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Input Filter Interactions with Switching Regulators
Christophe Basso Technical FellowChristophe Basso – Technical Fellow
IEEE Senior Member
Public Information
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions2 3/7/2017
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions3 3/7/2017
EMI Filter Interaction
A LC filter is inserted to prevent input line pollution
outVLinV outVLinV
gVloadRCgVloadRC
Switching converterSwitching converter ?inZ s
What load does the converter offer?
Public InformationChristophe Basso –Input Filter Interactions4 3/7/2017
C. Basso, “Designing Control Loops for Linear and Switching Power Supplies”, Artech House, 2012
A Negative Incremental Resistance Assume a 100%-efficient converter
out inP P in in out outI V I V
Infinite rejection
In closed-loop operation, Pout is constant, no link to Vin
outin in
in
PI V
V
P
in
For a constant Pout , if Vin increases, Iin drops11
2
out
in in in out
in in in
PddI V V P
dV dV V
0.6
0.8
AinI 0.6
0.8
AinI
10 15 20 25 300.2
0.4
10 15 20 25 300.2
0.4 The incremental input resistance is negative
2in
inout
VR
P
Public InformationChristophe Basso –Input Filter Interactions5 3/7/2017
VinV VinV
A Simple SPICE Simulation
A constant-power current source shows the negative resistance
bulk
Vin100
B1Current
bulkparameters
100Pout/V(bulk)Pout=50W
.TF V(bulk) Vin
***** SMALL SIGNAL DC TRANSFER FUNCTION
Infinite bandwidth
output_impedance_at_V(bulk) 0.000000e+000vin#Input_impedance -2.00000e+002Transfer function 1 000000e+000
Neg. resistance
Public InformationChristophe Basso –Input Filter Interactions6 3/7/2017
Transfer_function 1.000000e+000
A Simple LC Filter
The low-pass filter is built with L and C elements:
11 zs
H HoutV
Lr LCr
gV
10 2
0 0
1
zH s Hs sQ
R
1
1z
Cr C C 0
1 L
C
r Rr RLC
LC R
R
0 C
L C L C
LC r RQ
L C r r r R r R
A negative resistance cancels 0L C L CL C r r r R r R
A negative resistance cancels losses: poles become imaginary
If
C LL r r CRC r r
Q
Sustained oscillations
Public InformationChristophe Basso –Input Filter Interactions7 3/7/2017
C LC r r
A Negative Resistance Oscillator
If losses are compensated, the damping ratio is zero
0 21 zsH s H
1
2Q
If ohmic losses are gone the 0 2
20 0
1s sQ
2 If ohmic losses are gone, the
damping ratio is zero, Q is infinite.
Without precautions, instability can happen!
130
150
130
150 Input voltageLr
CrL outV
110
130 Output voltage
110
130
inVC
C
I (V)
70.0
90.0
70.0
90.0C
outPI Negative resistance(constant) 0 0 0
Public InformationChristophe Basso –Input Filter Interactions8 3/7/2017
3.64m 10.9m 18.2m 25.5m 32.7m3.64m 10.9m 18.2m 25.5m 32.7moutV Negative resistanceInfinite bandwidth
Filter Output Impedance
What is the output impedance of an LC filter? TV s
L Cr
loadR
TI s 0th
N sZ s R
D s
It follows the form
Lr C Ω ΩNo dimension
L
2
1 1||
1 ||
C outL
th L loadC load
Ls sr Cr
Z s r Rr RLs C r R r s LC
1 ||L load CL load L load
s C r R r s LCr R r R
All losses (ohmic, iron etc.) help decreasing Q
Public InformationChristophe Basso –Input Filter Interactions9 3/7/2017
( ) p g Q
Typical Filter Output Impedance
At low frequency, the inductive ohmic loss dominates
dBΩ °
90 0
180
10 0
20.0
