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]


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


Voltage=0No ChargeNo Current

1 2 3

Voltage build upCharge build upTransient Current

Voltage = 6VCharge=QNo Current

What is Capacitance?


Voltage=6VCharge=QNo Current

Voltage decayingCharge decayingTransient Current

Voltage=0No ChargeNo Current

4 5 6What is Capacitance?


( ) 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


Capacitance = Total charge

Voltage

CAPACITANCE


C = oAd

A : area

o : permittivity

Capacitance

d

V+

For more complex capacitance geometries, need to use numerical methods


( ) ( , ') ( ') '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


METHOD OF MOMENTSOperator equation

L(f) = g

L = integral or differential operator

f = unknown function

g = known function

Expand unknown function f

f nfnn


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


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

Copyright © by Jose E. Schutt‐Aine , All Rights ReservedECE 598‐JS, Spring 2012

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


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


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


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


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


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


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


Point matching

m m mw ( x x )( y y )

m

VV

g.

V

Conducting Plate


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


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


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


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


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


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


2

ˆ4Idl rdH

R

2

ˆ4Idl rH

R

Biot-Savart Law

Idl

R

dH


I

I

Magnetic Flux

S

B dS


Inductance

Inductance = Total flux linked

Current

z

rH

IdlR


dL Ndi

LI

2v

B HdvLI

Inductance


INDUCTANCE

L = N

I

N : number of turns

: flux per turn


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


( ) 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


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


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


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.


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


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


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


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


+-

+

-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


sp

Dielectrics

Charge density within volume is zeroSurface charge density is nonzero

D=o(1+e)E=E


1vLC

Dielectric Materials

tan

Germanium 2.2Silicon 0.0016Glass 10-10-10-14

Quartz 0.510-17

Material Conductivity (-1-m-1)


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


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

ece 598 js lecture 02 resistance, capacitance,...

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