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 and Telecommunication > Sem7 > Optical Communication and Networks

Marks: 10M

Year: May 2014

Given: η=65%=0.65

$E_p= 1.5 ×10^{-19}J$

$I_p= 2.5 µA$

To find: Operating wavelength

Incident optical power $P_in$

Solution:

$$R = \frac{ɳqλ}{hc}$$

$$hν = E_p \ \ \ and \ \ \ ν= \frac{c}λ$$

$$λ= \frac{hc}{E_p} = \frac{6.62×10^{-34}.3×10^8}{1.5 ×10^{-19} }=0.21µm$$

$$∴ λ=1.324 \ \ µm $$

$$R =\frac{ɳq}{hν}= \frac{ɳq}{E_p} = {0.65 × 1.6 ×10^{-19} }{1.5 ×10^{-19} }= 0.69$$

$$∴R=0.69$$

$$R = \frac{I_p}{P_{in}} = \frac{2.5×10^{-6}}{P_{in}} =0.69$$

$$∴ P_{in}=3.6 µW $$