Communications MCQ Set 1
1. A _________________ is a network connecting several regional or national networks together.
a) Long-haul network
b) Domain network
c) Short-haul network
d) Erbium network
Answer
Answer: a [Reason:] Long-haul networks are also called as core or backbone networks. These networks connect regional or national networks together on a large scale. This can be extended to extended long haul networks.
2. What is the range of transmission of extended long haul network?
a) 200-400 km
b) 600-1000 km
c) 1000-2000 km
d) 2000-4000 km
Answer
Answer: c [Reason:] Extended long haul networks comprise of DWDM links. The transmission ranges may vary depending upon the complexity of the network. The extended long haul network’s transmission range varies from 1000 to 2000 km.
3. What is the range of transmission of ultra-long haul network?
a) 200-400 km
b) 600-1000 km
c) 1000-2000 km
d) 2000-4000 km
Answer
Answer: d [Reason:] Ultra haul networks comprise of DWDM links which provides them maximum range. The transmission ranges may vary depending upon the complexity of the network. The ultra-long haul network’s transmission range varies from 2000 to 4000 km.
4. Which feature plays an important role in making the longer haul networks feasible?
a) Channeling
b) Forward error control
c) Backward error control
d) Interconnection
Answer
Answer: b [Reason:] The longer haul networks can be made feasible by improvements in the DWDM systems and using forward error control mechanism. Such networks operate at channel rates in G-bits.
5. Which of the following is not an element of a submerged cable system?
a) Repeater
b) Branching unit
c) Gain equalizer
d) Attenuator
Answer
Answer: d [Reason:] The submerged cable system consists of a dry and wet plant. The elements associated with it include a repeater, branching unit, gain equalizer and a line amplifier. Attenuator is not present in cable system.
6. ___________ provides interconnection between the United States and European countries.
a) TAT
b) WTE
c) PFE
d) POP
Answer
Answer: a [Reason:] TAT is abbreviated as transatlantic optical fiber cables. TAT-14 is the newest version which is used as a medium of interconnection between the countries.
7. TAT-14 employs a DWDM bidirectional ring configuration. State whether the given statement is true or false.
a) False
b) True
Answer
Answer: b [Reason:] TAT-14 was first used in the year 2000. Its transmission capacity is more that the previous TAT versions. It’s DWDM configuration enables it to connect the various countries of Europe with the United States.
8. A single fiber in TAT-14 can carry _________ wavelength channels.
a) One
b) Twelve
c) Sixteen
d) Ten
Answer
Answer: c [Reason:] TAT-14 ‘ s single fiber carries a total of 16 wavelength channels. Each channel can allow a transmission rate of 10 Gigabits. It possesses a high operational capacity.
9. Optical MAN’S are usually structured in _______ topologies.
a) Ring
b) Bus
c) Mesh
d) Star
Answer
Answer: a [Reason:] MAN’s are characterized by changing traffic patterns requiring the networks to be fast. Also, MAN must be cost effective in terms of both operation and maintenance. Hence, they are structured in ring topologies.
10. The ________ network is an element of public telecommunication network that connects access nodes to individual users or MAN’s.
a) Ring
b) Access
c) Mesh
d) Nodal
Answer
Answer: b [Reason:] Access network is usually a last link in the network. It provides the strategies to connect to end-point users on both sides of connection.
11. _____________ is a technique that combines two or more network resources for redundancy or higher throughput.
a) Signal bonding
b) Attenuation
c) Re-signaling
d) Channel bonding
Answer
Answer: d [Reason:] Channel bonding combines two interfaces. It increases the overall bandwidth by the number of channels bonded. The data-rates are similar in this technique.
12. The upstream traffic in EPON is managed by employing a TDM approach. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: a [Reason:] EPON upstream traffic is divided into time slots. The time slots are dedicated to each ONU(Optical network units) in order not to interfere with the data.
Communications MCQ Set 2
1. Optical fibre communications uses _______ dielectric waveguide structures for confining light.
a) Rectangular
b) Circular
c) Triangular
d) Planar
Answer
Answer: b [Reason:] The use of circular dielectric waveguide structures is universally acceptable. This has been a boon for optical fibre communications.
