Optical Detectors Questions and Answers

Q: The bandwidth of a photodetector is defined as :

The bandwidth is \([2\pi(\tau_{tr} + \tau_{RC})]^{-1}\).The response speed of a photodetector is limited by the transit time of carriers across the depletion region, \(\tau_{tr}\), and by the RC time constant of the diode capacitance and load, \(\tau_{RC}\). Treating their sum as a single time constant gives a first-order cut-off frequency\[ B = \frac{1}{2\pi(\tau_{tr} + \tau_{RC})} \]The other options are not a frequency: one subtracts 1 from a time, one is the time constant itself, and one adds a dimensionless 1 to a time before inverting.

Q: For a typical p-n photodiode the width of depletion layer (W) is 10 µm and drift velocity is \(10^{5}\) m/s, then what is the transit time ?

The transit time is 100 ps (printed 'PS').The transit time is the depletion width divided by the drift velocity:\[ \begin{aligned} t &= \frac{W}{v_d} = \frac{10 \times 10^{-6}}{10^{5}} \\ &= 10^{-10}\ \text{s} = 100\ \text{ps} \end{aligned} \]10 ps and 1000 ps come from slipping a power of ten in converting micrometres to metres or seconds to picoseconds, and 10000 ps is two powers of ten too large.

The bandwidth of a photodetector is defined as :

A.\([2\pi(\tau_{tr} + \tau_{RC})] - 1\)
B.\([2\pi(\tau_{tr} + \tau_{RC})]^{-1}\)
C.\(2\pi(\tau_{tr} + \tau_{RC})\)
D.\((2\pi(\tau_{tr} + \tau_{RC}) + 1)^{-1}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The bandwidth is \([2\pi(\tau_{tr} + \tau_{RC})]^{-1}\).

The response speed of a photodetector is limited by the transit time of carriers across the depletion region, \(\tau_{tr}\), and by the RC time constant of the diode capacitance and load, \(\tau_{RC}\). Treating their sum as a single time constant gives a first-order cut-off frequency

\[ B = \frac{1}{2\pi(\tau_{tr} + \tau_{RC})} \]

The other options are not a frequency: one subtracts 1 from a time, one is the time constant itself, and one adds a dimensionless 1 to a time before inverting.

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For a typical p-n photodiode the width of depletion layer (W) is 10 µm and drift velocity is \(10^{5}\) m/s, then what is the transit time ?

A.10 PS
B.10000 PS
C.100 PS
D.1000 PS

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The transit time is 100 ps (printed 'PS').

The transit time is the depletion width divided by the drift velocity:

\[ \begin{aligned} t &= \frac{W}{v_d} = \frac{10 \times 10^{-6}}{10^{5}} \\ &= 10^{-10}\ \text{s} = 100\ \text{ps} \end{aligned} \]

10 ps and 1000 ps come from slipping a power of ten in converting micrometres to metres or seconds to picoseconds, and 10000 ps is two powers of ten too large.

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Following are the detectors used in optical communication :

(i) Photo diode
(ii) Photo-transistor
(iii) Avalanche photodiode
(iv) Pin diode
Correct sequence of gain in decreasing order is given by

A.(iii), (iv), (ii), (i)
B.(iv), (iii), (ii), (i)
C.(i), (ii), (iii), (iv)
D.(iv), (ii), (iii), (i)

Exam Year(s)

UGC NET - 2012,

Answer: A

The key gives the detectors in decreasing order of gain as the avalanche photodiode, the PIN diode, the photo transistor and the photodiode, which is (iii), (iv), (ii), (i). The avalanche photodiode has internal gain from avalanche multiplication, so it is first. The other options place the photodiode or the PIN diode ahead of the avalanche device, or reverse the order, so they do not agree with the key. I could not check the relative gains of the PIN diode and the photo transistor in a standard text, and the order of these two rests on the key. The answer is (iii), (iv), (ii), (i).

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Direction for [ Question No: 4 To 4 ] :

Assertion – Reason type questions :
The following items consist of two statements, one labelled the ‘Assertion (A)’ and the other labelled the ‘Reason (R)’. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Codes :
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(C) (A) is true and (R) is false.
(D) (A) is false and (R) is true.

Assertion (A) : PIN diodes are used as photo detectors. This is achieved by doping lightly doped in material in between p and n junctions.

Reason (R) : These are used because higher frequency infra-red penetrate more deeply in the semiconductor material.

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true but (R) is not the correct explanation of (A).
C.(A) is true and (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2012,

Answer: C

The assertion is true: a PIN photodiode has a lightly doped (intrinsic) layer between the p and n regions, which widens the depletion region so that more light is absorbed and the response is faster. The key regards the reason as false. The reason says that infra-red penetrates more deeply at higher frequency, but light of longer wavelength, which is lower frequency, penetrates more deeply before it is absorbed, so the statement as written is not correct, and it is not the reason why the intrinsic layer is added. The answer is that (A) is true and (R) is false.

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Photo diode is reverse biased because

A.only one side is illuminated
B.majority carriers flow under these conditions.
C.reverse current is small as compared to photo current
D.reverse current is large as compared to photo current

Exam Year(s)

UGC NET - 2013,

Answer: C

A photodiode is operated in reverse bias because the reverse current is small compared with the photo current, option (C). In the dark only a small leakage current flows, so the current produced by incident light dominates and the device gives a clear signal that is nearly proportional to the light.

