# Fundamentals of Microelectronics

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Unformatted text preview: 2.1 (a) k = 8 . 617 × 10 − 5 eV / K n i ( T = 300 K) = 1 . 66 × 10 15 (300 K) 3 / 2 exp bracketleftbigg − . 66 eV 2 (8 . 617 × 10 − 5 eV / K) (300 K) bracketrightbigg cm − 3 = 2 . 465 × 10 13 cm − 3 n i ( T = 600 K) = 1 . 66 × 10 15 (600 K) 3 / 2 exp bracketleftbigg − . 66 eV 2 (8 . 617 × 10 − 5 eV / K) (600 K) bracketrightbigg cm − 3 = 4 . 124 × 10 16 cm − 3 Compared to the values obtained in Example 2.1, we can see that the intrinsic carrier concentration in Ge at T = 300 K is 2 . 465 × 10 13 1 . 08 × 10 10 = 2282 times higher than the intrinsic carrier concentration in Si at T = 300 K. Similarly, at T = 600 K, the intrinsic carrier concentration in Ge is 4 . 124 × 10 16 1 . 54 × 10 15 = 26 . 8 times higher than that in Si. (b) Since phosphorus is a Group V element, it is a donor, meaning N D = 5 × 10 16 cm − 3 . For an n-type material, we have: n = N D = 5 × 10 16 cm − 3 p ( T = 300 K) = [ n i ( T = 300 K)] 2 n = 1 . 215 × 10 10 cm − 3 p ( T = 600 K) = [ n i ( T = 600 K)] 2 n = 3 . 401 × 10 16 cm − 3 2.3 (a) Since the doping is uniform, we have no diffusion current. Thus, the total current is due only to the drift component. I tot = I drift = q ( nμ n + pμ p ) AE n = 10 17 cm − 3 p = n 2 i /n = (1 . 08 × 10 10 ) 2 / 10 17 = 1 . 17 × 10 3 cm − 3 μ n = 1350 cm 2 / V · s μ p = 480 cm 2 / V · s E = V/d = 1 V . 1 μ m = 10 5 V / cm A = 0 . 05 μ m × . 05 μ m = 2 . 5 × 10 − 11 cm 2 Since nμ n ≫ pμ p , we can write I tot ≈ qnμ n AE = 54 . 1 μ A (b) All of the parameters are the same except n i , which means we must re-calculate p . n i ( T = 400 K) = 3 . 657 × 10 12 cm − 3 p = n 2 i /n = 1 . 337 × 10 8 cm − 3 Since nμ n ≫ pμ p still holds (note that n is 9 orders of magnitude larger than p ), the hole concentration once again drops out of the equation and we have I tot ≈ qnμ n AE = 54 . 1 μ A 2.4 (a) From Problem 1, we can calculate n i for Ge....
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ch02sol - 2.1(a k = 8 617 × 10 − 5 eV K n i T = 300 K =...

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