Phase is positive
0
90.0
0
10.0
outZ f
-90.0-10.0 outZ f
r
CrPhase is negative
1 10 100 1k 10k 100k 1Megfrequency in hertz
-180-20.0Lr
Public InformationChristophe Basso –Input Filter Interactions10 3/7/2017
Negative Resistance at Low Frequency N i i b f f db k (P ) Neg. resistance exists because of feedback (Pout = constant )
3 18
L15u
5
R41.5m
PWMCM2_LX1L = 5u
VZin
2.70V2.70V2.70V
I d
4
Resr60m
7
Vin2.7
2
duty_cyclevout
c
cos
sy P
WM
it
ch C
M
p
duty
-cyc
le
SWD
L 5uFs = 1MegRi = -40mSe = 16.4k
I1AC = 1
L210GH
Rload
5.00V
2.70V
536mV
Average modelImpedancemeasurementsetup
16
Cout200uF
R91k 11
D1mbra210et3
13
Rdson140m
parameters
Rupper=10kC2C2
vca Lo
swi S 55.00V-455mV
-301mV
6
R1Rupper
9
14
R610k
R71k
Rupper=10kfc=8kpm=70Gfc=-20pfc=-101
G=10^(-Gfc/20)boost=pm-(pfc)-90pi=3.14159
15
R2R2
C1C1
C2
2.50V
2.50V
1.10V
99.9mV
2.50V
R310k
1
V22.5
Verr
X2AMPSIMPVHIGH = 100VLOW = 10m
SwitchMode Power Supplies: SPICESimulations and Practical Designs
pK=tan((boost/2+45)*pi/180)C2=1/(2*pi*fc*G*k*Rupper)C1=C2*(K^2-1)R2=k/(2*pi*fc*C1)
fp=1/(2*pi*R2*C2)fz=1/(2*pi*R2*C1)
G s
Public InformationChristophe Basso –Input Filter Interactions11 3/7/2017
SwitchMode Power Supplies: SPICE Simulations and Practical DesignsChristophe Basso - McGraw-Hill, 2014
Negative Resistance at Low Frequency Th i i l i 200 H The resistance is truly negative up to 200 Hz
1 2
24.0
12 0
0
12.0
(dBΩ) 1.02Ω
Well before crossover,
-24.0
-12.0 inZ f
In this region
e be o e c osso e ,the -180° argument is gone.
0
72.0
144
inZ f
(°)
In this regionPout constant
fc
-144
-72.0
180inZ
fc
Public InformationChristophe Basso –Input Filter Interactions12 3/7/2017
10 100 1k 10k 100k 1Meg
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions13 3/7/2017
A Filter Affects the Line Output Impedance Th li i d d i i h i l 0
thZ s inZ s ˆoutv 2H s
The line impedance driving the converter is no longer 0
1H s
ˆgv
V
Gd
inV s
G s
thV s iV s+
The system can be modeled by a minor loop
1
in thV s V s th inV s-
th inZ Z
1in th
th
in
Z sZ s
gain
Public InformationChristophe Basso –Input Filter Interactions14 3/7/2017
ga“Input Filter Considerations in Design and Application of Switching Regulators”, R. D. Middlebrook, IEEE Proceedings, 1976
Impedance Interactions Stability can be at stake when inserting the filter
inV s
thZ s thV s inZ s
in
Filter Switching Supply thZ s inZ s
1
The Nyquist criterion applies
C diti f 1thZ s
1
1in th
th
in
V s V sZ sZ s
1 and 180th thZ s Z sZ s Z s
Conditions foroscillations
1
inZ s
Public InformationChristophe Basso –Input Filter Interactions15 3/7/2017
in inZ s Z s
A Filter Modifies Converter’s Dynamics The EMI filter output impedance makes The converter input voltage is no longer zero in ac analysis
0thZ s
Z s ˆoutv H s
ˆoutv
0thZ ˆ 0inv H s
thZ s outout
d
d
0thZ
ˆ 0iv G s G s
outV sH s
D s
outV s
H sD s
0inv
Public InformationChristophe Basso –Input Filter Interactions16 3/7/2017
0inV sD s
0inV s
D s
Extra Element Theorem to Help It can be shown how an EMI filter affects the open-loop gain
V s H s
ˆoutv
0th
out
Z s
V sH s
D s
Gain without the filter
Z thZ s thZ s Filter output impedance
G sd
1 thZ s
Extra-Elementcorrection factor
1
0 0
1
1th th
th
out out N
thZ s Z s
V s V s Z sH s
