emlab 1 6. capacitance. emlab 2 contents 1.capacitance 2.capacitance of a two-wire line...
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EMLAB
1
6. Capacitance
EMLAB
2Contents
1. Capacitance
2. Capacitance of a two-wire line
3. Poisson’s and Laplace’s equations
4. Examples of Laplace’s equation
5. Examples of Poisson’s equation : p-n junction
EMLAB
3
Capacitance
EMLAB
4
0V
0Q
Due to Coulomb force, positive charges rush to the capacitor. As the amount of charges increases, the voltage increases.
If the voltage difference between the terminals of the capacitor is equal to the supply voltage, net flow of charges becomes zero.
Charging capacitor
EMLAB
5Potential distribution near parallel plates
EMLAB
6Capacitance
The magnitude of an electric field is proportional to charges, and voltages are pro-portional to electric field. Hence, charges are proportional to voltages. This propor-tionality constant is called capacitance.
CVQVQQEV
VV
d
d
d
V
QC SS
aE
sE
aE
EMLAB
7
V
QC
h
VolumeS
Capacitor
EMLAB
9
EMLAB
10
SS
S
d
d
SC
d
S
dz
da
dz
da
d
d
V
QC
d
S
SS
dS
S
S
S
S
00
1ˆˆ
)ˆ(ˆ
ˆ
zz
zz
sE
aE
zE
z
x
Example: Capacitance of a parallel plate capacitor
y
EMLAB
11Capacitance from electrostatic energy
22
22
2
2
1
2
1
2
1
2
1
2
1)2(
2
1
2
1)1(
sE
EE
EEED
d
d
V
WC
CVQC
QCVQVdVdVW
ddW
Ve
VV
e
VV
e
SS
S
dzx
d
SC
d
S
d
Sd
V
W2C
ddzˆˆV
Sd2
d2
1d
2
1W
ˆ
2
S
2S
2e
Sd
0
S
2S
V
2
S
V
e
S
zz
EE
zE
Example : parallel plate capacitor
EMLAB
12Capacitance of 2-wire transmission line
PEC
In this problem, capacitance between two parallel wires needed to be obtained whose radius is b. The distance between the center of the wires is 2h.
LρLρ
- +++++
++
+
+-------
-
With finite radii, the distribution of charges on the wires are non-uniform, which prevents the application of Gauss’ law.
-
To simplify the problem, the potential field is approximated by that of two wires with infini-tesimal radii.
--
-
--
- -
-
h2
b:radius
--
-
--
- -
- LρLρ
a2
0radius
EMLAB
14Capacitance : 2-wire line
LL
(a,0,0)
(-a,0,0)
P(x,y,0)
z
x
1
0
10101 ln
2,
2ˆ
charge) line one todue potential(
0r
r
rdV
rL
r
r
L
sErE
1r2r
VV
Equation for equi-potential surfaces
h2 )/(cosh)/(cosh2
)/(coshln2
10
1
0
1
0
22
0
222
1
0
bh
L
bh
L
V
L
V
QC
bhb
bhhV
b
bhheK
L
LL
LL
V
L
If the surfaces of wires are fitted to those equi-potential surfaces, bound-ary conditions on PEC’s are satisfied.
With the surfaces fitted to equi-potential contours, the relation between the volt-age and the charge density is
2
1
12
2
1
1
1
4
22
22
22
22
01
2
0
222
221
1
2
202
0
201
0
10
1
2
1
1
)(
)(
)(
)(log
4log
2
)(,)(
ln2
ln2
ln2
) wires two todue potential(
0
K
Kay
K
Kax
Keyax
yax
yax
yax
r
rV
yaxryaxr
r
r
rr
r
rr
r
rV
L
V
LL
LLL
a2
b:radius
b
bhhK
K
K
h
b
K
Kab
K
Kah
22
11
1
1
1
1
1
,1
2
1
2,
1
1
EMLAB
15Capacitance of a wire above a PEC plane
z
x
2r
LL
(a,0,0)
(-a,0,0)
z
x
1r2r
If image method is adopted, the PEC plane is replaced with a wire with negative charges. Then the problem geometry be-comes that of the previous example.
