Most Competitive Optical Communication MCQs – Optical Detectors ( Optical Communication ) MCQs
Latest Optical Communication MCQs
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Optical Communication MCQs – Optical Detectors ( Optical Communication ) MCQs
The most occurred mcqs of Optical Detectors ( ) in past papers. Past papers of Optical Detectors ( Optical Communication ) Mcqs. Past papers of Optical Detectors ( Optical Communication ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Optical Detectors ( Optical Communication ) Mcqs. The Important series of Optical Detectors ( Optical Communication ) Mcqs are given below:
Device Types
1. ____________ converts the received optical signal into an electrical signal.
a) Detector
b) Attenuator
c) Laser
d) LED
Answer: a
Explanation: A detector is an essential component of an optical fiber communication system. It dictates the overall system performance. Its function is to convert optical signal into an electrical signal. This electrical signal is then amplified before further processing.
2. The first generation systems of optical fiber communication have wavelengths between ___________
a) 0.2 and 0.3 μm
b) 0.4 and 0.6 μm
c) 0.8 and 0.9 μm
d) 0.1 and 0.2 μm
Answer: c
Explanation: The first generation systems operated at a bit-rate of 45 Mbps with repeater spacing of 10 km. It operates at wavelengths between 0.8 and 0.9μm. These wavelengths are compatible with AlGaAs laser and LEDs.
3. The quantum efficiency of an optical detector should be high.
a) True
b) False
Answer: a
Explanation: The detector must satisfy stringent requirements for performance and compatibility. The photo detector thus produces a maximum electrical signal for a given amount of optical power; i.e. the quantum efficiency should be high.
4. Which of the following does not explain the requirements of an optical detector?
a) High quantum efficiency
b) Low bias voltages
c) Small size
d) Low fidelity
Answer: d
Explanation: The size of the detector must be small for efficient coupling to the fiber. Also, ideally, the detector should not require excessive bias voltages and currents. The fidelity and quantum efficiency should be high.
5. How many device types are available for optical detection and radiation?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Two types of devices are used for optical detection and radiation. These are external photoemission and internal photoemission devices. External photoemission devices are too bulky and require high voltages for operation. Internal devices provide good performance and compatibility.
6. The ___________ process takes place in both extrinsic and intrinsic semiconductors.
a) Avalanche multiplication
b) External photoemission
c) Internal photoemission
d) Dispersion
Answer: c
Explanation: During intrinsic absorption, the received photons excite electrons from the valence band and towards the conduction band in the semiconductor. Extrinsic absorption involves impurity centers created with the material. Generally, intrinsic absorption is preferred for internal photoemission.
7. ____________ are widely used in first generation systems of optical fiber communication.
a) p-n diodes
b) 4-alloys
c) 3-alloys
d) Silicon photodiodes
Answer: d
Explanation: The first generation systems operates at wavelengths 0.8 and 0.9 μm. Silicon photodiodes have high sensitivity over the 0.8-0.9 μm wavelength band with adequate speed, long term stability. Hence, silicon photodiodes are widely used in first generation systems.
8. Silicon has indirect band gap energy of __________________
a) 1.2 eV
b) 2 eV
c) 1.14 eV
d) 1.9 eV
Answer: c
Explanation: Silicon’s indirect band gap energy of 1.14 eV gives a loss in response above 1.09μm. To avoid this, narrower bandgap materials are used. Hence, silicon’s usefulness is limited to first generation systems and not for second and third generation systems.
9. Which of the following detector is fabricated from semiconductor alloys?
a) Photoconductive detector
b) p-i-n detector
c) Photodiodes
d) Photoemission detectors
Answer: a
Explanation: The detectors fabricated from semiconductor alloys can be used for longer wavelengths. Photoconductive detector and hetero-junction transistor have found favor as a potential detector over a wavelength range of 1.1 to 1.6μm.
10. Silicon photodiodes provide high shunt conductance.
a) True
b) False
Answer: b
Explanation: Semiconductor photodiodes provide best solution for detection in optical fiber communications. Silicon photodiodes have high sensitivity, negligible shunt conductance and low dark current.
Optical Detection Principles
1. P-n photodiode is forward biased.
a) True
b) False
Answer: b
Explanation: p-n photodiode includes p and n regions. The electric field developed across the p-n junction sweeps holes and electrons to p and n regions respectively. P-n photodiode is thus reverse biased due to reverse leakage current.