Z sD s D sZ
th NZ s Z s
th DZ s Z s
Public InformationChristophe Basso –Input Filter Interactions17 3/7/2017
DZ s
What are ZD and ZN? ZD and ZN come from the Extra-Element Theorem, EET
0D i D sZ s Z s
Open-loop input impedance 0D s Open loop input impedance
ˆ 0outN i v
Z s Z s
Input impedance for a nulled outputIdeal input rejection
?DZ s ?NZ s Null
Ideal input rejection
ˆ 0outv ˆ 0outv
0 35%ˆ 0
D
d
0 35%
ˆ 0
D
d
ˆ 0outi ˆ 0outi
Ideal control
ˆ 0
Public InformationChristophe Basso –Input Filter Interactions18 3/7/2017
Open-loop input impedance Input impedance for ˆ 0outv
Consider Closed-Loop Input Impedance Th l h l d b d i d
Converter ZN(s) ZD(s) Ze(s)
These values have already been derivedZe(s) is the input impedance for a shorted output
Buck
22 21' 'L LCs s
D R D 2' 1 sLD R
L
2
sLD
2
2
1
1
Ls s LCR RD sRC
2
RD
Boost
Buck-Boost
2 22 ' ''1
D R DD RsRC
22' 1'
D RD R
sL
2
sLD
2
2 2
' 1'
D R sDLD D R
22 2 2
2
1' ' '1
L LCs sD R D R DD sRC
D
A converter closed-loop input impedance follows the form
1 1 1 11 1in N D
T sZ s Z s T s Z s T s
L i
Public InformationChristophe Basso –Input Filter Interactions19 3/7/2017
“Fundamentals of Power Electronics”, R. Erickson, D. Maksimovic, Springer, 2001Loop gain
The Output Impedance is Affected The filter degrades the closed-loop output impedance
1 thZ s No impact if
, ,0 01
th th
eout CL out CLZ s Z s
th
in CL
Z sZ s Z s
Z sZ s
th eZ s Z s
No impact if
,in CL
Ze(s) is the converter input impedance with a shorted output
eZ s 0outV 2H s
Short circuit
Public InformationChristophe Basso –Input Filter Interactions20 3/7/2017
Applying Dr. Middlebrook Criteria P i i id i hi ll l Previous equations consider switching cell alone Do not include the decoupling capacitor!
H H iZ s
ˆgvˆoutv
2H s 1H s inZ s
g
If possible, move the capacitor to the filter sidep , p
ˆoutv 2H s 1H s inZ s
ˆgv outv
Public InformationChristophe Basso –Input Filter Interactions21 3/7/2017
“Input Impedance and Filter Interactions Part I”, R. Ridley, ridleyengineering.com
At 1st-Order Check Crossover Points
dBΩdBΩ
Plot the converter input resistance value Plot the filter output impedance: check for overlaps: th inZ s Z s
65.0
36 dBΩ
undamped65.0
36 dBΩ
undamped2
in
out
VP
Overlap occurs?
1th
in
Z sZ s
5 00
35.036 dBΩ
5 00
35.036 dBΩ
1th
in
Z sZ s
Watch the
out in
-25.0
5.00
damped
3.6 /100 µF-25.0
5.00
damped
3.6 /100 µF
Watch the
argument of
th inZ Z
-55.0 outZ s-55.0 outZ s
th in
Output filter
Public InformationChristophe Basso –Input Filter Interactions22 3/7/2017
10 100 1k 10k 100k10 100 1k 10k 100k
Condition for Oscillations Plot the ratio of magnitude Plot the difference of arguments
th inZ Zarg argth inZ Z
Check phase margin at 0-dB points if any
60.00
° dBth inZ Z Conditions for
ill ti
30.0
60.0
-90.0
0 th in
1thZ sZ s
oscillations
safe
-30.0
0
-270
-180
and
180
in
th
Z s
Z s
problem
1 10 100 1k 10k 100k 1Meg
-60.0-360
arg argth inZ Z 180
inZ s
Public InformationChristophe Basso –Input Filter Interactions23 3/7/2017
1 10 100 1k 10k 100k 1Meg
Damping the Filter to Build Margin Damping the filter can provide phase margin at 0 dB The phase margin in this example is a bare 12°
020 0
dB °
Avoid magnitude overlaps by working on Zth and Zin!
