Package Inductance and RLC Circuit Analysis · Package Inductance and RLC Circuit Analysis FACTOR...
Transcript of Package Inductance and RLC Circuit Analysis · Package Inductance and RLC Circuit Analysis FACTOR...
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Package Inductance and RLC Circuit Analysis
FACTOR
Jawel, ik sta bekend als fenomeenTenminste onder mathematiciEn in wiskundig aangelegde kringen
Mijn naam, als u het weten wilt is iIk zit in allerlei berekeningenEn aan het einde hef ik mij weer op
U vraagt niet, denk ik, om verhandelingenMaar wenst dat ik mezelf nu eens ontpopWelnu - ik ben de wortel uit -1
Ik functioneer, ofschoon ik niet bestaDenk daar maar eens langdurig over na
(Uit: Drs. P en Marjolein Kool, Wis- en natuurlyriek met chemischsupplement, Amsterdam: Nijgh en van Ditmar, 2000)
Ch. 5
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New Today
Inductive effects, for switching (transient) behavior
Application of Complex Numbers
Relation between complex exponentials and harmonic(sinusoidal) functions
Damping factor, damped natural frequency, undamped natural frequency
Oscillatory behavior (opslingering)
Concepts of overdamped, critically damped, underdamped and undamped responses
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Package ParasiticsNeed transition from submicron on-chip world to PCB (printed Circuit Board) worldRelatively large ‘package parasitics’Parasitics = unwanted but non-avoidable electrical effects Inductance of package must often be accounted forOn-chip interconnectinductance also becoming important
Bull Microprocessor Chip
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Bondwires
Bondwires: connection from silicon chip to package pins
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Model for Invertor with Inductance
Cin
Rnvin
Rp
5 V
Lground
Lpower
voutCinCin
Rnvin
Rp
5 V
Lground
Lpower
vout
Many interesting properties
Can we determine speedand other properties of such systems?
Addition of L to R and C might result in oscillating behavior!
Need complex numbers to deal with it.
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Inductance (Inductantie)
L
i +
–v
dtdiLv =
C
i +
–v
dtdvCi =
L and C exhibit dual roles of iand vCauses many interesting and useful properties of electrical circuits and circuit analysisWe will apply this right now!
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DC Equivalent Circuit
DC value of capacitor current:
00 ==⇒= idt
dvcdt
dv cc
DC model for C is open circuit
00 ==⇒= vdtdiL
dtdi LL
DC value of inductor voltage:
DC model for L is
Dual!
short circuit
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First Order RL Circuits
0=++−dtdiLRiV L
Ls
( ) ( )( )
RVe
RVtiti stt
LR
sLL +⎟⎠⎞
⎜⎝⎛ −=
−− 00
VA
+
-Vs
t=t0
R
+
-L vL(t)
+
-
iL(t)
VA
+
-Vs
t=t0
R
+
-L vL(t)
+
-
iL(t)
1. Use KVL for circuit with t > t0 :
sLL V
Li
LR
dtdi 1
+−=
2. Rearrange:
3. Compare to RC case:
scc V
RCv
RCdtdv 11
+−=
5. Write down result (by inspection):
F(t) = (IV – FV) e-(t-t0)/τ + FV4. Remember general formula:
RC: τ = RC
RL: τ = L/R
F(t): waveform
IV: initial value
FV: final value
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Cascaded Invertors with Inductance
Rn
Rp
5V
Lground
Lpower
CinRn
Rp
5V
Lground
Lpower
Cin
Rint
+
-
Lpower
Package and power supply
5V
Driving gateRp
Rn
Lground
Cint
Interconnect Load gate
Cin
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Simplify Circuit
Rint
+
-
Lpower
Package and power supply
5V
Driving gateRp
Rn
Lground
Cint
Interconnect Load gate
Cin
+
-Vs C
Rt=0t=0
iL(t)
L vc(t)Prototypical RLC series circuit
Principal subject of this chapter