2. The depletion region must be ____________ to allow a large fraction of the incident light to be absorbed in the device(photodiode).
a) Thick
b) Thin
c) Long
d) Inactive
Answer: a
Explanation: In p-n photodiode, intrinsic conditions are created in the depletion region. The depletion region must be thick in order to achieve maximum carrier pair generation. Also, its width must be limited to enhance the speed of operation of the p-n photodiode.
3. The process of excitation of an electron from valence band to conduction band leaves an empty hole in the valence band and is called as ____________
a) Detection
b) Absorption
c) Degeneration of an electron-hole pair
d) Regeneration of an electron-hole pair
Answer: d
Explanation: A photon is incident in the depletion region of a device has an energy greater than or equal to the band gap energy of the fabricating material. This will cause excitation of an electron from valence to the conduction band. This creates an empty hole in valence band which is referred to as photo-generation of an electron-hole pair.
4. __________________ always leads to the generation of a hole and an electron.
a) Repulsion
b) Dispersion
c) Absorption
d) Attenuation
Answer: c
Explanation: Absorption affects the electron and excites it to some other level say conduction band. This is called as photo-generation as absorption always leads to the generation of hole and electron. This does not mean that both contribute to the electronic transport.
5. The electron hole pairs generated in a photodiode are separated by the ____________
a) Magnetic field
b) Electric field
c) Static field
d) Depletion region
Answer: b
Explanation: Electric field separates the electron-hole pairs in a photodiode. The electric field distribution is determined by an internal and an external field component. A reverse bias voltage is usually applied to the p-n photodiode.
6. Electric field in the depletion region should be high.
a) True
b) False
Answer: a
Explanation: The electric field in the depletion region is always kept high in order to extract all photogenerated carriers. Only the extracted electron hole pairs contribute to the overall photocurrent.
7. The photocurrent of an optical detector should be __________
a) Less
b) More
c) Linear
d) Non-linear
Answer: c
Explanation: A linear relationship must exist between the intensity of the incident light and the photocurrent. This makes the photodiode free of noise. It increases system performance.
8. How many types of optical detectors are available?
a) One
b) Four
c) Two
d) Three
Answer: d
Explanation: Three types of optical detectors are available. These are diodes, photoconductors and photo-transistors. Diodes include p-n photodiodes, p-i-n diodes, avalanche photodiodes and schottky diodes.
Absorption
1. The absorption of photons in a photodiode is dependent on __________
a) Absorption Coefficient α0
b) Properties of material
c) Charge carrier at junction
d) Amount of light
Answer: a
Explanation: Absorption in a photodiode is for producing carrier pans. Thus, photocurrent is dependent on absorption coefficient α 0of the light in semiconductor used to fabricate device.
2. The photocurrent in a photodiode is directly proportional to absorption coefficient.
a) True
b) False
Answer: a
Explanation: The absorption of photons produces carrier pairs. Thus, photocurrent is dependent on absorption coefficient and is given by
I = Po e(1-h)/hf(1-exp (-α rd))
Where r = Fresnel coefficient
D = width of absorption region.
3. The absorption coefficient of semiconductor materials is strongly dependent on __________
a) Properties of material
b) Wavelength
c) Amount of light
d) Amplitude
Answer: b
Explanation: In some common semiconductors, there is a variation in absorption curves for materials. It is found that they are each suitable for different wavelength and related applications. This is due to difference in band gap energies. Thus absorption coefficient depends on wavelength.
4. Direct absorption requires assistance of photon.
a) True
b) False
Answer: b
Explanation: Indirect absorption requires photon assistance resulting in conversation of energy and momentum. This makes transition probability less likely for indirect absorption than direct absorption where no photon is included.
5. In optical fiber communication, the only weakly absorbing material over wavelength band required is?
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: c
Explanation: The transition over wavelength band in silicon is due to indirect absorption mechanism. This makes silicon weakly absorbent over particular wavelength band.
6. The threshold for indirect absorption occurs at wavelength __________
a) 3.01 μm
b) 2.09 μm
c) 0.92 μm
d) 1.09 μm
Answer: d
Explanation: The band gap for silicon is 4.10 eV corresponding to threshold of 0.30 μm in ultraviolet. Thus it’s outside wavelength range is the one which is required.
7. The semiconductor material for which the lowest energy absorption takes place is?
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: d
Explanation: Germanium absorption is by indirect optical transition. The threshold for direct absorption is at 1.53μm. Below this, germanium becomes strongly absorbing to corresponding link.
