1 answer

Homework 5 Duc 02/21/20 by 7 pm Problem 2: Determining doping density in Si using the...

Question:

Homework 5 Duc 02/21/20 by 7 pm Problem 2: Determining doping density in Si using the bisection method (Chapra, Problem 5.16)

Homework 5 Duc 02/21/20 by 7 pm Problem 2: Determining doping density in Si using the bisection method (Chapra, Problem 5.16). The resistivity pof doped silicon is based on the charge q of an electron, the electron density, and the electron mobility u. The electron density is given in terms of the doping density and the intrinsic electron (carrier) density n. The electron mobility is described by the temperature T, the reference temperature To, and the reference mobility Ho. The equations required to compute the resistivity are p=qhy where n=:(w+ lw= +9mi) u = "» (?)** n = 4.7x10"T? exp(**, 24,1 [em] Determine N (in cm), given To = 300 K, T = 400 K, H = 1360 cm(V s)', q = 1.6022 x 109 C, k, = 8.62x107x and a desired resistivity p=4.5 x 102 cm. For Si, the energy bandgap (Eg) around that temperature is 1.1 eV. Start by using the incremental search method to find a physically reasonable bracket and then use the bisection method for finding the root. Take into consideration proper unit analysis for these calculations.

Answers

Here, μ0=1360cm2(Vs)-1

T=400 K

T0=300K

Determine the value of μ.

l = o(T) -2.42 = 1360 (40) -2.42 = 1360 (0.4985) = 677.934

ρ=4.5 x 102

q=1.6022 x 10-19 C

Determine the value of n.

p=quipe 4.5 x 102 = 1.6022x10-197(677.934) n = 4887.836 x 10-17 = 2.046 x 1013

kB=8.62 x 10-5 eV/K

Eg=1.1eV

Determine the value of ni

ni= 4.7 x 105Tie(3) = 4.7 x 1015 (400)} (.62x10-5}<a = 37.6 x 1018e- = 37.6 x 1018e-15.951 = 4.426 x 1012

Determine the value of N.

n=} (N + V N2 + 4n;2) 2.046 x 1013 = { (N+V N2 + 4(4.426 x 1012) 2) 4.091 x 1013 = N + N2 + 78.358 x 1024 (4.091 x 1013 – N)?

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