Diodes and Its Applications MCQ Quiz - Objective Question with Answer for Diodes and Its Applications - Download Free PDF

Last updated on Jun 12, 2025

Latest Diodes and Its Applications MCQ Objective Questions

Diodes and Its Applications Question 1:

Compared to Silicon, GaAs Semi-conductors offer advantage of

  1. Higher Electron Mobility
  2. Higher frequency operation
  3. Lower leakage current
  4. All of the above

Answer (Detailed Solution Below)

Option 4 : All of the above

Diodes and Its Applications Question 1 Detailed Solution

Explanation:

GaAs Semiconductors Compared to Silicon

Definition: Gallium Arsenide (GaAs) is a compound semiconductor composed of gallium (Ga) and arsenic (As). It is widely used in high-frequency and optoelectronic applications due to its superior electrical and physical properties compared to silicon (Si), which is the most commonly used semiconductor material.

Correct Option Analysis:

The correct option is:

Option 4: All of the above

This option is correct because GaAs semiconductors offer all the advantages mentioned in the other options, namely:

  • Higher Electron Mobility: GaAs has significantly higher electron mobility compared to silicon. Electron mobility is a measure of how quickly electrons can move through a material when subjected to an electric field. In GaAs, electrons can move more freely, resulting in faster signal transmission and better performance in high-speed electronic devices.
  • Higher Frequency Operation: Due to its higher electron mobility, GaAs is capable of operating at higher frequencies than silicon. This property makes it ideal for applications in radio frequency (RF) and microwave technologies, such as mobile phones, satellite communications, and radar systems.
  • Lower Leakage Current: GaAs has a wider bandgap than silicon, which reduces the leakage current in electronic devices. Lower leakage current improves the efficiency and reliability of GaAs-based components, especially in high-temperature and high-power applications.

Since all the advantages are valid, the correct answer is "All of the above."

Diodes and Its Applications Question 2:

Gunn Diode operated at a high frequency is a

  1. Negative resistance device
  2. Positive resistance device
  3. High noise device
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Negative resistance device

Diodes and Its Applications Question 2 Detailed Solution

Gunn Diode Operated at a high frequency is a Negative resistance device.

Explanation:

  • A Gunn diode is considered as a type of diode even though it does not contain any typical PN diode junction like the other diodes, but it consists of two electrodes.
  • It is also called a Transferred Electronic Device.
  • It is a negative differential resistance device, which is frequently used as a low-power oscillator to generate microwaves.
  • It consists of only an N-type semiconductor in which electrons are the majority charge carriers.
  • The most common materials used for the fabrication of Gunn diode are Gallium Arsenide (GaAs) and Indium Phosphide (InP)
  • For the construction of these diodes, only N-type material is used, which is due to the transferred electron effect applicable only to N-type materials and is not applicable to the P-type materials.

 

F1 J.P Madhu 15.06.20 D2

Advantages:

  • High bandwidth
  • High reliability
  • Low manufacturing cost
  • Fair noise performance (does not use avalanche principle).
  • Relatively low operating voltage

 

Disadvantages:

  • Low efficiency below 10 GHz
  • Poor stability – frequency varies with bias and temperature
  • FM noise high for some applications
  • Small tuning range

Diodes and Its Applications Question 3:

An ideal diode is connected in series with a 1 kΩ load resistor and the input voltage is given as V(t) = sin2(t) + cos2(t) V. What is the average output voltage across the load resistor?

  1. Average voltage cannot be determined
  2. +1/2 V
  3. 0 V
  4. +1 V

Answer (Detailed Solution Below)

Option 4 : +1 V

Diodes and Its Applications Question 3 Detailed Solution

Given:

  • Input voltage: V(t) = sin²(t) + cos²(t) V
  • Load resistor: 1 kΩ
  • Ideal diode (no voltage drop when conducting)

Key Analysis:

  1. Input Voltage Simplification:

    Using the trigonometric identity:

     sin2(t) + cos2(t)  = 1 

    Therefore, V(t) = 1 V (constant DC voltage at all times)

  2. Diode Behavior:

    Since the input is always positive (1V):

    • The diode remains continuously forward-biased
    • Acts as a short circuit (0V drop)
  3. Output Voltage:

    The entire input appears across the resistor:

    \( V_{out}(t) = V(t) = 1 \text{ V} \)

Average Voltage Calculation:

For a constant DC voltage:

\(V_{avg} = \frac{1}{T}\int_0^T 1\ dt = 1 \text{ V} \)


Final Answer:

4) +1 V

Diodes and Its Applications Question 4:

Which of the following statements is correct when comparing a bridge rectifier to a centre-tapped full-wave rectifier?

