lecture34

# lecture34 - Lecture 34 Wednesday April 29 EE 332...

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

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.

Unformatted text preview: Lecture 34 Wednesday April 29 EE 332 Semiconductor Devices Omit section 10.2 Problem Assignment Wednesday Æ 9.19 and 9.23 Friday Æ 10.1 and 10.3 Hybrid-Pi Circuit Model Identify each element with our model of a BJT AC Analysis and Modeling Write the total voltage (v EB ) as a sum of the DC component (V EB ) and an AC component (v eb ). eb EB EB v V v + = In the hybrid-pi model r π is given by: ( ) E kT v V q E pE E pE B W e p D qA I I eb EB / + = ≅ be b v i r ∂ ∂ ≅ 1 π The base current is approximately equal to I pE because: and α T =1 nE T pE B I I I ) 1 ( α − + = Finding r π The expression for r π in terms of the bias current becomes: C DC B B qI kT qI kT kT qI r β π = = ≅ 1 ( ) kT qI W e p D qA kT q v i B E kT v V q E pE E eb b eb EB = = ∂ ∂ + / The reciprocal of this equation is r π in terms of the device parameters as give in eq. 10.9 Finding r u CB C u v i r ∂ ∂ = 1 The denominator is change in current through the reverse biased collector junction with changing collector to base voltage,...
View Full Document

## This note was uploaded on 06/30/2010 for the course EECE 332 taught by Professor Constable during the Spring '08 term at Binghamton.

### Page1 / 4

lecture34 - Lecture 34 Wednesday April 29 EE 332...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online