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RLC Circuit Differential Equations
Lc i
dtdvC =
KCL @ vc
+
-Vs C
Rt=0t=0
iL(t)
L vc(t)+ vL -+ vR –
+ vC
–
iC=iLR L
0=+++− CL
Ls vdtdiLRiV
KVL around loop0=+++− CLRs vvvV
State variables: vC, iL
Note: KVL for inductor voltage is dual from KCL for capacitor current
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RLC Circuit Differential Equations
Lc i
dtdvC =
KCL & KVL
0=+++− CL
Ls vdtdiLRiV
+
-Vs C
Rt=0t=0
iL(t)
L vc(t)
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
LV
iv
LR
L
C
dtdidt
dv
sL
CL
C 01
10
Lc i
Cdtdv 1
=
LVi
LRv
Ldtdi s
LCL +−−=
1
Rewrite
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⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
LV
iv
LR
L
C
dtdidt
dv
sL
CL
C 01
10
Two coupled first order D.E.’s
Next: find vc(t) and iL(t)Satisfy D.E.’sSatisfy initial conditions vc(0), iL(0)
Procedure:Assume general solutionDetermine steady state solutionsFind natural frequencies (s1, s2)Determine coefficients of trans. part
General method identical to RC and RL case
General Procedure
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Assume General Solution( ) ( ) C
tstsCCC VeVeVVtvtv t ++=+= 21 21
( ) ( ) Ltsts
LtLL IeIeIItiti ++=+= 2211
+
-Vin C
Rt=0
iL(t)
L vc(t)
Determine Steady State Solutions
VC= vc(∞) =
IL= iL(∞) =
Vin
0
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Find Natural Frequencies
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
LV
iv
LR
L
C
dtdidt
dv
sL
CL
C 01
100.
01
1=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+
−
LRs
L
Cs
det1.
011=⎟
⎠⎞
⎜⎝⎛ −
−⎟⎠⎞
⎜⎝⎛ +
Cx
LLRss2.
012 =++LC
sLRs3. Characteristic Equation of series
RLC circuit
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D - discriminant
Sign of Discriminant012 =++
LCs
LRs
02 =++ cbsas
aacbbs
242 −±−
=⇒
Note: c≠C
LCc
LRba 1;;1 ===
Will be explained shortly
undampedRe(si) = 0R = 04underdampeds1 and s2∈ CD < 03critically dampeds1 = s2D = 02overdampedlike RC and RLD > 01
Four Cases depending on sign of discriminant
Next: Study cases 1, 2, 3 and 4
LCLR
LRs 1
22
2−⎟
⎠⎞
⎜⎝⎛±
−=⇒
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Damping
Many systems vibrate/oscillate/cycle(more or less) repetitive behaviormechanical systems, electrical, economic, …
Damping is the mechanism that tries to suppress oscillationsand force the system into stable steady-state
‘schokdempers’ – of a car‘dempers’ – of Erasmus bridge Bridges can oscillate!
Tacoma Narrows Bridge, November 7, 1940
Overdamped – critically damped – underdamped – undampedWhat does it mean?
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Tacoma Narrows Bridge
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Tacoma Narrows Bridge
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D>0
Case 1: Overdamped Series RLC Response
LCLR
LRs 1
22
21 −⎟
⎠⎞
⎜⎝⎛+−=
LCLR
LRs 1
22
22 −⎟
⎠⎞
⎜⎝⎛−−=
No extra difficulties compared to RC or RL case
LCLR
LRs 1
22
2−⎟
⎠⎞
⎜⎝⎛±
−=
Strictly decaying transient solution
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Solve for s
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
LV
iv
LR
L
C
dtdidt
dv
sL
CL
C 01
100.
01
1=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+
−
LRs
L
Cs
det1.
011 =⎟⎠⎞
⎜⎝⎛ −−⎟
⎠⎞
⎜⎝⎛ +
Cx
LLRss2.
012 =++LC
sLRs3. Characteristic Equation of series
RLC circuit
Two real natural frequencies
CLR
LCLR 2012
2>⇔>−⎟
⎠⎞
⎜⎝⎛
Now consider case of
§5.8.1
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Case 2: Critically Damped Response
Houston, we have a problem…
α−≡−==L
Rss221
α: damping factor
Then, V1 and V2 are not independent because they refer to same exponential term. Actually, V1+V2 can be replaced by V’.
