EE2403 Exam 1 Solutions.pdf - Exam 1 Solutions In prob 3 exam1a had ID3/ID1 and exam1b had ID5/ID1 1 a Charge neutrality requires xn n = xp p x n ND = x

# EE2403 Exam 1 Solutions.pdf - Exam 1 Solutions In prob 3...

• 3

This preview shows page 1 - 3 out of 3 pages.

Exam 1 Solutions In prob. 3, exam1a had I D3 /I D1 and exam1b had I D5 /I D1 . 1. a) Charge neutrality requires: x n n = x p p x n N D = x p N A 2 . · 10 17 = x p · 5 10 14 x p = 2 . 5 . · 10 3 = 400 μ m b) W total = 402 μ m. c) When voltage is applied. C = C 0 radicalbig 1 . - V a /V bi = C 0 radicalbig 1 . + 4 . 5 / 1 . 5 = C 0 2 W = K × (1 /C ) K is a proportionality constant. = 402 · 2 . = 804 μ m For exam1b, the answer is the same. 2. The number free carriers in a semiconductor obeys the mass action law independently on both sides of the junction. On the p side: p p n p = n 2 i p p · 10 8 = (1 . 5 · 10 10 ) 2 p p = 2 . 25 · 10 12 cm - 3 For exam1b, p p n p = n 2 i p p · 10 7 = (1 . 5 · 10 10 ) 2 p p = 2 . 25 · 10 13 cm - 3 1

Subscribe to view the full document.

3. I D3 = I S exp( V D3 /V T ) I D1 = I S exp( V D1 /V T ) I D3 I D1 = exp[( V D3 /V T ) - ( V D1 /V T )] = exp[ V D3 - V D1 ] /V T ] V D3 - V D1 V T = ln bracketleftbigg I D3 I D1 bracketrightbigg V D3 - V D1 = Δ V = V T ln bracketleftbigg I D3 I D1 bracketrightbigg = 0 . 259 × ln(3) = 28 . 6 mV For exam1b, the answer is: V D5 - V D1 = Δ V = V T ln bracketleftbigg I D5 I D1 bracketrightbigg numerically the answer would be 41.68 mV. 4. Change all dimensions to cm. 1 cm = 10 - 2 m. 1 μ m = 10 - 6 m. 1 μ m = 10
• Fall '13

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask 0 bonus questions You can ask 0 questions (0 expire soon) You can ask 0 questions (will expire )
Answers in as fast as 15 minutes