-90.0
0
10.0
20.0
arg argth inZ Z
168
-270
-180
-10 0
0
-360
-270
-20.0
-10.0
th inZ Z
Public InformationChristophe Basso –Input Filter Interactions24 3/7/2017
1 10 100 1k 10k 100k 1Meg
Preliminary Conclusion Y d i h i fil h i h h SMPS You design the input filter together with the SMPS
Converter loop gain T is unaffected th NZ s Z sWhen a filter is installed:
Converter output impedance is unaffected thZ s Z s
th DZ s Z sand
Converter output impedance is unaffected th eZ s Z sVout shorted, open loop
You design the input filter alone
andth th
in in
Z s Z sZ s Z s
If d i th
Don’t care about phase anymore
If damping thefilter guarantees
th in closed loopZ s Z s OK
Public InformationChristophe Basso –Input Filter Interactions25 3/7/2017
System is stable but overall performance may be altered
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions26 3/7/2017
Classical Circuits Analysis Techniques Apply classical Kirchhoff’s voltage and current laws The expression is correct but disorganized: high-entropy form
V response
What is the t f f ti ?
Lr 1LCr
R
outV
outV sH s
response
transfer function?
Apply brute2C
1RinV
in
H sV s
stimulus
1
2 1 1 2
1 ||CC
r RsC R sR r C
H s
Apply bruteforce algebra
stimulus
2 21 1 2 1 2 1 2 2 1 1 2 1
1 12
1 ||
C
L C L C L CC L
H sR r sL sC R r C R r s C r r s C L R s C L r s
r R r sLsC
No insight: poles? Zeros? Dc gain?
Public InformationChristophe Basso –Input Filter Interactions27 3/7/2017
o s g t po es e os c ga
Fast Analytical Circuits Techniques FACTs describe a set of tools to quickly write transfer functions
21 1 Csr CRH s
1 2 112 1 1 2
1 1
1 ||L CC L
L L
H sR r r RLs C r r R s L C
r R r R
Dc gain
1 s
Naturally showing gains, poles and zeros…
1 10
1 Cr R
0 2
0 0
1
zH s Hs sQ
2z
Cr C 0
11 2 Lr RL C
1 1Lr R
QL C r r R r r
0 0Q 01 2 1C L C LL C r r R r r
This is a low-entropy expression
Public InformationChristophe Basso –Input Filter Interactions28 3/7/2017
R. D. Middlebrook, “Methods of Design-Oriented Analysis: Low Entropy Expressions”, New Approaches to Undergraduate Education, July 1992
Two Different Stages
Consider dc and high-frequency states for L and C
C impedance 1 Dc state Z Cap is an open circuitC p 1CZ
sC Dc state
HF state
CZ 0CZ
Cap. is an open circuit
Cap. is a short circuit
impedance D 0Z I d i h i iLimpedance
LZ sL Dc state
HF state
0LZ
LZ Inductor is a short circuit
Inductor is an open circuit
Change the circuit depending on s
C L0s 0s s s
Public InformationChristophe Basso –Input Filter Interactions29 3/7/2017
Principles of FACTs: Time Constants
Determine time constants in two different conditions1. The excitation is set to zero (no excitation)2 The output is nulled (no response)2. The output is nulled (no response)
How do you determine a time constant?
Remove
capacitor
TI
TV1R
2R 3R1R
2R 3R 1Ccapacitor
||TVR R R R
Test generator
Remove L or C and look into its terminals: 2 3 1||T
T
R R R RI
Remove L or C and look into its terminals:
2 3 1 1||R R R C
Public InformationChristophe Basso –Input Filter Interactions30 3/7/2017
Excitation is set to 0In your head, imagine an ohm-meter placed across C1’s terminals
Turning the Excitation off – The Pole Turning the excitation off means A 0-V source becomes a short circuit A 0-A generator is an open circuit and disappears A 0 A generator is an open circuit and disappears
1RR
1L 1RR
Set source1L
0 V2RinV 2R
to 0 V?R
1 2R R
2R?R
2R
0 V
Set source
to 0 A1R
1LinI
?R
1R 1
1
LR
0 A
to 0 A
The inverse of the time constant in this case is a pole: 1p
Public InformationChristophe Basso –Input Filter Interactions31 3/7/2017
p p
Mathematical Definition of a Zero A zero is the root of the equation 0f x
15
2f10
2 4f x x
0f x 1x 2x
0
5f x( )
1 2x
1 2
4 2 0 2 45
x
2 2x
Transfer function zeros are the numerator roots
0N s 1 2, ...z zs s
Public InformationChristophe Basso –Input Filter Interactions32 3/7/2017
Nulling the Response If the numerator is 0, then the response is theoretically 0
HH((sszz))ˆ 0outv
Complex excitation
(( zz))
s s 0N s
Complex response
zs s 0zN s
What is happening in the box when ? zs s
The excitation cannot reach the output: the response is nulled
0out zV s ˆ 0out zv s
Public InformationChristophe Basso –Input Filter Interactions33 3/7/2017
How Does the Response Disappear? The signal is lost in the transformed network
1Rresponse response 1 zZ s
1Rp p 0out zV s 0out zV s
2 0zZ s in zV s in zV s1R
excitation excitations j
A series impedance A parallel impedance
s j s j
pbecomes infinite.
p pshorts the path to ground
What is a transformedtransformed network?
