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
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