Section 4 1 TRƯỜNG ĐIỆN TỪ

Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
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


4.1
Conductors
and Semiconductors


4.1-3

Material Media can be classified as
(1) Conductors
and Semiconductors
electric property
(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.


4.1-4

electron
cloud

free electrons
+ bound
elecrons

nucleus

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

J c E

Ohm’s Law
at a point


4.1-5

 conductivity (S/m)
 e Ne e

 
 h N h e   e N e e

conductors
semiconductor
s

 Mobility
Nh,e = Density of holes (h) or electrons (e)


4.1-6

Ohm’s Law
V  El

V
Jc = E =
l
A
I  Jc A 
V
l
l
V I
A
V  IR Ohm’s Law
l
R=
A

A

l



E, Jc

I
V


4.1-7

D4.1

I
0.1
Jc  
10 3 A m 2
A 10 –4

(a) For cu,

 5.8 10 7 S m
Jc
10 3
E 
17.24 V m

7
 5.8 10
(b)
 h  e  Ne e
1700  3600 10 4 2.5 1013 106
1.602 10 19
2.1229 S m
Jc
103
E 
471.1 V m
 2.1227


4.1-8

(c)

l
FromR 
1
A
l
1
10 6



Sm
–6

RA  10

Jc
10 3
E  6
3.14 mV m
 10 


4.1-9

Conductor in a static electric field

E

E


4.1-10

–

–

–

–

– S = –0E0
E0az

z=d

S0
z=d

z=0

z=0

+

+

+

+

+ S =  0 E 0

–

–

–

–

– S = –0E0

z=d

+

+

+

+

+

S = 0E0

z=0

–

–

–

–

–

S = –0E0

+

+

+

+

+ S = 0E0

E =0

+

+

+

+

+

–

–

–

–

–

E=–
–S0

S0
–
a z  E0 a z 0
0
S0  0 E0

S0
0 a z


4.1-11

P4.3
(a) S0  S1  S2
Ei = 0

S1

z

S2

S1
S2
Ei  –
az 
a z 0

2 0
2 0
1
  S1 S2   S0
2


4.1-12

(b)
E i1 = 0

S11
S12

Ei2 = 0

S21
S22

S11  S12  S1
S21  S22  S2
Write two more equations and solve for the four
unknowns.