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Diode Circuits and Applications Questions and Answers

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Read the passage below and answer the questions, that follows based on your understanding of the passage :

Many instruments require dc power of their operation. This power is available from a portable cell, however, the power output is low besides other problems. AC power is continuously and easily available, so it is convenient to use it. However, ac power must be converted into dc power.

The devices such as half-wave, full wave and bridge rectifiers convert bi-directional voltage to uni-directional voltage. However, the output has large ripple contents. Filters are used to reduce the ripple significantly. Various filters circuits and L, C, LC and CLC. The power supply with rectifier and filter is still unregulated.

The output of such power supplies varies with variations in a.c. mains voltage, load current and temperature.

In order to maintain the output at constant level, voltage or current regulation is necessary. This is obtained by using series or shunt feedback. In addition to regulation, the power supply also requires protection e.g. overload, over current or short circuit protection. Pre- regulation is also carried out to improve the stability.

Regulators can be designed by using discrete components. However, IC regulators are also available. These regulators have facilities like :

(i) voltage / current boosting
(ii) thermal shut-down
(iii) floating point to facilitate higher output voltage.
(iv) switching regulator to reduce the power dissipation.

1.

For a d.c. output of 30 V, from a full wave rectifier, the PIV rating of the diode should be minimum

Exam Year(s)

UGC NET - 2012,

Answer: C

For a full-wave rectifier the d.c. output is \(2V_m/\pi\), so a 30 V output needs a peak of 47 V, and for a centre-tapped full-wave rectifier the diode peak inverse voltage is \(2V_m\), about 94 V, so the diode must be rated at least 100 V (option C).

\[ V_m = \frac{30\pi}{2} = 47.1\ \text{V} \]

\[ PIV = 2V_m = 94.2\ \text{V} \]

The ratings 50 V and 75 V (A and B) are below 94 V and would break down, and 200 V (D) is more than the minimum that was asked.

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2.

The critical inductance in mains power supply ensures that the current through 'L' never becomes zero. For this, critical value of inductance is

Exam Year(s)

UGC NET - 2012,

Answer: D

For a full-wave rectifier with an inductor input filter the critical inductance is \(R_L/3\omega\), which is \(R_L/942\) for a 50 Hz supply (option D).

\[ L_c = \frac{R_L}{3\omega} \]

\[ L_c = \frac{R_L}{3 \times 2\pi \times 50} \]

\[ L_c = \frac{R_L}{942} \]

The other options, \(R_L/310\), \(R_L/450\) and \(R_L/620\), do not come from this relation at 50 Hz; \(R_L/310\) is not a value of \(R_L/3\omega\) for any usual supply frequency.

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Answer: A

The ideal model treats the diode as a perfect switch: no drop when conducting and no current when blocking, which is enough for deciding whether a diode conducts in a given circuit. The constant drop is the next model up, the piecewise-linear one. A current-dependent resistance belongs to the exponential law, and temperature-dependent leakage is a real effect outside all three models. Choosing the simplest model that answers the question is the whole skill in diode analysis, and the ideal model answers more questions than its crudeness suggests.

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Answer: B

It adds a constant forward drop, typically 0.7 V for silicon and 0.3 V for germanium, and often a small series resistance as well. The temperature coefficient, the exponential law and reverse breakdown are all refinements beyond it. The model is the workhorse of hand analysis because it turns a non-linear device into a battery and a switch, which any linear method can then handle.

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Answer: C

The thermal voltage is about 26 millivolts at room temperature, and it sets the steepness of the exponential characteristic - sixty millivolts per decade of current follows directly from it. The forward drop is a much larger quantity, breakdown is a reverse property, and thermal resistance is a packaging specification. A small voltage that governs an exponential is worth knowing numerically, because every logarithmic estimate in this topic depends on it.

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Answer: D

Because the saturation current rises steeply with temperature, less forward voltage is needed to carry the same current - about two millivolts less per degree. Bulk resistance, depletion width and mobility all change with temperature too, but the saturation current dominates by a wide margin. That coefficient is used deliberately in temperature sensors and has to be compensated in reference circuits, so its sign and size are both worth remembering.

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Answer: A

It is the small leakage that flows under reverse bias, nearly independent of the reverse voltage but strongly dependent on temperature, roughly doubling every ten degrees. The other three describe a forward characteristic feature, a rating and a breakdown condition. Its independence of voltage and dependence on temperature is what makes it the term that grows in the diode equation and the reason leakage matters in hot equipment. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

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Answer: B

At breakdown the reverse current rises very steeply, and unless it is limited the diode is destroyed by the power dissipated. It does not begin conducting forwards, and the forward drop and junction capacitance are unaffected. The same mechanism is used deliberately in a Zener diode, where the current is limited externally, which illustrates that a failure mode and a design feature can be the same physics viewed differently. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

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Answer: C

Under reverse bias the depletion capacitance dominates, and it falls as the reverse voltage widens the depletion layer - which is exactly the property a varactor exploits. Diffusion capacitance belongs to forward bias, where stored minority charge dominates. Package capacitance is small and fixed, and gate capacitance belongs to field effect devices. Which capacitance dominates depends on the bias, and that dependence is itself the basis of a whole class of tuning circuits.

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Answer: D

The minority carrier charge stored during conduction must be swept out before the diode can block, and that reverse recovery time sets the speed. Depletion formation is fast, the package's thermal constant is far slower but irrelevant to switching, and establishing the forward drop is quick. Stored charge is why Schottky diodes, which store almost none, switch so much faster, and it is the property to look for in any fast switching application. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

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