Diode Circuits and Applications Questions and Answers

Answer: A

The reason is true and is exactly why the assertion is false: because the voltage barely changes with current, nothing inside the diode limits that current, and something in series must. Without the resistor the diode is destroyed as soon as breakdown begins. A steep characteristic is what makes a good reference and what makes external current limiting mandatory, so the two facts are two faces of one property. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

Both hold and the second is the mechanism: reverse recovery time is the time taken to sweep out stored charge, and a metal-semiconductor junction stores almost none. The same physics gives the low forward drop, so the two advantages come together. Whenever a speed limit is traced to stored energy, the fast device is the one that stores less, which is a rule that holds well beyond diodes.

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.