# Optical Communications Questions and Answers – Semiconductor Photodiodes Without Internal Gain

This set of Optical Communications Multiple Choice Questions & Answers focuses on “Semiconductor Photodiodes Without Internal Gain”.

1. The width of depletion region is dependent on ___________ of semiconductor.
a) Doping concentrations for applied reverse bias
b) Doping concentrations for applied forward bias
c) Properties of material
d) Amount of current provided

Explanation: The depletion region is formed by immobile positively and immobile negatively charged donor and acceptor atoms in n- and p-type respectively. When carriers are swept towards majority side under electric field, lower the doping, wider the depletion region.

2. Electron-hole pairs are generated in ___________
a) Depletion region
b) Diffusion region
c) Depletion region
d) P-type region

Explanation: Photons are absorbed in both depletion and diffusion regions. The position and width of absorption region depends on incident photons energy. The absorption region may extend throughout device in weakly absorption of photons. Thus carriers are generated in both regions.

3. The diffusion process is _____________ as compared with drift.
a) Very fast
b) Very slow
c) Negligible
d) Better

Explanation: None.

4. Determine drift time for carrier across depletion region for photodiode having intrinsic region width of 30μm and electron drift velocity of 105 ms-1.
a) 1×10-10 Seconds
b) 2×10-10 Seconds
c) 3×10-10 Seconds
d) 4×10-10 Seconds

Explanation: The drift time is given by
tdrift = w/vd = 30×10-6/1×10-10 = 3×10-10 seconds.

5. Determine intrinsic region width for a photodiode having drift time of 4×10-10 s and electron velocity of 2×10-10ms-1.
a) 3×10-5M
b) 8×10-5M
c) 5×10-5M
d) 7×10-5M

Explanation: The drift time is given by
tdrift = w/vd
4×10-10 = w/2×105
= 4×10-10×2×105
= 8×10-5m.
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6. Determine velocity of electron if drift time is 2×10-10s and intrinsic region width of 25×10-6μm.
a) 12.5×104
b) 11.5×104
c) 14.5×104
d) 13.5×104

Explanation: The drift time is given by
tdrift = w/vd
vd = 25×10-6/2×10-10 = 12.5×104ms-1.

7. Compute junction capacitance for a p-i-n photodiode if it has area of 0.69×10-6m2, permittivity of 10.5×10-13Fcm-1 and width of 30μm.
a) 3.043×10-5
b) 2.415×10-7
c) 4.641×10-4
d) 3.708×10-5

Explanation: The junction capacitance is given by,
Cj = εsA/w = 10.5×10-13×0.69×10-6/30×10-13
= 2.415×10-7F.

8. Determine the area where permittivity of material is 15.5×10-15Fcm-1 and width of 25×10-6 and junction capacitance is 5pF.
a) 8.0645×10-5
b) 5.456×10-6
c) 3.0405×10-2
d) 8.0645×10-3

Explanation: The junction capacitance is given by,
Cj = εsA/ w = 5×10-12×25×10-6/15.5×10-15
= 8.0645×10-3m2.

9. Compute intrinsic region width of p-i-n photodiode having junction capacitance of 4pF and material permittivity of 16.5×10-13Fcm-1 and area of 0.55×10-6m2.
a) 7.45×10-6
b) 2.26×10-7
c) 4.64×10-7
d) 5.65×10-6

Explanation: The junction capacitance is given by,
Cj = εsA/ W
w = εsA/Cj
= 16.5×10-13 × 0.55×10-6/4×10-12
= 2.26×10-7.

10. Determine permittivity of p-i-n photodiode with junction capacitance of 5pF, area of 0.62×10-6m2 and intrinsic region width of 28 μm.
a) 7.55×10-12
b) 2.25×10-10
c) 5×10-9
d) 8.5×10-12

Explanation: The junction capacitance is given by,
Cj = εsA/ W
εs = Cj w/A = 5×10-12×28×10-6/0.62×10-6
= 2.25×10-10Fcm-1.

11. Determine response time of p-i-n photodiode if it has 3 dB bandwidth of 1.98×108Hz.
a) 5.05×10-6sec
b) 5.05×10-7Sec
c) 5.05×10-7sec
d) 5.05×10-8Sec

Explanation: The maximum response time is
Maximum response time = 1/Bm = 1/1.98×108 = 5.05×10-9sec.

12. Compute maximum 3 dB bandwidth of p-i-n photodiode if it has a max response time of 5.8 ns.
a) 0.12 GHz
b) 0.14 GHz
c) 0.17 GHz
d) 0.13 GHz

Explanation: The maximum response time is
Maximum response time = 1/Bm
= 1/5.8×10-9 = 0.17 GHz.

13. Determine maximum response time for a p-i-n photodiode having width of 28×10-6m and carrier velocity of 4×104ms-1.
a) 105.67 MHz
b) 180.43 MHz
c) 227.47 MHz
d) 250.65 MHz

Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = Vd/2ΠW = 4×10-4/2×3.14×28×10-6 = 227.47 MHz.

14. Determine carrier velocity of a p-i-n photodiode where 3dB bandwidth is1.9×108Hz and depletion region width of 24μm.
a) 93.43×10-5
b) 29.55×10-3
c) 41.56×10-3
d) 65.3×10-4

Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = Vd/2ΠW
Vd = Bm × 2Π × W
= 1.98×108×2Π×24×10-6
= 29.55×10-3.

15. Compute depletion region width of a p-i-n photodiode with 3dB bandwidth of 1.91×108and carrier velocity of 2×104ms-s.
a) 1.66×10-5
b) 3.2×10-3
c) 2×10-5
d) 2.34×104

Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = Vd/2ΠW
W = Vd/Bm2Π
= 2×10-5/1.91×108×2Π
= 1.66×10-5m.

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