Remember: for a second-order system, we actually need two independent exponential solutions and two constants to adjust for two initial conditions.
Similarly for iC(t)
Because s1 = s2 = – α we would have( ) ( ) C
tC
ttC VeVVVeVeVtv ++=++= −−− ααα
2121
CLR
LCLR 2012
2=⇔>=⎟
⎠⎞
⎜⎝⎛
Now consider case of
§5.8.2
D=0LCL
RLRs 1
22
2−⎟
⎠⎞
⎜⎝⎛±
−=
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Second Independent Solution to D.E.
tt eedtd αα α −− −=
=− ttedtd α
vc(t) = V1e-αt + V2 t e-αt + Vc iL(t) = I1e-αt + I2 t e-αt + ILModified solutions:
Product rule
dtduv
dtdvuvu
dtd
+=⋅
Derivative (almost) proportional to function
Modified solution compared to RC or RL caseOvershoot in transient solution
D = 0 ⇒ 1 natural frequency
More formal derivation in D.E. course
tttt etetdtdee
dtdt αααα α −−−− +−=+
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Critically Damped Waveform
vc(t) = V1e-αt + V2 t e-αt + Vc
V1 -5
V2 10
Vc 5
α 1
012345678
0 1 2 3 4 5
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Houston, we have another problem…
Case 3: Underdamped Response
We need square-root of negative number…
Solution: complex numbers
FACTOR
Jawel, ik sta bekend als fenomeenTenminste onder mathematiciEn in wiskundig aangelegde kringen
Mijn naam, als u het weten wilt is iIk zit in allerlei berekeningenEn aan het einde hef ik mij weer op
U vraagt niet, denk ik, om verhandelingenMaar wenst dat ik mezelf nu eens ontpopWelnu - ik ben de wortel uit -1
Ik functioneer, ofschoon ik niet bestaDenk daar maar eens langdurig over na
(Uit: Drs. P en Marjolein Kool, Wis- en natuurlyriek met chemischsupplement, Amsterdam: Nijgh en van Ditmar, 2000)
CLR
LCLR 2012
2<⇔><⎟
⎠⎞
⎜⎝⎛
Now consider case ofD<0
LCLR
LRs 1
22
2−⎟
⎠⎞
⎜⎝⎛±
−=
See TIO ‘Complexe Rekenwijze’
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Imaginary Numbers
We need square-root of negative number…
We offer number of which the square is -1
Conventionally denoted as i: i 2 = -1
In EE, i is reserved for current, use j instead (j2 = -1)
xjxjxjx ===− 22Hence:
Now, we have real numbers, e.g. 1, 2, 1.34132, …
And imaginary numbers, e.g. 3j
Their combination is called a complex number
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Complex NumbersComplex number is sum of real and imaginary number,
Addition keeps real and imaginary parts separate:
(1+3j) + (1+3j) = 1+1+j(3+3) = 2+6j
Multiplication uses j2 = -1:(1+3j) x (1+3j) = 1 + 3j + 3j + 9j2 = -8+6j
All normal rules for addition, subtraction, multiplication, division, but real and imaginary parts are kept separate, except for j2, which becomes -1
e.g. 1+3j
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Complex Plane and Polar Form
z = a + bjbjb = Im(z)
a = Re(z)real axis
imaginary axis
r = |z|ϕ = arg(z)
z=a+bj z = r(cosϕ + jsinϕ) = r·ejϕ
ejϕ = cosϕ + jsinϕ
|z1z2| = |z1||z2| arg(z1z2) = arg(z1) + arg(z2) (mod 2π)Multiplication often easier using polar co-ordinates:
Eulers identity will be important
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LRs2
)Re( =≡α Damping factor
20
2 ωαα −±−=s
α
Underdamped Response
20ω
LC1
0 ≡ω Undamped natural frequency
LCLR
LRs 1
22
2−⎟
⎠⎞
⎜⎝⎛±−=
Would-be natural frequency when R=0
D<0LCL
RLRs 1
22
2−⎟
⎠⎞
⎜⎝⎛±
−=
CLR
LCLR 2012
2<⇔><⎟
⎠⎞
⎜⎝⎛
Now consider case of
§5.8.3
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Damped natural frequency220)Im( α−ω=≡ω sd
2dω
22 ωαα −±−=s
< 0
( )( )2201 αωα −−±−=
220
2 αωα −±−= j 220 αωα −±−= j
> 0
Complex Natural Frequencies
220 αωα −±−= j
djs ωα +−=1 djs ωα −−=2
Main result: two complex natural frequencies
21 ss =
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Complex Arithmetic for Underdamped Response
s1 and s2 two complex natural frequenciesAlso need complex coefficients in linear combination of both independent solutions for vc(t) and iL(t)Result: Familiar general solution, but with complex arithmetic
( ) Ctsts
C Veetv +Φ+Φ= 21 21
( ) Ltsts
L Ieeti +Γ+Γ= 21 21
complexare,, iiis ΓΦ
But we can’t have complex voltages and currents, can’t we?
