Diode Circuits and Applications Questions and Answers

Answer: C

The output is pulsating DC.

A rectifier lets current through in one direction only. Its output therefore never reverses polarity, so it is DC, but it rises and falls with each half cycle of the input: a series of half sine waves. A filter capacitor or inductor is what smooths these pulses into nearly steady DC.

The output is not AC, since it no longer alternates. It is not smooth DC without a filter. A '25 Hz DC' does not describe it: a full wave rectifier on 50 Hz mains gives ripple at 100 Hz and a half wave rectifier at 50 Hz, never 25 Hz.

Answer: D

The fundamental ripple frequency is 100 Hz.

A full wave rectifier turns both half cycles of the input into positive pulses, so the output contains two pulses in every input cycle. With 50 Hz mains the output repeats every 10 ms, and the lowest frequency in the ripple is 2 × 50 = 100 Hz.

50 Hz is the ripple frequency of a half wave rectifier, which passes only one pulse per cycle. 25 Hz would need one pulse every two cycles, and 70.7 Hz looks like 50 × √2, mixing up an rms factor with a frequency.

Answer: B

The Zener current is 6 mA.

First check that the Zener conducts: without it the load would see \( 30 \times \frac{4}{2 + 4} = 20 \) V, above 12 V, so the Zener is in breakdown and holds the load at 12 V.

\[ \begin{aligned} I &= \frac{30 - 12}{2\ \text{k}\Omega} = 9\ \text{mA} \\ I_{L} &= \frac{12}{4\ \text{k}\Omega} = 3\ \text{mA} \\ I_{z} &= 9 - 3 = 6\ \text{mA} \end{aligned} \]

9 mA would be the answer if the load current were forgotten, and 4, 8 and 10 mA do not satisfy the two current balances above.

Answer: C

The tunnel diode is useful because it exhibits a negative resistance characteristic in the region between peak current I_P and valley current I_V.

A tunnel diode is very heavily doped, so its depletion layer is extremely thin and electrons can tunnel through the barrier. As forward voltage rises, current first rises to a peak and then falls to a valley as the bands move out of alignment. In that falling region, current decreases while voltage increases, a negative resistance, which can cancel the losses of a tuned circuit and sustain oscillation or provide gain at microwave frequencies.

The diode is heavily, not lightly, doped. A negative resistance supplies power to the circuit rather than absorbing it, and a positive resistance in that region would make it useless as an oscillator.

Answer: C

The PIV in a bridge rectifier is 0.5 times that in a centre-tap full wave rectifier.

In a centre-tap full wave rectifier, when one diode conducts, the other has the full secondary voltage across it: the peak of its half plus the peak on the conducting side, so its PIV is \(2V_{m}\), where \(V_{m}\) is the peak voltage of each half of the secondary.

In a bridge rectifier, the two non-conducting diodes are each connected across the secondary, through the conducting diodes, so each sees a PIV of only \(V_{m}\). The ratio is \(V_{m}/2V_{m} = 0.5\), one of the advantages of the bridge circuit. The answer 2 inverts the ratio.

Answer: B

The bulk resistance is 0.9 Ω.

The voltage across the diode is \(V = P/I = 5/2 = 2.5\) V. A silicon diode needs about 0.7 V to overcome its barrier potential; the rest drops across its bulk resistance:

\[ r_{B} = \frac{2.5 - 0.7}{2} = \frac{1.8}{2} = 0.9\ \Omega \]

The option 1.2 Ω uses V/I = 2.5/2 without subtracting the barrier potential (and rounds it down), and 2.5 Ω confuses volts with ohms.

Answer: B

The load current is nearly 18 mA.

In the centre-tapped full-wave rectifier, each half of the secondary gives 100 V r.m.s., a peak of \(V_{m} = 100\sqrt{2} = 141.4\) V. The average output of a full-wave rectifier is:

\[ \begin{aligned} V_{dc} &= \frac{2V_{m}}{\pi} = \frac{2 \times 141.4}{\pi} \\ &\approx 90\ \text{V} \end{aligned} \]

so \(I_{dc} = 90/5000 = 18\) mA. Using the r.m.s. value of 100 V would give 20 mA, and using the full 200 V across both halves would double the answer.

Answer: B

The zener current is 60 mA.

The zener holds the load at 10 V, so the series resistor drops the remaining 6 V. The currents are:

\[ \begin{aligned} I_{s} &= \frac{16 - 10}{60} = 0.1\ \text{A} = 100\ \text{mA} \\ I_{L} &= \frac{10}{250} = 0.04\ \text{A} = 40\ \text{mA} \\ I_{Z} &= I_{s} - I_{L} = 60\ \text{mA} \end{aligned} \]

The option 40 mA is the load current and 100 mA the total current from the battery; the zener carries the difference. 80 mA does not follow from the circuit values.

Answer: B

The average DC voltage is nearly 270 Volts.

With a 300-0-300 V centre-tapped transformer, each diode is fed by 300 V rms, whose peak is \(V_{m} = 300\sqrt{2} = 424\) V. A full-wave rectifier's average output is \(2V_{m}/\pi\):

\[ \begin{aligned} V_{dc} &= \frac{2V_{m}}{\pi} = \frac{2 \times 424.3}{3.1416} \\ &\approx 270\ \text{V} \end{aligned} \]

That is 0.9 times the rms value. 300 V confuses rms with average, 150 V is half the rms value (the half-wave average would be \(V_{m}/\pi = 135\) V), and the load resistance does not affect the average voltage when diode drops are neglected.

Answer: C

The diode needs a minimum rating of about 280 mW.

Replace the 33.4 V source with its two 60 \(\Omega\) resistors by a Thevenin equivalent seen by the diode branch:

\[ \begin{aligned} V_{TH} &= 33.4 \times \frac{60}{120} = 16.7\ \text{V} \\ R_{TH} &= 60 \parallel 60 = 30\ \Omega \\ I &= \frac{16.7 - 0.7}{30 + 10} = 0.4\ \text{A} \\ P_{D} &= 0.7 \times 0.4 = 0.28\ \text{W} \end{aligned} \]

The diode dissipates its forward drop times its current, 280 mW, so it must be rated for at least that. 240 mW and 180 mW would underrate it, and 350 mW, while safe, is not the minimum.