  1. The PIV of both rectifiers is the same.
  2. The transformer utilisation factor is the same for both circuits.
  3. A bridge rectifier has double the peak inverse voltage (PIV) compared to a centre-tapped rectifier.
  4. The transformer utilisation factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Answer (Detailed Solution Below)

Option 4 : The transformer utilisation factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Diodes and Its Applications Question 4 Detailed Solution

The Correct Answer is:  4) The transformer Utilisation Factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Explanation:
Transformer Utilisation Factor (TUF):
TUF indicates how efficiently the transformer is used in a rectifier circuit.

Bridge Rectifier TUF ≈ 0.812

Centre-Tapped Full-Wave Rectifier TUF ≈ 0.693

Therefore, a bridge rectifier uses the transformer more efficiently, making option 4 correct.

Additional Information 
1) The PIV of both rectifiers is the same
False — In a bridge rectifier, each diode withstands only Vm (peak voltage),
while in a centre-tapped rectifier, each diode must withstand 2Vm ⇒ PIV is higher.

2) The transformer utilisation factor is the same for both
 False — TUF is better in a bridge rectifier, not the same.

3) A bridge rectifier has double the PIV compared to the centre-tapped rectifier.
False — It's the centre-tapped rectifier that has higher PIV, not the bridge.

Final Answer:
4) The transformer utilisation factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Diodes and Its Applications Question 5:

A half-wave rectifier is designed using a transformer and a diode. The primary winding of the transformer, with N1 turns, is connected to a 240 sin(ωt) V supply. The secondary winding has N2 turns. What is the rectified DC output voltage (Vdc) if N1/N2 = 1 : 1?

  1. 240π
  2. 240/π
  3. 240
  4. 480/π

Answer (Detailed Solution Below)

Option 2 : 240/π

Diodes and Its Applications Question 5 Detailed Solution

Concept:

A half-wave rectifier converts AC to DC by allowing only one half-cycle of the input waveform to pass. The DC output voltage is the average value of the rectified waveform.

The average DC voltage for a half-wave rectifier is given by:

\(V_{dc} = \frac{V_m}{\pi} \)

where Vm is the peak voltage of the input AC signal.

Given:

  • Primary voltage: \(V_p = 240 \sin(\omega t) \; V\)
  • Turns ratio \( \frac{N_1}{N_2} = 1:1 \)
  • Rectifier type: Half-wave

Calculation:

  1. Determine Secondary Voltage:

    Since the turns ratio is 1:1, the secondary voltage equals the primary voltage:

    \(V_s = V_p = 240 \sin(\omega t) \text{ V (peak)}\)

  2. Calculate DC Output Voltage:

    For a half-wave rectifier:
    \( V_{dc} = \frac{V_m}{\pi} = \frac{240}{\pi} \text{ V} \)

Thus the correct answer is option 2

Top Diodes and Its Applications MCQ Objective Questions

The maximum efficiency of a half-wave rectifier is

  1. 33.3 %
  2. 40.6 %
  3. 66.6 %
  4. 72.9 %

Answer (Detailed Solution Below)

Option 2 : 40.6 %

Diodes and Its Applications Question 6 Detailed Solution

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Concept:

The efficiency of a rectifier is defined as the ratio of dc output power to input power.

The efficiency of a half-wave rectifier will be:

\(\eta = \frac{{{P_{dc}}}}{{{P_{ac}}}}\)

\(\eta= \frac{{\frac{{V_{dc}^2}}{{{R_L}}}}}{{\frac{{V_{rms}^2}}{{{R_L}}}}} \)

VDC = DC or average output voltage

RL = Load Resistance

For a half-wave rectifier, the output DC voltage or the average voltage is given by:

\(V_{DC}=\frac{V_m}{\pi}\)

Also, the RMS voltage for a half-wave rectifier is given by:

\(V_{rms}=\frac{V_m}{2}\)

Calculation:

The efficiency for a half-wave rectifier will be:

\(\eta= \frac{{{{\left( {\frac{{{V_m}}}{\pi }} \right)}^2}}}{{{{\left( {\frac{{{V_m}}}{{2 }}} \right)}^2}}} = 40.6\;\% \)

For Half wave rectifier maximum efficiency = 40.6%

NoteFor Full wave rectifier maximum efficiency = 81.2%

The direction of the arrow represents the direction of __________

When the diode is forward biased.

