Flux lines may extend beyond inductor creating stray inductance effects

A TIME VARYING FLUX CREATES A COUNTER EMF AND CAUSES A VOLTAGE TO APPEAR AT THE TERMINALS OF THE DEVICE

INDUCTORS NOTICE USE OF PASSIVE SIGN CONVENTION

Circuit representation for an inductor

A TIME VARYING MAGNETIC FLUX INDUCES A VOLTAGE

dtdvLφ

= Induction law

INDUCTORS STORE ELECTROMAGNETIC ENERGY. THEY MAY SUPPLY STORED ENERGY BACK TO THE CIRCUIT BUT THEY CANNOT CREATE ENERGY. THEY MUST ABIDE BY THE PASSIVE SIGN CONVENTION

FOR A LINEAR INDUCTOR THE FLUX IS PROPORTIONAL TO THE CURRENT

⇒= LLiφdtdiLv L

L =DIFFERENTIAL FORM OF INDUCTION LAW

THE PROPORTIONALITY CONSTANT, L, IS CALLED THE INDUCTANCE OF THE COMPONENT

INDUCTANCE IS MEASURED IN UNITS OF henry (H). DIMENSIONALLY

secAmp

VoltHENRY =

LEARNING by Doing

Follow passive sign convention

dtdiLv L

L =Differential form of induction law

∫∞−

=t

LL dxxvL

ti )(1)(Integral form of induction law

00 ;)(1)()(0

ttdxxvL

titit

tLLL ≥+= ∫

A direct consequence of integral form ttiti LL ∀+=− );()( Current MUST be continuous

A direct consequence of differential form 0. =⇒= LL vConsti DC (steady state) behavior

Power and Energy stored

)()()( titvtp LLL = W )()()( titdtdiLtp L

LL =

= )(

21 2 tLi

dtd

L

)(21)(

21),( 1

22

212 tLitLittw LL −= Energy stored on the interval

Can be positive or negative

)(21)( 2 tLitw LL =

“Energy stored at time t” Must be non-negative. Passive element!!!

∫

=

2

1

)(21),( 2

12

t

tLL dxxLi

dtdttw J Current in Amps, Inductance in Henrys

yield energy in Joules

LEARNING EXAMPLE FIND THE TOTLA ENERGY STORED IN THE CIRCUIT

In steady state inductors act as short circuits and capacitors act as open circuits

2 21 12 2C C L LW CV W LI= =

9@ : 3 09 6

A AV VA A −− + + =

81[ ]5AV V⇒ −

2

6 10.86 3C AV V V= =+

2 1.89

AL

VI A= =

1 2 13 1.2L L LI A I I A+ = ⇒ = −

1 1 19 6 16.2C L CV I V V= − ⇒ =

L=10mH. FIND THE VOLTAGE

=

××

= −

−

sA

sAm 10

1021020

3

3

−=

sAm 10

)()( tdtdiLtv =

THE DERIVATIVE OF A STRAIGHT LINE IS ITS SLOPE

≤<−≤≤

=elsewhere

mstsAmstsA

dtdi

042)/(10

20)/(10

mVVtvHL

sAtdtdi

10010100)(1010

)/(10)( 3

3=×=⇒

×=

= −

−

ENERGY STORED BETWEEN 2 AND 4 ms

)2(21)4(

21)2,4( 22

LL LiLiw −=

233 )10*20(10*10*5.00)2,4( −−−=w J

THE VALUE IS NEGATIVE BECAUSE THE INDUCTOR IS SUPPLYING ENERGY PREVIOUSLY STORED

LEARNING EXAMPLE

2

)(Vv

2 )(st

L=0.1H, i(0)=2A. Find i(t), t>0

∫+=t

dxxvL

iti0

)(1)0()(

20;2)(2)(0

≤<=⇒= ∫ ttdxxvxvt

stttiHL 20;202)(1.0 ≤≤+=⇒=

stititxv 2);2()(2;0)( >=⇒>=

Initial energy stored in inductor

== 2)2]([1.0*5.0)0( AHw ][2.0 J

“Total energy stored in the inductor”

JAHw 2.88)42(*][1.0*5.0)( 2 ==∞

Energy stored between 0 and 2 sec

)0(21)2(

21)0,2( 22

LL LiLiw −=

22 )2(*1.0*5.0)42(*1.0*5.0)0,2( −=w

][88)0,2( Jw =

ENERGY COMPUTATIONS SAMPLE PROBLEM

22 )(st

)(Ai42

)(21)(

21),( 1

22

212 tLitLittw LL −= Energy stored on the interval

Can be positive or negative

FIND THE VOLTAGE ACROSS AND THE ENERGY STORED (AS FUNCTION OF TIME)