2. __________ waveguide is formed when the film is sandwiched by layers of different refractive index.
a) Planar
b) Circular
c) Asymmetric
d) Symmetric
Answer
Answer: c [Reason:] When the film is sandwiched between layers of same refractive index, symmetric waveguide is formed. Owing to the different refractive index, asymmetry is observed and hence asymmetric waveguide is formed.
3. When the dimensions of the guide are reduced, the number of ___________ also decreases.
a) Propagating nodes
b) Electrons
c) Holes
d) Volume of photons
Answer
Answer: a [Reason:] The dimensions of the guide are directly proportional to the number of propagating nodes. As the dimensions are reduced, the number of propagating nodes also decreases.
4. What does hff stands for in the equation hff =h+x+x2?
a) Frequency of layer
b) Diameter of curve
c) Effective guide layer thickness
d) Space propagation
Answer
Answer: c [Reason:] In the above equation, h is the height, x and x2 are the evanescent field penetration depths. hff Denotes the effective guided layer thickness.
5. ___________ waveguides are plagued by high losses.
a) Circular
b) Planar
c) Depleted
d) Metal-clad
Answer
Answer: d [Reason:] All suitable waveguide materials are subject to limitations in the confinement. However, metal-clad waveguides are not so limited. Hence, they are plagued by high losses.
6. The planar waveguides may be fabricated from glasses and other isotropic materials such as ___________ and ______________
a) Octane and polymers
b) Carbon monoxide and diode
c) Fluorides and carbonates
d) Sulphur dioxide and polymers
Answer
Answer: d [Reason:] These materials are isotropic. However, their properties do not affect the fabrication of planar waveguides. Their properties cannot be controlled by external energy sources.
7. Which of the following devices are less widely used in the field of optical fibre communications?
a) Acousto-optic devices
b) Regenerators
c) Reflectors
d) Optical translators
Answer
Answer: a [Reason:] Acousto-optic devices are less widely used, mainly in the area of field deflection. Regenerators, reflectors form a base for the optical fibre communications.
8. Which of the following materials have refractive index near two?
a) GA As
b) Zinc
c) InP
d) AlSb
Answer
Answer: b [Reason:] Two basic groups are distinguished on the basis of the respective refractive indices near two and near three. GaAs, InP, AlSb have refractive indices near 3.
9. Passive devices are fabricated by __________ technique.
a) Fassbinder
b) High density integration
c) Radio-frequency sputtering
d) Lithium implantation
Answer
Answer: c [Reason:] Passive devices’ fabrication comes mainly from microelectronics industry. Radio frequency sputtering is used to deposit thin films of glass onto glass substrates.
10. Strip pattern in waveguide structures is obtained through ____________
a) Lithography
b) Cryptography
c) Depletion of holes
d) Implantation
Answer
Answer: a [Reason:] Field strength is an important aspect when it comes to strip patterns in waveguide structures. The electron and laser beam lithography is used to obtain stripe pattern in waveguide structures.
11. Propagation losses in slab and strip waveguides are smaller than the single mode fibre losses. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: b [Reason:] The losses are in the range of 0.1 to 0.3 dB/cm. In case of slab and stripe waveguides, the losses are much higher whereas in case of single-mode fibres, they are much less.
12. A passive Y-junction beam splitter is fabricated from __________
a) GaAs
b) ZnS
c) AlbS
d) LiNbO3
Answer
Answer: d [Reason:] A passive Y-junction splitter is used to combine signals from separate sources or to divide a signal into two or more channels. It is fabricated from the waveguide materials such as LiNbO3.
13. A passive Y-junction beam splitter is also used as a switch. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: a [Reason:] A passive junction beam splitter finds application where equal power division of the incident beam is required. It can be used as a switch if it is fabricated from an electro-optic material.
14. The linear variation of refractive index with the electric field is known as the ________
a) Linear implantation
b) Ionization
c) Koppel effect
d) Pockels effect
Answer
Answer: d [Reason:] The change in refractive index is related by the applied field via the linear and quadratic electro-optic coefficients. The variation of R.I with the electric field is known as Pockels effect.
15. Planar waveguides are used to produce _______ coupler.
a) MMI
b) CMI
c) Frequency
d) Differential
Answer
Answer: a [Reason:] MMI couplers are abbreviated as Multimode interference couplers. These are similar to fused fibre couplers. These are easily produced by using planar waveguides.
