{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Week7HWSolutionsV2

# Week7HWSolutionsV2 - Mark Lundstrom SOLUTIONS ECE 656...

This preview shows pages 1–3. Sign up to view the full content.

Mark Lundstrom 10/5/13 ECE-­‐656 Fall 2013 1 SOLUTIONS: ECE 656 Homework (Week 7) Mark Lundstrom Purdue University (Revised 10/30/13) 1) In Lecture 15, we derived a current equation for a 2D, n-­‐type conductor and wrote it as J n = σ S d F n q ( ) dx . Derive the corresponding equation for a p-­‐type semiconductor. Solution: I = 2 q h T E ( ) M V E ( ) f 1 f 2 ( ) dE −∞ E V (channels in the valence band are all below E = E V .) f 1 f 2 f 1 E qV T E ( ) = λ λ + L λ L I = 2 q h λ E ( ) M V E ( ) f 0 E −∞ E V dE qV L J px = I W qV = −Δ F p J px = 2 q h λ E ( ) M V E ( ) W ( ) f 0 E −∞ E V dE Δ F p L J px = 2 q h λ E ( ) M V E ( ) W ( ) f 0 E −∞ E V dE dF p dx = σ Sp dF p dx J px = σ Sp d F p q ( ) dx σ Sp = 2 q 2 h λ E ( ) M V E ( ) W ( ) f 0 E −∞ E V dE

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

View Full Document
Mark Lundstrom 10/5/13 ECE-­‐656 Fall 2013 2 2) In Lecture 15, we derived the drift-­‐diffusion equation for a 2D n-­‐type semiconductor with parabolic energy bands. Repeat the derivation for a 3D semiconductor with parabolic energy bands. Do not assume Maxwell-­‐Boltzmann statistics. Solution: Begin with: J nx = σ dF n q dx (i) n = N C F 1/2 η F ( ) η F = F n E C ( ) k B T N C = 1 4 2 m * k B T π 2 3/2 Now find the gradient of the electrochemical potential: dn dx = N C d d η F F 1/2 η F ( ) d η F dx = N C F
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}