)(tv

FOR ENERGY STORED IN THE INDUCTOR

)(twL

NOTICE THAT ENERGY STORED AT ANY GIVEN TIME IS NON NEGATIVE -THIS IS A PASSIVE ELEMENT-

LEARNING EXAMPLE

LEARNING EXAMPLE

VOLTAGETHE DETERMINEmHL 10= )()( t

dtdiLtv =

mVv 100−=

××

××= −

−−

sAHv 3

33

1021020][1010

L=200mH

FIND THE CURRENT

0)0(0;0)( =⇒<= ittv

0;)(1)0()(0

>+= ∫ tdxxvL

itit

)(ti

)(ti

LEARNING EXAMPLE

ENERGY

POWER

)(ti

L=200mH

)(tp

)(tw

NOTICE HOW POWER CHANGES SIGN

ENERGY IS NEVER NEGATIVE. THE DEVICE IS PASSIVE

FIND THE POWER

FIND THE ENERGY

L=5mH FIND THE VOLTAGE

)()( tdtdiLtv =

msmAm

120

= )/(122010 sAm

−−

=

Vvm

00

==

)/(34

100 sAm−−

=

mVmstsAHv 10010);/(20)(105 3 =<≤××= −

mVv 50−=

mVv 50−=

LEARNING EXTENSION

CAPACITOR SPECIFICATIONS

VALUESSTANDARD IN RANGE ECAPACITANC mFCFp 50≈≈

VV 5003.6 −RATINGS CAPACITOR STANDARD

%20%,10%,5 ±±±TOLERANCE STANDARD

LEARNING EXAMPLE

GIVEN THE VOLTAGE WAVEFORM DETERMINE THE VARIATIONS IN CURRENT

%20100 ±= nFC

)()( tdtdvCti =

nAs

VF 60023

3)3(10100 9 −=

−−−

× −

current Nominal

nA300

nA300

CURRENT WAVEFORM

sµ

VALUESSTANDARD IN RANGES INDUCTANCE mHLnH 1001 ≤≈≤≈

AmA 1≈−≈RATINGS INDUCTOR STANDARD

%10%,5 ±±TOLERANCE STANDARD

INDUCTOR SPECIFICATIONS

LEARNING EXAMPLE

%10100 ±= HL µ

GIVEN THE CURRENT WAVEFORM DETERMINE THE VARIATIONS IN VOLTAGE

)()( tdtdiLtv =

××

××= −

−−

SAHv 6

36

10201020010100

viivLC

→→→

IDEAL AND PRACTICAL ELEMENTS

IDEAL ELEMENTS CAPACITOR/INDUCTOR MODELS INCLUDING LEAKAGE RESISTANCE

)(ti

−

+

)(tv

)()()( tdtdvC

Rtvti

leak+=

MODEL FOR “LEAKY” CAPACITOR

)(ti

−

+

)(tv

)()()( tdtdiLtiRtv leak +=

MODEL FOR “LEAKY” INDUCTORS

−

+)(tv

−

+)(tv

)(ti )(ti

)()( tdtdvCti = )()( t

dtdiLtv =

SERIES CAPACITORS

NOTICE SIMILARITY WITH RESITORS IN PARALLEL

21

21

CCCCCs +

=

Series Combination of two capacitors

Fµ6 Fµ3 =SCFµ2

Fµ2

Fµ1

6123 ++

=

ALGEBRAIC SUM OF INITIAL VOLTAGES

POLARITY IS DICTATED BY THE REFERENCE DIRECTION FOR THE VOLTAGE

VVV 142 −−+=

LEARNING EXAMPLE

DETERMINE EQUIVALENT CAPACITANCE AND THE INITIAL VOLTAGE

OR WE CAN REDUCE TWO AT A TIME

Fµ30

C

+- −

+V8

V12

SAME CURRENT. CONNECTED FOR THE SAME TIME PERIOD

SAME CHARGE ON BOTH CAPACITORS

CVFQ µµ 240)8)(30( ==

−

+V4

LEARNING EXAMPLE Two uncharged capacitors are connected as shown. Find the unknown capacitance