Public InformationChristophe Basso –Input Filter Interactions34 3/7/2017
The Transformed Network
Reactances are replaced by their Laplace expression
1R 2R 1R 2R1R 2R
1C
1R
1
1sC
2R
1
1R
L L
1R2R 2R
2L 2sL
The circuit is then observed at the zero angular frequency
Public InformationChristophe Basso –Input Filter Interactions35 3/7/2017
Considering a Negative Angular Frequency F h RC i d i h i i For s = sz1, the RC impedance is a short circuit
Cr 1 sr C 11 0 ΩzZ s =0
Cr
1 1
11
1 Csr CZ s
sC
1
zs shunt
1sC 11
zCr C
For s = s 2, the RL impedance is infinite For s sz2, the RL impedance is infinite
2R
2 2sL RZ
22 ΩzZ s
2sL
2 22
2 2
Z sR sL
2
2z
Rs
L =0
Poles of the RL network become
Public InformationChristophe Basso –Input Filter Interactions36 3/7/2017
2LseriesPoles of the RL network become zeros of the transfer function.
Zeros by Inspection: Fastest Option! Id if f d i i / h i i Identify transformed open circuits/short circuits
1R 3R outV s Lrs Lr
Lr
1R2R
2sL
3R
Cr
out
2R
11
zsL 1
1z L
2R2
sL 1 inV s 2
2
2z
Rs
L
1
2
2
2z
RL
11sL1sC
31
1z
C
sr C
3
1
1z
Cr C
1 2 3
1 1 1z z z
s s sN s
Noequations!
Public InformationChristophe Basso –Input Filter Interactions37 3/7/2017
FACTs at Work in an Example
How would you calculate Vout / Vin?
1R 3R
Cr
outV s
2RC
C
4R inV s
1C
1. Transform the circuit with a Thévenin generator2. Apply impedance divider involving C1
Public InformationChristophe Basso –Input Filter Interactions38 3/7/2017
g 1
Apply Impedance Divider
Reduce circuit complexity with Thévenin
1 2 3||R R R1R 3R
R
outV s outV s1 2 3
Cr
1R
2R
3R
Cr thR s
thV s 2
1 2in
RV s
R R
1C
4R
1C
4R
1Z s
Apply impedance divider involving Z1 and Rth
1 2
1 1 2th
Z s RH s
Z s R s R R
1 41|| CZ s R rC
1 2 3||thR s R R R
“Who you
Public InformationChristophe Basso –Input Filter Interactions39 3/7/2017
1 41
C sC gonna call?”
High-Entropy Expression
H2 s( )R2 R4 C1 rC s 1
R1 R2 R1 R3 R1 R4 R2 R3 R2 R4 C1 R1 R2 R4 s C1 R1 R3 R4 s C1 R2 R3 R4 s C1 R1 R2 rC s C1 R1 R3 rC s C1 R1 R4 rC s C1 R2 R3 rC s C1 R2 R4 rC s
How do you make use of this result?
what is the pole/zero position? what affects the quasi-static gain for s = 0?q g
0
20 H f
You can plot the ac -20
-40
-60
0
50 You can plot the ac response but it yields no
insight on what drives
-80
-10010 100 1k 10k 100k 1Meg 10Meg 100Meg
-50 arg H f poles and zeros!
Public InformationChristophe Basso –Input Filter Interactions40 3/7/2017
10 100 1k 10k 100k 1Meg 10Meg 100Meg
Applying FACTs Now
What is the gain when Vin is a dc voltage?
tV s1R
R
3RCr
R
outV s
V2R
1C
4R inV s
The capacitor is open circuited, read the schematic! The capacitor is open circuited, read the schematic!
