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

Last updated on Jul 8, 2025

Latest Diodes and Its Applications MCQ Objective Questions

Diodes and Its Applications Question 1:

What is the output waveform of the circuit for the given input signal? Assume that the zener diodes are identical, amplitude of the input voltage Vin is twice the zener breakdown voltage, and RL = 10 R .

qImage686cf6c3ce33a9723ba0a44d

  1. image-Photoroom (9)
  2. image-Photoroom (10)
  3. qImage685de4a9f3aaadb5a0103d33
  4. qImage685de4aaf3aaadb5a0103d34

Answer (Detailed Solution Below)

Option 1 : image-Photoroom (9)

Diodes and Its Applications Question 1 Detailed Solution

Let V→ Breakdown, Vr = 0

Vin < Vz, D→ F.B D2 → R.F. (OFF) [Zenernot inbreak down region]

qImage6867967fb6b8e9b2e8b3de29

Vin > Vz, D1 → F.B D→ R.F. (ON) [Zener in break down region]

qImage6867967fb6b8e9b2e8b3de2a

ve half cycle: D→ R.B., D2 → F.B.

same logic can be applied with negative polarity.

Diodes and Its Applications Question 2:

A voltage regulator gives 5 V output when 10 V is applied to input. What is the input current when the load current of 0.75 A flows assuming an efficiency of 75% ?

  1. 1.5 A
  2. 2 A
  3. 0.5 A
  4. 0.75 A

Answer (Detailed Solution Below)

Option 3 : 0.5 A

Diodes and Its Applications Question 2 Detailed Solution

Explanation:

The efficiency of a voltage regulator is defined as the ratio of output power to input power, expressed as a percentage:

Efficiency (%) = (Output Power / Input Power) × 100

Rearranging this formula, we can calculate the input power:

Input Power = Output Power / Efficiency

Next, we use the relationship between power, voltage, and current to determine the input current:

Power = Voltage × Current

Rearranging this, the input current can be calculated as:

Input Current = Input Power / Input Voltage

Solution:

The output power is the product of the output voltage and the load current:

Output Power = Output Voltage × Load Current

Here, the output voltage is 5 V, and the load current is 0.75 A.

Output Power = 5 × 0.75 = 3.75 W

Using the efficiency formula, we calculate the input power. The efficiency is given as 75%, or 0.75 in decimal form:

Input Power = Output Power / Efficiency == 3.75 / 0.75 = 5 W.

Using the relationship between power, voltage, and current, we find the input current. The input voltage is 10 V:

Input Current = Input Power / Input Voltage = 5/10 = 0.5 A

Diodes and Its Applications Question 3:

In the figure, Vs is a square wave of levels ±10 V, frequency 100 Hz and duty cycle 50%. The diode is ideal. What is the average value of VL?

qImage684c3f0a82be0b8e0ca93c0a

  1. - 5 V
  2. - 2.5 V
  3. 0 V
  4. + 2.5 V

Answer (Detailed Solution Below)

Option 4 : + 2.5 V

Diodes and Its Applications Question 3 Detailed Solution

Concept:

We are given a square wave input VS of ±10 V, 100 Hz frequency, and 50% duty cycle. The diode is ideal.

We need to find the average voltage across VL, which is the voltage across the load resistor RL.

Step-by-step Analysis:

When VS = +10 V (50% of the time):

  • The diode is forward-biased.
  • Current flows through the upper 100 Ω resistor, diode, and lower 100 Ω + RL.
  • It acts like a voltage divider between two 100 Ω resistors.
  • So, VL=102=+5 V

When VS = -10 V (other 50% time):

  • The diode is reverse-biased, so it acts as an open circuit.
  • No current flows; hence, VL=0 V

Average value over full cycle:

Vavg=12(+5 V)+12(0 V)=+2.5 V

Diodes and Its Applications Question 4:

In the Zener diode voltage regulator shown, the 7 V Zener has knee current of 5mA. What is the minimum value of RL to get regulated output voltage at VL?

qImage684c3e6803ea9cea6c89e397

  1. 70 Ω
  2. 93.3 Ω
  3. 100 Ω
  4. 150 Ω 

Answer (Detailed Solution Below)

Option 3 : 100 Ω

Diodes and Its Applications Question 4 Detailed Solution

Concept:

A Zener diode regulator maintains a constant output voltage (VZ) across the load resistor (RL) as long as the Zener diode current is greater than or equal to its knee current (minimum operating current).

