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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Test  >  GATE ECE (Electronics) Mock Test Series 2027  >  Test: Electronic Devices - 3 - Electronics and Communication Engineering (ECE) MCQ

GATE ECE (Electronics) Test: Electronic Devices - 3 Free Online Test 2026


MCQ Practice Test & Solutions: Test: Electronic Devices - 3 (10 Questions)

You can prepare effectively for Electronics and Communication Engineering (ECE) GATE ECE (Electronics) Mock Test Series 2027 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Electronic Devices - 3". These 10 questions have been designed by the experts with the latest curriculum of Electronics and Communication Engineering (ECE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 10

Sign up on EduRev for free to attempt this test and track your preparation progress.

Test: Electronic Devices - 3 - Question 1
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The open-circuit p-n junction diode without any Biasing is shown if NA = 2 × 1017/cm3, ND = 5 × 1016/cm3 then the depletion region width (W) is ______μm

Detailed Solution: Question 1

For the p-n junction diode, the depletion region width is given by

W = 10 μm

Test: Electronic Devices - 3 - Question 2
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Which of the following statements is TRUE for the function f(z) = z sin(z) / (z - π)²?

Detailed Solution: Question 2

f(z) = z sin(z) / (z - π)² has a pole of order 2 at z = π

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Test: Electronic Devices - 3 - Question 3
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The correct output waveform for the circuit shown if the input is a sinusoidal signal of maximum Amplitude Vmax is 
 

Detailed Solution: Question 3

The circuit shown in a half-wave voltage double During Positive half cycle C1 and D1 Conducts and C is charged to Vmax During Negative half cycle C2 and D2conducts and

 Vout = Vmax + VCl

= 2 Vmax

*Answer can only contain numeric values
Test: Electronic Devices - 3 - Question 4
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In the circuit shown, the forward biased LED has a voltage drop of 1.5 volts. If the battery voltage is 6V. Then the power displaced in the resistor R in milliwatts is _________mW. Current through LED is 15mA.


Detailed Solution: Question 4

If r is the internal resistance of LED

Total current = 

As the drop across LED = 1.5 V

Internal resistance r = 
The external resistance to be connected

R = 400 – 100 = 300Ω

Power dissipated in the resistor

I2R = 67.5 mW

Test: Electronic Devices - 3 - Question 5
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The voltage regulator circuit using a Zener diode is shown. The Zener diode current is limited in the range 5 ≤ iz ≤ 100 mA 


The range of load resistance is

Detailed Solution: Question 5


 

IZ = (125 - IZ)mA

Case A:

IZ = 5 mA

IZ = 120 mA

= 0.12 A


 

Case B:

IZ = 100 mA

IZ = (125 – 100) mA

= 25 mA

= 0.025 A

*Answer can only contain numeric values
Test: Electronic Devices - 3 - Question 6
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For a P±n Si junction the reverse current at room temperature is 0.9 nA/cm2. If donor density is 1015 cm-3 and intrinsic carrier concentration is 1.05 × 1010. The minority carrier life time is ________ n sec.

[Assume μP = 450 cm2/V-sec, kT = 25mV ] 


Detailed Solution: Question 6

For P±n junction the current density is given by

Substituting

DP = 0.025 × 450

= 11.25

τp = 4.32 × 10-9 sec

Test: Electronic Devices - 3 - Question 7
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The sketch of output voltage Vo vs. input voltage VI is of from

Detailed Solution: Question 7

For voltages -5V < VI < 5 V both the diodes are off since the diodes are reverse biased.


The output voltage Vo = VI

For VI ≤ -5V

The diode D1 is forward biased, while D2 is reverse biased



For VI ≥ 5 V

The diode D1 is reverse biased, while D2 is forward biased


Drawing sketch of Vo Vs VI

*Answer can only contain numeric values
Test: Electronic Devices - 3 - Question 8
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Consider a pn junction at zero bias with an electric distribution as sketched below. Calculate the built-in-potential


Detailed Solution: Question 8

The built in potential is the integral of election field at zero bais:

⇒ 75 × 10-2 V

⇒ 0.75 V

Test: Electronic Devices - 3 - Question 9
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The correct waveform of the output of the given circuit is

Assume the Zener diodes are ideal with threshold voltage Vk = 0.7

Detailed Solution: Question 9

For positive cycle when VI = 0.7 V, the 6 V Zener diode is forward biased while 4 V is reverse biased

For negative cycle VI <-0.7 V, the 6 V Zener diode is reverse biased 4 V Zener diode is forward biased


For -0.7 < VI < 0.7 both the diodes are reverse biased hence Vout = VI

The output waveform will be of form

Test: Electronic Devices - 3 - Question 10
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The correct statement regarding depletion and diffusion capacitance is

i) Depletion capacitance is dominant in reverse-bias voltage

ii) Diffusion capacitance is dominant in reverse-bias voltage

iii) The diffusion capacitance is due to a stored charge of minority electrons and minority holes near the depletion region

iv) Depletion capacitance is directly proportional to the width of the depletion region

Detailed Solution: Question 10

Diffusion capacitance is due to transfer of minority carries during forward bias. The minority carries diffuse from one end of junction to other, causing variation of charge with applied voltage. This leads to capacitance, it is present only in forward bias and is significantly higher than depletion capacitance in forward bias.

Depletion capacitance is due to storage of charges in reverse bias junction, which acts like parallel plate capacitance with forward voltage, the depletion width decreases increasing depletion capacitance. It is less than diffusion capacitance in forward bias mode.

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About this Electronics and Communication Engineering (ECE) Practice Test

This test contains 10 MCQs on Electronic Devices - 3 with detailed solutions for each question. It is part of the GATE ECE (Electronics) Mock Test Series 2027 course on EduRev, designed to align with the current Electronics and Communication Engineering (ECE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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