Diode Circuits and Applications Questions and Answers

Q: The current 'I' supplied by the power supply in the given circuit is ...............

The current is 0.1 A.Conventional current leaves the positive (left) plate of the cell, passes the 20 Ω resistor and reaches the left junction. There it meets two diodes. The middle diode points from left to right, so it is forward biased and, taken as ideal, acts as a short. The top diode points from right to left, so it is reverse biased and blocks its branch, 50 Ω resistor and all.The current therefore flows through the middle diode and the 30 Ω resistor back to the cell:\[ \begin{aligned} I &= \frac{5\ \text{V}}{20\ \Omega + 30\ \Omega} \\ &= 0.1\ \text{A} \end{aligned} \]0.05 A is what one gets by sending the current through the 50 Ω branch as well (5/100), which the reversed top diode does not allow. 1 mA and 5 mA are far too small for a 5 V source driving 50 Ω.

Q: A displacement transducer with a shaft stroke of 100 mm is applied to the circuit shown below. The total resistance is 10 kΩ and voltage is 5 V. If the position of wiper is at a distance of 40 mm from the end A then the value of output voltage V0 = ..................

The output voltage is 3 V.The transducer is a potentiometer: 5 V is spread uniformly along the resistance element between A and B, and the wiper picks off a fraction of it in proportion to its position. The output is taken between B and the wiper (via C). With the wiper 40 mm from A on a 100 mm stroke, it is 60 mm from B:\[ V_{0} = \frac{60}{100} \times 5\ \text{V} = 3\ \text{V} \]The total resistance of 10 kΩ does not enter as long as the output is not loaded, because only the ratio of the two parts matters.2 V is the voltage between A and the wiper (40% of 5 V), measured from the wrong end. 2.5 V is the mid-stroke value, and 1.5 V has no basis.

Q: For a given full wave bridge rectifier the input line voltage has nominal value of 120V. The actual voltage coming out of the power outlet vary from 105 to 125V and corresponding load voltage as 8.8V and 12.8V respectively. Therefore, the line regulation is .................

The line regulation is 45%.Line regulation measures how much the load voltage moves when the line voltage swings from its low to its high value. With the definition used in standard electronics texts, the change in load voltage is expressed as a percentage of the low-line load voltage:\[ \begin{aligned} \text{LR} &= \frac{V_{HL} - V_{LL}}{V_{LL}} \times 100\% \\ &= \frac{12.8 - 8.8}{8.8} \times 100\% \approx 45\% \end{aligned} \]Such a large figure is typical of an unregulated rectifier, whose output simply follows the line.22.5% is half the value, and 4.5% and 2.25% are ten times smaller; none of them follows from the 4 V change on an 8.8 V base.

Q: In the filtered half wave rectifier circuit as shown below, the value of the DC output voltage is ............ .

The DC output voltage is 16 V.The transformer steps 240 V rms down by 20 : 1, and the capacitor charges to the peak of the secondary voltage less one diode drop:\[ \begin{aligned} V_{2,\text{rms}} &= 240/20 = 12\ \text{V} \\ V_{2,\text{peak}} &= 12\sqrt{2} \approx 16.97\ \text{V} \\ V_{dc} &\approx 16.97 - 0.7 \approx 16.3\ \text{V} \end{aligned} \]With 500 mF across 1 kΩ the ripple is negligible (the time constant is 500 s against a 20 ms cycle), so the output sits at about 16 V.17 V is the ideal-diode peak, which ignores the diode drop. 12 V is the rms value, and 18 V exceeds the peak of the secondary altogether.

Q: In a silicon diode, the potential barrier at room temperature is approximately

The barrier potential of a silicon diode is about 0.7 V.The built-in potential across the depletion region of a silicon PN junction at room temperature is about 0.6-0.7 V. So a silicon diode conducts noticeably only once the forward bias reaches roughly that cut-in voltage.0.3 V is the figure for germanium. 1.1 V (more exactly 1.12 eV) is the band-gap energy of silicon, not its barrier potential. 0.1 V is too low for any common semiconductor junction.

Q: The reverse saturation current of a PN junction diode

The reverse saturation current increases with temperature.The reverse saturation current comes from minority carriers generated thermally near the junction. Their number rises steeply with temperature, so the current roughly doubles for every 10 degrees Celsius rise in silicon and germanium diodes. This is why diodes and transistors are prone to thermal drift.It therefore does not decrease with temperature and is not independent of it. By definition it is the current under reverse bias, so it does not depend on the forward voltage.

Q: PN junction diode can be used as a/an

A PN junction diode can be used as a rectifier.The junction conducts easily when forward biased and blocks when reverse biased, so on an alternating input it passes current in one direction only, which turns AC into pulsating DC.An oscillator and an amplifier need gain, which a two-terminal passive junction cannot give. A diode is also a conductor in one direction, so it is not an insulator.Rectifiers made with such diodes are used in power supplies, either as a half wave circuit with one diode or as a full wave bridge with four. A reverse biased diode passes only a very small leakage current, so the one-way property is nearly ideal for small voltages.

