solution 4 21, 22 ,23, 24

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21. A Fabry Perot laser active region has an index of 3.55 and a cavity length of 1.65 mm. If the laser has a dominant mode output of 9mW and a most dominant secondary mode output of 1.5mW, calculate the mode separation and the mode suppression ratio. Index, n= 3.55 cavity length, L= 1.65mm Dominant mode output, Pm=9mW Ps = 1.5mW i) Mode separation ∆V= c 2 nL ¿ 310 8 23.55 1.6510 3 ¿ 25.61 GHz ii) Mode suppression ratio ¿ 10 log Pm / Ps ¿ 10log9 m / 1.5 m ¿ 7.78 dB

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Page 1: SOLUTION 4 21, 22 ,23, 24

21. A Fabry Perot laser active region has an index of 3.55 and a cavity length of 1.65 mm. If the laser has a dominant mode output of 9mW and a most dominant secondary mode output of 1.5mW, calculate the mode separation and the mode suppression ratio.

Index, n= 3.55 cavity length, L= 1.65mm Dominant mode output, Pm=9mW

Ps = 1.5mW

i) Mode separation

∆V= c2nL

¿ 3∗108

2∗3.55∗1.65∗10−3

¿25.61GHz

ii) Mode suppression ratio

¿10 log Pm/Ps¿10 log 9m /1.5m¿7.78dB

Page 2: SOLUTION 4 21, 22 ,23, 24

22. Define external power efficiency. Give the corresponding formula and explain it briefly.

The external power efficiency of the device ηep in converting electrical input to optical output is given by:

ƞep=PePX 100 %

= PeVIX 100 % = ƞT

EgVX 100 %

Where P=IV is the d.c. electrical input power and Pe = power emitted

23. What is the spectral bandwidth of an LED? Draw the spectral bandwidth curve and

explain.

Spectral bandwidth is the wavelength interval in which a radiated spectral quantity is not less

than half its maximum value. It is a measure of the extent of the spectrum. For

a Light Source typical spectral widths are 20 to 60 nm for a LED . LED’s are narrow band

emitters. Typically, the bandwidth is 30nm wide. ILT Light Sources LED’s have spectral peaks

from 430nm up to 700nm in the visible light range and 880nm and 940nm in the near infrared

range. The spectral properties of an LED are important to aid manufacturers in their design

Page 3: SOLUTION 4 21, 22 ,23, 24

efforts and process control. End-users use these values in determining the correct LED for their

application.

23. When a LED has a 2 V applied to its terminals, it draws 100 mA and produce 2 mW of optical power. What is the LED’s conversion efficiency from electrical to optical power?

Given V = 2V

I = 100mA

P0 = 2mW

LED conversion efficiency, η from electrical PE to optical power PO

η=PoPEx 100 %

PE=I x V

¿100mA x2V

¿0.2W

η=2mW0.2W

x100 %

¿1 %

Page 4: SOLUTION 4 21, 22 ,23, 24

24. A simple system consist of a transmitter with a source responsivity of 0.0125 W/A, a

fiber with a loss of 8.0 dB, and a receiver. What is the power at the receiver in dBm for a

12-mA diode drive current?

R=0.00123WA

L=8 dB

I=12mA

R=ρ0

I

0.00125=ρ0

12m

ρ0=0.00125×12m

¿150 μW

Power Receive in dBm,

ρ0=10 logρ0

¿10 log150 μ

¿−8.239dBm