Question 1

Linearizing circuit techniques are used for LEDs.

Accepted Answer

Linearizing circuit techniques are used for LEDs to ensure that their brightness can be controlled smoothly and predictably, as the relationship between current and light output is nonlinear.

Question 2

The internal quantum efficiency of LEDs decreasing _______________ with ________________ temperature.&#10;A)exponentially, decreasing&#10;B)exponentially, increasing&#10;C)linearly, increasing&#10;D)linearly, decreasing

Accepted Answer

The internal quantum efficiency of LEDs decreases exponentially with increasing temperature due to increased non-radiative recombination processes at higher temperatures.

Question 3

To utilize _____________________ of SLDs at elevated temperatures, the use of thermoelectric coolers is important.&#10;A)low-internal efficiency&#10;B)high-internal efficiency&#10;C)high-power potential&#10;D)low-power potential

Accepted Answer

Thermoelectric coolers are used to manage heat in devices with high-power potential, ensuring they operate efficiently at elevated temperatures.

Question 4

For particular materials with smaller bandgap energies operating in _____________ wavelength, the linewidth tends to ______________&#10;A)2.1 to 2.75 ?m, increase&#10;B)1.1 to 1.7 ?m, increase&#10;C)2.1 to 3.6 ?m, decrease&#10;D)3.5 to 6 ?m, decrease

Accepted Answer

Materials with smaller bandgap energies operating in the 1.1 to 1.7 μm wavelength range tend to have their linewidth increase due to the higher energy transitions involved, which are more sensitive to temperature variations and other factors that can broaden the linewidth.

Question 5

The active layer composition must be adjusted if a particular center wavelength is desired.

Accepted Answer

The active layer composition in devices like lasers or LEDs determines the bandgap energy, which directly affects the center wavelength of the emitted light. Adjusting the composition allows for tuning of the center wavelength.

Question 6

In optical fiber communication, the electrical signal dropping to half its constant value due to modulated portion of optical signal corresponds to _______&#10;A)6 db&#10;B)3 db&#10;C)4 db&#10;D)5 db

Accepted Answer

When an electrical signal drops to half its power, it corresponds to a 3 dB decrease. This is a standard reference in signal processing and telecommunications, indicating a halving of power.

Question 7

The optical 3 dB point occurs when currents ratio is equal to&#10;A)?&#10;B)?&#10;C)&#189;&#10;D)&#190;

Accepted Answer

The optical 3 dB point occurs when the current ratio is equal to ½. This is because a 3 dB decrease corresponds to a power reduction to half its original value, and since power is proportional to the square of the current, the current ratio must be ½ to achieve this power reduction.

Question 8

The optical bandwidth is _____________ the electrical bandwidth.&#10;A)smaller&#10;B)greater&#10;C)same as&#10;D)zero with respect to

Accepted Answer

The optical bandwidth is typically greater than the electrical bandwidth because optical signals can carry more information due to their higher frequencies, allowing for a wider bandwidth.

Question 9

When a constant d.c. drive current is applied to device, the optical o/p power is 320 ?m. Determine optical o/p power when device is modulated at frequency 30 MHz with minority carrier recombination lifetime of LED i.e. 5ns.

A)4.49*10^-12
B)6.84*10^-9
C)1.29*10^-6
D)2.29*10^-4

Accepted Answer

The answer of When a constant d.c. drive current is...

Question 10

The optical power at 20 MHz is 246.2 ?W. Determine dc drive current applied to device with carrier recombination lifetime for LED of 6ns. A)3.48*10^-4 B)6.42*10^-9 C)1.48*10^-3 D)9.48*10^-12

Accepted Answer

The optical power (P) at a given frequency can be related to the dc drive current (I) through the efficiency of the device and the energy of the photons being emitted. Given the carrier recombination lifetime (τ) for the LED, one can use the formula P = (I*q)/(2*τ), where q is the charge of an electron (approximately 1.6*10^-19 C). Rearranging for I and substituting the given values (P = 246.2 μW = 246.2*10^-6 W, τ = 6 ns = 6*10^-9 s), we can solve for I. The correct answer must account for these values and units correctly, leading to an answer in the order of magnitude of microamperes (μA), which is option A.

Question 11

Determine the 3 dB electrical bandwidth at 3 dB optical bandwidth Bopt of 56.2 MHz.&#10;A)50.14&#10;B)28.1&#10;C)47.6&#10;D)61.96

Accepted Answer

The answer of Determine the 3 dB electrical bandwidth at...

Question 12

The 3 dB electrical bandwidth B is 42 MHz. Determine 3dB optical bandwidth Bopt:&#10;A)45.18&#10;B)59.39&#10;C)78.17&#10;D)94.14

Accepted Answer

The answer of The 3 dB electrical bandwidth B is...

Question 13

Determine degradation rate ?rif constant junction temperature is 17 degree celsius. A)7.79*10^-11 B)7.91*10^-11 C)6.86*10^-11 D)5.86*10^-11

Accepted Answer

The answer of Determine degradation rate ?rif constant junction temperature...

Question 14

Determine CW operating lifetime for LED with ?rt = -0.58 and degradation rate ?r= 7.86*10^-11 h-1. A)32.12 B)42 C)22.72 D)23.223

Accepted Answer

The answer of Determine CW operating lifetime for LED with...

Deck 18: Optical Communication and Led Performance Analysis