Diode Circuits and Applications Questions and Answers

Q: What does the ideal diode model assume?

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.

Q: What does the piecewise-linear model add to the ideal model?

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.

Q: Which quantity does the diode equation's thermal voltage represent?

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.

Q: Why does a silicon diode's forward drop fall as temperature rises?

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.

Q: What is reverse saturation current?

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.

Q: What happens at reverse breakdown in an ordinary rectifier diode?

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.

Q: Which capacitance dominates in a reverse biased junction?

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.

Direction for [ Question No: 1 To 1 ] :

Assertion - Reason type questions : The following items consist of two statements, one labelled the 'Assertion (A)' and the other labelled the 'Reason (R)'. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Assertion (A) : Tunnel diode represents negative resistance characteristics and when operated in this region may be used as an oscillator.
Reason (R) : Tunnel diode is heavily doped p-n junction having an extremely narrow junction which electrons are able to tunnel through it.

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.(A) is true and (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2013,

Answer: A

A tunnel diode shows a negative resistance region in its current-voltage curve, and biased there it can be used as an oscillator, so (A) is true. It is made of a heavily doped p-n junction with a very narrow depletion region, through which electrons tunnel, so (R) is true. The tunnelling is what causes the negative resistance, so (R) explains (A). The peak current is reached at a low forward voltage, after which the current falls as the voltage rises, which is the negative resistance used with a resonant circuit. The answer is both are true and (R) is the correct explanation of (A).

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Direction for [ Question No: 2 To 3 ] :

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.

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

A.50 V
B.75 V
C.100 V
D.200 V

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

A.\(\frac{R_L}{310}\)
B.\(\frac{R_L}{450}\)
C.\(\frac{R_L}{620}\)
D.\(\frac{R_L}{942}\)

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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What does the ideal diode model assume?

A.Zero forward drop and infinite reverse resistance
B.A constant forward drop of about seven tenths of a volt
C.A forward resistance that varies with the applied current
D.A reverse leakage that doubles for every ten degrees of rise

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.

Discuss in Forum

What does the piecewise-linear model add to the ideal model?

A.A finite reverse breakdown voltage
B.A fixed forward voltage drop
C.A temperature coefficient for the drop
D.An exponential relation between current and voltage

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.

Discuss in Forum

Which quantity does the diode equation's thermal voltage represent?

A.The reverse voltage at which avalanche breakdown starts
B.The temperature rise per milliwatt the package dissipates
C.Roughly 26 millivolts at room temperature
D.The forward drop at which conduction begins in silicon

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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Why does a silicon diode's forward drop fall as temperature rises?

A.The series resistance of the bulk material decreases
B.The depletion region narrows with increasing temperature
C.The carrier mobility rises and so lowers the drop
D.The reverse saturation current increases strongly

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.

Discuss in Forum

What is reverse saturation current?

A.The small current flowing when the diode is reverse biased
B.The current at which the forward characteristic becomes linear
C.The maximum current the diode may carry continuously
D.The current at which reverse breakdown becomes destructive

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.

Discuss in Forum

What happens at reverse breakdown in an ordinary rectifier diode?

A.The junction capacitance rises without limit as bias increases
B.Current rises sharply for little further voltage
C.The diode reverts to conducting in the forward direction
D.The forward drop falls to a much lower value than usual

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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10.

Which capacitance dominates in a reverse biased junction?

A.Stray package capacitance
B.Gate-to-channel capacitance
C.Depletion capacitance
D.Diffusion capacitance

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.

Discuss in Forum

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Diode Circuits and Applications Questions and Answers

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