ECE 598 JS Lecture 02 Resistance, Capacitance,...
Transcript of ECE 598 JS Lecture 02 Resistance, Capacitance,...
Copyright © by Jose E. Schutt‐Aine , All Rights ReservedECE 598‐JS, Spring 2012 1
ECE 598 JS Lecture 02Resistance, Capacitance, Inductance
Spring 2012
Jose E. Schutt-AineElectrical & Computer Engineering
University of [email protected]
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Faraday’s Law of Induction
Ampère’s Law
Gauss’ Law for electric field
Gauss’ Law for magnetic field
Constitutive Relations
BEt
H J D
t
D
0B
B H
D E
MAXWELL’S EQUATIONS
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Voltage=0No ChargeNo Current
1 2 3
Voltage build upCharge build upTransient Current
Voltage = 6VCharge=QNo Current
What is Capacitance?
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Voltage=6VCharge=QNo Current
Voltage decayingCharge decayingTransient Current
Voltage=0No ChargeNo Current
4 5 6What is Capacitance?
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( ) dv dQi t Cdt dt
Relation: Q = Cv
Q: charge stored by capacitorC: capacitancev: voltage across capacitori: current into capacitor
0
1( ) ( )t
v t i dC
Capacitance
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Capacitance = Total charge
Voltage
CAPACITANCE
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C = oAd
A : area
o : permittivity
Capacitance
d
V+
For more complex capacitance geometries, need to use numerical methods
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( ) ( , ') ( ') 'r g r r r dr
( ) ( )r potential known
( , ') ' (known)g r r Green s function
( ') arg ( )r ch e distribution unknown
Q=CV
Once the charge distribution is known, the total charge Q can be determined. If the potential =V, we have
To determine the charge distribution, use the moment method
Capacitance Calculation
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METHOD OF MOMENTSOperator equation
L(f) = g
L = integral or differential operator
f = unknown function
g = known function
Expand unknown function f
f nfnn
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Matrix equation
n nn
( f ) g L
n m n mn
w , f w ,g L
mn n mg l
in terms of basis functions fn, with unknown coefficients n to get
Finally, take the scalar or inner product with testing of weighting functions wm:
METHOD OF MOMENTS
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1 1 1 2
2 1 2 2mn
w , f w , f ...w , f w , f ............. ............. ...
l
L LL L
n 1
2
.
gm w1,gw2,g
.
Solution for weight coefficients
1n nm mg l
METHOD OF MOMENTS
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Green’s function G: LG =
D
E
2
L (x,y,z) (x' ,y' ,z' )4R dx' dy' dz'
R (x x' )2 (y y' )2 (z z' )2
Moment Method Solution
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Subdomain bases
P(xn ) 1 xn 2 x xn
2
0 otherwise
x1 x2 xn
T(xn ) 1 x xn x xn 0 otherwise
x1 x2 xn
Testing functions often (not always) chosen same as basis function.
Basis Functions
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2a
2b
2b
2aSn
Z
Y
X
conducting plate
(x,y,z) dx'a
a dy' (x' ,y' )
4Ra
a
= charge density on plate
Conducting Plate
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for |x| < a; |y| < a
R (x x' )2 (y y' )2 (z z' )2
V dx'a
a dy' (x' , y' )
4 (x x' )2 (y y' )2a
a
1 a a
a a
qC dx' dy' ( x, y )V V
Setting = V on plate
Capacitance of plate:
Conducting Plate
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Basis function Pn
2 21
2 20
m m
n m n
n n
s sx x xP ( x , y )
s sy y y
otherwise
1
N
n nn
( x, y ) f
Representation of unknown charge
Conducting Plate
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Matrix equation:
1
N
mn nn
V f
l
2 2
14n n
mn x ym m
dx' dy'( x x') ( y y')
l
1
1
1 N
n n mn nn mn
C s sV
l
Matrix element:
Conducting Plate
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Operator form(x,y,z) dx'
a
a dy'
b
b
(x' ,y' )4R
f(x, y) (x,y)
g(x, y) V | x |< a; | y |< a
2 24
a a
a a
f ( x', y')( f ) dx' dy'( x x') ( y y')
L
a a
a af ,g dx dy f ( x, y )g( x, y )
Conducting Plate
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Point matching
m m mw ( x x )( y y )
m
VV
g.
V
Conducting Plate
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Calculation of matrix Elements
2 2
14
b b
nn b bdx dy
x y
l
2 1 2b ln( ) 2 0 8814b ( . ) 22 as b
N
(1) Self-term (diagonal element)
where
(2) Off-diagonal terms
2 2
4n
mnmn
2
m n m n
s m nR
b=( x x ) ( y y )
l
Conducting Plate
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2a
2a
z
y
x
d
tt tb
bt bb
[ ] [ ][ ]
[ ] [ ]
l lll l
Using N unknowns per plate, we get 2N 2N matrix equation:
Subsript ‘t’ for top and ‘b’ for bottom plate, respectively.
