Diode Circuits and Applications Questions and Answers

Answer: C

A Schottky is the choice, because its fast recovery and low drop keep switching and conduction losses down at the tens or hundreds of kilohertz such supplies run at. A Zener regulates rather than rectifies, a varactor tunes, and a tunnel diode oscillates. Matching the device to the duty means asking what the circuit demands most - here speed and efficiency together - and then finding the diode built for exactly that. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: D

A varactor gives a capacitance controlled by a voltage, which is precisely what electronic tuning needs and is how every phase-locked loop's oscillator is steered. A Schottky rectifies quickly, a light emitting diode indicates, and a Zener regulates. The fact that the same physical effect - a bias-dependent depletion width - is a nuisance in a rectifier and the whole point of a varactor is a good illustration of how devices are specialised. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: A

Subtracting the forward drop before dividing:

\[ I = \frac{5 - 0.7}{1000} = 4.3 \ \mathrm{mA} \]

giving 4.3 mA. Five milliamps ignores the drop, which is the ideal-diode answer and about sixteen per cent out here. The remaining options subtract or add the wrong quantity. Whether the drop matters depends on how it compares with the supply, and at five volts it plainly does.

Answer: B

Two drops must be subtracted:

\[ I = \frac{5 - 1.4}{1000} = 3.6 \ \mathrm{mA} \]

so 3.6 mA. 4.3 mA counts one drop and 5 mA none. Halving the single-diode answer, which gives 1.8 mA, would be right for two resistors but not for two diodes, since a diode is a voltage drop rather than a resistance. Which model a component obeys decides whether series elements add voltages or resistances.

Answer: C

The exponential law gives about sixty millivolts per decade:

\[ \begin{aligned} \Delta V &= 2.3 V_{T} \log_{10}(10) \\ &\approx 60 \ \mathrm{mV} \end{aligned} \]

so the drop rises to about 0.76 V. Assuming it stays at 0.7 V is the piecewise-linear approximation, adequate for many purposes. Seven volts would follow from treating the diode as a resistance. The sixty-millivolts-per-decade figure is worth memorising, since it makes every such estimate a matter of counting decades.

Answer: D

Two diodes conduct in series on each half cycle:

\[ V_{dc} \approx 20 - 2(0.7) = 18.6 \ \mathrm{V} \]

so about 18.6 V before allowing for ripple and regulation. Twenty volts ignores the drops, 19.3 V counts only one, and 12.7 V is the root-mean-square value of an unfiltered sine, which a capacitor filter does not produce. Counting how many junctions the current passes through is a small point that changes the answer by a volt and a half. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: A

The capacitor discharges at the load current for about one ripple period:

\[ \begin{aligned} \Delta V &= \frac{I t}{C} \\ &= \frac{0.1 \times 0.01}{10^{-3}} = 1 \ \mathrm{V} \end{aligned} \]

giving about 1 volt. The remaining options misplace a decade. The relation says ripple is proportional to load current and inversely to capacitance, which is why a supply's ripple specification is always quoted at a stated load.

Answer: B

Ripple is inversely proportional to capacitance:

\[ \Delta V \propto \frac{1}{C} \]

so doubling the capacitor halves the ripple to about 0.5 volts. Two volts inverts the relationship and a quarter of a volt would need four times the capacitance. Because the improvement is only linear in capacitance, getting ripple down by a large factor with brute capacitance quickly becomes expensive, which is the argument for an active regulator after the filter. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: C

More load current means a faster discharge, so the capacitor sags further before the next peak refills it, and the average output falls - transformer and diode resistances add to the effect. The voltage rating is a protection matter, the ripple frequency is fixed by the supply, and a turns ratio is fixed by construction. Regulation and ripple in a capacitor filter have the same cause, which is why improving one usually improves the other. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: D

Both are true and the second explains the first: the bridge's idle diodes share the reverse voltage, so each sees only the secondary peak, while the centre-tapped circuit's idle diode sees twice the half-winding peak. That is the bridge's real advantage, alongside needing no centre tap, and it outweighs the extra two devices. Counting components is a poor guide to cost when the ratings differ.