8. The wavelength range of interest for Germanium is __________
a) 0.8 to 1.6 μm
b) 0.3 to 0.9 μm
c) 0.4 to 0.8 μm
d) 0.9 to 1.8 μm
Answer: a
Explanation: Germanium is used in fabrication of detectors over the whole wavelength range i.e. first and second generation 0.8 to 1.6 μm while specially taking into consideration that indirect absorption will occur up to a threshold of 1.85 μm.
9. A photodiode should be chosen with a ________________ less than photon energy.
a) Direct absorption
b) Band gap energy
c) Wavelength range
d) Absorption coefficient
Answer: d
Explanation: A photodiode selection must be made by choosing that diode having band gap energy less than photon energy corresponding to longest operating wavelength. This provides high absorption coefficient which ensures a good response and limits the thermally generated carriers to obtain low dark current with no incident light.
10. ________________ photodiodes have large dark currents.
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: c
Explanation: Germanium photodiodes provide narrow band gaps as compared to other semiconductor materials. This is main disadvantage with use of germanium photodiodes at shorter wavelength and thus they have large dark current.
11. For fabrication of semiconductor photodiodes, there is a drawback while considering _________________
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: d
Explanation: Due to drawback with germanium to be used as fabricating material, there
is an increased investigation of direct band gap III and V alloys for longer wavelength region.
12. _________________ materials are potentially superior to germanium.
a) GaAs
b) Silicon
c) GaSb
d) III – V alloys
Answer: d
Explanation: The band gap energies for III – V alloys materials can be tailored to required wavelength. This can be achieved by changing relative concentration of their constituents which results in low dark currents. Thus, III – V alloys are superior potentially to germanium.
13. ____________ alloys such as InGaAsP and GaAsSb deposited on InP and GaSb substrate.
a) Ternary
b) Quaternary
c) Gain-guided
d) III – V alloys
Answer: a
Explanation: Ternary alloys are used to fabricate photodiodes for longer wavelength band. Thus, these alloys such as InGaAsP and GaAsSb are deposited on InP and GaSb substrates.
14. _________________ alloys can be fabricated in hetero-junction structures.
a) InGaSb
b) III – V alloys
c) InGaAsP
d) GaAsSb
Answer: b
Explanation: III – V alloys enhances the high speed operations of hetero-junction structures. Thus these structures can be fabricated with III-V alloys.
15. The alloys lattice matched to InP responds to wavelengths up to 1.7μm is?
a) InAsSb
b) III – V alloys
c) InGaSb
d) InGaAs
Answer: d
Explanation: Although there were difficulties in growth of IOnGaAs alloys, the problems are now reduced. These alloys lattice matched to InP responding to wavelength around 1.7 μmare widely utilized for fabrication of photodiodes operating around 1.7μm.
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
Answer: a
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
Answer: b
Explanation: While considering intrinsic absorption process, the energy of incident photon must be greater than band gap energy of material fabricating photo detector.
Optical Source : The Laser And Light – Emitting Diode MCQs
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
Answer: b
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
Answer: d
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
Answer: a
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.
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%
Answer: b
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
Answer: b
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
Answer: c
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
Answer: a
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
Answer: d
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
Answer: c
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
Answer: b
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
Answer: d
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
Answer: c
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
Answer: b
Explanation: The Responsivity of photodiode is given by
R = Ip/Po
Ip = R*Po
= 0.70*3.51*10-6
= 2.457μm.
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
Answer: a
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
Answer: c
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
Answer: b
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
Answer: c
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
Answer: b
Explanation: The drift time is given by
tdrift = w/vd
4×10-10 = w/2×105
= 4×10-10×2×105
= 8×10-5m.
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
Answer: a
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
Answer: b
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
Answer: d
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
Answer: b
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
Answer: b
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
Answer: c
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
Answer: c
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
Answer: c
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
Answer: b
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
Answer: a
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.
Semiconductor Photodiodes With Internal Gain
1. ___________ has more sophisticated structure than p-i-n photodiode.
a) Avalanche photodiode
b) p-n junction diode
c) Zener diode
d) Varactor diode
Answer: a
Explanation: Avalanche photodiode is second major type of detector in optical communications. This diode is more sophisticated so as to create a much higher electric field region.
