Question 1

Determine the internal quantum efficiency generated within a device when it has a radiative recombination lifetime of 80 ns and total carrier recombination lifetime of 40 ns.&#10;A)20 %&#10;B)80 %&#10;C)30 %&#10;D)40 %

Accepted Answer

The internal quantum efficiency ($\eta$) can be calculated using the formula $\eta = \frac{\tau_{total}}{\tau_{radiative}}$, where $\tau_{total}$ is the total carrier recombination lifetime and $\tau_{radiative}$ is the radiative recombination lifetime. Plugging in the given values, $\eta = \frac{40 ns}{80 ns} = 0.5$ or 50%. However, since 50% is not an option and the correct calculation should actually be $\eta = \frac{\tau_{radiative}}{\tau_{total}} = \frac{80 ns}{40 ns} = 2$ or 200%, which indicates a mistake in the calculation. The correct approach is indeed $\eta = \frac{\tau_{radiative}}{\tau_{total}}$, but correctly interpreted, it should yield a value that represents the efficiency in terms of how much of the total recombination is radiative. The correct formula and approach, considering the options, suggest a misunderstanding in the explanation provided. The correct calculation for internal quantum efficiency, considering the typical formula $\eta = \frac{\tau_{total}}{\tau_{radiative}}$ and the options given, does not match the provided options, indicating a mistake in the explanation process. The correct answer should reflect the proportion of radiative recombination to total recombination, which is not accurately captured in the response.

Question 2

Compute power internally generated within a double-heterojunction LED if it has internal quantum efficiency of 64.5 % and drive current of 40 mA with a peak emission wavelength of 0.82 ?m.&#10;A)0.09&#10;B)0.039&#10;C)0.04&#10;D)0.06

Accepted Answer

0.039&#10;

Question 3

The Lambertian intensity distribution __________ the external power efficiency by some percent.&#10;A)reduces&#10;B)does not affects&#10;C)increases&#10;D)have a negligible effect

Accepted Answer

The Lambertian intensity distribution reduces the external power efficiency because it spreads the emitted light over a wide angle, leading to a portion of the light not being effectively utilized or captured, especially in applications requiring directed or focused light output.

Question 4

A planar LED fabricated from GaAs has a refractive index of 2.5. Compute the optical power emitted when transmission factor is 0.68.&#10;A)3.4 %&#10;B)1.23 %&#10;C)2.72 %&#10;D)3.62 %

Accepted Answer

The optical power emitted, often referred to as the extraction efficiency, can be calculated using the formula for the transmission factor at the interface between two media, which is given by $T = \frac{4n}{(n+1)^2}$, where $n$ is the refractive index of the material. Rearranging this formula to solve for $T = 0.68$, we get $0.68 = \frac{4 \times 2.5}{(2.5+1)^2}$, which simplifies to $0.68 = \frac{10}{3.5^2}$, or $0.68 = \frac{10}{12.25}$, which does not directly give us the percentage. To find the correct answer, we use the given transmission factor directly as it represents the fraction of optical power emitted. Since the transmission factor is 0.68, and we want to express this as a percentage, we multiply by 100 to get $0.68 \times 100 = 68\%$. However, the correct calculation method involves understanding that the transmission factor given does not directly translate to the optical power emitted without considering the critical angle and total internal reflection phenomena in planar LEDs. The correct approach involves more complex calculations involving the refractive index and the geometry of the LED. Given the options and the misunderstanding in the calculation, the correct answer based on the provided options and assuming a direct correlation would be to find the closest match to the theoretical calculation of emission efficiency based on the refractive index, which is not straightforwardly provided in the question. The correct answer choice, therefore, is based on typical extraction efficiencies for LEDs with high refractive indices without direct calculation from the given transmission factor.

Question 5

A planar LED is fabricated from GaAs is having a optical power emitted is 0.018% of optical power generated internally which is 0.018% of optical power generated internally which is 0.6 P. Determine external power efficiency.

A)0.18%
B)0.32%
C)0.65%
D)0.9%

Accepted Answer

The answer of A planar LED is fabricated from GaAs...

Question 6

For a GaAs LED, the coupling efficiency is 0.05. Compute the optical loss in decibels.&#10;A)12.3 db&#10;B)14 db&#10;C)13.01 db&#10;D)14.6 db

Accepted Answer

The optical loss in decibels can be calculated using the formula: Loss (dB) = -10 log(coupling efficiency). For a coupling efficiency of 0.05, the calculation is: Loss = -10 log(0.05) ≈ 13.01 dB.

Question 7

In a GaAs LED, compute the loss relative to internally generated optical power in the fiber when there is small air gap between LED and fiber core. (Fiber coupled = 5.5 * 10-4Pint)&#10;A)34 db&#10;B)32.59 db&#10;C)42 db&#10;D)33.1 db

Accepted Answer

The loss relative to internally generated optical power in the fiber, when there is a small air gap between the LED and the fiber core, can be calculated using the formula for optical power loss in decibels (dB): Loss (dB) = -10 * log10(Pout/Pin), where Pout is the output power and Pin is the input power. Given that the fiber-coupled power is 5.5 * 10^-4 times the internally generated power (Pint), we can set Pout = 5.5 * 10^-4 * Pint and Pin = Pint. Substituting these values into the formula gives Loss = -10 * log10(5.5 * 10^-4), which calculates to approximately 32.59 dB.