Communications MCQ Set 3
1. Which equations are best suited for the study of electromagnetic wave propagation?
a) Maxwell’s equations
b) Allen-Cahn equations
c) Avrami equations
d) Boltzmann’s equations
Answer
Answer: a [Reason:] Electromagnetic mode theory finds its basis in electromagnetic waves. Electromagnetic waves are always represented in terms of electric field E, magnetic field H, electric flux density D and magnetic flux density B. These set of equations are provided by Maxwell’s equations.
2. When λ is optical wavelength in vacuum, k is given by k=2Π/λ. What does k stand for in the above equation?
a) Phase propagation constant
b) Dielectric constant
c) Boltzmann’s constant
d) Free-space constant
Answer
Answer: a [Reason:] In the above equation, k= 2Π/λ, also termed as wave equation, k gives us the direction of propagation and also the rate of change of phase with distance. Hence it is termed as phase propagation constant.
3. Constructive interference occurs when total phase change after two successive reflections at upper and lower interfaces is equal to? (Where m is integer)
a) 2Πm
b) Πm
c) Πm/4
d) Πm/6
Answer
Answer: a [Reason:] The component of phase waves which is in x direction is reflected at the interference between the higher and lower refractive index media. It is assumed that such an interference forms a lowest order standing wave, where electric field is maximum at the center of the guide, decaying towards zero.
4. When light is described as an electromagnetic wave, it consists of a periodically varying electric E and magnetic field H which are oriented at an angle
a) 90 degree to each other
b) Less than 90 degree
c) Greater than 90 degree
d) 180 degree apart
Answer
Answer: a [Reason:] In case of electromagnetic wave which occur only in presence of both electric and magnetic field, a particular change in magnetic field will result in a proportional change in electric field and vice versa. These changes result in formation of electromagnetic waves and for electromagnetic waves to occur both fields should be perpendicular to each other in direction of wave travelling.
5. A monochromatic wave propagates along a waveguide in z direction. These points of constant phase travel in constant phase travel at a phase velocity Vp is given by
a) Vp=ω/β
b) Vp=ω/c
c) Vp=C/N
d) Vp=mass/acceleration
Answer
Answer: a [Reason:] Velocity is a function of displacement. Phase velocity Vp is a measure of angular velocity.
6. A most important velocity in the study of transmission characteristics of optical fiber is
a) Phase velocity
b) Group velocity
c) Normalized velocity
d) Average velocity
Answer
Answer: b [Reason:] Group velocity is much important in relation to transmission characteristics of optical fiber. This is because the optical wave propagates in groups or form of packets of light.
7. Refraction is the
a) Bending of light waves
b) Reflection of light waves
c) Diffusion of light waves
d) Refraction of light waves
Answer
Answer: a [Reason:] Unlike reflection, refraction involves penetration of a light wave from one medium to another. While penetrating, as it passes through another medium it gets deviated at some angle.
8. The phenomenon which occurs when an incident wave strikes an interface at an angle greater than the critical angle with respect to the normal to the surface is called as
a) Refraction
b) Partial internal reflection
c) Total internal reflection
d) Limiting case of refraction
Answer
Answer: c [Reason:] Total internal reflection takes place when the light wave is in the more dense medium and approaching towards the less dense medium. Also, the angle of incidence is greater than the critical angle. Critical angle is an angle beyond which no propagation takes place in an optical fiber.
Communications MCQ Set 4
1. ____________ results from small lateral forces exerted on the fiber during the cabling process.
a) Attenuation
b) Micro-bending
c) Dispersion
d) Stimulated Emission
Answer
Answer: b [Reason:] Optical fibers must be designed so that the transmission characteristics of the fiber are maintained after the cabling process. The main problem which occurs in the cabling process is the meandering of the axis of the fiber core on a microscopic scale within the cable form. This phenomenon is called as micro-bending.
2. Microscopic meandering of the fiber core axis that is micro-bending is caused due to
a) Environmental effects
b) Rough edges of the fiber
c) Large diameter of core
d) Polarization
Answer
Answer: a [Reason:] Micro-bending can be generated at any stage during manufacturing process, cable installation process or during service. This is mainly due to environmental effects, mainly varying temperatures causing differential expansion or contraction.