2. The phenomenon leading to avalanche breakdown in reverse-biased diodes is known as _______
a) Auger recombination
b) Mode hopping
c) Impact ionization
d) Extract ionization
Answer: c
Explanation: In depletion region, almost all photons are absorbed and carrier pairs are generated. So there comes a high field region where carriers acquire energy to excite new carrier pairs. This is impact ionization.
3. _______ is fully depleted by employing electric fields.
a) Avalanche photodiode
b) P-I-N diode
c) Varactor diode
d) P-n diode
Answer: a
Explanation: APD is fully depleted by electric fields more than 104V/m. This causes all the drifting of carriers at saturated limited velocities.
Optical Fiber Connections : Joints, Couplers And Isolators MCQs
4. At low gain, the transit time and RC effects ________
a) Are negligible
b) Are very less
c) Dominate
d) Reduce gradually
Answer: c
Explanation: Low gain causes the dominance of transit time and RC effects. This gives a definitive response time and thus device obtains constant bandwidth.
5. At high gain, avalanche buildup time ________
a) Is negligible
b) Very less
c) Increases gradually
d) Dominates
Answer: d
Explanation: High gain causes avalanche buildup time to dominate. Thus the bandwidth of device decreases as increase in gain.
6. Often __________ pulse shape is obtained from APD.
a) Negligible
b) Distorted
c) Asymmetric
d) Symmetric
Answer: c
Explanation: Asymmetric pulse shape is acquired from APD. This is due to relatively fast rise time as electrons are collected and fall time dictated by transit time of holes.
7. Fall times of 1 ns or more are common.
a) False
b) True
Answer: b
Explanation: The use of suitable materials and structures give rise times between 150 and 200 ps. Thus fall times of 1 ns or more are common which in turn limits the overall response of device.
8. Determine Responsivity of a silicon RAPD with 80% efficiency, 0.7μm wavelength.
a) 0.459
b) 0.7
c) 0.312
d) 0.42
Answer: a
Explanation: The Responsivity of a RAPD is given by-
R = ηeλ/hc A/w where, η=efficiency, λ = wavelength, h = Planck’s constant.
9. Compute wavelength of RAPD with 70% efficiency and Responsivity of 0.689 A/w.
a) 6μm
b) 7.21μm
c) 0.112μm
d) 3μm
Answer: c
Explanation: The wavelength can be found from the Responsivity formula given by-
R = ηeλ/hc. The unit of wavelength isμm.
10. Compute photocurrent of RAPD having optical power of 0.7 μw and responsivity of 0.689 A/W.
a) 0.23 μA
b) 0.489 μA
c) 0.123 μA
d) 9 μA
Answer: b
Explanation: The photocurrent is given byIP=P0R. Here IP = photocurrent, P0=Power, R = responsivity.
11. Determine optical power of RAPD with photocurrent of 0.396 μAand responsivity of 0.49 A/w.
a) 0.91 μW
b) 0.32 μW
c) 0.312 μW
d) 0.80 μW
Answer: d
Explanation: The photocurrent is given by IP = P0R. Here IP = photocurrent, P0 = Power, R = responsivity.
P0 = IP/R gives the optical power.
12. Determine the Responsivity of optical power of 0.4μW and photocurrent of 0.294 μA.
a) 0.735
b) 0.54
c) 0.56
d) 0.21
Answer: a
Explanation: The photocurrent is given by IP = P0R. Here IP = photocurrent, P0 = Power, R = responsivity.
R = IP/P0 gives the responsivity.
13. Compute multiplication factor of RAPD with output current of 10 μAand photocurrent of 0.369μA.
a) 25.32
b) 27.100
c) 43
d) 22.2
Answer: b
Explanation: The multiplication factor of photodiode is given by-
M = I/IP where I = output current, IP = photocurrent.
14. Determine the output current of RAPD having multiplication factor of 39 and photocurrent of 0.469μA.
a) 17.21
b) 10.32
c) 12.21
d) 18.29
Answer: d
Explanation: The multiplication factor of photodiode is given by-
M = I/IP where I = output current, IP = photocurrent. I = M*IP gives the output current inμA.
15. Compute the photocurrent of RAPD having multiplication factor of 36.7 and output current of 7μA.
a) 0.01 μA
b) 0.07 μA
c) 0.54 μA
d) 0.9 μA
Answer: a
Explanation: The multiplication factor of photodiode is given by-
M = I/IP where I = output current, IP = photocurrent. IP = I/M Gives the output current inμA.
