Analog Electronics MCQ Quiz - Objective Question with Answer for Analog Electronics - Download Free PDF

Last updated on Jul 8, 2025

Latest Analog Electronics MCQ Objective Questions

Analog Electronics 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 .

Answer (Detailed Solution Below)

Option 1 :

Analog Electronics Question 1 Detailed Solution

Let V→ Breakdown, Vr = 0

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

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

ve half cycle: D→ R.B., D2 → F.B.

same logic can be applied with negative polarity.

Analog Electronics Question 2:

A bipolar junction transistor having common emitter current gain of 100 is operating in active region with collector current of 2 mA. Assuming that the thermal voltage VT is 25 mV, what is the input impedance of the transistor?

  1. 2.5 k Ω
  2. 2 k Ω
  3. 1.25 k Ω
  4. 1 k Ω

Answer (Detailed Solution Below)

Option 3 : 1.25 k Ω

Analog Electronics Question 2 Detailed Solution

Concept:

The input impedance (rπ) of a BJT in the active region is given by:

Given:

Common emitter current gain,

Thermal voltage,

Collector current,

Calculation:


Answer:

Option 3) 1.25 kΩ

Analog Electronics Question 3:

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

Analog Electronics Question 3 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

Analog Electronics Question 4:

The current flow in an Enhancement P-type MOSFET is primarily due to 

  1. Electron flow from source to drain 
  2. Hole flow from source to drain
  3. Minority carriers in source region
  4. Carrier injection from gate to substrate

Answer (Detailed Solution Below)

Option 2 : Hole flow from source to drain

Analog Electronics Question 4 Detailed Solution

Explanation:

Current Flow in Enhancement P-Type MOSFET:

Definition: An Enhancement P-type MOSFET is a type of Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET) where the channel is formed by applying a positive voltage to the gate terminal. It is designed to operate in enhancement mode, meaning the device is normally off, and current flows only when a sufficient voltage is applied to the gate to create a conductive channel.

Working Principle: In an Enhancement P-type MOSFET, the source terminal is connected to a higher potential, and the drain terminal is connected to a lower potential. The gate voltage influences the creation of a p-type channel between the source and drain terminals. When a positive gate voltage is applied relative to the source, it repels electrons and attracts holes (the majority carriers in a p-type semiconductor), forming a conductive channel of holes. The flow of current is primarily due to the movement of these holes from the source to the drain.

Advantages:

  • High input impedance due to the insulated gate structure.
  • Low power consumption as no gate current flows under steady-state conditions.
  • High switching speeds suitable for digital and analog applications.

Disadvantages:

  • Vulnerable to static discharge due to the thin insulating layer between the gate and the channel.
  • Complex fabrication process compared to BJTs (Bipolar Junction Transistors).

Correct Option Analysis:

The correct option is:

Option 2: Hole flow from source to drain.

This option correctly describes the primary mechanism of current flow in an Enhancement P-type MOSFET. In a p-type device, the majority carriers are holes. When the gate voltage is applied, it creates a conductive channel of holes between the source and drain terminals. The holes, being positively charged, move from the source (higher potential) to the drain (lower potential), constituting the current flow.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Electron flow from source to drain.

This option is incorrect because it describes the operation of an n-type MOSFET, where the majority carriers are electrons. In an Enhancement P-type MOSFET, the majority carriers are holes, not electrons. Therefore, current flow due to electrons does not apply in this case.

Option 3: Minority carriers in source region.

This option is misleading because the current flow in an Enhancement P-type MOSFET is primarily due to the majority carriers (holes) in the source region. Minority carriers do not play a significant role in the current conduction process in this device.

Option 4: Carrier injection from gate to substrate.

This option is incorrect as the gate in a MOSFET is insulated from the substrate by a thin oxide layer. No carriers (electrons or holes) are injected from the gate to the substrate. Instead, the gate voltage influences the electric field, which modulates the channel formation in the semiconductor material.

Conclusion:

Understanding the operation of Enhancement P-type MOSFETs is crucial for correctly identifying the mechanism of current flow. The device operates by forming a p-type channel when a positive gate voltage is applied, allowing holes to move from the source to the drain. This simplicity and efficiency make it widely used in various applications, including amplifiers, switches, and integrated circuits.

Analog Electronics Question 5:

In the OP AMP circuit shown below, what is the input voltage if output voltage Vout = -0.62 V?

  1. -20 mV
  2. +20 mV
  3. -30 mV
  4. +30 mV

Answer (Detailed Solution Below)

Option 1 : -20 mV

Analog Electronics Question 5 Detailed Solution

Concept:

An inverting amplifier has the configuration shown below. The voltage gain of an inverting OP-AMP is given by:

Given:

Feedback resistor, , Input resistor, , Output voltage,

Calculation:

Gain of the amplifier:

Using the gain formula:

Since it is an inverting amplifier, input voltage must be negative:

Answer:

Option 1) -20 mV

Top Analog Electronics 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 %

Analog Electronics 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:

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:

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

Calculation:

The efficiency for a half-wave rectifier will be:

For Half wave rectifier maximum efficiency = 40.6%

NoteFor Full wave rectifier maximum efficiency = 81.2%

A transistor can be made to operate as a switch by operating it in which of the following regions?

