330hw2-solutions.pdf

# D p 10 8 cm 2 s 1 kt q μ p 12 in a region of a si

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D P = 10 . 8 cm 2 s - 1 = kT q μ p 12. In a region of a Si sample the doping concentration as a function of position is N D = 5 × 10 16 cm - 3 - 4 × 10 14 cm - 3 μ m - 1 x for 0 x 100 μ m. (a) Find the carrier concentrations n and p as functions of x . 330hw2-solutions.tex Winter 2018

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P. R. Nelson 9 (b) Find an expression for E F - E i as a function of x and plot the band diagram showing all important values. (The plot is best done with a tool such as excel, matlab, mathematica or other program. It is diﬃcult to do with a sketch). (c) Find expressions for the diffusion current densities J N d iff and J P diff as functions of x and plot the results. (d) Find an expression for V as a function of x and plot the result. (e) Find an expression for the electric field as a function of x and plot the result. (f) Find expressions for the drift current densities J N drift and J P drift as func- tions of x and plot the result. (g) Find expressions for the total current densities J N and J P . (h) What would you expect the total current density to be? Do your results agree? Explain your reasoning. The python 3 code in jupyter notebook used to generate the plots is available as a separate file. (a) Let N D = c 1 - c 2 x . n = 5 × 10 16 cm - 3 - 4 × 10 14 cm - 3 μ m - 1 = N D p = n 2 i c 1 - c 2 x = n 2 i n (b) E F - E i = kT ln ( c 1 - c 2 x n i ) = kT ln ( n n i ) 330hw2-solutions.tex Winter 2018
P. R. Nelson 10 (c) Assume that the mobilities are constant, even though they are somewhat a function of dopant concentration. Approximate the average dopant concentration as 2 × 10 16 cm - 3 . Then from the information on page 80 of the text, μ n 1165 cm 2 V - 1 s - 1 and μ p 419 cm 2 V - 1 s - 1 . J N diff = qD N dn dx = - q ( kT q μ n ) c 2 = - 19 . 3 A cm - 2 J P diff = - qD P dp dx = - q ( kT q μ p ) n 2 i c 2 ( c 1 - c 2 x ) 2 10 12 A cm - 2 Note that the total diffusion current is due to electrons, and that it is constant to a very good approximation. The plot is thus irrelevant. (d) Let V ( x = 0) = 0. Then V = - E i ( x ) q + E i ( x = 0) q = - kT q [ ln ( c 1 - c 2 x n i ) - ln ( c 1 n i )] = - kT q ln ( 1 - c 2 c 1 x ) 0 V 42 mV (e) E = - dV dx = kT/q c 1 /c 2 - x 2 . 07 V cm - 1 ≤ E ≤ 10 . 4 V cm - 1 330hw2-solutions.tex Winter 2018

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P. R. Nelson 11 (f) J N drift = - qn ( x ) μ n E = q ( kT q μ n ) c 2 J P drift = qp ( x ) μ p E = q ( kT q μ p ) n 2 i c 2 ( c 1 - c 2 x ) 2 The drift currents are the exact negative of the diffusion currents.
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• Winter '08
• Liu,J
• P. R. Nelson

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