Mid Infrared and Far Infrared Photodiodes
1. In the development of photodiodes for mid-infrared and far-infrared transmission systems, lattice matching has been a problem when operating at wavelengths ____________
a) 1 µm
b) Greater than 2 µm
c) 2 µm
d) 0.5 µm
Answer: b
Explanation: Lattice matching for alloy materials is obtained at wavelengths above 2 µm. For example, a lattice-matched alloy material system (GaSb) was utilized in a p-i-n photodiode for high speed operation at wavelengths up to 2.3 µm.
2. What is generally used to accommodate a lattice mismatch?
a) Alloys
b) Attenuator
c) Graded buffer layer
d) APD array
Answer: c
Explanation: The use of indium alloy cause inherent problems of dislocation-induced junction leakage and low quantum efficiency. To avoid these problems, a compositionally graded buffer layer is used to accommodate a lattice mismatch.
3. HgCdTe material system is utilized to fabricate long-wavelength photodiodes.
a) True
b) False
Answer: a
Explanation: HgCdTe family alloys allow resonant characteristics via hole ionization. Its band gap energy variation enables optical detection to far-infrared. Thus, this material can be used for fabrication of long-wavelength photodiodes.
4. Avalanche photodiodes based on HgCdTe are used for ______________ in both the near and far infrared.
a) Dispersion
b) Dislocation
c) Ionization
d) Array applications
Answer: d
Explanation: Avalanche photodiodes based on HgCdTe are used for array applications. The materials of APDs based on HgCdTe possess uniform avalanche gain across an array. This variation in gain is variation in gain is lower in HgCdTe as compared with silicon.
5. The detection mechanism in ____________ relies on photo excitation of electrons from confined states in conduction band quantum wells.
a) p-i-n detector
b) Quantum-dot photo detector
c) p-n photodiode
d) Avalanche photodiodes
Answer: b
Explanation: Quantum-dot photo detector’s detection mechanism involves photo excitation of electrons. This process of photo excitation in photo detectors is similar to that in the Quantum-dot semiconductor optical amplifier. The dots-in-well in Quantum-dot detector is called as DWELL structure.
6. When determining performance of a photo detector ___________ is often used.
a) No. of incident photon
b) No. of electrons collected
c) Responsivity
d) Absorption coefficient
Answer: c
Explanation: The expression for quantum efficiency does not include photon energy. Thus for characterizing performance of photo detector, Responsivity is used.
7. The important parameter for exciting an electron with energy required from valence band to conduction band is?
a) Wavelength
b) Absorption coefficient
c) Responsivity
d) Band gap energy
Answer: a
Explanation: As wavelength of incident photon becomes longer, the photon energy is less than energy required to excite electron. Mostly parameters of photodiode are dependent on wavelength.
8. __________ is less than or unity for photo detectors.
a) Absorption coefficient
b) Band gap energy
c) Responsivity
d) Quantum efficiency
Answer: d
Explanation: Quantum efficiency determines the absorption coefficient of semiconductor material of photo detector. It is not all incident photons are absorbed to create electron-hole pairs. Thus quantum efficiency must be less than unity.
9. There must be improvement in __________ of an optical fiber communication system.
a) Detector
b) Responsivity
c) Absorption Coefficient
d) Band gap energy
Answer: a
Explanation: If proper and improved and highly efficient detector is utilized, it will then reduce the repeated stations. It will also lower down both capital investment and maintenance cost.
Phototransistors and Metal – Semiconductor – Metal Photodetectors
1. The _____________ is photosensitive to act as light gathering element.
a) Base-emitter junction
b) Base-collector junction
c) Collector-emitter junction
d) Base-collector junction and Base-emitter junction
Answer: a
Explanation: Base-collector junction is photosensitive in n-p-n phototransistor and act as light gathering element. This light absorbed affects the base current and gives multiplication of primary photocurrent in device.
2. A large secondary current _________________ in n-p-n InGaAs phototransistor is achieved.
a) Between base and collector
b) Between emitter and collector
c) Between base and emitter
d) Plasma
Answer: b
Explanation: The photo-generated holes are swept to the base. This increases the forward bias device. This generates secondary current between emitter and collector.
3. _______ emitter-base and collector-base junction capacitances is achieved by use of hetero-structure along with _________ base resistance.
a) Low, high
b) High, low
c) Low, low
d) High, negligible
Answer: c
Explanation: In hetero-structure, there is low doping level in emitter and collector which is coupled with heavy doping base. This is due low emitter-base and collector-base junction capacitance and low base resistance. This allows large current gain.
