Diode Circuits and Applications Questions and Answers

Answer: C

The reason is true but the assertion is false: an inductor input filter delivers roughly the average of the rectified wave rather than its peak, so its output is lower. What it buys is a smoother current and hence far smaller peak diode currents, which is why it suits heavy loads. Trading output voltage for current waveform quality is the exchange these two filters offer, and only the load decides which is worth having. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: D

The first two are correct and the third is not: a half-wave rectifier's diode stands off the input peak, not twice it, since no current flows on the reverse half cycle. Twice the peak is the centre-tapped full-wave figure. Keeping the peak inverse voltages of the three topologies straight is worth the effort, because it is the rating that most often decides whether a diode survives.

Answer: A

The first and third are correct; the second is backwards, since a heavier load discharges the capacitor faster and increases the ripple. The third is the hidden cost of a large capacitor: it refills in a shorter burst and so demands a higher peak current from the diodes and transformer. A change that improves one specification while worsening another is the normal case, and noticing the second effect is what makes a choice informed. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: B

The first is correct and the second is not: a photodiode is normally reverse biased, so that the light-generated current flows against a wide depletion region and the response is fast and linear. Both devices exploit properties of the reverse-biased junction - one its capacitance, the other its light-dependent leakage. Grouping the special diodes by which bias region they use is a compact way to keep them straight. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: C

The first is true and the second is false: above five volts avalanche multiplication dominates and the coefficient turns positive, which is exactly why a device near 5.6 V, where the two mechanisms overlap, drifts hardly at all. The crossover is the whole reason that particular voltage appears so often in reference circuits, and it is a good example of a design value chosen because two effects cancel there. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: D

The pairing is A-P, B-Q, C-R. Each device is an ordinary junction pushed until one property dominates, and knowing which property that is settles every question about where the device belongs: the Schottky in a switching rectifier, the varactor in a tuned circuit, and the tunnel diode in an oscillator or amplifier that needs negative resistance.

Answer: A

The correct pairing is A-P, B-Q, C-R, and the centre-tapped case is the one that catches people out because its idle diode sees both half-windings at once. The bridge and the half-wave circuit both need only the single peak. Working out what the non-conducting device stands off, rather than what the conducting one carries, is the habit these ratings reward.

Answer: B

The order is 1, 2, 3, 4. Each position follows from function: the transformer sets the amplitude, the rectifier makes the current unidirectional, the filter removes what remains of the alternating component, and the load is served last. A regulator, when fitted, sits between filter and load, which is worth remembering because its input ripple is what the filter has to deliver.

Answer: C

The first three are correctly paired and the fourth is not, so three match. A voltage doubler doubles the output voltage rather than the frequency, by charging two capacitors on alternate half cycles and adding them in series. Each of these circuits uses a diode and a reference and they are easily confused, so the distinction between altering the shape, the position and the amplitude is worth fixing firmly. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: D

Using the discharge relation with a ten millisecond ripple period:

\[ \begin{aligned} \Delta V &= \frac{0.2 \times 0.01}{2.2 \times 10^{-3}} \\ &\approx 0.9 \ \mathrm{V} \end{aligned} \]

so about 0.9 volts, roughly five per cent of the output. The other options misplace a decade or halve the result. Reading a specification table means identifying which three of its numbers the calculation actually needs. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.