To maintain regulation, the Zener must conduct at least the knee current (IZK = 5 mA).

Given:

Input Voltage (Vin) = 14.5 V

Zener Voltage (VZ) = 7 V

Series Resistance Rs = 100 Ω

Zener Knee Current IZK = 5 mA

Calculation:

Total current through Rs:

Itotal=VinVZRs=14.57100=0.075 A=75 mA

Minimum current required by Zener = 5 mA

So, maximum load current:

IL=ItotalIZK=755=70 mA

Using Ohm’s law to find minimum RL:

RL=VZIL=70.07=100 Ω

Final Answer:

Option 3) 100 Ω 

Diodes and Its Applications Question 5:

In an ideal half-wave rectifier with zero mean sine wave input, the output is

  1. non zero during only one half cycle of input 
  2. non zero during both half cycles of input
  3. clipped form of input during both half cycles
  4. clamped form of input during both half cycles

Answer (Detailed Solution Below)

Option 1 : non zero during only one half cycle of input 

Diodes and Its Applications Question 5 Detailed Solution

Explanation:

Ideal Half-Wave Rectifier

Definition: A half-wave rectifier is an electronic circuit that converts an alternating current (AC) input signal into a pulsating direct current (DC) output signal. It achieves this by allowing current to flow through the load only during one half-cycle of the input AC waveform, while blocking the current during the other half-cycle. This operation is typically achieved using a single diode in the circuit.

Working Principle: In a half-wave rectifier, the diode is forward-biased during one half of the AC cycle (positive half-cycle for a typical configuration), allowing current to pass through. During the other half of the AC cycle (negative half-cycle), the diode becomes reverse-biased and blocks the current. As a result, the output voltage is non-zero only during one half of the AC input cycle and zero during the other half.

Correct Option Analysis:

The correct option is:

Option 1: Non-zero during only one half cycle of input.

This option is correct because the primary characteristic of an ideal half-wave rectifier is that it allows current to flow through the load only during one half of the AC input cycle. For a zero mean sine wave input, this means the rectifier will produce a pulsating output that corresponds to the positive or negative half-cycles of the input waveform, depending on the orientation of the diode. The output will remain zero during the other half-cycle when the diode is reverse-biased.

Mathematical Analysis:

Let the input AC signal be represented as:

vin(t) = Vmsin(ωt), where:

  • Vm = Peak amplitude of the sine wave.
  • ω = Angular frequency of the AC signal.

During the positive half-cycle of the input (0 ≤ ωt ≤ π), the diode is forward-biased, and the output voltage is:

vout(t) = Vmsin(ωt)

During the negative half-cycle of the input (π ≤ ωt ≤ 2π), the diode is reverse-biased, and the output voltage is:

vout(t) = 0

Thus, the output of an ideal half-wave rectifier is non-zero only during one half-cycle of the input AC signal

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

Download Solution PDF

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:

η=PdcPac

η=Vdc2RLVrms2RL

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:

VDC=Vmπ

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

Vrms=Vm2

Calculation:

The efficiency for a half-wave rectifier will be:

η=(Vmπ)2(Vm2)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

Download Solution PDF
  • 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

Download Solution PDF

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

Download Solution PDF

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=AϵW

  • 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

Download Solution PDF
  • 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

Download Solution PDF
  • 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

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 12 Detailed Solution

Download Solution PDF

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

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 13 Detailed Solution

Download Solution PDF

Concept

The value of Zener current is given by:

IZ=ISIL

IZ=VSVZRSVZRL

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

IZ=30200.1201

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 14 Detailed Solution

Download Solution PDF

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

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

Download Solution PDF
  • 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
Get Free Access Now
Hot Links: teen patti boss teen patti joy 51 bonus teen patti royal - 3 patti