  1. Active region
  2. active region, cut-off region
  3. Active region, saturation region
  4. Saturation region, cut-off region

Answer (Detailed Solution Below)

Option 4 : Saturation region, cut-off region

Analog Electronics Question 7 Detailed Solution

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Mode

EB Biasing

Collector Base Biasing

Application

Cut off

Reverse

Reverse

OFF switch

Active

Forward

Reverse

Amplifier

Reverse Active

Reverse

Forward

Not much Useful

Saturation

Forward

Forward

On Switch

Find the approximate collector current in the given transistor circuit. (Take current gain, β = 100)

  1. 10 mA
  2. 1.25 mA
  3. 1 mA
  4. 11.5 mA

Answer (Detailed Solution Below)

Option 3 : 1 mA

Analog Electronics Question 8 Detailed Solution

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Concept:

For a transistor, the base current, the emitter current, and the collector current are related as:

IE = IB + IC

where IC = β IB

β = Current gain of the transistor

Typical base-emitter voltages, VBE for both NPN and PNP transistors are as follows:

  • If the transistor is made up of a silicon material, the base-emitter voltage VBE will be 0.7 V.
  • If the transistor is made up of a germanium material, the base-emitter voltage VBE will be 0.3 V.
     

Application:

From the given figure, Apply KVL

10 - I× RB - VBE = 0

Let us assume VBE = 0.7 V

10 - IB (1 × 106) - 0.7 = 0

IB = 9.3 μA

We know that,

IC = β IB

Where,

IC  & IB = collector current and base current

Therefore,

IC = 100 × 9.3 μA

= 930 μA

= 0.93 mA

1 mA

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

Analog Electronics Question 9 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
    • 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.

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

Answer (Detailed Solution Below)

Option 2 : 0 V

Analog Electronics Question 10 Detailed Solution

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Concept: 

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

Calculation:

Given,

Zener voltage Vz = 10 V

Ein = 6 V ⇒ Ein z

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

Analog Electronics Question 11 Detailed Solution

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Varactor diode:

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

  • 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

​           

  • 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

Diodes

Application

 Schottky diode

rectifying circuits requiring high switching rate

Varactor diode

Tuned circuits

PIN diode 

High-frequency switch

Zener diode

voltage regulation

State the correct condition for transistor to operate in cut-off region.

  1. Emitter base junction: forward bias
    Collector base junction: forward bias
  2. Emitter base junction: reverse bias
    Collector base junction: forward bias
  3. Emitter base junction: forward bias
    Collector base junction: reverse bias
  4. Emitter base junction: reverse bias
    Collector base junction: reverse bias

Answer (Detailed Solution Below)

Option 4 : Emitter base junction: reverse bias
Collector base junction: reverse bias

Analog Electronics Question 12 Detailed Solution

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BJT Amplifier:

  • Transistors biasing is done to keep stable DC operating conditions needed for its functioning as an amplifier.
  • A properly biased transistor must have it's Q-point (DC operating parameters like IC and VCE) at the center of saturation mode and cut-off mode i.e. active mode.
  • In the active mode of transistor operation, the emitter-base junction is forward biased and the collector-base junction is reverse biased.​
  • In the cut-off mode of transistor operation, the emitter-base junction is reverse biased and the collector-base junction is reverse biased.​

Different modes of BJT operations are:

Mode

Emitter-base

 Junction

Collector-Base

 Junction

Cut off

Reverse

Reverse

Active

Forward

Reverse

Reverse Active

Reverse

Forward

Saturation

Forward

Forward

The early effect in BJT is related to

  1. Base narrowing
  2. Avalanche breakdown
  3. Zener breakdown
  4. Thermal runaway

Answer (Detailed Solution Below)

Option 1 : Base narrowing

Analog Electronics Question 13 Detailed Solution

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Early Effect:

  • large collector base reverse bias is the reason behind the early effect manifested by BJTs.
  • As reverse biasing of the collector to base junction increases, the depletion region penetrates more into the base, as the base is lightly doped.
  • This reduces the effective base width and hence the concentration gradient in the base increases.
  • This reduction in the effective base width causes less recombination of carriers in the base region which results in an increase in collector current. This is known as the Early effect.
  • The decrease in base width causes ß to increase and hence collector current increases with collector voltage rather than staying constant.
  • The slope introduced by the Early effect is almost linear with IC and the common-emitter characteristics extrapolate to an intersection with the voltage axis VA, called the Early voltage.

 

This is explained with the help of the following VCE (Reverse voltage) vs IC (Collector current) curve:

 

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

Analog Electronics Question 14 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.
  • 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.

For a bipolar junction transistor, the common base current gain is 0.98 and the base current is 120 μA. Its common-emitter current gain will be:

  1. 98
  2. 56
  3. 49
  4. 118

Answer (Detailed Solution Below)

Option 3 : 49

Analog Electronics Question 15 Detailed Solution

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Concept:

Where β = common-emitter current gain

α = Common base current gain

Calculation:

Common base current gain = α = 0.98

Note:  & 

Where IC = Collector current

IE = Emitter current

IB = Base current

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