2 40
1 2 1 2 3 4||R R
HR R R R R R
Public InformationChristophe Basso –Input Filter Interactions41 3/7/2017
Determine the First Time Constant
Look at the resistance driving the storage element1. When the excitation is turned off, Vin = 0 V
1R 3RCr
2RC
4R
?R
1 1 2 3 4 1|| ||Cr R R R R C
Public InformationChristophe Basso –Input Filter Interactions42 3/7/2017
1 1 2 3 4 1|| ||C Denominator calculation
Determine the Second Time Constant
Inspect the circuit and find the transformed short circuit
0tV s 1R
R
3RCr
R
0out zV s
V2R
1
1sC
4R in zV s
1Z1sC1
If Z is equal to 0 the output is nulled
1
1 0Cr s C
If Z1 is equal to 0, the output is nulled
1
1z
C
sr C
Public InformationChristophe Basso –Input Filter Interactions43 3/7/2017
1zs C 1Cr CNumerator calculation
Assemble the Terms
You immediately have a low-entropy form
R R
1 s
2 40
1 2 1 2 3 4||R R
HR R R R R R
1 0
1
zH s Hs
1
1 2 3 4 1
1|| ||p
Cr R R R R C
p
1
1z
Cr C
Way cool!
We did not write a single line of algebra!
Public InformationChristophe Basso –Input Filter Interactions44 3/7/2017
Use Mathcad® to Check ResultsR1 1k R2 22k rC 0.1 R3 150 R4 100
|| x y( )x y
x y C1 1F
40
20
0
50
20 log H1 i 2 fk 10 arg H1 i 2 fk 180
H s( )
R4 rC1
s C1
R4 rC1
s C1
R4 rC1
R2R1 R2 80
60
40
50
0
20 log H1 i 2 fk 10 20 log H2 i 2 fk 10
1
arg H2 i 2 fk 180
R4 rC s C1
R4 rC1
s C1
R1 R2
R1 R2 R3
H2 s( )R2 R4 C1 rC s 1
R R R R R R R R R R C R R R C R R R C R R R C R R C R R C R R C R R C R R
10 100 1 103 1 104 1 105 1 106 1 107 1 108100
fk
2 R1 R2 R1 R3 R1 R4 R2 R3 R2 R4 C1 R1 R2 R4 s C1 R1 R3 R4 s C1 R2 R3 R4 s C1 R1 R2 rC s C1 R1 R3 rC s C1 R1 R4 rC s C1 R2 R3 rC s C1 R2 R4 rC s
2 C1 rC R1 || R2 R3 || R4 91.812 s
1 C1 rC 100 nsSuperimposing both transfer functions,
1 C1 rC 100 ns
H0R4
R4 R1 || R2 R3
R2R1 R2 0.079
H1 s( ) H01 s 1
matching should be perfect. If not, there is mistake.
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H1 s( ) H0 1 s 2
Fractions and Dimensions
A 1st-order system follows the form
N11 a s
f t i
0 1
0 1
N s a a sH s
D s b b s
0 0
10
0
1
a aH s
bb sb
factoring
A leading term (if any) carries the unit0b
a 11a
s a
1
00
1
1
1
a sa
Z s Rb
0
1 sa
b
1
0
s Naa
time
1
0
1 sb
Unitless
1
0
1b
sb
Unitless
1
0
s Dbb
time
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Unitless Unitless
2nd-Order System
A 2nd-order system follows the form
2 2
2
0 1 22
0 1 2
s sH s
s s
21 2
0 21 2
11
a s a sH s H
b s b s
Factoring 0Unitless
Factoring 0
The second fraction is unitless
Carries the unit
11 1 2
0
s N Na
2 1 222 1 2 2 1
0
s orN N N Na
sum product
time²
11 1 2
0
s D Db
1 2
2 1 222 1 2 2 1
0
s orD D D Db
time²
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reactance 1 reactance 2
Alternating the Reactance States In a 1st order circuit there is one reactance In a 1st-order circuit, there is one reactance it is either in a high-frequency state or in a dc state
0s s
In a 2nd-order circuit, there are two reactances we can consider individual states
0s L s 0H H1L2C 1L2C
1L
2C
0
?R ?R
21
12
1L2C1L2C
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?R ?R
Introducing the Notation Set one reactance into its high-frequency state
1 Reactance 1 is in its high-frequency state
2?R
What resistance drives reactance 2?
?R
2 Reactance 2 is in its high-frequency state
1?R
What resistance drives reactance 1?
