ECE474S09_Lecture31 - = 0.607 eV Neutral region p-side...

Info iconThis preview shows pages 1–20. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 474: Principles of Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu
Background image of page 1

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

View Full DocumentRight Arrow Icon
Lecture 31: Chp. 05: Current State-of-Art Transistors Introduction to pn junctions pn junction no battery: No battery V0: OFF Calculate V0 (3 ways!) Calculate maximum E-field E0 Calculate depletion region width W (xp0, xn0) pn junction with battery: Battery in Forward bias Vf: ON Battery in Reverse bias Vr: really good OFF
Background image of page 2
pn junction: another Built-in E-field E F p 0 = 10 13 cm -3 Junction Depletion region W n 0 = 10 17 cm -3 B B B- B- E F P+ P+ P P P P P P P Neutral region p-side Neutral region n-side E i E i E (x)
Background image of page 3

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

View Full DocumentRight Arrow Icon
pn junction: new: “Energy Barrier” E F p 0 = 10 13 cm -3 Junction Depletion region W n 0 = 10 17 cm -3 B B B- B- E F P+ P+ P P P P P P P qV 0 = 0.2 eV + 0.407 eV
Background image of page 4
Background image of page 5

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

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

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

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

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

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

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

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

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

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

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

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

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

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19

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

View Full DocumentRight Arrow Icon
Background image of page 20
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 0.607 eV Neutral region p-side Neutral region n-side E i E i E (x) pn junction: Evaluate the Energy Barrier E F E i E V E C E gap = 1.1 eV E i E V E C E gap = 1.1 eV 0.407 eV 0.2 eV p = 10 13 cm-3 n = 10 17 cm-3 E F qV = 0.2 eV + 0.407 eV = 0.607 eV Therefore: V = 0.607 V contact potential by definition an equilibrium quantity p. 172 Another way to evaluate V : Another way to evaluate V : n(x) Another way to evaluate V : Another way to evaluate V : n(x) Example problem V0: Built in E-field E(x): Built in E-field E(x): Built in E-field E(x): Built in E-field E(x): -x p0 Built in E-field E(x): Built in E-field E(x): |E(x ) |-magnitude changes across the depletion region E(x )-direction is right to left...
View Full Document

Page1 / 20

ECE474S09_Lecture31 - = 0.607 eV Neutral region p-side...

This preview shows document pages 1 - 20. Sign up to view the full document.

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