Section 4 1 TRƯỜNG ĐIỆN TỪ

No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]

to supplement “Elements of Engineering

Electromagnetics, Sixth Edition”

by

Nannapaneni Narayana Rao

Edward C Jordan Professor of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign, Urbana, Illinois, USA

Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India

Conductors and Semiconductors

Material Media can be classified as

(1) Conductors

and Semiconductors

(2) Dielectrics

(3) Magnetic materials – magnetic property

Conductors and Semiconductors

Conductors are based upon the property of

conduction, the phenomenon of drift of free

electrons in the material with an average drift velocity proportional to the applied electric field

electric property

electron cloud nucleus

free electrons + bound

elecrons

In semiconductors, conduction occurs not only by electrons but also by holes – vacancies created by detachment of electrons due to breaking of

covalent bonds with other atoms

The conduction current density is given by

Jc E Ohm’s Lawat a point

 conductivity (S/m)

 e N e e

h N h e  e N e e







conductors semiconductor s

N h,e = Density of holes (h) or electrons (e)

V  El

l

V  I l

 A

V  IR

R = l

 A

Ohm’s Law

Ohm’s Law

l A

I

V

E, Jc



(a) For cu,

(b)

3

5.8 107 17.24 V m

19

3

1.602 10 2.1229 S m

2.1227

c

N e

J E









(c) FromR  l

A 1

  l

1

 10–6 10

6

 S m

3

Conductor in a static electric field

E

E

– S0

0 az  E0az 0

S0 0E0

S = –0E0

S =  0E0

S = –0E0

S =  0E0

S = – 0E0

S =  0E0

+ + + +

– – – –

z = d

z = 0

E0az

+

–

z = d

z = 0

+ + + + +

+ + + + +

– – – – –

– – – – –

E = 0

S0

–S0

E = – S0

 0 az

z = d

z = 0

+ –

+ –

+ –

+ – +

–

(a) S0 S1  S2

Ei – S1

20 az 

S2

20 az 0

 S1 S2  1

2S0

Ei = 0 S1 z

S2

Ei1 = 0

S11

S12

Ei2 = 0 S21

S22

S11  S12 S1

S21  S22 S2

Write two more equations and solve for the four unknowns.