3. How many forms of modal power distribution are considered?
a) One
b) Two
c) Three
d) Four
Answer
Answer: b [Reason:] Two forms of modal power distribution are considered. The first form is seen when a fiber is excited by a diffuse Lambertian source, and is called as fully filled mode distribution. The second form occurs when, due to mode coupling and attenuation, the distribution of optical power becomes invariant with the distance of propagation along the fiber, and is called as steady-state mode distribution.
4. What does micro-bending losses depend on?
a) Core material
b) Refractive index
c) Diameter
d) Mode and wavelength
Answer
Answer: d [Reason:] Micro-bending losses cause differential expansion or contraction. These losses are mode dependent. The number of modes is a function inverse to the wavelength of the transmitted light and thus micro-bending losses are wavelength dependent.
5. The fiber should be________________ to avoid deterioration of the optical transmission characteristics resulting from mode-coupling-induced micro-bending.
a) Free from irregular external pressure
b) Coupled with plastic
c) Large in diameter
d) Smooth and in a steady state
Answer
Answer: a [Reason:] Micro-bending losses results from environmental effects such as temperature variation. The irregular external pressure deteriorates the quality of transmission through the fiber. Thus, controlled coating and cabling of the fiber is essential in order to reduce the cabled fiber attenuation.
6. The diffusion of hydrogen into optical fiber affects the ______________
a) Transmission of optical light in the fiber
b) Spectral attenuation characteristics of the fiber
c) Core of the fiber
d) Cladding of the fiber
Answer
Answer: b [Reason:] The hydrogen absorption by an optical fiber increases optical fiber losses. It forms absorption peaks where the hydrogen diffuses into interstitial spaces in the glass. At high temperatures, these losses can increase and reduced if the hydrogen source is removed.
7. __________ can induce a considerable amount of attenuation in optical fibers.
a) Micro-bending
b) Dispersion
c) Diffusion of hydrogen
d) Radiation Exposure
Answer
Answer: d [Reason:] The optical transmission characteristics of the fiber cables can be degraded by exposure to nuclear radiation. The nature of this attenuation depends upon fiber structures, optical intensity, wavelength etc. The radiation-induced attenuation comprises both permanent and temporary components which makes the exposure irreversible and reversible respectively.
8. The radiation-induced attenuation can be reduced through photo-bleaching. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: a [Reason:] Photo-bleaching can be exploited to study the diffusion of molecules. It is used to remove the radiation exposure by quenching auto-fluorescence. It helps to increase signal-to-noise ratio of the fiber and thus reduces attenuation.
9. The losses due to hydrogen absorption and reaction with fiber deposits can be temporary. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: b [Reason:] Hydrogen absorption occurs in two mechanisms. First phenomenon affects silica-based glass fibers whereas the second one occurs when hydrogen reacts with the fiber deposits to give P-OH, Ge-OH absorption. These losses are permanent.
10. The losses caused due to hydrogen absorption mechanisms are in the range of
a) 20 dB/km to25 dB/km
b) 10 dB/km to15 dB/km
c) 25 dB/km to50 dB/km
d) 0 dB/km to5 dB/km
Answer
Answer: c [Reason:] The diffusion of hydrogen into optical fiber leads to increase in optical fiber losses, causing damage to spectral loss characteristics. This phenomenon gets vibrant at higher temperatures. The losses caused due to such absorption are greater than25 dB/km.
Communications MCQ Set 5
1. A multimode step index fiber has source of RMS spectral width of 60nm and dispersion parameter for fiber is 150psnm-1km-1. Estimate rms pulse broadening due to material dispersion.
a) 12.5ns km-1
b) 9.6ns km-1
c) 9.0ns km-1
d) 10.2ns km-1
Answer
Answer: c [Reason:] The RMS pulse broadening per km due to material dispersion is given by
σm(1 km) = σλLM
= 60*1* 150pskm-1
= 9.0nskm-1
Where σλ= rms spectral width
L= length of fiber
M = dispersion parameter.