There is redundancy: pick the simplest result1
2 1 2b 22 2 1b
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2 1 2 2 2 1
Example with Capacitors Assume the following 2-capacitor circuit
R
outV s
R1RCr
2C inV s
0s 1RCr
0 1H
1C
Determine the two time constants while Vin is 0 V
R ?R C R b 1RCr
?R
1 1 1CC r R
2 2 1C R
1 1 2
1 1 1 2 1C
bb C r R C R
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Determining the Higher-Order Term Place C1 in its high-frequency state and look into C2
1R r ?R1Cr ?R
12 1 2 1 1 2 1 ||C Cb C r R C R r 0 VinV
1C 2C
12 1 2|| CR r C
Place C2 in its high-frequency state and look into C1
1
2 g q y 1
1RCr
2
?R 21 1Cr C
22 2 1 2 1 1 Cb C R C r 0 VinV
1C2C
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1 1Cr C
Denominator is Completed The denominator can be assembled
2 21 2 1 1 2 1 2 1 11 1 C CD s b s b s C r R C R s C R C r s
Is there a zero in this network?
0out zV s
1RCr 1
1 0r 1
1
1z
C
sr C
1sC
2sC in zV s1
0Cr sC
11
1z
Cr C
Z1sC
If Z1 becomes a transformed short, the response disappears
1Z
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Final Expression and Conclusion
12
11
Csr CH s
C R C R C R C
Gather the pieces to form the transfer function
2
1 1 2 1 2 1 11 C CC r R C R s C R C r s
1 s
0 2
0 0
1
zH s Hs sQ
No algebra!
0 0Q
This expression was determined in a flashing time We did not use KVL or KCL: inspection is easy
Fast Analytical Circuits Techniques for Electrical and Electronic Circuits – Vatché Vorpérian – Cambridge Press 2002
Linear Circuit Transfer Function – A Tutorial Introduction to Fast Analytical Techniques –Christophe Basso – Wiley & Sons 2016
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Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
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A Buck Converter in Voltage Mode Replace the diode and the switch by the PWM switch
caVoutV
1LLr
Cr
loadRinV
Control
p2C
..
Small-signal model
V Vorpérian "Simplified Analysis of PWM Converters using Model of PWM Switch parts I and II"
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V. Vorpérian, Simplified Analysis of PWM Converters using Model of PWM Switch, parts I and IIIEEE Transactions on Aerospace and Electronic Systems, Vol. 26, NO. 3, 1990
Buck Input Impedance Inductance LoL lets you sweep the input to have Zin
L1 rLLoLVZin B2
Voltage
10 4
L1100uH
5
rL10m
rC30m
a c
LoL1GH
a
B1CurrentI(Vc)*V(d)
3
V(a,p)*V(d)/V(D0)
B3CurrentI(Vc)*V(D0)
11
B4VoltageV(3,p)*V(D0)
Vc
c
D0
12.0V
5.00V 5.00V 4.99V12.0V 5.00V
417mV12.0V
Ac block
VinVin parameters
Vin=12D=0.417
6
C247uF
R35
I1AC = 1
pp
V4D
d
V5AC = 0
4.99V-582nV
0V
In this mode, is equal to zerod
ˆ 0
in
in d
V sI s
Source B2 and B1 are zero
, inV a p VNode p is ground Simplify schematic
Check ac response
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0in d
inpInput impedance
Simplifying and Rearranging is Key Install the dc transformer to obtain Zin
CLr 1LIExcitation
Cr
2C
loadR
L 1TI
TV 1 0D
Tin
in
V sZ s
I s
Excitation
R
Reflect elements to the primary side
Response
p y
22 0C D
l dR2
0
LrD
12
0
LDTI
1 0D20
loadRD
20
CrD
TV
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Dc input resistance
Start with s = 0 – Draw Circuit in dc Short the inductor, open the capacitor
LrI 1L
20
loadRD
2Cr
D
20DTI
TV 0 20
L loadr RR
D
2C
20D
For the time constants, suppress the excitation, IT = 0
l dR
20
LrD
l dR
20
LrD ?R
2C
1L
2C
1L
20
loadRD
20
CrD
?R
22 2 0 2
0
C Loadr RC D
D
20
loadRD
20
CrD
11 2
0
LD
2C 2C
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Input impedance
Higher Order Coefficients Avoid indeterminacy with : use instead Determine
1 221
2L High frequency state21
?R2
0
LrD
1L
2C
High-frequency state
?R
2 1L ?R
20
loadRD
2Cr
D
1 20D
0D
2 2 12 2 1 2 0 2 2
0 0
0C Loadr R Lb C D
D D
2 12 0 22 2
0 0
1 1C LoadC Load
r R LD s C D s sC r R
D D
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Input impedance
The Numerator is of 2nd-Order Type Null the response across the current source
Degenerate case, short the generator’s terminals!