)b/h(cosh
L2
)b/h(cosh2
L
V
L
V
QC
)b/h(cosh2b
bhhln
2V
b
bhheK
b
bhhK,
1K
K2
h
b
1K
Ka2b,
1K
1Kah
10
1
0
L
LL
1
0
L22
0
L
22V2
1
22
11
1
1
1
1
1
L
0
In calculating capacitance, it should be noted that the voltage difference is V as compared with 2V of the previous example.
L
(a,0,0)
1r
EMLAB
16
Poisson’s and Laplace’s equations
EMLAB
17Derivation of Poisson’s & Laplace’s equations
EDD ,
)equationsPoisson'(2
V
VE
)()( VE
VV 2
for homogeneous medium
2
2
2
2
2
2
ˆˆˆˆˆˆ
ˆˆˆ
z
V
y
V
x
V
z
V
y
V
x
V
zyx
z
V
y
V
x
VV
zyxzyx
zyx
Laplace operator has different forms for different coordinate systems.
EMLAB
18Laplace’s equations
The differential equation for source-free region becomes a Laplace equation.
)equations'Lapace(0V2
2
2
2
2
22 11
z
VVVV
2
2
2
2
2
22
z
V
y
V
x
VV
2
2
2222
22
sin
1sin
sin
11
V
r
V
rr
Vr
rrV
(rectangular coordinate)
(cylindrical coordinate)
(spherical coordinate)
EMLAB
19Uniqueness theoremSolution of a differential equation
The solution that satisfies the differential equation and its boundary condition is unique regardless of the solution proce-
dure.
conditionboundary
),( yxVb
)y,x(Vinterior)(
)()boundary(,0
)()boundary(,0
21
222
112
VV
fVV
fVV
r
r
022
2 VV
e ddW EE
boundary:S To prove the uniqueness theorem, we assume that there exist two distinct solutions that satisfy the same boundary condition.
Then, the difference of the those two solutions will have zero value at the boundary and will have non-zero value in the inte-rior region.
0)interior(
0)boundary(,02
21
VV
0(interior)
This situation is contradictory to the original assumption that there exist two distinct solution that satisfy the same boundary condition.
0constant(interior)0(interior)
)0(02
)(2
)(22
22
Sondd
dd
SV
VV
a
EMLAB
20Example 1 : Laplace eqs.
Sd
zx
0V
V0
Unlike the procedures in the previous chapters, the potential V is first obtained solving Laplace equation. Then, using the potential, E, D, , Q, C are obtained.
02
2
2
2
2
2
2
22
zzyx
d
S
dV
SV
V
QC
d
SVSdaQ
d
Vd
Vd
Vd
V
zd
Vz
s
S
s
s
s
0
0
0
0
0
0
0
0
)6(
)5(
ˆˆbottom)(
ˆˆtop)()4(
ˆ)3(
ˆ)2(
)()1(
DzDn
DzDn
zED
zE
If the plates are wide enough to ignore the variation of electric field along x and y directions
0,0
yx
constant),( BABAzAz
zd
Vz
d
VA
BVBAdd
00
0
)(
0)0(,)(
Using the boundary conditions on the two plates,
EMLAB
21Example 2
)/ln(
)/ln()(
)/ln(
ln,
)/ln(
ln)(,0ln)(
,conditionsboundary theUsing
constants),(ln,0
0,0
symmetricaxially anddirectionz in the infinite be toassumed iscylinder The
0111
000
0
2
2
2
2
2
2
2
2
22
ab
bVV
ba
bVB
ba
VA
VBaAaVBbAbV
BABAVAVV
z
VV
V
z
VVVV
)/ln(
2)6(
2)/ln(
1)5(
)/ln(
1ˆˆ)(
)/ln(
1ˆˆ)()4(