2. A multimode fiber has RMS pulse broadening per km of 12ns/km and 28ns/km due to material dispersion and intermodal dispersion resp. Find the total RMS pulse broadening.
a) 30.46ns/km
b) 31.23ns/km
c) 28.12ns/km
d) 26.10ns/km
Answer
Answer: a [Reason:] The overall dispersion in multimode fibers comprises both chromatic and intermodal terms. The total RMS pulse broadening σT is given by
Where σm = RMS pulse broadening due to material dispersion
σi= RMS pulse broadening due to intermodal dispersion.
3. Γg= dβ / C*dk. What is β in the given equation?
a) Attenuation constant
b) Propagation constant
c) Boltzmann’s constant
d) Free-space
Answer
Answer: b [Reason:] Above given equation is an equation of transit time or a group delay(Γg) for a light pulse. This light pulse is propagating along a unit length of a single mode fiber.
4. Most of the power in an optical fiber is transmitted in fiber cladding. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: b [Reason:] Most of the power in optical fiber is transmitted in fiber core. This is because in multimode fibers, majority of modes propagating in the core area are far from cutoff. Hence more power is transmitted.
5. A single mode fiber has a zero dispersion wavelength of 1.21μm and a dispersion slope of 0.08 psnm-2km-1. What is the total first order dispersion at wavelength 1.26μm.
a) -2.8psnm-1 km-1
b) -3.76psnm-1 km-1
c) -1.2psnm-1 km-1
d) 2.4psnm-1 km-1
Answer
Answer: b [Reason:] The total first order dispersion for fiber at two wavelength is obtained by
DT(1260 nm) = λS0/4 [1-(λ0/λ)4]
= (1260*0.08*10-12)/4* (1-[1550/1260] 4)
= -3.76psnm-1km-1
Where
λ0= zero dispersion wavelength
λ= wavelength
S0=dispersion slope
DT= total first order dispersion.
6. The dispersion due to material, waveguide and profile are -2.8nm-1km-1, 20.1nm-1km-1 and 23.2nm-1km-1respectively. Find the total first order dispersion?
a) 36.2psnm-1 km-1
b) 38.12psnm-1 km-1
c) 40.5psnm-1 km-1
d) 20.9psnm-1 km-1
Answer
Answer: c [Reason:] The total dispersion is given by
DT= DM+ DW+ DP(psnm-1km-1)
Where
DW = waveguide dispersion
DM= Material dispersion
DP= profile dispersion.
7. Dispersion-shifted single mode fibers are created by
a) Increasing fiber core diameter and decreasing fractional index difference
b) Decreasing fiber core diameter and decreasing fractional index difference
c) Decreasing fiber core diameter and increasing fractional index difference
d) Increasing fiber core diameter and increasing fractional index difference
Answer
Answer: c [Reason:] It is possible to modify the dispersion characteristics of single mode fibers by tailoring of some fiber parameters. These fiber parameters include core diameter and relative index difference.
8. An alternative modification of the dispersion characteristics of single mode fibers involves achievement of low dispersion gap over the low-loss wavelength region between –
a) 0.2 and 0.9μm
b) 0.1 and 0.2μm
c) 1.3 and 1.6μm
d) 2 and 3μm
Answer
Answer: c [Reason:] Dispersion characteristics can be altered by changing fiber parameters and wavelength. The achievement of low dispersion gap over the region 1.3 and 1.6μm modifies the dispersion characteristics of single mode fibers.
9. The fibers which relax the spectral requirements for optical sources and allow flexible wavelength division multiplying are known as-
a) Dispersion-flattened single mode fiber
b) Dispersion-enhanced single mode fiber
c) Dispersion-compressed single mode fiber
d) Dispersion-standardized single mode fiber
Answer
Answer: a [Reason:] The dispersion-flattened single mode fibers (DFFS) are obtained by fabricating multilayer index profiles with increased waveguide dispersion. This is tailored to provide overall dispersion say 2psnm-1km-1 over the wavelength range 1.3 to 1.6μm.
10. For suitable power confinement of fundamental mode, the normalized frequency v should be maintained in the range 1.5 to 2.4μm and the fractional index difference must be linearly increased as a square function while the core diameter is linearly reduced to keep v constant. This confinement is achieved by-
a) Increasing level of silica doping in fiber core
b) Increasing level of germanium doping in fiber core
c) Decreasing level of silica germanium in fiber core
d) Decreasing level of silica doping in fiber core
Answer
Answer: b [Reason:] The tailoring of fiber parameters provides suitable power confinement. These parameters may be diameter, index-difference, frequency etc. The doping level of germanium contributes to the tailoring of fiber parameters; which in turn provides suitable power confinement.