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( ) Ctsts
C Veetv +Φ+Φ= 2211
( ) Ltsts
L Ieeti +Γ+Γ= 2211
Ltdjttdjt Iee +Γ+Γ= −−+− ωαωα
21
( ) ( )( ) Lddddt Itjttjte +−Γ++Γ= − ωωωωα sincossincos 21
( ) ( )( ) Lddt Itjte +Γ−Γ+Γ+Γ= − ωωα sincos 2121
Apply Euler’s Identity
Dilemma can be solved by exploiting Euler’s Identity
ejϕ = cosϕ + jsinϕ
Similar for vC(t)
djs ωα +−=1
djs ωα −−=2
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( ) ( )( ) Lddt
L Itjteti +Γ−Γ+Γ+Γ= − ωωα sincos)( 2121
Looks complicated complexBut physical signals are simply realSolution: Γ1 and Γ2 can not be chosen independentlyIm(Γ1 + Γ2 ) = 0 Im( j (Γ1 - Γ2 )) = 0
Γ1 and Γ2 must be complex conjugates:
Re(Γ1) = Re(Γ2) and Im(Γ1) = - Im(Γ2)
Γ1 = Γ2
Real or Complex?
Like s1 and s2
Then: Γ1 + Γ2 = 2a and j(Γ1 – Γ2) = 2b
Let: Γ1 = a - jb and Γ2 = a + jb
Clearly Real!Finally, let: I1 = 2a and I2 = 2b
( ) Lddt
L ItItIeti ++=⇒ − ωωα sincos)( 21 Similar for vC(See book)
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12 Γ=+=Γ jba
If we wouldn’t know that Γ1 and Γ2 are complex conjugates, we can ‘discover’ it as follows:
Γ1 and Γ2 are Complex Conjugates
jba −=Γ1 jdc −=Γ2
)(21 dbjca +−+=Γ+Γ
bd −=⇔=Γ+Γ 0)Im( 21
)(21 dbjca −−−=Γ−Γ
( ) caca =⇔=−=Γ−Γ 0Re 21
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Underdamped Response
( ) ( ) Cddt
C VtsinVtcosVetv ++= − ωωα21
( ) ( ) Lddt
L ItsinItcosIeti ++= − ωωα21
Exponentially damped
Sinusoidal Waveform
Steady State
Vi and Ii again follow from initial conditionand derivatives at t=0 (our method) (or substitution of above general solution in original DE (book))
Damped natural frequencyDamping factor
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Underdamped Waveforms
-6
-4
-2
0
2
4
6
8
0 1 2 3 4 5
( ) ( ) Cddt
C VtsinVtcosVetv ++= − ωωα21
αωdV1V2Vc
αωdV1V2Vc
0.52
-505
0.52
-505
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Case 4: Undamped Series RLC Response
01 ωj
LCs ±=−±=
( ) LL ItsinItcosIti ++= 0201 ωω
When R = 0 ⇒ α = 0, ωd=ω0
Oscillating behavior, without damping, oscillation continues forever
Perpetuum Mobile?
