Answer to Question #214976 in Optics for Soham Samanta

Question

Photons of wavelength 0.8 µm are incident on a p-n photodiode at a rate of 5 x 

10¹⁰ s-1 and on average the electrons are collected at the terminals of the diode at 

the rate of 1.5 x 10¹⁰ s-1. Calculate (a) The quantum efficiency (%) (b) band gap (in 

eV) of the semiconductor material and (c) the mean output photo current for 12 

µW incident optical power. 

Answer 1

Answer:-

a)$\eta=\frac{r_e}{r_p}=\frac{1.5\times10^{10}}{5\times10^{10}}$ =0.3 = 30%

b)$E_g=1k(log^p_e)(\frac{1}{t})\\$

$=\times 8.6\times10^{-15}\times2.30\times 10^3log^p_{10}(\frac{1}{t})\\ =2\times 8.6\times 10^{-5}\times2.3\times 10^3\times slope$

k=Boltzmann constant

T = temperature in kelvin

Slope = $log^p_{10}(\frac{10^5}{t})\\$

$p=(0.213\times v )(I)$ where I is current

c) $R=\frac{Photo \ current }{incident \ optical power }=\frac{I}{12\mu W}$