11. Any amount of stress occurring at the core-cladding interface would be reduced by grading the material composition. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: a [Reason:] A problem arises with that of simple step index approach to dispersion shifting is high. The fibers produced exhibit high dopant-dependent losses at operating wavelengths. These losses are caused by induced-stress in the region of core-cladding interface. This can be reduced by grading the material composition of the fiber.
12. The variant of non-zero-dispersion-shifted fiber is called as
a) Dispersion flattened fiber
b) Zero-dispersion fiber
c) Positive-dispersion fiber
d) Negative-dispersion fiber
Answer
Answer: d [Reason:] The dispersion profile for non-zero dispersion shifted fiber is referred to as bandwidth non-zero-dispersion-shifted fiber. It was introduced to provide wavelength division multiplexed applications to be extended into the s-band. The variant of non-zero-dispersion-shifted fiber can also be referred to as dispersion compensating fiber.
13. Non-zero-dispersion-shifted fiber was introduced in the year 2000. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: b [Reason:] Non-zero-dispersion-shifted fiber was introduced in mid-1990s to provide wavelength division multiplexing applications. In the year 2000, the dispersion profile for non-zero-dispersion-shifted fiber was introduced.
Communications MCQ Set 6
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
Answer: a [Reason:] 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
Answer: c [Reason:] 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
Answer: b [Reason:] None.
4. Determine drift time for carrier across depletion region for photodiode having intrinsic region width of 30μmand 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
Answer: c [Reason:] The drift time is given by
tdrift= w/vd = 30×10-6/1×10-10= 3×10-10seconds.
5. Determine intrinsic region width for a photodiode having drift time of 4×10-10s and electron velocity of2×10-10ms-1.
a) 3×10-5M
b) 8×10-5M
c) 5×10-5M
d) 7×10-5M
Answer
Answer: b [Reason:] 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
Answer: a [Reason:] 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-1and width of 30μm
a) 3.043×10-5
b) 2.415×10-7
c) 4.641×10-4
d) 3.708×10-5
Answer
Answer: b [Reason:] 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 is15.5×10-15Fcm-1and 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
Answer: d [Reason:] 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-1and 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
Answer: b [Reason:] 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 0.62×10-6m2and 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
Answer: b [Reason:] 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) 5.05×10-7 sec
d) 5.05×10-8Sec
Answer
Answer: c [Reason:] 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
Answer: c [Reason:] 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 28×10-6mand carrier velocity of 4×104ms-1.
a) 105.67 MHz
b) 180.43 MHz
c) 227.47 MHz
d) 250.65 MHz
Answer
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: a [Reason:] 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.
Communications MCQ Set 7
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
Answer: a [Reason:] 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. State whether the given statement is true or false?
a) True
b) False
Answer
Answer: a [Reason:] 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
Answer: b [Reason:] 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. State whether the given statement is true or false.
a) True
b) False
Answer
Answer: b [Reason:] 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
Answer: c [Reason:] 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
Answer: d [Reason:] 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
Answer: d [Reason:] 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
Answer: a [Reason:] 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
Answer: d [Reason:] 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
Answer: c [Reason:] 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
Answer: d [Reason:] 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
Answer: d [Reason:] 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
Answer: a [Reason:] 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
Answer: b [Reason:] 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.
a) InAsSb
b) III – V alloys
c) InGaSb
d) InGaAs
Answer
Answer: d [Reason:] 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.
Communications MCQ Set 8
1. Determine dispersion equalization penalty if total RMS pulse broadening is 4.8ns, BT is 25 Mbits/s.
a) 0.03 dB
b) 0.08 dB
c) 7 dB
d) 0.01 dB
Answer
Answer: a [Reason:] Dispersion equalization penalty is denoted by DL. It is given by-
DL=2 (2σTBT√2)4.Here σT=RMS pulse broadening.
2. Determine RMS pulse broadening with mode coupling if pulse broadening is 0.6 over 8km.
a) 1.6ns
b) 1.7ns
c) 1.5ns
d) 1.4ns
Answer
Answer: b [Reason:] Total RMS pulse broadening with mode coupling is given by-
σT=σ√L. Here σT=RMS pulse broadening, L =length of the fiber.
