FE Electrical Electronics Practice Problems

Answer: A) \(-5\) V

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Operational Amplifiers, p. 386 (PDF viewer: page 392). The page states the two ideal-op-amp rules — the input currents are zero, and in linear operation \(v_2 - v_1 = 0\) — and then gives the two-source result

$$v_0 = -\frac{R_2}{R_1}v_a + \left(1+\frac{R_2}{R_1}\right)v_b$$

with the explicit note: “If \(v_b = 0\), we have an inverting amplifier with \(v_0 = -(R_2/R_1)v_a\).” In the Handbook's notation \(v_a\) is the source that feeds the inverting input through \(R_1\) and \(v_b\) the one at the non-inverting input; the Handbook calls the feedback resistor \(R_2\), which is the \(R_f\) of this figure.

Step 1 – Read which branch is which. The figure shows \(V_{in}\) entering through \(R_1\) to the inverting terminal, and the non-inverting terminal tied to ground. So \(v_a = V_{in} = 0.5\) V and \(v_b = 0\), and the second term of the Handbook's formula vanishes. That single reading is what separates the answer from two of the distractors.

Step 2 – Apply the inverting-amplifier result.

$$V_{out} = -\frac{R_f}{R_1}V_{in} = -\frac{100\;\text{k}\Omega}{10\;\text{k}\Omega}(0.5) = -10(0.5) = -5\;\text{V}$$

Answer: (A).

Where that formula comes from, in two lines, using only the two rules on the same Handbook page. The non-inverting input sits at 0 V, so by \(v_2 - v_1 = 0\) the inverting input sits at 0 V too — a virtual ground. No current enters the op amp, so everything arriving through \(R_1\) must continue through \(R_f\):

$$\frac{V_{in} - 0}{R_1} = \frac{0 - V_{out}}{R_f} \quad\Longrightarrow\quad V_{out} = -\frac{R_f}{R_1}V_{in}$$

The current here is \(0.5\;\text{V}/10\;\text{k}\Omega = 50\;\mu\)A, and it drops \(50\;\mu\text{A} \times 100\;\text{k}\Omega = 5\) V across \(R_f\), starting from 0 V at the virtual ground — hence \(-5\) V at the output.

What the gain does and does not depend on. Only the ratio \(R_f/R_1\) appears. Doubling both resistors changes nothing; the op amp's own gain \(A\) has dropped out entirely, which is the whole point of negative feedback and why \(A > 10^4\) is enough to be called “infinite.”

Step 3 – Eliminate the distractors:

(B) \(+5.5\) V is \(\left(1+R_f/R_1\right)V_{in} = 11(0.5)\): the non-inverting gain, the second term of the Handbook's two-source formula. It applies when the signal arrives at the + terminal. Here the figure grounds that terminal, so the term is zero. This is the most common op-amp error there is, and the figure is the only thing that tells you which formula to use.

(C) \(-50\) V is a decade slip on the input resistor: reading \(R_1\) as 1 kΩ gives a gain of \(-100\). The gain is the ratio of the two resistances, \(100/10 = 10\), and a factor of ten in \(R_1\) moves the answer by a factor of ten. Reading the two resistor labels off the figure together, rather than one at a time, is what catches it.

(D) \(-0.5\) V takes the gain as \(-1\), i.e. \(R_f = R_1\). The figure shows \(R_f\) ten times larger.

Answer: A) 0.105 mA

The Handbook's Shockley equation on p. 390 is \(i_D \approx I_s\left(e^{v_D/(\eta V_T)} - 1\right)\), with \(\eta = 1\) here and the thermal voltage \(V_T = kT/q \approx 0.026\) V at 300 K (p. 387). The exponent is

$$\dfrac{V_D}{V_T} = \dfrac{0.6}{0.026} = 23.08$$

$$e^{23.08} = 1.052 \times 10^{10}$$

$$I = 10^{-14}\left(1.052 \times 10^{10} - 1\right) = 1.052 \times 10^{-4}\text{ A} = 0.105\text{ mA}$$

The \(-1\) is negligible against \(10^{10}\): in forward bias the diode equation is effectively \(I_s e^{V_D/V_T}\). Ten orders of magnitude of exponential gain is why a silicon diode goes from leakage to milliamperes over a few tenths of a volt, and why the exponent must be carried to two decimals: an error of 0.1 in \(V_D/V_T\) is a 10% error in the current. Keeping the exponent as 23.08 and evaluating \(e^{23}e^{0.08} = (9.74 \times 10^{9})(1.083)\) reproduces \(1.05 \times 10^{10}\) without a calculator overflow.

0.265 mA uses \(V_T = 25\) mV, so the exponent becomes 24 and \(e^{24} = 2.65 \times 10^{10}\); the stem fixes \(V_T\) at 26 mV.

1.05 mA is the correct \(1.05 \times 10^{-4}\) A misconverted by one decade when going to milliamperes.

4.93 mA substitutes the 0.7 V rule of thumb for the given 0.6 V; the exponent rises to 26.92 and the current by a factor of 47.

