Diode Circuits and Applications Questions and Answers

Answer: C

A clipper cuts off whatever lies beyond a reference level, leaving the rest untouched, which is how a waveform is limited or a peak protected. Shifting the direct level is a clamper's job, multiplying the peak is a voltage multiplier's, and pulse generation is something else again. Clipping and clamping are constantly confused because both use a diode and a reference, so it is worth fixing which one alters the shape and which one alters the position.

Answer: D

A clamper adds a direct offset so that the positive or negative peak rests at a chosen level, leaving the waveform's shape and amplitude intact. Removing part of the waveform is clipping, amplitude reduction is attenuation, and keeping one polarity is rectification. The distinction matters in practice because a clamper is used to restore a direct level that capacitive coupling has removed, which no clipper could do. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: A

The series capacitor charges to the offset needed and holds it, while the diode provides the fast path that sets that charge on the first peak. The load resistance sets how quickly the charge leaks away and hence the low-frequency limit, and the reference merely shifts the level at which clamping occurs. A circuit that has to remember something always contains a storage element, and identifying it is the quickest route to understanding how the circuit behaves over time.

Answer: B

The time constant of the capacitor and its discharge path must be long compared with the signal period, or the clamped level sags within each cycle and the waveform tilts. Recovery time matters at high frequencies rather than low ones, the voltage rating is a protection matter, and the reference impedance is usually negligible. A storage element's time constant against the signal period is the standard question to ask of any circuit that holds a value between events.

Answer: C

A doubler charges one capacitor on each half cycle and puts the two in series, so the output approaches twice the input peak without any change to the transformer. Doubling the turns ratio would work but is a different circuit and needs a bigger transformer. Diode drops subtract rather than add, and rectifying twice with one diode gains nothing. Getting a higher voltage by storing energy at different times and then adding it is the principle behind every multiplier and every charge pump.

Answer: D

Multipliers have high source impedance, so the output falls markedly as soon as current is drawn - which is why they suit high-voltage, low-current duties like photomultiplier supplies. The peak inverse voltage per diode stays modest, the ripple frequency is not reduced, and they work perfectly well at high frequencies, which is why switching supplies use them. A topology that trades regulation for voltage is fine wherever the load is light and steady. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: A

A Zener works in reverse breakdown, where its voltage is nearly independent of current - the property that makes it a reference. Forward biased it behaves as an ordinary diode with a 0.7 V drop and no useful regulation. Below breakdown it merely leaks. And the forward knee is not a Zener property. Using deliberately the region that destroys an ordinary rectifier is what the device is designed for, and the current limiting is the designer's responsibility.

Answer: B

The series resistor drops the surplus voltage and limits the current, without which the diode would be destroyed the moment breakdown began. A capacitor may be added for noise but is not essential, an inductor plays no part, and a bleeder is not required. Every device operated in a steep region of its characteristic needs something in series to set the operating current, and that is as true of a Zener as of a light emitting diode. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: C

All the resistor's current passes through the diode:

\[ I = \frac{15-10}{500} = 10 \ \mathrm{mA} \]

so 10 mA, and the diode must be rated for the 100 milliwatts this dissipates. The other values follow from using the wrong voltage across the resistor. The no-load case is always the worst case for the diode, because every milliamp the load does not take is a milliamp the diode must.

Answer: D

The resistor can supply only ten milliamps in total:

\[ \begin{aligned} I_{R} &= 10 \ \mathrm{mA} \\ I_{Z} &= 10 - 50 < 0 \end{aligned} \]

so the diode stops conducting and regulation is lost, with the output falling below ten volts. This is the design check that matters most: the series resistor has to pass the maximum load current plus a minimum keep-alive current for the diode.