EEE340 Lecture 18 1

0 Y - Y"2).

0 X X" 1).

Let us assume 20

From 1).

xgxg

eCeCxX

XX

sc

xjxi

sincos

)(

0"

10

2

EEE340 Lecture 18 2

Boundary conditions:

xb

ngx

b

n

nb

bbgX

gggX

n

sbx

cscx

sin)(X

4,... 3, 2, 1,n

0sin0sin0

00010

n

0

0 b

a

x

yVo

EEE340 Lecture 18 3

yhy

Y

s

sinhcoshh Y(y)

0- Y").2

c

2

Boundary conditions:

0000 cscoy hhhY

Hence, yb

nsinhh)y(Y s

π

or ... 2, 1,n yb

nsinhh)y(Y nn

π

EEE340 Lecture 18 4

yhy

Y

s

sinhcoshh Y(y)

0- Y").2

c

2

Boundary conditions:

00000 cscy hhhY

Hence, y

b

nsinhh)y(Y s

π

or

... 2, 1,n yb

nsinhh)y(Y nn

π

0 b

a

x

Vo

EEE340 Lecture 18 5

b

ynsinh

b

xnsinhg

)y(Y)x(X)y,x(V

nn

nnn

ππ

2. Find unknown constant Cn. 1.) Superposition (Linear Space)

b

ynsinh

b

xnsinC)y,x(V

1nn

ππ

nnn ghC Let

2.) Boundary condition (to find Cn)

b

an

b

xnC

VyxV

nn

ay

sinhsin

),(

1

0

(1)

EEE340 Lecture 18 6

3.) Orthogonality of the Fourier SeriesFourier invented his series and his transform from his work of the PDE for heat transfer.

nm 0nm 10

{sinsin2

dxb

nx

b

mx

b

b

Multiplying (1) by

xb

msindx

b

0

π

0b00

b

0 b

xmcos

m

bV

b

xmsin Vdx.S.H.L

ππ

π

)cos1(0

mm

bV

EEE340 Lecture 18 7

)sinh(22

)sinh(

2

2cos1

sinh

sinsinh...

0

0

2

b

ambCb

b

amC

dxx

bm

b

amC

dxb

xm

b

amCSHR

mm

b

m

b

m

i.e.,

1,3,5,...m 4

... 6, 4, 2,m 0

0

0

{

)cos1(2

sinh

m

V

m mm

V

b

amC

EEE340 Lecture 18 8

evenn 0

oddn

ban

sinhn

V4

C

o

n

ππ

,...5,3,1

0

sinh

sinhsin4),(

n

ban

n

ybn

bxn

VyxV

Hence,

Substituting Cn into (1), we obtain

(4-113)

(4-114)

EEE340 Lecture 18 9

EEE340 Lecture 18 10

Chapter 5: Steady Electric CurrentsCharges in motion constitute current flow.

Electric current consists of three types:

Conduction current: current in a wireOhm’s law

Convection Current:

Displacement Current:

5-2: Current Density and Ohm’s LawTotal current flowing through an arbitrary surface S

usually, we choose a cross-section

EJVI R1

σ

uJ v

ρ

t

DJ

S

)A( sdJI

Vacuum tube

Capacitor

(5.5)

EEE340 Lecture 18 11

Example 5-1: Vacuum-tube diode

Electron cloud

where (y) is negative.

The velocity is

Newton’s law

Hence

where

)y(u)y(yJ ρ

)y(uyu

dy

dVe)y(eE

dt

dum

2

2

1mu

dy

d

dy

dumu

dt

dy

dy

dumLHS

eVmu2

1 2

0V and 0u0y0y

(5.9)

(5.11)

(5.8)

y

Cathode

Anode

JE

0

(5.10)

EEE340 Lecture 18 12

Therefore

On the other hand, from Poisson’s Equation,

where

Therefore

o

vV

2

21

2 V

e

mJ

u

Jv

21

22

2 V

e

mJ

dy

Vd

oo

v

(5.14)

21

Vm

e2u

(5.12)

(5.13)

EEE340 Lecture 18 13

Solving (5.14), one obtains

Or

Child-Langmuir law.

The I-V curve can be plotted (non-linear)

de2

mJ2V

3

4 41

43

oo

ε

2o2o

mA V

m

e2

d9

4J 2

3ε (5.17)