  1. P-type material
  2. N-type material
  3. P-N Junction
  4. Conventional current flow

Answer (Detailed Solution Below)

Option 4 : Conventional current flow

Diodes and Its Applications Question 7 Detailed Solution

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  • A diode is an electronic device allowing current to move through it only in one direction.
  • Current flow is permitted when the diode is forwaforward-biased
  • Current flow is prohibited when the diode is reversed-biased.
  • The direction of the arrow represents the direction of conventional current flow when the diode is forward biased

F1 U.B. Nita 11.11.2019 D 4

  • In the figure given above, the symbol represents the circuit symbol of a semiconductor junction diode.
  • The ‘P’ side of the diode is always positive terminal and is designated as anode for forward bias.
  • Another side that is negative is designated as cathode and is the ‘N’ side of diode.

Find the output voltage of the given network if Ein = 6 V and the Zener breakdown voltage of the Zener diode is 10 V.

F1 Koda Raju 12.4.21 Pallavi D2

  1. 4 V
  2. 0 V
  3. 10 V
  4. 6 V

Answer (Detailed Solution Below)

Option 2 : 0 V

Diodes and Its Applications Question 8 Detailed Solution

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Concept: 

The working of the Zener diode is explained in the below figures.

F1 S.B 1.9.20 Pallavi D28 (1)

Calculation:

Given,

Zener voltage Vz = 10 V

Ein = 6 V ⇒ Ein < Vz

Hence zener will be reverse biased and get open-circuited.

Output voltage E0 = 0 V

Which of the following diodes is also known as a ‘voltacap’ or ‘voltage-variable capacitor diode’?

  1. Varactor diode
  2. Step recovery diode
  3. Schottky diode
  4. Gunn diode

Answer (Detailed Solution Below)

Option 1 : Varactor diode

Diodes and Its Applications Question 9 Detailed Solution

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Varactor diode:

  • It is represented by a symbol of diode terminated in the variable capacitor as shown below:

hark3

  • Varactor diode refers to the variable Capacitor diode, which means the capacitance of the diode varies linearly with the applied voltage when it is reversed biased.
  • The junction capacitance across a reverse bias pn junction is given by

​           \(C=\frac{A\epsilon}{W}\)

  • As the reverse bias voltage increases, the depletion region width increases resulting in a decrease in the junction capacitance.
  •  Varactor diodes are used in electronic tuning systems to eliminate the need for moving parts
  • Varactor [also called voltacap, varicap, voltage-variable capacitor diode, variable reactance diode, or tuning diode] diodes are the semiconductor, voltage-dependent, variable capacitors
  • Varactors are used as voltage-controlled capacitors and it operated in a reverse-biased state

26 June 1

Diodes

Application

 Schottky diode

rectifying circuits requiring high switching rate

Varactor diode

Tuned circuits

PIN diode 

High-frequency switch

Zener diode

voltage regulation

A limiter circuit is also known as a:

  1. clamp circuit
  2. chopping circuit
  3. clipper circuit
  4. chopper circuit

Answer (Detailed Solution Below)

Option 3 : clipper circuit

Diodes and Its Applications Question 10 Detailed Solution

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  • A limiter circuit is also known as a clipper circuit.
  • A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal.
  • The clipping (removal) of the input AC signal is done in such a way that the remaining part of the input AC signal will not be distorted
  • In the below circuit diagram, the positive half cycles are removed by using the series positive clipper.

F1 U.B 10.4.20 Pallavi D 5

NoteA Clamper circuit can be defined as the circuit that consists of a diode, a resistor, and a capacitor that shifts the waveform to the desired DC level without changing the actual appearance of the applied signal.

The following symbol is used for __________.