Q: Zener diode is a

A Zener diode is a reverse biased diode in its normal use.It is designed to work in the breakdown region, where the voltage across it stays almost constant while the current changes. For this it is connected with the reverse polarity, with the cathode towards the positive supply, and so it is used as a voltage regulator.It is not a variable voltage source, as its voltage is nearly fixed. It is not a constant current source either. A forward biased Zener behaves like an ordinary diode and gives no regulation.

141.

The current 'I' supplied by the power supply in the given circuit is ...............

Circuit: a 5 V cell in series with a 20 ohm resistor feeds two parallel branches, the top a 50 ohm resistor with a diode pointing left and the middle a diode pointing right, which rejoin and return through a 30 ohm resistor

A.1 mA
B.0.05 A
C.0.1 A
D.5 mA

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The current is 0.1 A.

Conventional current leaves the positive (left) plate of the cell, passes the 20 Ω resistor and reaches the left junction. There it meets two diodes. The middle diode points from left to right, so it is forward biased and, taken as ideal, acts as a short. The top diode points from right to left, so it is reverse biased and blocks its branch, 50 Ω resistor and all.

The current therefore flows through the middle diode and the 30 Ω resistor back to the cell:

\[ \begin{aligned} I &= \frac{5\ \text{V}}{20\ \Omega + 30\ \Omega} \\ &= 0.1\ \text{A} \end{aligned} \]

0.05 A is what one gets by sending the current through the 50 Ω branch as well (5/100), which the reversed top diode does not allow. 1 mA and 5 mA are far too small for a 5 V source driving 50 Ω.

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142.

A displacement transducer with a shaft stroke of 100 mm is applied to the circuit shown below. The total resistance is 10 kΩ and voltage is 5 V. If the position of wiper is at a distance of 40 mm from the end A then the value of output voltage V₀ = ..................

Displacement transducer: a 5 V supply across the ends A and B of a resistance element D; a wiper W on the shaft slides along the element and touches a conducting strip leading to terminal C; Vout is taken between B and C

A.3V
B.2V
C.1.5V
D.2.5V

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The output voltage is 3 V.

The transducer is a potentiometer: 5 V is spread uniformly along the resistance element between A and B, and the wiper picks off a fraction of it in proportion to its position. The output is taken between B and the wiper (via C). With the wiper 40 mm from A on a 100 mm stroke, it is 60 mm from B:

\[ V_{0} = \frac{60}{100} \times 5\ \text{V} = 3\ \text{V} \]

The total resistance of 10 kΩ does not enter as long as the output is not loaded, because only the ratio of the two parts matters.

2 V is the voltage between A and the wiper (40% of 5 V), measured from the wrong end. 2.5 V is the mid-stroke value, and 1.5 V has no basis.

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143.

For a given full wave bridge rectifier the input line voltage has nominal value of 120V. The actual voltage coming out of the power outlet vary from 105 to 125V and corresponding load voltage as 8.8V and 12.8V respectively. Therefore, the line regulation is .................

A.4.5%
B.45%
C.22.5%
D.2.25%

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The line regulation is 45%.

Line regulation measures how much the load voltage moves when the line voltage swings from its low to its high value. With the definition used in standard electronics texts, the change in load voltage is expressed as a percentage of the low-line load voltage:

\[ \begin{aligned} \text{LR} &= \frac{V_{HL} - V_{LL}}{V_{LL}} \times 100\% \\ &= \frac{12.8 - 8.8}{8.8} \times 100\% \approx 45\% \end{aligned} \]

Such a large figure is typical of an unregulated rectifier, whose output simply follows the line.

22.5% is half the value, and 4.5% and 2.25% are ten times smaller; none of them follows from the 4 V change on an 8.8 V base.

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144.

In the filtered half wave rectifier circuit as shown below, the value of the DC output voltage is ............ .

Half wave rectifier: Vin = 240 V rms 50 Hz on a 20 to 1 transformer, a diode in series, then a capacitor C = 500 mF in parallel with a load RL = 1 kilohm across which Vdc is taken

A.12V
B.16V
C.18V
D.17V

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The DC output voltage is 16 V.

The transformer steps 240 V rms down by 20 : 1, and the capacitor charges to the peak of the secondary voltage less one diode drop:

\[ \begin{aligned} V_{2,\text{rms}} &= 240/20 = 12\ \text{V} \\ V_{2,\text{peak}} &= 12\sqrt{2} \approx 16.97\ \text{V} \\ V_{dc} &\approx 16.97 - 0.7 \approx 16.3\ \text{V} \end{aligned} \]

With 500 mF across 1 kΩ the ripple is negligible (the time constant is 500 s against a 20 ms cycle), so the output sits at about 16 V.

