Diode Circuits and Applications Questions and Answers

Answer: B

Load regulation is the output change per unit load change, and with the load current diverted into or out of the diode, a shallow slope means a smaller voltage shift. A larger series resistor would worsen the maximum load rather than help, dissipation is a separate matter, and the temperature coefficient is unrelated. When a specification is a slope, the component whose own slope is in the path is the one that determines it. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: C

Germanium conducts at about 0.3 V, against silicon's 0.7 V, because its smaller band gap needs less forward bias. That lower drop is why germanium survived in small-signal detection long after silicon took over elsewhere, and it is matched by a much larger leakage current and a lower maximum temperature. 1.2 V is typical of an infrared emitter and 2.0 V of a visible one. Band gap decides forward drop, leakage and temperature limit together, so the three always move as a set.

Answer: D

A red emitter drops about 1.8 V, and the drop rises through the colours to well over three volts for blue and white, because the photon energy and hence the band gap increase. 0.7 V and 0.3 V are silicon and germanium rectifiers. Knowing that the drop tracks the colour is what lets a series resistor be sized without a data sheet, and it explains why a blue emitter will not light from a single cell.

Answer: A

Peak inverse voltage is the reverse blocking rating, and exceeding it drives the diode into breakdown where it usually fails. Surge current is a forward rating for brief overloads, recovery time a switching specification, and junction temperature the thermal limit. A device is specified by one rating per way it can be destroyed, and matching each rating to the corresponding worst case in the circuit is the whole of selecting a component. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: B

The surge rating covers the large brief current that flows when a discharged filter capacitor is first connected, which can be many times the steady peak. Continuous temperature, reverse transients and the normal peak are all covered by other ratings. Switch-on is a distinct worst case from steady operation, and a supply that runs perfectly can still destroy its rectifier on the first switch-on if this rating was ignored. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: C

The four diodes steer whichever end of the winding is positive to the load, so a single winding serves both half cycles. The diodes conduct in pairs but carry the full load current, not half. The doubled ripple frequency and the lower voltage rating are both real advantages but neither explains the absence of a centre tap. A switching arrangement that can reverse a connection removes the need for a second source, which is the same idea as an H-bridge in a motor drive.

Answer: D

Each diode conducts for half the cycle, working with its diagonal partner, which is why the average current per diode is half the load current. With a capacitor filter the conduction narrows to a burst near each peak, so the last option describes a real effect but not the resistive-load case the question asks about. Establishing the resistive-load behaviour first and then adding the filter's effect is the reliable way through these circuits. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

Each diode carries the full load current, but only during its own half cycle, so its average is half. It is a duty-cycle effect rather than a current-division one: the diodes are in series with the load, not in parallel with each other. The transformer supplies everything the load takes, and the ripple frequency is a consequence rather than a cause. Distinguishing average from peak current is essential in rectifier ratings, since the two differ by a large factor.

Answer: B

An ideal full-wave rectifier converts about 81 per cent of the input power into direct power, twice the half-wave figure of 40.6 per cent. It is not a hundred per cent because the remainder appears in the ripple rather than being lost as heat - a point worth being clear about, since the figure measures conversion into the wanted component rather than efficiency in the thermal sense. Knowing what a stated efficiency is a ratio of prevents a great deal of confusion.

Answer: C

Current flows in the secondary only on one half cycle, so a net direct component magnetises the core and pushes it towards saturation, wasting capability and increasing loss. The primary's ripple, the available voltage and the ripple frequency are all separate matters. Any winding carrying unidirectional current biases its core, which is why full-wave and bridge arrangements are preferred at anything above the smallest powers. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.