ˆ)/ln(
1)3(
ˆ)/ln(
1)2(
)/ln(
)/ln()()1(
0
0
0
0
0
0
ab
L
V
QC
aLaba
VdaQ
abb
Vb
aba
Va
ab
V
ab
VV
ab
bVV
S
s
s
s
DρDn
DρDn
ρED
ρE
x
y
r
a
b
V00V
EMLAB
22
2tanln
2tanln
)(0,
2tanln
02
,2
tanln)(
,conditionsboundary theUsing
2tanln
2tan
2sec
21
2cos
2cos
2sin2
2cos
2sin2sin
) ,(sin
,sin0sin
0,0
symmetricaxially are anddirection radial thealonginfinity toextend surfaces The
0sinsin
1
sin
1sin
sin
11
00
0
2
2
2
2
2
2
2
2222
22
VVBV
A
BVVBAV
Ad
AdAdAdA
BABdA
VAVV
V
r
V
V
r
V
r
V
rr
Vr
rrV
상수는
0V
V0
2cotln
r2
V
QC)6(
dr
2cotln
V2
sinr
drdsinr
2cotln
VdaQ)5(
2cotlnsinr
Vˆˆ)a()4(
ˆ
2tanlnsinr
V)3(
ˆ
2tanlnsinr
VˆV
r
1V)2(
2tanln
2tanln
V)(V)1(
1
r
0
0
r
0
2
0
0
S
s
0s
0
0
0
11
DθDn
θED
θθE
Example 3
EMLAB
23
Scanning tunneling microscope probeBi-conical antenna
Examples
EMLAB
24Intrinsic semiconductor
EMLAB
25Semiconductor doping
EMLAB
26
EMLAB
27Example : Poisson eq. Diode (simple model)
V
P
NNxPx
P-type N-type VV 2
V
dx
Vd
2
2
1Cxdx
dV V
02
2
dx
Vd02 C
dx
dV03 CV
1Cxdx
dVE P
x
)(,
0)(
1
1
pP
xpP
pP
px
xxExC
Cxdx
dVxE
)( pP xx
dx
dV
22)(
2)( CxxxV p
P
0)( 2 CxV p
2)(2
)( pP xxxV
xE
PN
NxPx
)( pN xx
dx
dV
22)(
2)( CxxxV N
N
2
)(2
)(2
)0(
22
2
22
2
PPNN
PP
NN
xxC
xCxxV
2)(
2)(
222 PPNN
NN xx
xxxV
V
NxPx
P-type region
N-type region
EMLAB
28Method of separation of variables
)()()(),,(
02
2
2
2
2
2
zZyYxXzyx
zyx
0)(1
011
0111
0111
,by Divided
0
222
222
2
2
22
22
2
22
2
2
22
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
Zdz
Zd
dz
Zd
Z
Ydy
Yd
dz
Zd
Zdy
Yd
Y
Xdx
Xd
dx
Xd
Xdz
Zd
Zdy
Yd
Y
dz
Zd
Zdy
Yd
Ydx
Xd
X
XYZ
dz
ZdXY
dy
YdXZ
dx
XdYZ
zyx
s.conditioinboundary thefrom fixed becan ,,,,,
coshsinh)(
cossin)(
cossin)(
0)(
0
0
0111
2222
222
2
22
2
22
2
2
2
2
2
2
2
FEDCBA
zFzEzZ
xDxCyY
xBxAxX
Zdz
Zd
Ydy
Yd
Xdx
Xd
dz
Zd
Zdy
Yd
Ydx
Xd
X
•The original equation on the left is split into three equations containing single variable. The equa-tions are related to one another through variables and .• , can be fixed using boundary conditions.
• If the boundary condition becomes complex, three dimensional Laplace equa-tion is very difficult to solve. This is because the Laplace equation is a partial differential equation that contains three variables x, y and z.
• If the boundary is parallel to coordinate axes, the solution to Laplace equation can be represented as a multiplication of three functions that contain only one variable.
• This method is called as a method of separation of variables.
EMLAB
29Example : Separation of variables
0VV0
V0
V0 a
b
),( yxV
)()(),(
02
2
2
2
yYxXyxV
y
V
x
V
If the boundary extends to infinity in
the z-direction, the derivative with re-spect to z becomes zero.