3. Determine dispersion equalization penalty with mode coupling of 1.7ns if BT is 25 Mbits/s.
a) 4.8 * 104dB
b) 4 * 104dB
c) 4.2 * 104dB
d) 3.8 * 104dB
Answer
Answer: c [Reason:] Dispersion equalization penalty is denoted by DL.With mode coupling, it is given by-
DL=2 (2σTBT√2)4. Here σT=RMS pulse broadening.
4. Determine dispersion equalization penalty without mode coupling if BTis 150 Mbits/s and total rms pulse broadening is 4.8ns.
a) 34 dB
b) 33 dB
c) 76.12 dB
d) 34.38 dB
Answer
Answer: d [Reason:] Dispersion equalization penalty is denoted by DL(WM). It is given by-
DL(WM)=2 (2σTBT√2)4.Here σT=RMS pulse broadening, (WM) =without mode coupling.
5. Determine ratio of SNR of coaxial system to SNR of fiber system if peak output voltage is 5V, quantum efficiency of 70%, optical power is 1mW, wavelength of 0.85μm.
a) 1.04 * 104
b) 2.04 * 104
c) 3.04 * 104
d) 4.04 * 104
Answer
Answer: a [Reason:] Ratio of SNR of coaxial system to SNR of fiber system is given by-
Ratio =V2hc/2KTZ0ηPiλ. Here, η=quantum efficiency, Pi = 0ptical power in mW, V=optical output voltage.
6. Determine the peak output voltage if efficiency is 70%, wavelength is 0.85μmand output power is 1mW.
a) 7V
b) 8V
c) 5V
d) 6V
Answer
Answer: Peak output voltage is given by-
V2= (2KTZ0ηPiλ * Ratio)/hc. Hereη=quantum efficiency, Pi=0ptical power in mW, V=optical output voltage.
7. Determine the efficiency of a coaxial cable system at 17 degree Celsius with peak output voltage 5V, 0.85μmwavelength and SNR ratio of 1.04 * 104.
a) 80%
b) 70%
c) 40%
d) 60%
Answer
Answer: b [Reason:] The efficiency of a coaxial cable system is η=V2hc/2KTZ0ηPiλ * Ratio. Hereη=quantum efficiency, Pi=0ptical power in mW, V=optical output voltage.
8. Determine the wavelength of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 100Ω impedance, optical power of 1mW, 70% quantum efficiency.
a) 0.39μm
b) 0.60μm
c) 0.85μm
d) 0.98μm
Answer
Answer: c [Reason:] The wavelength can be determined by –
λ= V2hc/2KTZ0ηPi * Ratio. Hereη=quantum efficiency, Pi =0ptical power in mW, V=optical output voltage.
9. Determine the impedance of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 0.85μmwavelength, optical power of 1mW, 70% quantum efficiency and SNR ratio of 1.04 * 104.
a) 80Ω
b) 50Ω
c) 90Ω
d) 100Ω
Answer
Answer: d [Reason:] The impedance is given by-Z0=V2hc/2KTPi * Ratio. Hereη=quantum efficiency, Pi = Optical power in mW, V=optical output voltage.
10. The 10-90% rise times for components used in D-IM analog optical link is given.
LED=10ns
Intermodal=9ns/km
Chromatic=2ns/km
APD = 3ns
Link is of 5km. Determine the total rise time.
a) 62ns
b) 53ns
c) 50ns
d) 52ns
Answer
Answer: d [Reason:] Total rise time is given by-
Tsyst=1.1[Ts2+Tn2+Tc2+TD2]1/2. Here Ts =rise time,Tn =intermodal time,Tc =Chromatic time.
11. The 10-90% rise times for components used in D-IM analog optical link is given.
LED=10ns
Intermodal=9ns/km
Chromatic=2ns/km
APD = 3ns
Link is of 5km. It has an optical bandwidth of 6MHz. Determine maximum permitted system rise time.
a) 58.3ns
b) 54ns
c) 75ns
d) 43.54ns
Answer
Answer: a [Reason:] The maximum permitted system rise time is given by-
Tsyst(Max)=0.35/Bopt. Here, Bopt=Optical Bandwidth.