Parallel Plates
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2
2 2 2
tbmn
m n m n
b( x x ) ( y y ) d
l
20 282 2 14 2
tbnn
. d d( b )b b
l
gm gm
t gm
b
Right hand side
Symmetry of matrix element blocks
tt bb l l l
tb bt l l
Approximate expressions for matrix elements
Parallel Plates
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where
t bm m
VV
g g..
top plate at constant potential V and bottom at -V
Taking advantage of symmetry, the coefficients of top and bottom plates can be related as
t tn n
n b tn n
Parallel Plates
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lmntt lmn
tb nt gm
t Matrix equation becomes
Solution: 1t tt tb tm nmn
l l g
12
tn n
top
charge on top plateC2V
sV
tg V
122 tt tb
mnmn
C b
l l
Capacitance
Using s=4b2 and all elements of
Parallel Plates
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2
1 ˆ4 o
qqF RR
q q'
0
R
rr'
F q U B
+
movingcharge q
Direction ofmagnetic field
B
F
v
Magnetic ForceElectric Force
Fields
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2
ˆ4Idl rdH
R
2
ˆ4Idl rH
R
Biot-Savart Law
Idl
R
dH
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I
I
Magnetic Flux
S
B dS
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Inductance
Inductance = Total flux linked
Current
z
rH
IdlR
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dL Ndi
LI
2v
B HdvLI
Inductance
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INDUCTANCE
L = N
I
N : number of turns
: flux per turn
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Radius ad/2
ba
1cosh ( / )2 2
oL d H ml a
ln ( / )2
oL b H ml a
Wire above ground plane Coaxial Line
Inductance Calculation
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( ) di dv t Ldt dt
Relation: = Li: flux stored by inductorL: inductancei: current through inductorv: voltage across inductor
0
1( ) ( )t
i t v dL
Inductance
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ˆˆ 0t iz n V
ˆ sri t i
o
qn V
ˆˆ 0t ziz n A
1ˆ t zi o zri
n A J
CV Q LI
Electrostatics Magnetostatics
2-D Isomorphism
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I
I
I
akI
a
1 1loop
a a
L B da A daI I I
QUESTION: Can we associate inductance with piece of conductor rather than a loop? PEEC Method
Loop Inductance
Inductance Calculation
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Package Inductance & Capacitance
Component Capacitance Inductance(pF) (nH)
68 pin plastic DIP pin† 4 3568 pin ceramic DIP pin†† 7 2068 pin SMT chip carrier † 2 7
† No ground plane; capacitance is dominated by wire-to-wire component.
†† With ground plane; capacitance and inductance are determined by the distance between the lead frame and the ground plane, and the lead length.
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Component Capacitance Inductance(pF) (nH)
68 pin PGA pin†† 2 7256 pin PGA pin†† 5 15Wire bond 1 1Solder bump 0.5 0.1
† No ground plane; capacitance is dominated by wire-to-wire component.
†† With ground plane; capacitance and inductance are determined by the distance between the lead frame and the ground plane, and the lead length.
Package Inductance & Capacitance
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Metallic Conductors
Length
Area
Re sist an ce : R
Package level:W=3 milsR=0.0045 /mm
R = Le ng th Are a
Submicron level:W=0.25 micronsR=422 /mm
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Metal Conductivity -1 m 10-7)
Silver 6.1Copper 5.8Gold 3.5Aluminum 1.8Tungsten 1.8Brass 1.5Solder 0.7Lead 0.5Mercury 0.1
Metallic Conductors
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3.00 m 0.03 1 102.00 m 0.04 2 151.00 m 0.09 4 300.75 m 0.12 5 400.50 m 0.18 8 600.25 m 0.36 16 120
Wint Aluminum WSi2 Polysilicon( = 3-cm ) ( =130-cm) ( =1000-cm)
sq intR . = int
intH
intH = 3intW
Sheet Resistance of Interconnections
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+-
+
-E
Dielectrics contain charges that are tightly bound to the nucleiCharges can move a fraction of an atomic
distance away from equilibrium positionElectron orbits can be distorted when an
electric field is applied
Dielectrics
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sp
Dielectrics
Charge density within volume is zeroSurface charge density is nonzero
D=o(1+e)E=E
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1vLC
Dielectric Materials
tan
Germanium 2.2Silicon 0.0016Glass 10-10-10-14
Quartz 0.510-17
Material Conductivity (-1-m-1)
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Dielectric Materials
Polyimide 2.5-3.5 16-19Silicon dioxide 3.9 15Epoxy glass (FR4) 5.0 13Alumina (ceramic) 9.5 10
Material r v(cm/s)
r
1vLC
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Conductivity of Dielectric Materials
Material Conductivity ( -1 m-1)
Germanium 2.2
Silicon 0.0016
Glass 10-10 - 10-14
Quartz 0.5 x 10-17
Loss TANGENT : tan
r j i