Answer: C) 44 kHz

The Handbook's statement of Nyquist's (Shannon's) sampling theorem on p. 230 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) \(-96\)

Each transistor carries half the tail current at balance, \(I_C = I_{EE}/2 = 1\) mA, so by p. 391 of the Handbook

$$g_m = \dfrac{I_{CQ}}{V_T} = \dfrac{1 \times 10^{-3}}{0.026} = 38.5\text{ mS}$$

The gain comes from the pair's own equation on p. 388, \(i_{C1} = \alpha I/(1 + e^{(v_{B2} - v_{B1})/V_T})\). Its slope at \(v_d = v_{B1} - v_{B2} = 0\) is \(\alpha I/(4V_T)\), which with \(g_m = \alpha I/(2V_T)\) is \(g_m/2\): a differential input \(v_d\) splits equally, \(v_d/2\) across each base-emitter junction, so the signal current in one collector is \(g_m v_d/2\). The voltage at that collector is therefore

$$A_{d,se} = \dfrac{v_{c1}}{v_d} = -\dfrac{g_m R_C}{2} = -\dfrac{(0.0385)(5000)}{2} = -96$$

Taking the output between the two collectors doubles it, because \(v_{c2}\) swings the opposite way: \(A_d = -g_m R_C = -192\). That is the single fact to keep straight in differential-pair problems: the single-ended gain is half the differential-output gain, and the Handbook's transfer curve on p. 389 shows the two collector currents moving in opposite directions around \(I/2\).

\(-192\) is \(-g_m R_C\), the gain with the output taken between the collectors; the stem asks for one collector to ground.

\(-385\) computes \(g_m\) from the whole 2 mA tail current instead of the 1 mA in each transistor and also omits the one half.

\(-9.6\) reads the collector resistor as 500 Ω.

Answer: D) 324 V

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Power Conversion, p. 389 (PDF viewer: page 395): “For an \(n\)-pulse rectifier with a line-to-line RMS input voltage of \(V_{rms}\) and no output filter, the average output voltage is”

$$V_{dc} = V_{rms} \times \frac{n\sqrt{2}}{\pi}\sin\frac{\pi}{n}$$

The formula is short, and the entire difficulty of this question is deciding what to put into it.

Step 1 – How many pulses? Three diodes, one per phase, all cathodes tied to a common bus: whichever phase is highest at each instant takes the load. Each phase conducts once per cycle, so the output has three pulses per cycle. \(n = 3\).

Step 2 – Which voltage does this circuit actually rectify? This is the step that decides the answer. Follow the load current in the figure: it leaves the common cathode, goes down through \(R_L\), and returns on the neutral line \(N\). So each diode presents its own phase voltage — line-to-neutral — to the load, and the rectified waveform is the upper envelope of the three phase voltages.

The Handbook writes “line-to-line” because the two rectifiers it has in view, the single-phase bridge and the three-phase bridge, do rectify terminal-to-terminal voltages. \(V_{rms}\) in that expression is really the rms of whatever waveform the rectifier is rectifying. Here that is the phase voltage:

$$V_{rms} = V_{phase} = \frac{V_{LL}}{\sqrt{3}} = \frac{480}{\sqrt{3}} = 277.1\;\text{V}$$

Step 3 – Substitute.

$$V_{dc} = 277.1 \times \frac{3\sqrt{2}}{\pi}\sin 60^\circ = 277.1\left(1.3505\right)\left(0.8660\right) = 324\;\text{V}$$

Answer: (D).

Cross-check without the Handbook. Averaging the peaks of three phase voltages over a \(2\pi/3\) conduction window gives the classical 3-pulse result:

$$V_{dc} = \frac{3\sqrt{3}}{2\pi}V_{m,phase} = 0.827\left(\sqrt{2}\times277.1\right) = 0.827(391.9) = 324\;\text{V}\;\checkmark$$

Two independent routes to the same number is worth the thirty seconds it costs.

The shape of the formula. Read it as \(\left(n/\pi\right)\sin\left(\pi/n\right)\) applied to the peak of whatever is being rectified. As \(n\) grows, \(\left(n/\pi\right)\sin\left(\pi/n\right) \to 1\): more pulses per cycle means the output hugs the peak more closely, because each diode is handed over to the next before its own waveform has fallen far. That single reading tells you why more pulses always means a higher and smoother dc output, without memorising a table of constants.

Why nobody builds this. A three-phase half-wave rectifier draws dc current through each phase in one direction only, which saturates supply transformers and pushes current into the neutral. It appears on exams as the clean illustration of \(n\)-pulse rectification, and almost nowhere else; the six-pulse bridge does the same job with no neutral current and far less ripple — a ripple factor near 4% against roughly 18% here.

Step 4 – Eliminate the distractors:

(A) 561 V is the Handbook's formula with \(n = 3\) fed the line-to-line 480 V, taking the words of the caption rather than the circuit in front of you. It is the answer for a hypothetical 3-pulse rectifier of line voltages; this one rectifies phase voltages, as the neutral return in the figure shows.

(B) 648 V is \(n = 6\): the three-phase full bridge, \(1.35\,V_{LL}\). That circuit needs six diodes and no neutral. The figure has three diodes and a neutral.

(C) 432 V is \(n = 2\): \(0.900\,V_{LL}\), the single-phase bridge factor applied to a three-phase source.