EE331HW2solutions - EE-331 Devices and Circuits 1 Prof R B...

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n 200K 8.797 10 10 × cm 3 = n 200K n i 200 K () 2 N A := n 300K 7.539 cm 3 = n 300K n i 300 K 2 N A := For both 300K and 200K, the majority carrier hole concentration is just equal to N A = 6 x 10 18 cm -3 . However, the minority carrier electron concentrations will depend upon n i (T): n i 200 K ( ) 7.265 10 4 × cm 3 = n i 300 K ( ) 6.725 10 9 × cm 3 = n i T ( ) 1.08 10 31 K 3 cm 6 T 3 exp 1.12 eV k B T 0.5 := (a) This is p-type silicon; holes are the majority carrier; electrons are the minority carrier. (b,c) For this problem, we need the intrinsic carrier concentration at two different temperatures. The more precise formula for n i (T) in Eq. (2.1) will be used to compute this: (Boron) N A 610 18 cm 3 := Problem 3: Problem 2.27(a,b,c) of Jaeger and Blalock. J max 1.602 10 5 × amp cm 2 = J max qv dmax N D := Since N D >> n i , this will be n-type material and electron conduction will dominate over hole conduction. The maximum drift current density (magnitude) is then simply N D 10 17 cm 3 := v dmax 10 7 cm sec 1 := Problem 2: Problem 2.24 of Jaeger and Blalock. (a) When Ge from group IV replaces In from group III, it has one too many outer shell electrons, so it behaves like an donor. (b) When Ge from group IV replaces P from group V, it has one too few outer shell electrons, so it behaves like a acceptor. Germanium is known as an amphoteric dopant in InP, because it can function as either a donor or an acceptor, depending upon where in the crystal lattice it is located. Problem 1: Problem 2.22(a,b) of Jaeger and Blalock. h 6.626 10 34 joule sec := q 1.602 10 19 coul := c 2.998 10 10 cm sec 1 := k B 8.62 10 5 eV K 1 := eV 1.602 10 19 joule := µ m1 0 6 m :=
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EE331HW2solutions - EE-331 Devices and Circuits 1 Prof R B...

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