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ECE250 HW2S

# ECE250 HW2S - ECE 250 HW#2 Solution Define units and...

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ECE 250 HW #2 Solution Define units and constants eV 1.602 10 19 coul volt := K B 86 10 6 eV K := Specify some constants for silicon E g_Si 1.12 eV := B Si 5.23 10 15 1 cm 3 K 1.5 := q 1.602 10 19 coul := Problem 3.109 Length 10 μ m := Voltage 5 V := E field Voltage Length := E field 51 0 3 × V cm = μ n 1350 cm 2 Vs := μ p 480 cm 2 := v dn μ n E field := v dn 6.75 10 4 × m s = v dp μ p E field := v dp 2.4 10 4 × m s = Problem 3.111

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μ n 1350 cm 2 Vs := Voltage 1 V := Length 10 μ m := E field Voltage Length := E field 11 0 3 × V cm = J n 1 mA μ m 2 := For a highly doped n-type semiconductor, assume that the current density is carried mostly by the electrons. n1 0 16 1 cm 3 := Initial Guess given J n qn ⋅μ n E field = n find n () := n 4.624 10 17 × 1 cm 3 = Check the current density due to holes. n i 1.5 10 10 1 cm 3 := p n i 2 n := p 486.608 1 cm 3 = μ p 480 cm 2 := J p qp p E field := J p 3.742 10 16 × mA μ m 2 = Pretty small - our approximation was a good one.
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ECE250 HW2S - ECE 250 HW#2 Solution Define units and...

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