HW SOLN 5

# HW SOLN 5 - ECE331 Homework#5 Solution 1 Streetman 3.2 The...

This preview shows pages 1–4. Sign up to view the full content.

1 ECE331 Homework #5 Solution 1. Streetman, 3.2 The probability that a state at E above E f is occupied is ) / exp( 1 1 ) / ) exp(( 1 1 ) ( kT E kT E E E E E f f f f Δ + = Δ + + = Δ + The probability that a state at E below E f is empty is ) / exp( 1 1 ) / ) exp(( 1 1 1 ) ( 1 kT E kT E E E E E f f f f Δ + = Δ + = Δ

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

View Full Document
2 2. Since Ec-Ed >> kT, not all donors are ionized. The ionization probability is 1-f(Ed). Then the ionized donor ion concentration is 3 14 15 10 733 . 8 ) 0259 . 0 / 05 . 0 exp( 1 1 1 10 ) / ) exp(( 1 1 1 ) ( 1 ( + × = + = + = = cm kT E E N E f N N f d d d d d For carrier concentrations, we have kT E E N n F C C ) ( exp 0 = kT E E N p V F V ) ( exp 0 = So, 10 0 0 10 15 . 1 ) 0259 . 0 / ) 25 . 0 1 . 1 ( exp( ) 0259 . 0 / 25 . 0 exp( ) / ) ( exp( ) / ) ( exp( × = = = kT E E kT E E p n V F F C Based on the space charge neutrality, we have n 0 = p 0 + N d + So, we can solve the carrier concentrations based on the above two equations, n 0 = 8.73×10 14 cm -3 p0 = 7.59×10 4 cm -3 So the intrinsic carrier concentration, 3 9 4 14 0 0 10 14 . 8 10 59 . 7 10 73 . 8 × = × × × = = cm p n n i 3. Since ) exp( 0 kT E E n n i F i = , so Ef-Ei = kTln(n0/ni) = 0.0259×ln(10 16 /1.5×10 10 ) = 0.347 eV 4.
3 5. This material has N c =10 19 cm -3 and N v =5X10 18 cm -3 , and E g =2eV at 300 K. At T = 900 K (627°C), From 2 / 3 2 * 2 2 = h kT m N n c π , we have 3 19 2 / 3 19 2 / 3 1 2 1 2 10 20 . 5 ) 300

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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