01
1
22
2
22
2
2
2
2
2
sinsinh),(
sinsinh),(
),3,2,1(
0sinsinh),(
0
0)coshsinh()0,(
0
0)cossin(),0(
)cossin(
)coshsinh(),(
cossin
coshsinh
0
00
11
Vb
yn
b
anAyaV
b
yn
b
xnAyxV
nb
nnb
bxAbxV
D
DxBxAxV
B
yDyCByV
yDyC
xBxAyxV
yDyCY
xBxAX
Ydy
Yd
Xdx
Xd
dy
Yd
Ydx
Xd
X
nn
nn
10
00
0
0
0
00
01
)12(sin
)12(sinh)12(
)12(sinh
4),(
;sinh
4
sinh
])1(1[2
])1(1[cossin2
sinh
~0 sin
sinsinh),(
n
m
m
mb
b
m
nn
b
yn
ban
n
bxn
VyxV
m
bam
m
V
bamm
VA
m
bV
b
ym
m
bVdy
b
ymV
b
b
amA
bb
ym
Vb
yn
b
anAyaV
홀수은
적분하면까지곱하고를양변에
nmforb
nmfor
b
ymn
mnb
ymn
mn
b
dyb
ymn
b
ymndy
b
ym
b
yn
b
bb
2
0
)(sin
1)(sin
1
2
)(cos
)(cos
2
1sinsin
0
00
EMLAB
30Result-Matlab code
% potential_demo.m% 변수 분리 방법에 의해 푼 전압을 그림 .
clear,clf;hold on;imax = 30;jmax = 30;a=2.5; b = 1;for i=1:imax+1 for j=1:jmax+1 x(i,j) = (i-1)*a/imax; y(i,j) = (j-1)*b/jmax; z(i,j) = Vseries(x(i,j),y(i,j)); endend
colormap jet;surf(x,y,z);shading interp;%caxis([0,100]);colorbar('vert');view([0,90]);
%Vseries.m
% 변수 분리 방법에 의한 전압 계산function f=Vseries(x,y)
V=100;f=0;a=2.5;b=1;
for i=1:10 n=2*i-1; f=f+sinh(n*pi*x/b)/sinh(n*pi*a/b)*sin(n*pi*y/b)*(1/n);endf=f*4*V/pi;
EMLAB
31
)(0
0
,,
0,0
2
2
2
2
2
2
2
LaplacianV
z
V
y
V
x
V
z
VE
y
VE
x
VE
z
E
y
E
x
E
zyx
zyx
E
Solving Laplace equation through Finite difference method
43210
204321
2
2
2
2
24002
2
2
23001
2
2
3001
4
1
04
,
VVVVV
h
VVVVV
y
V
x
V
h
VVVV
h
xV
xV
y
V
h
VVVV
hxV
xV
x
V
h
VV
x
V
h
VV
x
V
db
ca
ca
0V 1V
2V
3V
4V
acd
b
0V
0V
0V
10V Electrostatic potentials can be found from the Laplace equation.
43210 4
1VVVVV Numerical Laplace equation
EMLAB
32
0
0
0
0
0
0
10
10
10
V
V
V
V
V
V
V
V
V
410100000
141010000
014001000
100410100
010141010
001014001
000100410
000010141
000001014
0V0V0V4:V
0VVVVV4:V
0V100VV4:V
0V10VVV4:V
0V10V0V4:V
3,3
2,3
1,3
3,2
2,2
1,2
3,1
2,1
1,1
1,1N2,N1,N1,N
1j,i1j,ij,1ij,1ij,ij,i
3,22,13,13,1
2,23,11,12,12,1
1,22,11,11,1
1,1V 2,1V 3,1V
1,2V 2,2V
10
0
The unknown voltages on the lattice points are set to V(i,j). Using the numerical Laplace equation, the un-knowns are related to one another. If as many equations as the number of unknowns are generated, a set of si-multaneous equations is formed that has unique solu-tion.
Matrix formulation
04: 1,1,,1,1,, jijijijijiji VVVVVV
][]][[ SVVA