4. A ________ is created by hetero-junction at collector-base junction.
a) Potential barrier
b) Depletion region
c) Parasitic capacitance
d) Inductance
Answer: a
Explanation: Potential barrier is created at emitter-base junction by hetero-junction. This eliminates hole junction from base. This is achieved when junction is forward-biased and provides good emitter-base efficiency.
5. Phototransistors based on hetero-junction using _________ material are known as waveguide phototransistors.
a) InGaP
b) InGaAs
c) InGaAsP/ InAlAs
d) ErGaAs
Answer: c
Explanation: Phototransistor using InGaAsP/ InAlAs are known as waveguide phototransistors. They function as waveguide phototransistors. They function as high performance photo-detectors at 1.3 micro-meter wavelength. They utilize a passive waveguide layer under active transistor region.
6. A phototransistor has collector current of 18 mA, incident optical power of 128 μW with a wavelength of 1.24 μm. Determine an optical gain.
a) 1.407 *102
b) 19.407 *102
c) 2.407 *102
d) 3.407 *102
Answer: a
Explanation: The optical gain is given by-
G0=hcIc/λeP0, where h=Planck’s constant, Ic=collector current, λ=wavelength, P0=incident optical power.
7. For a phototransistor having gain of 116.5, wavelength of 1.28 μm, optical power 123μW. Determine collector current.
a) 0.123 mA
b) 0.0149 mA
c) 1.23 mA
d) 0.54 mA
Answer: b
Explanation: The collector current is given by-
Ic= G0λeP0/ hc, where h=Planck’s constant, Ic=collector current, λ=wavelength, P0=incident optical power.
8. The detection mechanism in the ____________ photo-detector includes inter sub-band transitions.
a) Dwell
b) Set
c) Avalanche
d) Futile
Answer: a
Explanation: The inter sub-band transitions are also known as type-2 transitions. It comprises of mini-bands within a single energy band, The detection mechanism in DWELL photo-detector includes inter sub-band transitions.
9. Which of the following is the difference between the n-p-n and conventional bipolar transistor?
a) Electric property
b) Magnetic property
c) Unconnected base
d) Emitter base efficiency
Answer: c
Explanation: The n-p-n bipolar transistor differs in the following ways: base is unconnected, base-collector junction is photosensitive as a light gathering element.
10. The n-p-n hetero-junction phototransistor is grown using ______________
a) Liquid-phase tranquilizers
b) Liquid-phase epistaxis
c) Solid substrate
d) Hetero poleax
Answer: b
Explanation: The technique LPE consists of a thin layer of n-type collector based on a p-type base layer. Liquid phase epistaxis is used in hetero-junction technology.
11. The _____________ at emitter-base junction gives good emitter base injection efficiency.
a) Homo-junction
b) Depletion layer
c) Holes
d) Hetero-junction
Answer: d
Explanation: The hetero-junction at the emitter-base junction effectively eliminates hole injection from the base when the junction is forward biased. This gives good emitter-base injection efficiency.
12. Waveguide phototransistors utilize a ___________ waveguide layer under the _________ transistor region.
a) Active, passive
b) Passive, active
c) Homo, hetero
d) Hetero, homo
Answer: b
Explanation: Waveguide phototransistors are based on hetero-junction structure. They function as high-performance photo-detectors and thus utilize a passive waveguide layer under the active transistor region.
13. What is the main benefit of the waveguide structure over conventional hetero-junction phototransistor?
a) High depletion region
b) Depletion width
c) Increased photocurrent, responsivity
d) Low gain
Answer: c
Explanation: Waveguide structure offers increased photocurrent. Photocurrent is directly proportional to the responsivity; thus in turn increases responsivity.
14. Waveguide structure provides high quantum efficiency.
a) True
b) False
Answer: b
Explanation: Responsivity and quantum efficiency follow a different path. They are indirectly proportional to each other. Thus, in waveguide structure, as the responsivity increases, quantum efficiency remains low.
15. Metal-semiconductor-metal (MSM) photo-detectors are photoconductive detectors.
a) True
b) False
Answer: a
Explanation: MSM photo-detectors are the simplest of photo-detectors. It provides the simplest form of photo-detection within optical fiber communications and are photoconductive.