1 ENE 325 Electromagnetic Fields and Waves Lecture 5 Conductor, Semiconductor, Dielectric and...

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1 ENE 325 ENE 325 Electromagnetic Electromagnetic Fields and Waves Fields and Waves Lecture 5 Lecture 5 Conductor, Conductor, Semiconductor, Dielectric and Semiconductor, Dielectric and Boundary Conditions Boundary Conditions

Transcript of 1 ENE 325 Electromagnetic Fields and Waves Lecture 5 Conductor, Semiconductor, Dielectric and...

Page 1: 1 ENE 325 Electromagnetic Fields and Waves Lecture 5 Conductor, Semiconductor, Dielectric and Boundary Conditions.

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ENE 325ENE 325Electromagnetic Electromagnetic Fields and WavesFields and Waves

Lecture 5Lecture 5 Conductor, Conductor, Semiconductor, Dielectric and Semiconductor, Dielectric and Boundary ConditionsBoundary Conditions

Page 2: 1 ENE 325 Electromagnetic Fields and Waves Lecture 5 Conductor, Semiconductor, Dielectric and Boundary Conditions.

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Review (1)Review (1)

The electric potential difference Vba is a work done by an external force to move a charge from point a to point b in an electric field divided by the amount of charge moved.

The electric potential is the same no matter which routes are used. Only displacement distance (shortest route) matters

b

baa

WV E dL

Q ����������������������������

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Review (2)Review (2) Conductors and Ohm’s law

Current, I is defined as the amount of charge that passes through a reference plane in a given amou

nt of time. Ampere.

Current density, J is defined as the amount of cu rrent per unit area A/m2

the relationship between I and J , convection current conduction current

dQI

dt

I J S ����������������������������

sI J S

����������������������������

vJ v����������������������������

J E����������������������������

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OutlineOutline

Conductor and boundary conditions Semiconductor and insulator

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Conductors and boundary Conductors and boundary conditionsconditions

charge density is zero inside a conductor. Surface charge density DS is on the

conductor surface. An electric field inside a conductor is zero.

outside charges cause an electric field. Ds=0

E=0

++

+

+

++

+

+

+

+

+

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Tangential and normal fields. Tangential and normal fields.

The electric field on the surface can be divided into two components. tangential electric field, Et, = 0 for an

equipotential surface normal electric field, En

EnEt

E

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Boundary conditions (1) Boundary conditions (1)

Consider a conductor-free space boundary

From

0E dl ����������������������������

a b

cd

h

w

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Boundary conditions (3) Boundary conditions (3)

Consider Gauss’s law

From

D dS Q ����������������������������

s

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Boundary conditions (3) Boundary conditions (3)

For conductor-free space boundary conditions (B.Cs.)

Dt = Et = 0

Dn = 0

En = s

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Ex1Ex1 Let V Let V and let a point and let a point PP(0.1, (0.1, /12, /12, /24) /24) locate at the conductor-free space locate at the conductor-free space boundary. At point boundary. At point PP, determine, determine

a) V

b) E

5100 sin(3 )cos(4 )xV e y z

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c) En

d) Et

e) S

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SemiconductorsSemiconductors

Electron and hole currents conductivity mobility is 10-100 times higher than

conductor. electron and hole density depend on

temperature. Doping is the process of adding impurities

to a semiconductor to alter the polarity.

e e h h

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Dielectric or insulator Dielectric or insulator

no free charge microscopic electric dipoles energy stored capability polar and non-polar molecules

+

-+ - +

+

+

+

+

+

+

-

-

-- -

polar molecules non-polar molecules

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Alignment of dipoles with E Alignment of dipoles with E field field

+

-

+

-

+

-

+

-

+

-

+

-+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

E

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Polarization Polarization

Each dipole has its dipole moment,

where Q is the positive one of the two bound charges is the vector from the negative to the positive charge.

where n = number of dipoles per volume.

