EE331W11E1solutions

EE331W11E1solutions - EE-331 Devices and Circuits 1 Prof....

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
V bi 0.811 volt = W 2 ε r ⋅ε o V bi v D () q 1 N A 1 N D + := W 2.046 10 5 × cm = W 0.205 µ m = (b) depletion capacitance: Ad h := A 1.00 10 2 × cm 2 = C j ε r ε o A W := C j 506.4 pF = (c) reverse saturation current: I Sp qA V T µ p N D L p n i 2 := I Sp 3.204 10 16 × amp = (holes) I Sn V T µ n N A L n n i 2 := I Sn 4.005 10 13 × amp = (electrons) I S I Sp I Sn + := I S 4.008 10 13 × amp = Electrons dominate the current flow. (d) diode current: i D I S exp v D V T 1 := i D 78.5 mA = EE-331 Devices and Circuits 1 Prof. R. B. Darling Exam # 1 Solutions Problem 1: pn-Junction Diode Calculations. eV 1.602 10 19 joule := µ m1 0 4 cm := ps 10 12 sec := n i 1.0 10 10 cm 3 := N A 5.0 10 15 cm 3 := N D 2.5 10 18 cm 3 := V T 25 mV := q 1.602 10 19 coul := ε r 11.7 := ε o 8.854 10 14 farad cm 1 := k B 1.38 10 23 joule K 1 := µ n 500 cm 2 volt 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

EE331W11E1solutions - EE-331 Devices and Circuits 1 Prof....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online