Ask
Login
0
4.0kviews
Incident optical power
written 10.1 years ago by gravatar for teamques10 teamques10 ★ 70k •   modified 10.1 years ago

A photodiode has quantum efficiency of 65% when a photon of energy of $1.5 x 10^{-19}J$ are incident upon it. At what wavelength is the photodiode operating?

Calculate the incident optical power required to operate photo current of 2.5 µA when the photodiode is operating as above.

Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication

Marks: 10M

Year: May 2014
optical fiber communication
ADD COMMENT EDIT
1 Answer
0
34views
written 10.1 years ago by gravatar for teamques10 teamques10 ★ 70k

Given: $η=65\%=0.65$

$$E_p= 1.5 ×10^{-19}J \\ \; \\ I_p= 2.5 µA \\ \; \\ $$

To find: Operating wavelength λ

Incident optical power ${P_{in}}$

Solution:

$R = \frac{ɳqλ}{hc} \\ \; \\ hν = E_p and ν= \frac{c}λ \\ \; \\ λ= \frac{ hc}{E_p} = \frac{6.62×10^{-34}∙3×10^8}{1.5 ×10^{-19} }=0.21µm \\ \; \\ ∴ λ=1.324 µm \\ \; \\ R =\frac{ɳq}{hν}= \frac{ɳq}{E_p} = \frac{0.65 × 1.6 ×10^{-19} }{1.5 ×10^{-19} }= 0.69 ∴ R=0.69 \\ \; \\ R = \frac{I_p}{P_{in}} = \frac{2.5×10^{-6}}P_{in} =0.69 \\ \; \\ ∴ P_{in}=3.6 µW \\ \; \\ $

ADD COMMENT EDIT
Please log in to add an answer.