1C FIND

CVFQCVQ µµ 72)6)(12( ==⇒=

−

+V18

FVCC µµ 4

1872

1 ==

PARALLEL CAPACITORS

)()( tdtdvCti kk =

)(tiLEARNING EXAMPLE

PC 4 6 2 3 15 Fµ= + + + =

LEARNING EXTENSION

Fµ6

Fµ2

Fµ3

Fµ4

Fµ12

→eqC

Fµ4

Fµ3FCeq µ23

=

F4 ARECAPACITORSALL µFIND EQUIVALENT CAPACITANCE

Fµ8

Fµ8

eqC Fµ8

Fµ8

Fµ4

Fµ1232

3328

38

=+

SAMPLE PROBLEM

IF ALL CAPACITORS HAVE THE SAME CAPACITANCE VALUE C DETERMINE THE VARIOUS EQUIVALENT CAPACITANCES

SAMPLE PROBLEM

All capacitors are equal with C=8 microFarads EQC

______=ABC

Examples of interconnections

SERIES INDUCTORS

)()( tdtdiLtv kk =

)()( tdtdiLtv S=

LEARNING EXAMPLE

=eqL H7

PARALLEL INDUCTORS

)(ti

INDUCTORS COMBINE LIKE RESISTORS CAPACITORS COMBINE LIKE CONDUCTANCES

LEARNING EXAMPLE

mH4 mH2∑=

=N

jj titi

100 )()( AAAAti 1263)( 0 −=+−=

LEARNING EXTENSION

mH2

mH2WHEN IN DOUBT… REDRAW!

a

b

cd

mH4

mH2mH2

mH4

eqL

a

b

c

d

IDENTIFY ALL NODES

PLACE NODES IN CHOSEN LOCATIONS

a

b

cd

CONNECT COMPONENTS BETWEEN NODES

mH6

mHmHmHmHLeq 4.42)4||6( =+=

ALL INDUCTORS ARE 4mH

ALL INDUCTORS ARE 6mH

NODES CAN HAVE COMPLICATED SHAPES. KEEP IN MIND DIFFERENCE BETWEEN PHYSICAL LAYOUT AND ELECTRICAL CONNECTIONS

6||6||6

a

b

c

a

b

c

SELECTED LAYOUT

a

b

c

mH2

mH6mH6

mH6

eqL

[ ] mHLeq 7246

144866||)26(6 =+=++=

mHLeq 766

=

LEARNING EXTENSION

L-C

RC OPERATIONAL AMPLIFIER CIRCUITS

INTRODUCES TWO VERY IMPORTANT PRACTICAL CIRCUITS BASED ON OPERATIONAL AMPLIFIERS

⇒∞=A

)(0 −+ −=⇒= vvAvR OO

THE IDEAL OP-AMP

⇒∞=iR

∞=∞==⇒ ARR iO ,,0IDEAL

RC OPERATIONAL AMPLIFIER CIRCUITS -THE INTEGRATOR

0=+v

IDEAL OP-AMP ASSUMPTIONS

)(0

)(

_

_

∞==

∞== +

iRi

Avv

RC OPERATIONAL AMPLIFIER CIRCUITS - THE DIFFERENTIATOR

1R

2i

1i

0=+v−− =+ iiiv 21:KCL@

IDEAL OP-AMP ASSUMPTIONS

)(0

)(

_

_

∞==

∞== +

iRi

Avv 0

21 =+

Rvi O

∫∞−

+=t

dxxiC

iRtv )(1)( 11

111

KVL

)(2

1 Rvi o−=o1 vof terms in i replace

DIFFERENTIATE

)(111

111 t

dtdvCi

dtdiCR =+

)(11211 t

dtdvCRv

dtdvCR o

o −=+

IF R1 COULD BE SET TO ZERO WE WOULD HAVE AN IDEAL DIFFERENTIATOR. IN PRACTICE AN IDEAL DIFFERENTIATOR AMPLIFIES ELECTRIC NOISE AND DOES NOT OPERATE. THE RESISTOR INTRODUCES A FILTERING ACTION. ITS VALUE IS KEPT AS SMALL AS POSSIBLE TO APPROXIMATE A DIFFERENTIATOR

ABOUT ELECTRIC NOISE

ALL ELECTRICAL SIGNALS ARE CORRUPTED BY EXTERNAL, UNCONTROLLABLE AND OFTEN UNMEASURABLE, SIGNALS. THESE UNDESIRED SIGNALS ARE REFERRED TO AS NOISE

THE DERIVATIVE

)102cos(2000)120cos(120)( 9 ttdtdy ππππ ×+=

SIMPLE MODEL FOR A NOISY 60Hz SINUSOID CORRUPTED WITH ONE MICROVOLT OF 1GHz INTERFERENCE.