Question 8

Determine coupling efficiency into the fiber when GaAs LED is in close proximity to fiber core having numerical aperture of 0.3&#10;A)0.9&#10;B)0.3&#10;C)0.6&#10;D)0.12

Accepted Answer

The coupling efficiency is determined by the numerical aperture of the fiber and the acceptance angle of the LED. The numerical aperture is given as 0.3, which means that the maximum acceptance angle of the fiber is 17.5 degrees. The acceptance angle of the LED is not given, but it is typically around 10 degrees. Therefore, the coupling efficiency is approximately 0.9, which is the ratio of the acceptance angle of the fiber to the acceptance angle of the LED.

Question 9

If a particular optical power is coupled from an incoherent LED into a low-NA fiber, the device must exhibit very high radiance .

Accepted Answer

The radiance of an optical source is defined as the optical power emitted per unit area per unit solid angle. For a low-NA fiber, the solid angle of acceptance is small, so the radiance of the source must be high in order to couple a significant amount of optical power into the fiber.

Question 10

The amount of radiance in planer type of LED structures is&#10;A)low&#10;B)high&#10;C)zero&#10;D)negligible

Accepted Answer

Planar type LED structures typically have low radiance due to their design, which does not efficiently direct light out of the device.

Question 11

In optical fiber communication, ______________ major types of LED structures are used&#10;A)2&#10;B)4&#10;C)6&#10;D)3

Accepted Answer

The answer of In optical fiber communication, ______________ major types...

Question 12

As compared to planar LED structure, Dome LEDs have ______________ External power efficiency, ____________ effective emission area and _____________ radiance.&#10;A)greater, lesser, reduced&#10;B)higher, greater, reduced&#10;C)higher, lesser, increased&#10;D)greater, greater, increased

Accepted Answer

The answer of As compared to planar LED structure, Dome...

Question 13

The techniques by Burros and Dawson in reference to homo structure device is to use an etched well in GaAs structure. Determine the given statement is True or false.

Accepted Answer

The answer of The techniques by Burros and Dawson in...

Question 14

In surface emitter LEDs, more advantage can be obtained by using&#10;A)bh structures&#10;B)qc structures&#10;C)dh structures&#10;D)gain-guided structure

Accepted Answer

The answer of In surface emitter LEDs, more advantage can...

Question 15

Internal absorption in DH surface emitter Burros type LEDs is&#10;A)cannot be determined&#10;B)negligible&#10;C)high&#10;D)very low

Accepted Answer

The answer of Internal absorption in DH surface emitter Burros...

Question 16

DH surface emitter generally give&#10;A)more coupled optical power&#10;B)less coupled optical power&#10;C)low current densities&#10;D)low radiance emission into-fiber

Accepted Answer

The answer of DH surface emitter generally give&#10;A)more coupled optical...

Question 17

A DH surface emitter LED has an emission area diameter of 60?m. Determine emission area of source A)1.534*10^-6 B)5.423*10^-3 C)3.564*10-2 D)2.826*10^-9

Accepted Answer

The answer of A DH surface emitter LED has an...

Question 18

Estimate optical power coupled into fiber of DH SLED having emission area of 1.96*10^?5, radiance of 40 W/rcm2, numerical aperture of 0.2 and Fresnel reflection coefficient of 0.03 at index matched fiber surface.

A)5.459*10^?5
B)1.784*10^-3
C)3.478*10²
D)9.551*10^?5

Accepted Answer

The answer of Estimate optical power coupled into fiber of...

Question 19

In a multimode fiber, much of light coupled in the fiber from an LED is&#10;A)increased&#10;B)reduced&#10;C)lost&#10;D)unaffected

Accepted Answer

The answer of In a multimode fiber, much of light...

Question 20

Determine the overall power conversion efficiency of lens coupled SLED having forward current of 20 mA and forward voltage of 2 V with 170 ?Wof optical power launched into multimode step index fiber. A)1.256*10^?5 B)4.417*10² C)4.25*10^-3 D)2.14*10^-3

Accepted Answer

The answer of Determine the overall power conversion efficiency of...

Question 21

The overall power conversion efficiency of electrical lens coupled LED is 0.8% and power applied 0.0375 V. Determine optical power launched into fiber&#10;A)0.03&#10;B)0.05&#10;C)0.3&#10;D)0.01

Accepted Answer

The answer of The overall power conversion efficiency of electrical...

Question 22

Mesa structured SLEDs are used&#10;A)to reduce radiance&#10;B)to increase radiance&#10;C)to reduce current spreading&#10;D)to increase current spreading

Accepted Answer

The answer of Mesa structured SLEDs are used&#10;A)to reduce radiance&#10;B)to...

Question 23

The InGaAsP is emitting LEDs are realized in terms of restricted&#10;A)length strip geometry&#10;B)radiance&#10;C)current spreading&#10;D)coupled optical power

Accepted Answer

The answer of The InGaAsP is emitting LEDs are realized...

Question 24

The active layer of E-LED is heavily doped with&#10;A)zn&#10;B)eu&#10;C)cu&#10;D)sn

Accepted Answer

The answer of The active layer of E-LED is heavily...

Question 25

&#160;Intrinsically _________________ are a very linear device.&#10;A)injection lasers&#10;B)dh lasers&#10;C)gain-guided&#10;D)leds

Accepted Answer

The answer of &#160;Intrinsically _________________ are a very linear device.&#10;A)injection...

Deck 17: Led Efficiency and Power Generation in Optical Communication