Optical Communications Questions and Answers – LED Power and Efficiency

This set of Optical Communications Multiple Choice Questions & Answers (MCQs) focuses on “LED Power and Efficiency”.

1. The absence of _______________ in LEDs limits the internal quantum efficiency.
a) Proper semiconductor
b) Adequate power supply
c) Optical amplification through stimulated emission
d) Optical amplification through spontaneous emission
View Answer

Answer: c
Explanation: The ratio of generated electrons to the electrons injected is quantum efficiency. It is greatly affected if there is no optical amplification through stimulated emission. Spontaneous emission allows ron-radiative recombination in the structure due to crystalline imperfections and impurities.

2. The excess density of electrons Δnand holes Δpin an LED is ____________
a) Equal
b) Δpmore than Δn
c) Δn more than Δp
d) Does not affects the LED
View Answer

Answer: a
Explanation: The excess density of electrons ΔnandΔp (holes) is equal. The charge neutrality is maintained within the structure due to injected carriers that are created and recombined in pairs. The power generated internally by an LED is determined by taking into considering the excess electrons and holes in p- and n-type material respectively.

3. The hole concentration in extrinsic materials is _________ electron concentration.
a) much greater than
b) lesser than
c) equal to
d) negligible difference with
View Answer

Answer: a
Explanation: In extrinsic materials, one carrier type will be highly concentrated than the other type. Hence in p-type region, hole concentration is greater than electron concentration in context of extrinsic material. This excess minority carrier density decays with time.

4. The carrier recombination lifetime becomes majority or injected carrier lifetime.
a) True
b) False
View Answer

Answer: b
Explanation: The initial injected excess electron density and τrepresents the total carrier recombination time. In most cases, Δnis a small fraction of majority carriers and contains all minority carriers. So in these cases, carrier recombination lifetime becomes minority injected carrier lifetime τi.

5. In a junction diode, an equilibrium condition occurs when ____________
a) Δngreater than Δp
b) Δnsmaller than Δp
c) Constant current flow
d) Optical amplification through stimulated emission
View Answer

Answer: c
Explanation: The total rate at which carriers are generated in sum of externally supplied and thermal generation rates. When there is a constant current flow in this case, an equilibrium occurs in junction diode.
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6. Determine the total carrier recombination lifetime of a double heterojunction LED where the radioactive and nonradioactive recombination lifetime of minority carriers in active region are 70 ns and 100 ns respectively.
a) 41.17 ns
b) 35 ns
c) 40 ns
d) 37.5 ns
View Answer

Answer: a
Explanation: The total carrier recombination lifetime is given by
τ = τrτnrrnr = 70× 100/70 + 100 ns = 41.17 ns
τr = radiative recombination lifetime of minority carriers
τnr = nonradioactive recombination lifetime of minority carriers.

7. Determine the internal quantum efficiency generated within a device when it has a radiative recombination lifetime of 80 ns and total carrier recombination lifetime of 40 ns.
a) 20 %
b) 80 %
c) 30 %
d) 40 %
View Answer

Answer: b
Explanation: The internal quantum efficiency of device is given by
ηint = τ/τr = 40/80 ×100 = 80%
τ = total carrier recombination lifetime
τr = radiative recombination lifetime.

8. Compute power internally generated within a double-heterojunction LED if it has internal quantum efficiency of 64.5 % and drive current of 40 mA with a peak emission wavelength of 0.82 μm.
a) 0.09
b) 0.039
c) 0.04
d) 0.06
View Answer

Answer: b
Explanation: The power internally generated within device i.e. double-heterojunction LED can be computed by
Pint = ηint hci/eλ = 0.645×6.626×10-34×3×108×40×10-3/ 1.602×10-19 × 0.82 × 10-6
= 0.039 W
ηint = internal quantum efficiency
h = Planck’s constant
c = velocity of light
i = drive current
e = electron charge
λ = wavelength.

9. The Lambertian intensity distribution __________ the external power efficiency by some percent.
a) Reduces
b) Does not affects
c) Increases
d) Have a negligible effect
View Answer

Answer: a
Explanation: In Lambertian intensity distribution, the maximum intensity I0is perpendicular to the planar surface but is reduced on the sides in proportion to the cosine of θ i.e. viewing angle as apparent area varies with this angle. This reduces the external power efficiency. This is because most of the light is tapped by total internal refraction when radiated at greater than the critical angle for crystal air interface.

10. A planar LED fabricated from GaAs has a refractive index of 2.5. Compute the optical power emitted when transmission factor is 0.68.
a) 3.4 %
b) 1.23 %
c) 2.72 %
d) 3.62 %
View Answer

Answer: c
Explanation: The optical power emitted is given by
Pe = PintFn2/4nx2 = Pint (0.680×1/4×(2.5)2) = 0.0272 Pint.
Hence power emitted is only 2.72 % of optional power emitted internally.
Fn2 = transmission factor
nx = refractive index.

11. A planar LED is fabricated from GaAs is having a optical power emitted is 0.018% of optical power generated internally which is 0.018% of optical power generated internally which is 0.6 P. Determine external power efficiency.
a) 0.18%
b) 0.32%
c) 0.65%
d) 0.9%
View Answer

Answer: d
Explanation: Optical power generated externally is given by
ηcp = (0.018Pint/2Pint)*100
Pint = power emitted
ηcp = external power efficiency.

12. For a GaAs LED, the coupling efficiency is 0.05. Compute the optical loss in decibels.
a) 12.3 dB
b) 14 dB
c) 13.01 dB
d) 14.6 dB
View Answer

Answer: c
Explanation: The optical loss in decibels is given by-
Loss = -10log10 ηc
ηc = coupling efficiency.

13. In a GaAs LED, compute the loss relative to internally generated optical power in the fiber when there is small air gap between LED and fiber core. (Fiber coupled = 5.5 * 10-4Pint)
a) 34 dB
b) 32.59 dB
c) 42 dB
d) 33.1 dB
View Answer

Answer: b
Explanation: The loss in decibels relative to Pint is given by-
Loss = -10log10Pc/Pint
Pc = 5.5 * 10-4Pint.

14. Determine coupling efficiency into the fiber when GaAs LED is in close proximity to fiber core having numerical aperture of 0.3.
a) 0.9
b) 0.3
c) 0.6
d) 0.12
View Answer

Answer: a
Explanation: The coupling efficiency is given by
ηc = (NA)2 = (0.3)2 = 0.9.

15. If a particular optical power is coupled from an incoherent LED into a low-NA fiber, the device must exhibit very high radiance.
a) True
b) False
View Answer

Answer: a
Explanation: Device must have very high radiance specially in graded index fiber where Lambertian coupling efficiency with same NA is about half that of step-index fibers. This high radiance is obtained when direct bandgap semiconductors are fabricated with DH structure driven at high current densities.

Sanfoundry Global Education & Learning Series – Optical Communications.

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

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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