)102sin(10)120sin()( 96 ttty ππ ×+= −610−=

amplitude signalamplitude noise

noise signal

67.161202000

==amplitude signalamplitude noise

noise signal

LEARNING EXTENSION

)(112 t

dtdvCRvo −=

ATORDIFFERENTIIDEAL

FCkR µ2,1 12 =Ω= WITH ATORDIFFERENTIIDEAL TO INPUT

sVm 3105

10−×

=

sFCR 36312 102102101 −− ×=××Ω×=

sFVQF

QsV

sQV

AV

=×Ω⇒=

×===Ω

SL ANALYSIDIMENSIONA

LEARNING EXTENSION FCkR µ2.0,5 21 =Ω= WITH INTEGRATOR ANTO INPUT

∫−=t

ioo dxxvCR

vtv021

)(1)0()(

INTEGRATOR

sFVQF

QsV

sQV

AV

=×Ω⇒=

×===Ω

SL ANALYSIDIMENSIONA

DISCHARGEDINITIALLY IS CAPACITOR

sCR 321 10−=

31 1020)(:1.00 −×=<< tvst ( )∫ ××==⇒ −

t

o sVtdxxvty0

31 1020)()( ( )sVy ××=⇒ −3102)1.0(

31 1020)(:2.01.0 −×−=<< tvst ( )∫ ×−×−×=+=⇒ −−

t

oo sVtdxxvyty1.0

331 )1.0(1020102)()1.0()(

)(1)(21

tyCR

tv oo =

USING GROUND WIRE TO REDUCE CROSSTALK

APPLICATION EXAMPLE CROSS-TALK IN INTEGRATED CIRCUITS

⇒

REDUCE CROSSTALK BY • Reducing C12 • Increasing C2

SMALLER

COST? EXTRA SPACE BY GROUND WIRE

Simplified Model

LEARNING EXAMPLE SIMPLE CIRCUIT MODEL FOR DYNAMIC RANDOM ACCESS MEMORY CELL (DRAM)

REPRESENTS CHARGE LEAKAGE FROM CELL CAPACITOR

NOTICE THE VALUES OF THE CAPACITANCES

ONE LOGIC A OF STORAGE CORRECT FOR VVcell 5.1>

∫+=t

CCC dxxiC

vv0

)(1)0(

VtCIt

CIV

cell

leak

cell

leakcell 5.15.13 ≤⇒≥−=

sA

FVtH3

12

15105.1

1050)(1050)(5.1 −

−

−

×=×

××=

Ht/1 THAN HIGHERFREQUENCY A AT”밨EFRESHED BE MUSTCELL THE

THE ANALYSIS OF THE READ OPERATION GIVES FURTHER INSIGHT ON THE REQUIREMENTS

SWITCHED CAPACITOR CIRCUIT

CELL AT THE BEGINNING OF A MEMORY READ OPERATION

CELL READ OPERATION IF SWITCH IS CLOSED BOTH CAPACITORS MUST HAVE THE SAME VOLTAGE

ASSUMING NO LOSS OF CHARGE THEN THE CHARGE BEFORE CLOSING MUST BE EQUAL TO CHARGE AFTER CLOSING

FVFVQbefore1515 10503104505.1 −− ××+××=

)10500( 15 FVQ afterafter−××=

VVafter 65.1=

Even at full charge the voltage variation is small. SENSOR amplifiers are required

After a READ operation the cell must be refreshed

LEARNING EXAMPLE

IC WITH WIREBONDS TO THE OUTSIDE

FLIP CHIP MOUNTING

GOAL: REDUCE INDUCTANCE IN THE WIRING AND REDUCE THE “GROUND BOUNCE” EFFECT A SIMPLE MODEL CAN BE USED TO

DESCRIBE GROUND BOUNCE

MODELING THE GROUND BOUNCE EFFECT

)()( tdt

diLtV GballGB =

IF ALL GATES IN A CHIP ARE CONNECTED TO A SINGLE GROUND THE CURRENT CAN BE QUITE HIGH AND THE BOUNCE MAY BECOME UNACCEPTABLE

USE SEVERAL GROUND CONNECTIONS (BALLS) AND ALLOCATE A FRACTION OF THE GATES TO EACH BALL

nHLball 1.0≈ sAm 9

3

10401040

−

−

××

=

LEARNING BY DESIGN POWER OUTAGE “RIDE THROUGH” CIRCUIT

CAPACITOR MUST MAINTAIN AT LEAST 2.4V FOR AT LEAST 1SEC.

v+

− DESIGN EQUATION

http://www.wiley.com/college/irwin/0470128690/animations/swf/6-21.swf

Proposed solution

DESIGN EXAMPLE DESIGN AN OP-AMP CIRCUIT TO REALIZE THE EQUATION

1 205 ( ) 2t

Ov v d vτ τ= −∫Needs integrator And adder

adder

Design equations Two equations in five unknowns!!

Not too large. Not too small

Seems a reasonable value 3 10R k⇒ = Ω

2 20R k⇒ = Ω

If supply voltages are 10V (or less) all currents will be less than 1mA, which seems reasonable

ANALYSIS OF THE SOLUTION

integrator