LCLR
LRs 1
22
2−±−=
LR2
−≡α LC1
0 ≡ω
( ) CC VtsinVtcosVtv ++= 0201 ωω
( ) ( ) Cddt
C VtsinVtcosVetv ++= − ωωα21
α = 0, ωd=ω0
§5.8.4
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The natural frequencies can be shown in complex plane
Im
Re
Overdamped
Im
Re
Underdamped
Im
Re
UndampedCritically Damped
Im
Re(2x)
Review of Natural Frequencies
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Digital System Switching Speed
§5.9
Compare switching speed with and without supply line inductance
Completely worked example
But also see other examples in book
Fully worked example 5.7
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Digital System Switching Speed
Determine high-to-low waveform here
Assume initially high for long time
Compare RC and RLC case
Rout = 100Ω
Cin = 1pFL = 10nH
Vs = 5V
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Compare RC and RLC Case
Without L
t=0 1pF
100Ω v ?With L
1pFt=0
100Ω v ? 10nH
Rout = 100Ω
Cin = 1pFL = 10nH
Vs = 5V
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Without L
τ =
v(0) =
v(∞) =
v(t) = FV + (IV – FV)e-t/τ
te10105 −=
1pFt=0
100Ω
5
0
100Ω x 1pF = 0.1 nS
-2.00
-1.00
0.00
1.00
2.00
3.00
4.00
5.00
6.00
0.00 0.20 0.40 0.60 0.80 1.00
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Strategy: Solving Second Order RLC circuits
1. Write down differential equations (determine state vars)2. Write down characteristic equation and solve for s3. Assume general solution, depending on si, as in scheme
below (A, B and C can be current or voltage or …)
4. Solve for steady state solutions, and constants (use I.C. and time derivatives at t=0)
5. Check solution!
CBeAe tsts ++ 21Overdamped: two different, real s
CBteAe tt ++ −− ααCritically damped:s1 = s2 = –α
( ) ( )( ) CtcosBtcosAe ddt ++− ωωα
Underdamped: s1 and s2 complex conjugate (undamped when α = 0)
( ) ( )sImsRe d == ωα
CBeAe tsts ++ 21Overdamped: two different, real s
CBteAe tt ++ −− ααCritically damped:s1 = s2 = –α
( ) ( )( ) CtcosBtcosAe ddt ++− ωωα
Underdamped: s1 and s2 complex conjugate (undamped when α = 0)
( ) ( )sImsRe d == ωα
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Determine (Complex) Natural Frequencies
t=0 1pF
100Ω vC ?10nH
iL+
-
LCLR
LRs 1
22
2−⎟
⎠⎞
⎜⎝⎛±−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
LV
iv
LR
L
C
dtdidt
dv
sL
CL
C 01
10
20
2
99 101
1020100
1020100
−−− −⎟⎟⎠
⎞⎜⎜⎝
⎛
×±
×−=
j99 1066.8105 ×±×−= Underdamped !
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α = 5 Damping factor
Write Assumed Solution
js 99 1066.8105 ×±×−= Unit: 1/s
js 66.85 ±−= Unit: 1/ns
Change seconds into nano seconds (for convenience)
( ) ( ) ( )( ) Cddt
C VtsinVtcosVetv ++= − ωωα21
( ) ( ) ( )( ) Lddt
L ItsinItcosIeti ++= − ωωα21
ωd = 8.66 Damped Natural Frequency
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Steady State and Initial Values
vC(∞) = 0 iL(∞) = 0
vC(0) = iL(0) =5 0
t=0 1pF
100Ω vC ?10nH
iL+
-
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Determine Constants
( ) ( ) ( )( ) Cddt
C VtsinVtcosVetv ++= − ωωα21
( ) ( ) ( )( ) Lddt
L ItsinItcosIeti ++= − ωωα21
Next: Determine constants (V1, V2, I1, I2)From initial conditions
From derivatives at t=0
(book uses substitution in D.E.)