Polarization is dipole moment per unit volume,

p��������������

p Qd C m ����������������������������

1

n v

itotali

p p

��������������

2

0 1

1lim /

n v

iv i

P p C mv

��������������

d��������������

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Equivalent polarization Equivalent polarization

The movement of bound charges when induced by an electric field causes changes in surface and volume charge densities.1. Equivalent polarization surface charge density, s

2. Equivalent polarization volume charge density, v

2/ns P a C m ��������������

+-

+-

+-

+-

+-

+-

+-

+-

+-

+-+-

+-+-

+-+-

E

+-

+-+-

+-+-+-

+-+-+-

+-

+-+-+-

+-

+-

an

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Electric flux density in a Electric flux density in a dielectric material (1)dielectric material (1) can be calculated from free and bound charges.can be calculated from free and bound charges.

Net bound charges flowing out of the closed surfaces,

bS

Q P dS C ����������������������������

Let

where QT = total charge Q = free charge

T bQ Q Q

+

-+

-

+

-

D��������������

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Electric flux density in a Electric flux density in a dielectric material (2)dielectric material (2)

From

then

Let

we can write

where is an electric flux density in a dielectric material.

0Ts

Q E dS ����������������������������

0( ) .T bs

Q Q Q E P dS ������������������������������������������

0D E P ������������������������������������������

sQ D dS

����������������������������

D��������������

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Equivalent divergence Equivalent divergence relationshipsrelationships

From

use a divergence theorem, we have

Since

then

We can also show that

therefore

,nbs

Q P a dS ����������������������������

.s vP dS Pdv ������������������������������������������

b vQ dv

.v bP ��������������

0 ,TE ��������������

.T b vD ��������������

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Electric flux density in Electric flux density in dielectric medium (1) dielectric medium (1)

If the dielectric material is linear and isotropic,the polarization is proportional to the electric

field . P��������������

E��������������

0 eP E ����������������������������

where e is an electric susceptibility. Then 0 0 .

������������������������������������������eD E E

Let 1r e

where r is a relative permittivity or a dielectric constant.

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Electric flux density in Electric flux density in dielectric medium (2) dielectric medium (2)

So we can write

0rD E ����������������������������

o r

D E����������������������������

where F/m. 0 r.

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Ex2Ex2 A dielectric material has an A dielectric material has an electric susceptibility electric susceptibility ee =0.12 and =0.12 and has a uniform electric flux density has a uniform electric flux density DD = 1.6 nC/m = 1.6 nC/m22 , determine , determine

.

a) E

b) P

c) Average dipole moment if there are210x 19 dipoles/m3

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Boundary conditions for dielect Boundary conditions for dielect ric materials ric materials: tangential fields: tangential fields

It is useful to determine the electric field in a dielectric medium.

Dielectric-Dielectric

w

h

1 2

Et1 Et2

Et1 =Et2

and 1 1

2 2

t

t

D

D

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Boundary conditions for dielect Boundary conditions for dielect ric materials ric materials: normal fields: normal fields

Dielectric-Dielectric

s

Dn1 = Dn2

and

1

En1 = 2

En2.

1 221 ( ) Sa D D

����������������������������

For a free of charge boundary,

21ais the unit vector pointing from

medium 2 to medium 1.

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Use B.C.s to determine Use B.C.s to determine magnitude of magnitude of and and

��������������D

��������������E

Dn1

Dn2

Dt1q1

q2

D t2

D2

D1

12

22 22

2 1 1 121

cos sinD D

q q

22 21

2 1 1 122

sin cosE E

q q

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Ex3Ex3 The isotropic dielectric medium The isotropic dielectric medium with with r1r1 = 3 and = 3 and r2r2 = =22 is connected as s is connected as s

hown. Given hown. Given V/m, determine and its m V/m, determine and its m

agnitude, and its magnitude, agnitude, and its magnitude, qq11

, and , and qq

22..

1 5 4x y zE a a a

��������������2E

��������������

2D��������������

Z

-Z

r1 3

r2 2x-y

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Boundary conditions for diel Boundary conditions for dielectricectric-conductor-conductor

h

wEt

En

conductor

dielectric

Dt = Et =0Dn = En = s

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ExEx44 - Between a dielectric conductor inter - Between a dielectric conductor inter face has a surface charge density of face has a surface charge density of ss = =

2 10x2 10x -9-9 C/m C/m22 . Given . Given V/m, determine V/m, determine ..

1 30 50 70

��������������x y zE a a a 2E

��������������