Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  Electronic Devices  >  Test: Photo Diode - Electronics and Communication Engineering (ECE) MCQ

Test: Photo Diode - Electronics and Communication Engineering (ECE) MCQ


Test Description

10 Questions MCQ Test Electronic Devices - Test: Photo Diode

Test: Photo Diode for Electronics and Communication Engineering (ECE) 2024 is part of Electronic Devices preparation. The Test: Photo Diode questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Photo Diode 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: Photo Diode below.
Solutions of Test: Photo Diode questions in English are available as part of our Electronic Devices for Electronics and Communication Engineering (ECE) & Test: Photo Diode solutions in Hindi for Electronic Devices course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt Test: Photo Diode | 10 questions in 30 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study Electronic Devices for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
Test: Photo Diode - Question 1

The quantum efficiency η for the photodetector is -
Where Iph = average photo current
Po = Average incident optical power
hc/λ = incident photon energy

Detailed Solution for Test: Photo Diode - Question 1

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:

Mathematically, Quantum efficiency will be:

Special Note:
Quantum efficiency is also defined in terms of responsivity (R).
Responsivity (R): It is a measure of the sensitivity of a photodetector to light. It is defined as the ratio of photocurrent Iph to the incident optical power P0 for a particular wavelength.
Responsivity 
Responsivity depends on:

  • The wavelength of the incident light.
  • Applied reverse bias.
  • Temperature.

Hence,

E = Energy of photon given as:

Test: Photo Diode - Question 2

______ is used to detect the optical signal.

Detailed Solution for Test: Photo Diode - Question 2
  • A photodiode is indeed used to detect optical signals.
  • A photodiode is a type of photodetector that converts light energy into an electrical current. When photons (light particles) strike the photodiode, they generate electron-hole pairs, resulting in a measurable current flow.
  • This current can then be amplified and processed to detect and interpret the optical signal.
  • Photodiodes are commonly used in various applications, including optical communication systems, light sensors, barcode readers, optical switches, and many other devices that require the detection of optical signals.
1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Photo Diode - Question 3

Which one of the following can be used as a photo detector in Fiber Optic Communication?

Detailed Solution for Test: Photo Diode - Question 3
  • PIN photodiodes are used in fibre optic network cards and switches.
  • They are designed to detect photons and are used in optical detection.
  • Reverse current flows through the photodiode when it is sensing light. 
  • When photons excite carriers in a reverse-biased p-n junction, a very small current proportional to the light intensity starts to flow.
  • The intensity depends on the wavelength of the light.
  • Photodiodes convert incident light into electric energy, so they are used as optical detectors.
Test: Photo Diode - Question 4

___ current is the leakage current that flows through a photo diode with no input used in as light detectors.

Detailed Solution for Test: Photo Diode - Question 4
  • In physics and in electronic engineering, the dark current is the relatively small electric current that flows through photosensitive devices such as a photomultiplier tube, photodiode, or charge-coupled device even when no photons are entering the device;
  • it consists of the charges generated in the detector when no outside radiation is entering the detector (the current the photodiode produces with no light hitting the diode).
  • It is referred to as reverse bias leakage current in non-optical devices and is present in all diodes.
  • Physically, the dark current is due to the random generation of electrons and holes within the depletion region of the device.
*Answer can only contain numeric values
Test: Photo Diode - Question 5

A light detector circuit using an ideal photo-diode is shown in the figure. The sensitivity of the photo-diode is 0.5 μA/μW. With Vr = 6 V, the output voltage Vo = −1.0 V for 10 μW of incident light. If Vr is changed to 3 V, keeping all other parameters the same, the value of Vo in volts is __________V.


Detailed Solution for Test: Photo Diode - Question 5

The function of battery Vr is to reverse bias the photocurrent is independent of reverse bias voltage
Ip = sensitivity × Incident power
Vout = -Ip (R)
Application:
Since Vout = -IpR does not contain any function of Reverse bias voltage.
Vout = -1V

Test: Photo Diode - Question 6

In the given circuit as the intensity of light falling on photodiode increases, the output voltage.

Detailed Solution for Test: Photo Diode - Question 6

The output voltage is given by V­out = Vcc - IcRc
Vout = 5 - Ic (1k) p Ic is in mA]
Also
Ic = β IB
As the Intensity of light falling on the photodiode increases, the Base current IB increase
as IB increases IC Increases
Hence
Output voltage which is given by Vout = 5 - Ic(1) decreases. 

*Answer can only contain numeric values
Test: Photo Diode - Question 7

The figure shows a spot of light of uniform intensity 50 W/m2 and size 10 mm × 10 mm incident at the exact center of a photo-detector, comprising two identical photo-diodes D1 and D2. Each diode has a sensitivity of 0.4 A/W and is operated in the photoconductive mode. If the spot of light is displaced upwards by 100 μm, the resulting difference between the photocurrents generated by D1 and D2 in micro amperes, is __________ μA.


Detailed Solution for Test: Photo Diode - Question 7

Photocurrent = Sensitivity × intensity × Area
After displacing by 100 μm

Current form the upper part:

IL = 80 μA
Difference = (120 – 80) μA
= 40 μA

*Answer can only contain numeric values
Test: Photo Diode - Question 8

The photo diode in the figure below has an active sensing area of 10 mm2, a sensitivity of 0.5 A/W and a dark current of 1 µA. The i-to-v converter has a sensitivity of 100 mV/µA. For an input light intensity of 4 W/m2, the output VO in volt is ____.


Detailed Solution for Test: Photo Diode - Question 8

The output voltage of I-V converter is

Photo current generate
Ip = Area × sensitivity × Intensity

Ip = 20 μA
Output of I-V converter

= 2000 mV
= 2 V

Test: Photo Diode - Question 9

A 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, is

Detailed Solution for Test: Photo Diode - Question 9

The whole power of 100 W will not fall on the active area
Intensity of light at a distance ‘d’ from an isotropic source is given by

Here the photodetector is at 1m distance
So, d = 1m

Power falling on photodetector = Intensity x active area

Output current = Power on photodetector x Responsivity

Hence, the photo-current generated in the detector is 1 mA

*Answer can only contain numeric values
Test: Photo Diode - Question 10

Quantum efficiency of a photodiode (ratio between the number of liberated electrons and the number of incident photons) is 0.75 at 830 nm. Given Planck’s constant h = 6.624 × 10-34 J-sec, the charge of an electron e = 1.6 × 10-19 C and the velocity of light in the photodiode Cm = 2 × 108 m/s. For an incident optical power of 100 μW at 830 nm, the photocurrent in μA is __________.


Detailed Solution for Test: Photo Diode - Question 10

Given:
η = 0.75
λ = 830 nm
h = 6.626 x 10-34 J-sec
e = 1.6 x 10-19 C
cm = 2 x 108 m/s
Pin = 100 μW
Concept:
The ratio of output current to input power is given by Responsivity

Calculation:

I = 75.18 μA

21 docs|29 tests
Information about Test: Photo Diode Page
In this test you can find the Exam questions for Test: Photo Diode solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Photo Diode, EduRev gives you an ample number of Online tests for practice

Top Courses for Electronics and Communication Engineering (ECE)

Download as PDF

Top Courses for Electronics and Communication Engineering (ECE)