7.5 Consider a 100 m thick p-doped c-Si wafer illuminated with a monochromatic l

Question

7.5 Consider a 100 m thick p-doped c-Si wafer illuminated with a monochromatic light at a wavelength of 650 nm as illustrated in Figure 7.8. The optical complex refractive index (n = n-ik) of the c-Si at 650 nm is n 3.84-0.0151. The incident irradiance is 1,000 W/m2. The absorption coefficient is given by 4k/A. Calculate: (a) The absorption coefficient at 650 nm. (b) The reflectance at interface air/Si (assume Hair 1) (c) The photon flux after reflection at x 0 and x = 50 m. (d) The generation rate GL at x = 50 m. (e) The excess of minority carriers,an at x = 50pm in the p-doped wafer. Assume the following steady-state conditions: sample is uniformly illuminated along y dir- ection as shown in the figure; dominant thermal recombination and generation process and condition of low injection level and finally there is no current flowing through the wafer, which means: an =0. Ot diff Ot drift

Explanation / Answer

Answer:-a) The value of absorption coefficient at lambda = 650 x 10-9 m is = 4x3.14x0.015 / 650x10-9 = 2889993.168

b) We have nair = 1 and |nsi| = | 3.84 - 0.015i | = 3.84, So reflectance at interface is = [ (3.84 - 1) / (3.84 + 1) ]2 = 0.344. It means 34.4% power is reflected.

c) photon flux N0 is the ratio of spectral irradiance to the photon energy. Photon energy is equal to hc/lambda. So photon energy = (6.626x10-34 x 3 x 108) / 650 x 10-9 = 3.0581x10-19 joule. Hence at x=0 the photon flux is(after reflection) = ( (1-0.344) x (1000) / (3.0581 x 10-19 ) ) = 2.145x1021 photon m-2 s-1 .

At x = 50 x 10-6 m, photon flux = 2.145x1021 x e-(alpha x 0.00005) = 1.08 x 1015 photon m-2 s-1 .

d) The genertion rate is = alpha x photon flux = 2889993.168 x 1.08 x 1015 = 3.138 x 1020 m-3 .

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