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Test: Photodetectors - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test - Test: Photodetectors

Test: Photodetectors for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Test: Photodetectors questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Photodetectors MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Photodetectors below.
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Test: Photodetectors - Question 1

The quantum efficiency (η) and responsivity (R) at a wavelength λ (in μm) in a p-i-n photodetector are related by

Detailed Solution for Test: Photodetectors - Question 1

Responsivity (R) is defined as the ratio of the photon current to the optical power, i.e.

If the electron generation rate is re and photon incident rate is rp, then Responsivity will be:

where η is the quantum efficiency defined as:

Test: Photodetectors - Question 2

The quantum efficiency η for the photo-detector is:


 

where: Iph = Average photocurrent

Po = Average incident optical power

hc/λ = incident photon energy

Detailed Solution for Test: Photodetectors - Question 2

The Quantum efficiency of a photodetector is defined as the fraction of the incident photons that are absorbed by the photoconductor to the generated electrons that are collected at the detector terminal.
Or Quantum efficiency is defined as the fraction of the number of electrons generated that contributes to the photocurrent to the total number of incident photons.
It is dented by ‘η’ and is given by:

Hence,

Mathematically, Quantum efficiency will be:

 

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*Answer can only contain numeric values
Test: Photodetectors - Question 3

When the optical power incident on a photodiode is 10 μW and the responsivity is 0.8 A/W, the photocurrent generated (in μA ) is


Detailed Solution for Test: Photodetectors - Question 3

Concept:
In a photodiode, the responsivity is given by:
R = I/p
I = photocurrent generated
P = Optical power incident
R = Responsitivity

Calculation:
Given:
R = 0.8 A/W and P = 10 μW

Test: Photodetectors - Question 4

Opto-Isolator device consists of the following device/devices _______.

Detailed Solution for Test: Photodetectors - Question 4

  • An optocoupler (also called an opto-isolator, photocoupler, or optical isolator) is an electronic component that transfers electrical signals between two isolated circuits by using light.
  • Opto-isolators prevent high voltages from affecting the system receiving the signal.
Test: Photodetectors - Question 5

_____________ is a semiconductor device with a P-N junction that converts light into electrical current.

Detailed Solution for Test: Photodetectors - Question 5

Photodiode:

  • A photodiode is a PN-junction diode that consumes light energy to produce an electric current. They are also called a photodetector, light detectors, and photo sensors.
  • Photodiodes are designed to work in reverse bias conditions. Typical photodiode materials are Silicon, Germanium, and Indium gallium arsenide.

Symbol of Photodiode:

Applications of Photodiode:

  • Photodiodes are used in solar cell panels.
  • Photodiodes are used in logic circuits.
  • Photodiodes are used in the detection circuits.
  • Photodiodes are used in character recognition circuits.
  • Photodiodes are used for the exact measurement of the intensity of light in science and industry.
  • Photodiodes are faster and more complex than normal PN junction diodes and hence are frequently used for lighting regulation and optical communication.
Test: Photodetectors - Question 6

Which of the following statement is true for a wavelength of 1000 nm. A p-n photodiode is fabricated from a semiconductor with band gap of 1.9 eV.

Detailed Solution for Test: Photodetectors - Question 6

Concept
A p-n photodiode can only detect those signals whose excitation energy is greater than the band gap energy of the p-n photodiode.
Energy associated with a wave is calculated by the formula

where, 
h = plank's constant
c = velocity of the wave i.e. light
λ = wavelength of the signal

Calculation
Given λ = 1000nm = 10-6m
h = 6.6 x10-34 m2kg/s
c = 3x108 m/s

Now we know that 1eV = 1.6 x 10-19J

Since the energy of the signal is 1.23eV less than the energy band gap (1.9eV) of the semiconductor, Hence the signal cannot be detected by the p-n photodiode.

Test: Photodetectors - Question 7

Direction: The following item consists of two statements, one labelled as ‘Statement (I)’ and the other as ‘Statement (II)’. Examine these two statements carefully and select the answers to these items using the code given below:

Statement (I): PIN photodiodes are preferred to PN diodes.
Statement (II): PIN photodiode is a constant voltage generator at a given light level.

Detailed Solution for Test: Photodetectors - Question 7

PIN (p-type, intrinsic, and n-type) structure is mostly used for constructing the photodiode instead of p-n (p-type and n-type) junction structure because the PIN structure provides fast response time. 

PIN photodiodes are mostly used in high-speed applications.

(Statement I is true)

 

In absence of right:
The entire bias voltage drop across the photodiode and VRL = 0

When Light falls
light power = P,
Photocurrent generated = pP
Where p = Responsivity
Voltage drop across RL = pPRL

Beyond this photodiode is saturated
The maximum voltage drop across RL can be bias voltage. (Statement II is true.​)
Hence, Both Statement I and Statement II are individually true but Statement II is not the correct explanation of Statement I

Test: Photodetectors - Question 8

Swept-out voltage in PIN diode happens when the PIN diode is

Detailed Solution for Test: Photodetectors - Question 8

PIN Diode:
It’s a special type of diode that contains an undoped intrinsic semiconductor between the p-type and n-type regions. The intrinsic layer does not conduct electric current well.

  • PIN Diode is an inferior rectifier but widely suitable for switches, attenuators, etc.

PIN diode with different biasing:

1. Forward Bias:

  • The Depletion region reduces, resulting in the flow of current.
  • Acts as a variable resistance in the forward bias.
  • A High electric field developed across the junction and this speeds up the transport of carriers.

2. Reverse Bias:

  • Width of the depletion region increases.
  • At a certain voltage, the entire intrinsic layer will be swept out of charge carriers. This is called

“Swept out” voltage.

Test: Photodetectors - Question 9

When the light increases, the reverse current in a photo-diode

Detailed Solution for Test: Photodetectors - Question 9

Photo-Diode:

  • A photodiode is a semiconductor p–n junction device that converts light into an electrical current
  • The current is generated when photons are absorbed in the photodiode.
  • The photodiode is a special type of diode which operates in reverse bias conditions.
  • When the incident light on the photodiode increases, the reverse current in a photo-diode also increases.
  • Diode current flows from n to p.
  • Net current flowing is due to only minority charge carriers.
  • The current flowing in the photodiode is due to diffusion current only.
Test: Photodetectors - Question 10

In a photodiode, carriers are generated in the ________.

Detailed Solution for Test: Photodetectors - Question 10

In a Photodiode, carriers are generated in the depletion region because the width of the depletion region increased when the photodiode operates in reverse biased.

So option (2) is correct.

The basic working model of photo-diode is as shown:

Here,
e- = electrons
O = holes
⊕ = positive ions
⊖ = negative ions

  • Carriers are generated in the depletion region mainly due to the incident photons on the surface of semiconductor material.
  • A photodiode is used to convert light signals into electrical signals.
  • In photodiode, the current called photo-current is proportional to the intensity of the light incident on the surface.
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