HW 8 Solution

HW 8 Solution - HW 8 Solution Problem 1: a) Photoconductor:...

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HW 8 Solution Problem 1: a) Photoconductor: first consider the dark current. For a p-type sample, we have a resistivity ρ = 1/ (q μ P p) and a resistance R = ρ L/A Using the stated parameters, along with the Einstein relationship (D = kT/q μ), we can calculate μ P = 400 cm2/Vsec, ρ = 1.563 ohm cm and R = 78.1 K ohm. For a bias of 1V, the dark current is 12.8 μA. For the light current, we should consider the resistivity of the sample under illumination. The procedure is the same as in the dark, with the exception that the resistivity will now be ρ = 1/ q (μ P (p 0 + p) + μ N n p ) Using our special-case solution of minority carrier diffusion equation: n p = G L τ n = G L L n 2 / D N Which yields n p = 5.5 x 10 12 cm-3. Note for this steady-state, uniform illumination, we also have p p = n p . The new rho is 1.546 ohm cm and R = 77.3 kohm. Therefore, the ratio between light and dark currents is 1.01 . b) Diode: again, we consider the dark current first. For the specified reverse bias point, we have I 0 = q A ni 2 ( D N /L N N A + D P /L P N
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HW 8 Solution - HW 8 Solution Problem 1: a) Photoconductor:...

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