Illumination of only one side (option A) is not a reason, and the majority carriers flow in forward bias, not in reverse bias (option B). Option D says the reverse current is large as compared with the photo current, which would bury the signal in noise and is the opposite of the actual situation.

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The carrier velocity in a silicon p-i-n photodiode with 25 µm depletion layer width is \(3 \times 10^{4}\) m/s. The maximum response time for the pin diode is given by :

A.52 ns
B.2.5 ns
C.62 ns
D.5.2 ns

Exam Year(s)

UGC NET - 2014,

Answer: D

The keyed answer is 5.2 ns. The transit time across the depletion layer is the width divided by the carrier velocity:

\[ t_{tr} = \frac{25 \times 10^{-6}}{3 \times 10^{4}} \]

\[ t_{tr} \approx 0.83\ \text{ns} \]

The official key takes the response time as \(2\pi\) times this transit time:

\[ 2\pi \times 0.83\ \text{ns} \approx 5.2\ \text{ns} \]

Of the four options only 5.2 ns is on the scale of this calculation. The values 52 ns and 62 ns are about ten times too large, and 2.5 ns does not follow from the figures given. The printed velocity is used as it stands.

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Direction for [ Question No: 7 To 7 ] :

Read the paragraph and answer the questions :

p-i-n photodiode contains a layer of intrinsic semi-conductor material sandwiched between p-and-n regions. The depletion layer is wholly contained within the i region. Thickness of the intrinsic region can be adjusted to produce device with optimum sensitivity and frequency response. P-i-n photodiode is most common type of depletion layer photodiode. The other class of photodiodes, avalanche photodiodes, are reverse-biased p-n junction diodes that are operated at voltages above the breakdown voltage. Current multiplication of electron-hole pairs generated by the incident electromagnetic radiation, occurs due to avalanche process. The photomultiplication factor Mpn is defined as ratio of the multiplied photocurrent Ipm to the photocurrent Ipho at voltage below breakdown where no avalanche multiplication takes place.

Photo-diode is reverse biased because

A.only one side is illuminated
B.majority swept are reverse biased across the function
C.reverse current is small as compared to photo current
D.reverse current is large as compared to photocurrent

Exam Year(s)

UGC NET - 2013,

Answer: C

A photodiode is operated in reverse bias because the reverse current is small compared with the photo current, so that the light-generated current can be measured easily above the dark current. The drift of carriers in the strong field also increases the response speed. The other options give a wrong reason, such as illumination of one side only or a large reverse current. So the third option is correct.

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Direction for [ Question No: 8 To 8 ] :

Assertion-Reason type questions Q. 41 to 50 : The following items consist of two statements, one labelled as 'Assertion (A)' and the other labelled the 'Reason (R)'. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Assertion (A) : PIN diodes have wider depletion region, in order to allow light to penetrate more deeply in the semiconductor material.

Reason (R) : III-V semiconductor materials are used to fabricate in order to allow shorter wavelength in the region of 400-700 nm.

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.(A) is true but (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2013,

Answer: C

The Assertion is true and the Reason is false, option (C). A PIN diode has an intrinsic layer, which gives a wider depletion region and allows light to penetrate more deeply into the semiconductor.

The Reason is wrong, because III-V materials are used to make devices for longer wavelengths in the infrared, which silicon does not cover, and not for shorter wavelengths of 400 to 700 nm. Options A and B need a true Reason and option D a false Assertion.

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Direction for [ Question No: 9 To 10 ] :

Read the following paragraph and answer the questions :

Gunn effect is instrumental in the generation of microwave oscillations in bulk semiconductor material. The effect was exhibited by gallium arenide. The effect is a bulk property of semiconductors. If a small dc voltage is placed across a thin slice of germanium arsanide a negative resistance will manifest itself under certain conditions. The voltage gradient across the slice is increase of about 3300 v/cm, oscillations will then occur at microwave frequencies if the slice is connected to a suitably tuned circuit. The Gunn effect is independent of total voltage or current and is not effected by magnetic field, it occurs in n-type of materials only. The PIN diode consists of a narrow layer of p-type semiconductor separated from an equally narrow layer of n type material. In PIN diode silicon tend to be the main material. The PIN diode is used for microwave power switching, limiting and modulation.

Response time of PIN diode is of the order of

A.0.1 ns
B.1 ns
C.10 ns
D.1 milli.sc

Exam Year(s)

UGC NET - 2013,

Answer: B

The response time of a PIN diode is of the order of 1 ns, option (B), which is the keyed value. The thick intrinsic layer limits the switching speed, which is why such a diode is suited to microwave switching.

0.1 ns (option A) is faster than a typical PIN diode, and 10 ns (option C) is slower. A response time of 1 millisecond (option D) is far too slow for microwave switching, which is what the passage describes as its use.

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10.

PIN diode is used as

A.sine wave modulator
B.triangular wave modulator
C.square wave modulator
D.low frequency rectifier

Exam Year(s)

UGC NET - 2013,

Answer: D

The keyed answer is option (D), low frequency rectifier. At low frequencies a PIN diode behaves like an ordinary rectifying junction, because the stored charge in its intrinsic layer has time to follow the signal, whereas at microwave frequencies it acts as a variable resistor.

The passage says the PIN diode is used for microwave power switching, limiting and modulation, but the options give sine wave, triangular wave and square wave modulators, which are waveform names and not these uses. The key accepts D, and a reviewer may wish to check this item.

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Optical Detectors Questions and Answers

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