17 V is the ideal-diode peak, which ignores the diode drop. 12 V is the rms value, and 18 V exceeds the peak of the secondary altogether.

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145.

In a silicon diode, the potential barrier at room temperature is approximately

A.0.7V
B.0.3V
C.0.1V
D.1.1V

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

The barrier potential of a silicon diode is about 0.7 V.

The built-in potential across the depletion region of a silicon PN junction at room temperature is about 0.6-0.7 V. So a silicon diode conducts noticeably only once the forward bias reaches roughly that cut-in voltage.

0.3 V is the figure for germanium. 1.1 V (more exactly 1.12 eV) is the band-gap energy of silicon, not its barrier potential. 0.1 V is too low for any common semiconductor junction.

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146.

The reverse saturation current of a PN junction diode

A.Decreases with temperature
B.Increases with temperature
C.Is independent of temperature
D.Depends on the applied forward voltage

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

The reverse saturation current increases with temperature.

The reverse saturation current comes from minority carriers generated thermally near the junction. Their number rises steeply with temperature, so the current roughly doubles for every 10 degrees Celsius rise in silicon and germanium diodes. This is why diodes and transistors are prone to thermal drift.

It therefore does not decrease with temperature and is not independent of it. By definition it is the current under reverse bias, so it does not depend on the forward voltage.

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147.

Match the given list of devices with their respective application :

List-I	List-II
(a) Diode	(i) Rectification
(b) Tunnel diode	(ii) Microwave switching
(c) Zener diode	(iii) Oscillator
(d) PIN diode	(iv) Voltage regulator

A.a-i, b-iii, c-iv, d-ii
B.a-i, b-iv, c-iii, d-ii
C.a-i, b-ii, c-iii, d-iv
D.a-i, b-iv, c-ii, d-iii

Exam Year(s)

UGC NET - 2012,

Answer: A

The matching is a-i, b-iii, c-iv, d-ii. A diode is used for rectification. A tunnel diode shows negative resistance and is used in oscillators. A Zener diode keeps a steady voltage in reverse breakdown, so it serves as a voltage regulator. A PIN diode has a wide intrinsic region and works as a switch at microwave frequencies. Option (B) gives the tunnel diode as a regulator and the Zener as an oscillator. Option (C) gives the Zener diode as the oscillator. Option (D) gives the tunnel diode as a regulator, the Zener diode as a microwave switch and the PIN diode as an oscillator. Only (A) is right for all four.

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148.

PN junction diode can be used as a/an

A.Oscillator
B.Amplifier
C.Insulator
D.Rectifier

Exam Year(s)

UGC NET - 2013,

Answer: D

A PN junction diode can be used as a rectifier.

The junction conducts easily when forward biased and blocks when reverse biased, so on an alternating input it passes current in one direction only, which turns AC into pulsating DC.

An oscillator and an amplifier need gain, which a two-terminal passive junction cannot give. A diode is also a conductor in one direction, so it is not an insulator.

Rectifiers made with such diodes are used in power supplies, either as a half wave circuit with one diode or as a full wave bridge with four. A reverse biased diode passes only a very small leakage current, so the one-way property is nearly ideal for small voltages.

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149.

Zener diode is a

A.reverse biased diode
B.variable voltage source
C.constant current source
D.forward biased diode

Exam Year(s)

UGC NET - 2013,

Answer: A

A Zener diode is a reverse biased diode in its normal use.

It is designed to work in the breakdown region, where the voltage across it stays almost constant while the current changes. For this it is connected with the reverse polarity, with the cathode towards the positive supply, and so it is used as a voltage regulator.

It is not a variable voltage source, as its voltage is nearly fixed. It is not a constant current source either. A forward biased Zener behaves like an ordinary diode and gives no regulation.

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150.

List - I	List - II
a. Varistor	i. Nonlinear resistance
b. Varactor diode	ii. Nonlinear reactance
c. Gunn diode	iii. Negative resistance
d. PIN diode	iv. Controllable impedance type

Codes : a b c d

A.i ii iii iv
B.ii iii iv i
C.iii iv i ii
D.iv i ii iii

Exam Year(s)

UGC NET - 2013,

Answer: A

The correct matching is a-i, b-ii, c-iii, d-iv, which is code (A).

A varistor has a non-linear resistance (i), a varactor diode a voltage dependent, non-linear reactance (ii), a Gunn diode a negative resistance region (iii) and a PIN diode a controllable impedance at microwave frequencies (iv).

Codes (B), (C) and (D) shift these properties to the wrong devices.

The device behaviours differ in kind. The varistor and varactor have non-linear characteristics of resistance and reactance respectively, the Gunn diode shows a region where current falls as voltage rises, and the PIN diode behaves as a variable resistance at radio and microwave frequencies depending on its bias current.

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