Diode Circuits and Applications Questions and Answers

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.

Answer: A

This is a clamper, which adds a direct offset while leaving the shape and amplitude untouched. A clipper removes part of the waveform instead, a rectifier passes one polarity, and a multiplier raises the output voltage. The reason these are so often confused is that all four are a diode, a capacitor or a resistor, and a reference, arranged differently - so the effect on the waveform, not the parts list, is what identifies them. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

This is the reverse saturation current, so named because it is almost independent of the reverse voltage applied. Recovery describes the time taken to stop conducting, breakdown current flows only beyond the breakdown voltage, and leakage inductance belongs to transformers. Its strong temperature dependence is why leakage matters in hot equipment and why it is the term that grows in the diode equation. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

Voltage gain is not among them: a diode is a two-terminal device with no control terminal, so it can steer, drop, clip and store but never amplify. Rectification, regulation and clamping are all standard diode applications. The absence of a third terminal is the fundamental limit that separates diodes from transistors, and it explains why every amplifying circuit in this subject needs a device with one. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: D

Increasing the load current increases the ripple, since the capacitor discharges faster between peaks. More capacitance, full-wave rectification and a following inductor all reduce it, the second by halving the discharge interval as well as raising the average. A proposed remedy that acts in the wrong direction on the governing relation can be ruled out without any arithmetic, which is what makes the discharge relation worth carrying around.

Answer: A

When the coil's current is interrupted, its inductance produces a large reverse voltage, and the diode provides a path for that current to decay harmlessly instead of destroying the switching device. It is not rectifying, regulating or indicating. Any inductive load switched by a semiconductor needs somewhere for its stored energy to go, and a single diode is the cheapest possible answer.

Answer: B

Each diode conducts when its input is above the output, so the output follows the highest input less a forward drop. Inversion and amplification both need an active device, and there is no storage in the arrangement. The accumulating forward drop is exactly why diode logic cannot be cascaded far, and why transistors were needed to build real logic families - a useful illustration of what a passive device cannot do. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: C

Because every stage costs 0.7 V, the logic levels converge until the following stage cannot distinguish them. No inversion occurs, delay is not doubled by the topology, and fan-out is limited by the resistor rather than halved per stage. Signal levels that degrade monotonically with each stage are the general problem restoration solves, and a transistor stage is the cheapest restorer available.