverse-biased base-collector junction sweeps almost all of them into the collector before they can recombine. That is what makes \(i_C = \beta i_B\) and \(i_C = \alpha i_E\) hold, with \(\alpha = \beta/(\beta + 1)\) close to one, and it is the only region in which the transistor behaves as a controlled current source and can amplify. For an NPN device the conditions read \(V_{BE} \approx 0.7\) V and \(V_{CB} > 0\), i.e. \(V_{CE}\) above roughly 0.2 V; for a PNP the polarities reverse.

Both junctions forward biased is saturation: the collector also injects carriers, \(V_{CE}\) collapses to about 0.2 V and \(i_C\) falls below \(\beta i_B\); this is the on state of a switch, not an amplifier.

Neither junction forward biased means both reverse biased, which is cutoff: no injection, \(i_C \approx 0\), the off state of a switch.

Base-emitter reverse and base-collector forward biased is the reverse-active mode, the transistor run backwards; it conducts, but with a very small current gain because the collector is lightly doped and makes a poor emitter, so it is not used for amplification.

Answer: A) \(\beta/g_m\)

In the low-frequency small-signal model on p. 387 of the Handbook the base and emitter are joined by \(r_\pi\), and the controlled source \(g_m v_{be}\) sits between collector and emitter. With the emitter grounded, a test voltage at the base sees only \(r_\pi\), so the input resistance is

$$R_{in} = r_\pi \approx \dfrac{\beta}{g_m} = \dfrac{\beta V_T}{I_{CQ}}$$

using the Handbook's \(g_m \approx I_{CQ}/V_T\). The same result comes from currents: \(i_b = i_c/\beta = g_m v_{be}/\beta\), so \(v_{be}/i_b = \beta/g_m\). At \(I_C = 1\) mA and \(\beta = 100\) this is \(100 \times 0.026/0.001 = 2.6\) kΩ, the moderate input resistance that characterises the common-emitter stage; it rises with \(\beta\) and falls as the bias current increases. The collector resistor does not enter because, with \(r_o\) very large, nothing connects the collector node back to the base.

\(1/g_m\) is the resistance seen looking into the emitter, the low input resistance of the common-base stage, smaller than \(r_\pi\) by the factor \(\beta\).

\(\beta R_C\) multiplies the current gain by the load, which has the units of a resistance but corresponds to no node of the model.

\(R_C \| r_o\) is the resistance seen looking into the collector, the output resistance of the stage, not its input resistance.

Answer: A) 10 V

This is the Handbook's two-source configuration on p. 382, \(v_0 = -(R_2/R_1)v_a + (1 + R_2/R_1)v_b\), where \(v_a = V_1\) and \(v_b\) is the voltage at the non-inverting terminal. Since the ideal op amp draws no input current, \(R_3\) and \(R_4\) form a plain divider:

$$v_b = V_2\dfrac{R_4}{R_3 + R_4} = 5 \times \dfrac{50}{60} = 4.167\text{ V}$$

$$v_0 = -\dfrac{50}{10}(3) + \left(1 + \dfrac{50}{10}\right)(4.167) = -15 + 25 = 10\text{ V}$$

Because the ratios match, \(R_2/R_1 = R_4/R_3 = 5\), the two terms combine into the difference form \(v_0 = (R_2/R_1)(V_2 - V_1) = 5(5 - 3) = 10\) V, and the common part of the inputs cancels exactly: 3 V and 5 V give the same output as 103 V and 105 V would, which is what the circuit is for. A mismatch in the ratios lets a fraction of the common-mode voltage through, and the CMRR of the finished amplifier is then set by resistor tolerance rather than by the op amp.

25 V is the non-inverting term alone, \((1 + R_2/R_1)v_b\), with the contribution of \(V_1\) through \(R_1\) omitted.

12 V applies the non-inverting gain of 6 to the difference, \(6(5 - 3)\); the difference is multiplied by \(R_2/R_1\), not by \(1 + R_2/R_1\).

40 V multiplies the sum of the inputs, \(5(5 + 3)\), instead of their difference.

Answer: C) 44 kHz

The Handbook's statement of Nyquist's (Shannon's) sampling theorem on p. 224 is that the sample rate must be larger than twice the highest frequency contained in the measured signal, \(f_s > 2f_N\) with \(f_N\) that highest frequency:

$$2f_{max} = 2 \times 22\text{ kHz} = 44\text{ kHz}$$

So 44 kHz is the Nyquist rate, the limiting value that the sampling frequency must exceed. Each cycle of the highest component must be sampled more than twice, otherwise the sequence of samples could equally have come from a lower frequency; the Handbook calls the impostors alias frequencies. In practice the rate is set somewhat above the limit and an anti-aliasing low-pass filter removes whatever lies above \(f_s/2\) before sampling, since a real signal is not band-limited exactly at 22 kHz. Compact-disc audio, sampled at 44.1 kHz, is this very calculation with a margin of 0.1 kHz for the filter's transition band.

22 kHz samples at the signal's own highest frequency, one sample per cycle, which cannot distinguish that component from dc.

11 kHz halves instead of doubling; it is the highest frequency that a 22 kHz sampler could capture, the relation read backwards.

88 kHz is twice the Nyquist rate, a comfortable choice but not the theoretical minimum asked for.

Answer: C) 0.75

The Handbook's Power Conversion section on p. 385 gives the voltage gain of an ideal boost converter as \(1/(1 - D)\), with \(D\) the duty ratio, the fraction of the switching period during which the switch is on.

$$\dfrac{V_{out}}{V_{in}} = \dfrac{1}{1 - D} \quad\Rightarrow\quad 1 - D = \dfrac{V_{in}}{V_{out}} = \dfrac{12}{48} = 0.25$$

$$D = 1 - 0.25 = 0.75$$

Check: \(12/(1 - 0.75) = 12/0.25 = 48\) V. The switch is closed for 75% of each period, during which the inductor charges from the 12 V source, and open for 25%, during which the inductor releases its stored energy in series with the source into the 48 V output. Volt-second balance on the inductor, \(V_{in}DT = (V_{out} - V_{in})(1 - D)T\), is where the gain formula comes from and gives the same 0.75 directly. A gain of 4 is near the practical limit of a boost stage: as \(D \to 1\) the ideal gain grows without bound but conduction losses in the switch and inductor take over.

0.25 is \(1 - D\), the off-time fraction, reported as the duty ratio.

0.80 comes from the buck-boost gain \(D/(1 - D) = 4\); that topology inverts the output and is not a boost converter.

0.20 is \(V_{in}/(V_{in} + V_{out})\), the complement of the buck-boost answer.