# Optical Communications Questions and Answers – Quantum Efficiency, Responsivity and Long – Wavelength Cut-Off

This set of Optical Communications Mcqs focuses on “Quantum Efficiency , Responsivity and Long – Wavelength Cut-Off”.

1. The fraction of incident photons generated by photodiode of electrons generated collected at detector is known as ___________________
a) Quantum efficiency
b) Absorption coefficient
c) Responsivity
d) Anger recombination

Explanation: Efficiency of a particular device is obtained by ratio of input given to that of output obtained. Thus, similarly, in photodiode, input i.e. incident photon and output generated electrons and their ratio is quantum efficiency.

2. In photo detectors, energy of incident photons must be ________________ band gap energy.
a) Lesser than
b) Greater than
c) Same as
d) Negligible

Explanation: While considering intrinsic absorption process, the energy of incident photon must be greater than band gap energy of material fabricating photo detector.

3. GaAs has band gap energy of 1.93 eV at 300 K. Determine wavelength above which material will cease to operate.
a) 2.431*10-5
b) 6.424*10-7
c) 6.023*103
d) 7.234*10-7

Explanation: The long wavelength cutoff is given by
λc = hc/Eg = 6.6268*10-34*2.998*108/1.93*1.602*10-19
= 6.424*10-7μm.

4. The long cutoff wavelength of GaAs is 0.923 μm. Determine bandgap energy.
a) 1.478*10-7
b) 4.265*10-14
c) 2.784*10-9
d) 2.152*10-19

Explanation: Long wavelength cutoff of photo detector is given by
λc = hc/Eg
Eg = hc/λc = 6.6268*10-34*2.998*108/0.923*10-6
= 2.152*10-19eV.

5. Quantum efficiency is a function of photon wavelength.
a) True
b) False

Explanation: Quantum efficiency is less than unity as all of incident photons are not absorbed to create electrons holes pairs. For example quantum efficiency of 60% is equivalent to 60% of electrons collected per 100 photons. Thus efficiency is a function of photon wavelength and must be determined at a particular wavelength.
Note: Join free Sanfoundry classes at Telegram or Youtube

6. Determine quantum efficiency if incident photons on photodiodes is 4*1011 and electrons collected at terminals is 1.5*1011?
a) 50%
b) 37.5%
c) 25%
d) 30%

Explanation: Quantum efficiency is given by
Quantum Efficiency = No. of electrons collected/No. of incident photons
= 1.5*1011/4*1011
= 0.375 * 100
= 37.5%.

7. A photodiode has quantum efficiency of 45% and incident photons are 3*1011. Determine electrons collected at terminals of device.
a) 2.456*109
b) 1.35*1011
c) 5.245*10-7
d) 4.21*10-3

Explanation: Quantum efficiency is given by
Quantum efficiency = No. of electrons collected/No. of incident photons
Electrons collected = Quantum efficiency * number of incident photons
= 45/100 * 3*1011
= 1.35*1011.

8. The quantum efficiency of photodiode is 40% with wavelength of 0.90*10-6. Determine the responsivity of photodiodes.
a) 0.20
b) 0.52
c) 0.29
d) 0.55

Explanation: Responsivity of photodiodes is given by
R = ηe λ/hc
= 0.4*1.602*10-19 * 0.90*10-6/6.626*10-34 * 3*108
= 0.29 AW-1.

9. The Responsivity of photodiode is 0.294 AW-1at wavelength of 0.90 μm. Determine quantum efficiency.
a) 0.405
b) 0.914
c) 0.654
d) 0.249

Explanation: Responsivity of photodiode is
R = ηe λ/hc
η = RXhc/eλ
= 0.294*6.626*10-34*3*108/ 1.602*10-19*0.90*108
= 0.405 AW-1.

10. Determine wavelength of photodiode having quantum efficiency of 40% and Responsivity of 0.304 AW-1.
a) 0.87 μm
b) 0.91 μm
c) 0.88 μm
d) 0.94 μm

Explanation: The Responsivity of photodiode is
R = ηe λ/hc
λ = Rhc/ηe
= 0.304*6.626*10-34*3*108/0.4*1.602*10-19
= 0.94 μm.

11. Determine wavelength at which photodiode is operating if energy of photons is 1.9*10-19J?
a) 2.33
b) 1.48
c) 1.04
d) 3.91

Explanation: To determine wavelength,
λ = hc/t
= 6.626*10-34*3*108/1.9*10-19
= 1.04 μm.

12. Determine the energy of photons incident on a photodiode if it operates at a wavelength of 1.36 μm.
a) 1.22*10-34J
b) 1.46*10-19J
c) 6.45*10-34J
d) 3.12*109J

Explanation: The wavelength of photodiode is given by
λ = hc/t
E = hc/λ
= 6.626*10-34*3*108/1.36*10-6
= 1.46*10-19J.

13. Determine Responsivity of photodiode having o/p power of 3.55 μm and photo current of 2.9 μm.
a) 0.451
b) 0.367
c) 0.982
d) 0.816

Explanation: The Responsivity of photodiode is
R = Ip/Po
= 2.9*10-6/3.55*10-6
= 0.816 A/W.

14. Determine incident optical power on a photodiode if it has photocurrent of 2.1 μA and responsivity of 0.55 A/W.
a) 4.15
b) 1.75
c) 3.81
d) 8.47

Explanation: The Responsivity of photodiode is
R = Ip/Po
Po = Ip/R
= 2.1*10-6/0.55
= 3.81 μm.

15. If a photodiode requires incident optical power of 0.70 A/W. Determine photocurrent.
a) 1.482
b) 2.457
c) 4.124
d) 3.199

Explanation: The Responsivity of photodiode is given by
R = Ip/Po
Ip = R*Po
= 0.70*3.51*10-6
= 2.457μm.

Sanfoundry Global Education & Learning Series – Optical Communications.

To practice MCQs on all areas of Optical Communications, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]