F1 Jai.P 29-12-20 Savita D4

  1. Tunnel diode
  2. Varactor diode
  3. Zener diode
  4. Photo diode

Answer (Detailed Solution Below)

Option 1 : Tunnel diode

Diodes and Its Applications Question 11 Detailed Solution

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  • A tunnel diode is a highly doped semiconductor diode.
  • The p-type and n-type semiconductor is heavily doped in a tunnel diode due to a greater number of impurities. Heavy doping results in a narrow depletion region.
  • When compared to a normal p-n junction diode, tunnel diode has a narrow depletion width.
  • The Fermi level moves in the conduction band on the n-side and inside the valence band on the p-side.
  • Below the Fermi level, all states are filled and above the Fermi level all states are empty

 

The tunnel diode is represented by the symbol

 

F1 P.Y Madhu 9.03.20 D10

The symbols of different diodes are given below.

Diode

Symbol

Tunnel diode

F1 P.Y Madhu 9.03.20 D10

Varactor diode

hark3

Zener diode

06.11.2018.001..08

Schottky diode

F1 S.B Pallavi 09.11.2019 D 3

Photo diode

F1 J.S 30.3.20 Pallavi D9

The value of Iz shown in the given circuit is ________.

F2 Vilas Engineering 8.12.2022 D2

  1. 0.02 A
  2. 0.02 mA
  3. 0.08 mA
  4. 0.08 A

Answer (Detailed Solution Below)

Option 4 : 0.08 A

Diodes and Its Applications Question 12 Detailed Solution

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Concept

The value of Zener current is given by:

\(I_Z=I_S-I_L\)

\(I_Z={V_S-V_Z\over R_S}-{V_Z\over R_L}\)

where, Iz = Zener current

Vz = Zener voltage

Vs = Source voltage

Rs = Source resistance

RL = Load resistance

Calculation

Given, Vz = 20 V

Vs = 30 V

Rs = 100 Ω = 0.1 kΩ 

RL = 1 kΩ

\(I_Z={30-20\over 0.1}-{20\over 1}\)

IZ = 100 - 20 mA = 0.08 A

Which of the following statements is FALSE about LED lamps?

  1. An LED is doped with silicon and germanium like all semiconductor devices. 
  2. An LED works in forward biased condition only.
  3. An LED is a semiconductor device.
  4. The cathode region of an LED would have a slight bent in the structure in comparison with the anode portion.

Answer (Detailed Solution Below)

Option 1 : An LED is doped with silicon and germanium like all semiconductor devices. 

Diodes and Its Applications Question 13 Detailed Solution

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Light Emitting Diode (LED)

  • A light-emitting diode (LED) is a semiconductor device that emits light when an electric current flows through it.
  • When current passes through an LED, the electrons recombine with holes emitting light in the process.
  • LEDs allow the current to flow in the forward direction and blocks the current in the reverse direction.
  • Light-emitting diodes are heavily doped p-n junctions made of a semiconductor material such as gallium and arsenide.
  • Based on the semiconductor material used and the amount of doping, an LED will emit colored light at a particular spectral wavelength when forward-biased.

F1 J.S 30.3.20 Pallavi D8

Identify the device in the following symbol.

F1 Raju Madhuri 13.04.2021 D 5

  1. Zener diode
  2. Varactor diode
  3. Tunnel diode
  4. Photo diode

Answer (Detailed Solution Below)

Option 2 : Varactor diode

Diodes and Its Applications Question 14 Detailed Solution

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Symbols of diodes:

Zener Diode
F1 Harish Batula 13.5.21 Pallavi D7
Varactor Diode

F1 Harish Batula 13.5.21 Pallavi D19

Tunnel Diode
F1 Harish Batula 13.5.21 Pallavi D20
Photo Diode

F1 Harish Batula 13.5.21 Pallavi D21

Among these alternatives, the PIV rating of which diode is lower than that of equivalent vacuum diode?

  1. PN junction Diode
  2. Crystal diode
  3. Tunnel diode
  4. Small single diode

Answer (Detailed Solution Below)

Option 2 : Crystal diode

Diodes and Its Applications Question 15 Detailed Solution

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  • The maximum value of the reverse voltage that a PN junction or diode can withstand without damaging itself is known as its Peak Inverse Voltage
  • This rating of Peak Inverse Voltage (PIV) is given and described in the datasheet provided by the manufacturer
  • PIV rating of Crystal diode is lower than that of the equivalent vacuum diode
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