This set of Optical Communications Multiple Choice Questions & Answers (MCQs) focuses on “Receiver Noise”.

1. Which are the two main sources of noise in photodiodes without internal gain?
a) Gaussian noise and dark current noise
b) Internal noise and external noise
c) Dark current noise & Quantum noise
d) Gaussian noise and Quantum noise

Explanation: The two main sources of noise in photodiodes without internal gain are dark current noise and quantum noise. They are regarded as shot noise on the photocurrent. These noise are together called as analog quantum noise.

2. The dominating effect of thermal noise over the shot noise in photodiodes without internal gain can be observed in wideband systems operating in the range of ________
a) 0.4 to 0.5 μm
b) 0.8 to 0.9 μm
c) 0.3 to 0.4 μm
d) 0.7 to 0.79 μm

Explanation: When the photodiode is without internal avalanche gain, the detector load resistor and active elements’ thermal noise in the amplifier tends to dominate. It is seen in wideband systems operating in the 0.8 to 0.9 μmwavelength band. This is because the dark currents in the silicon diodes can be made very small.

3. A silicon p-i-n photodiode incorporated in an optical receiver has following parameters:

```Quantum efficiency = 70%
Wavelength = 0.8 μm
Dark current = 3nA
Incident optical power = 150nW.
Bandwidth = 5 MHz
```

Compute the photocurrent in the device.
a) 67.7nA
b) 81.2nA
c) 68.35nA
d) 46.1nA

Explanation: The photocurrent is given by
Ip = ηP0eλ/hc
Where η = Quantum efficiency
P0 = Incident optical power
e = electron charge
λ = Wavelength
h = Planck’s constant
c = Velocity of light.

4. In a silicon p-i-n photodiode, if load resistance is 4 kΩ, temperature is 293 K, bandwidth is 4MHz, find the thermal noise in the load resistor.
a) 1.8 × 10-16A2
b) 1.23 × 10-17A2
c) 1.65 × 10-16A2
d) 1.61 × 10-17A2

Explanation: The thermal noise in the load resistor is given by –
it2 = 4KTB/RL
Where T = Temperature
B = Bandwidth
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5. ________________ is a combination of shunt capacitances and resistances.
a) Attenuation
b) Shunt impedance
d) Thermal capacitance

Explanation: Admittance is a measure of how easily a circuit will allow a current to flow. It is the inverse of impedance and is measured in Siemens. It is a combination of shunt capacitances and resistances.

6. ______________ is used in the specification of optical detectors.
a) Noise equivalent power
b) Polarization
c) Sensitivity
d) Electron movement

Explanation: Noise equivalent power is defined as the amount of incident optical power per unit bandwidth required to produce an output power equal to detector output noise power.
Noise equivalent power is the value of incident power which gives an output SNR of unity.

7. A photodiode has a capacitance of 6 pF. Calculate the maximum load resistance which allows an 8MHz post detection bandwidth.
a) 3.9 kΩ
b) 3.46 kΩ
c) 3.12 kΩ
d) 3.32 kΩ

Explanation: The load resistance is given by-
RL = 1/2πCdB
Where
B = Post detection bandwidth
Cd = Input capacitance

8. The internal gain mechanism in an APD is directly related to SNR. State whether the given statement is true or false.
a) True
b) False

Explanation: The internal gain mechanism in an APD increases the signal current into the amplifier. This improves the SNR because the load resistance and amplifier noise remains unaffected.

9. ____________ is dependent upon the detector material, the shape of the electric field profile within the device.
a) SNR
b) Excess avalanche noise factor
d) Noise power

Explanation: Excess avalanche noise factor is represented as F (M). Its value depends upon the detector material, shape of electric field profile and holes and electrons inclusion. It is a function of multiplication factor.

10. For silicon APDs, the value of excess noise factor is between _________
a) 0.001 and 0.002
b) 0.5 and 0.7
c) 0.02 and 0.10
d) 1 and 2

Explanation: The excess noise factor (K) is same as that of the multiplication factor. In case of holes, the smaller values of K produce high performance and therefore the performance is achieved when k is small. For silicon APDs, k = 0.02 to 0.10.

11. __________ determines a higher transmission rate related to the gain of the APD device.
a) Attenuation
b) Gain-bandwidth product
c) Dispersion mechanism
d) Ionization coefficient

Explanation: Gain-bandwidth product is defined as Gain multiplied by the bandwidth. Gain is a dimensionless quantity but the gain-bandwidth product is therefore measured in the units of frequency.

12. _________________ APDs are recognized for their high gain-bandwidth products.
a) GaAs
c) Germanium
d) Silicon

Explanation: Silicon APDs possess a large asymmetry of electron and hole ionization coefficient. Thus, they possess high gain-bandwidth products. These APDs do not operate at high transmission rates.

13. APDs do not operate at signal wavelengths between 1.3 and 1.6μm.
a) True
b) False

Explanation: APDs having high gain-bandwidth products do not operate at signal wavelengths between 1.3 and 1.6 μm.Hence, these APDs are not prefered for use in receivers operating at high transmission rates.

Sanfoundry Global Education & Learning Series – Optical Communications.

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