Exam-1_Student_S07_Sol

Exam-1_Student_S07_Sol - ECE 271 Exam-I (02/26/2007)...

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ECE 271 Exam-I (02/26/2007) Solutions Prof. Hanafi (1) (15 points) Use voltage and current division to find V 1 , V 2 , I 2 , and I 3 in the circuit in Fig. (1) if V = 18 V, R 1 = 39 k Ω , R 2 = 43 k Ω , and R 3 = 11 k Ω . Figure (1) Solution. V = 18 V, R 1 = 39 k Ω , R 2 = 43 k Ω and R 3 = 11 k Ω . V + - V 1 V 2 + - R 1 R 2 R 3 I 3 I 2 V 1 = 18 V 39 k Ω 39 k Ω+ 43 k Ω 11 k Ω () = 14.7 V V 2 = 18 V 43 k Ω 11 k Ω 39 k 43 k Ω 11 k Ω = 3.30 V I 2 = I 1 11 k Ω 43 k 11 k Ω = 18 V 39 k 43 k Ω 11 k Ω 11 k Ω 43 k 11 k Ω = 76.7 μ A I 3 = I 1 43 k Ω 43 k 11 k Ω = 18 V 39 k 43 k Ω 11 k Ω 43 k Ω 43 k 11 k Ω = 0.300 mA
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(2) (10 points) The maximum drift velocity for holes in silicon is 10 7 cm/s. If the hole density in a sample is 10 18 / cm 3 , what is the maximum hole current density? If the sample has a cross section of 2 μ m × 25 μ m, what is the maximum current? Solution j p = qpv p = 1.602 x 10 19 C () 10 18 cm 3 10 7 cm s ⎟ = 1.60 x 10 6 A cm 2 i p = j p A = 1.60 x 10 6 C A cm 2 2 x 10 4 cm 25 x 10 4 cm = 0.80
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This note was uploaded on 04/07/2008 for the course ECE 271 taught by Professor Hanafi during the Spring '05 term at NJIT.

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Exam-1_Student_S07_Sol - ECE 271 Exam-I (02/26/2007)...

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