VC=vC(∞) =0 IL=iL(∞) =0
First: steady-state values
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Use Initial Condition(and Steady-State Value)
( ) ( ) ( )( ) Cddt
C VtsinVtcosVetv +ω+ω= α−21
( ) ( ) ( )( ) CC VsinVcosVev ++== 0050 210
( ) 121 0011 VVV =+⋅+⋅= ⇒ V1 = 5
( ) 121 0011 III =+⋅+⋅= ⇒ I1 = 0
( ) ( ) ( )( )tsinVtcosetv ddt
C ω+ω= α−25
( ) ( )tIeti dt
L ω= α− sin2
3.
( ) ( ) ( )( ) Lddt
L ItsinItcosIeti +ω+ω= α−21
( ) ( ) ( )( ) LL IsinIcosIei ++== 0000 210
2.
1.
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Analysis Tools Reminder
Combined: ( ) ( ) ( )tetetedtd ttt ωα−ωω=ω α−α−α− sincossin3.
Chain Rule: dxdu
dudy
dxdy
⋅=
( ) ( ) ( )xdxdusin
dudxsin
dxd ωω ⋅=
( ) xxu ⋅= ω
( ) ω⋅= ucos( )xcos ωω=
2.
Example:
AnalysisProduct Rule: dtduv
dtdvuvu
dtd
+=⋅
Example: tsinedtd tα− tt e
dtdtsintsin
dtde αα −− +=
tsinetcose tt αα α −− −=
1.
Review!Determine ( )te
dtd t ωα− cos
Review!Determine ( )te
dtd t ωα− cos
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Use Time-Derivative at t=0 for Assumed Solution
( ) ( ) ( )( )tVtetvdtd
ddt
C ω+ωα−= α− sincos5 2
( ) ( )( )tVte ddddt ωω+ωω−+ α− cossin5 2
( ) ( ) ( )( )tsinVtcosetv ddt
C ω+ω= α−25
( ) dt
C Vtvdtd
ω+α−==
20
5Hence:
cos(0) = 1 sin(0) = 0 e0 = 1Note:
cos(0) = 1 sin(0) = 0 e0 = 1Note:Use:
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( ) 0)0(1)0(00
++×==
LCt
C iC
vtvdtd
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
LV
iv
LR
L
C
dtdidt
dv
sL
CL
C 01
10
Use Time-Derivative at t=0 for D.E.
Known!
00150 =×+×=C
( ) 00
==t
C tvdtd
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( ) ( ) ( )( )t.sin.t.cosetv tC 66889266855 += −
Final result (t in ns):
Combine Time-Derivative Info at t=0
( ) dt
C Vtvdtd
ω+α−==
20
5
( ) 00
==t
C tvdtd
α=ω 52 dV
89.22 =V
( )t.sine.i tL 66805770 5−−=
Use the preceding procedure for iLUse the preceding procedure for i L
α = 5ωd = 8.66earlier:
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( ) ( ) ( )( )t.sin.t.cosetv tC 66889266855 += −
( )t.sine.i tL 66805770 5−−=
Final result (t in ns):
Inverter Pair Waveforms
t=0 1pF
100Ω v ?
t=0 1pF
100Ω v ?
-2.00
-1.00
0.00
1.00
2.00
3.00
4.00
5.00
6.00
0.00 0.20 0.40 0.60 0.80 1.00
vC(t)
when L = 010nH
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Summary: Solving Second Order RLC circuits
1. Write down differential equations (determine state vars)2. Write down characteristic equation and solve for s3. Assume general solution, depending on si, as in scheme
below (A, B and C can be current or voltage or …)
4. Solve for steady state solutions, and constants (use I.C. and time derivatives at t=0)
5. Check solution!
CBeAe tsts ++ 21Overdamped: two different, real s
CBteAe tt ++ −− ααCritically damped:s1 = s2 = –α
( ) ( )( ) CtcosBtcosAe ddt ++− ωωα
Underdamped: s1 and s2 complex conjugate (undamped when α = 0)
( ) ( )sImsRe d == ωα
CBeAe tsts ++ 21Overdamped: two different, real s
CBteAe tt ++ −− ααCritically damped:s1 = s2 = –α
( ) ( )( ) CtcosBtcosAe ddt ++− ωωα
Underdamped: s1 and s2 complex conjugate (undamped when α = 0)
( ) ( )sImsRe d == ωα