2Lr
D Use Fast 2
2 0C D12
LD
20
loadRD
Cr
20D
TI
Use Fast Analytical Circuits
Techniques!0TV
20D
20D
212 1 21 || C load
L load CL load L load
r RLN s s C r R r s L Cr R r R
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C. Basso, “Introduction to Fast Analytical Techniques, Application to Small-Signal Modeling”, APEC 2016 Professional Seminar
Assemble the Pieces
2
1 s sQ
R1
The transfer function dimension is ohm
0 0, 0
1in OL
p
QZ s R s
0 2
0
L loadr RR
D
2
1p
C Loadr R C
1 2 0
1 2
C load
L C L load C load
L C r RQ
L C r r r R r R
,in OLZ f
100°100
01 L loadr R
C load C load
Z f01 2 C loadr RL C ,in OLZ f
-90°
29.2 dBΩdBΩ
0
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10 Hz 10 MHz
Closed-Loop Input Impedance
H s V s
We want the input impedance once the loop is closed
outV s
D s
Control to output
Z
PWMG2C
1C 2R3R
1R
inZ s
3C
tV s V s G s
G s err outV s V s G s
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Stabilize the Buck with a Type 3 A 10 kHz crossover is selected for the compensation A 10-kHz crossover is selected for the compensation
d
a c1 2 7
L2100u
R120m v outparameters
Vout=5Pout=50
20.0V 5.20V
260mV
5.20V
5.00V
1
2
1 1z
z
ss
G G
d
PWM switch VM p4
X1PWMVML = 100uFs = 100kG
AIN
5
XPWMGAINK = 1/Vp
8
R250m
C1470u
R4Rload
Pout=50Rload=Vout^2/PoutRupper=10kAOL=1TVp=2fc=10kVin=20Gfc=-22
V3Vin
521mV
5.00V
2
1 2
0
1 1p p
G s Gs s
Neg. sign
18060.0
dB ° T f
T f
5
C2C1
3
C3C2
R7R2 R9
R3
v outPM=70PS=-119boost=PM-PS-90
G=10^(-Gfc/20)pi=3.14159
fz1=540
AC = 1V1
v out
Vin
2.50V5.00V
0
90.0
0
30.0
9
10
R5Rupper 12
R3
C5C3
E1AOL
5 0fz2=230fp1=6.8kfp2=fp12
C1=1/(2*pi*fz1*R2)C2=C1/(C1*R2*2*pi*fp1-1)C3=(fp2-fz2)/(2*pi*Rupper*fp2*fz2)R3=Rupper*fz2/(fp2 fz2)
2.50V
5.00V
10 kHzcf
m
cf
-180
-90.0
-60.0
-30.0Rlower10k
11
Vref2.5
R3=Rupper*fz2/(fp2-fz2)
a=sqrt((fc^2/fp1^2)+1)b=sqrt((fc^2/fp2^2)+1)c=sqrt((fz1^2/fc^2)+1)d=sqrt((fc^2/fz2^2)+1)
R2=((a*b/(c*d))/(fp1-fz1))*Rupper*G*fp1
2.50V
T s H s G s
60c
m
f
cf
G s
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1 10 100 1k 10k 100kG0=((R2*C1)/(Rupper*(C1+C2)))*c*d/(a*b) G s
Use Large-Signal Model for Reference U h PWM i h h k h
VZin
aL2100u
R110m vout
5.00V
Use the PWM switch to check the responseH
L11T
d
a c
PWM switch VM p
4
2
5
R250
20.0V5.10V
255mV
5.10V
5.00V
10
Vin
I1AC = 1
X1PWMVML = 100uFs = 100k
GAI
N XPWMGAINK = 1/Vp
7
50m
C1
R4Rload
20.0V5.00V
d f
Vin
3
K 1/Vp C1470u
511mV
G f errV fSetup for input
impedancemeasurement.
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compensator
Loop Gain Decreases as f Increases I t i d i ti t l f i l
55.0 ,in CLZ f
Input impedance is negative at low frequencies only
25 0
35.0
45.0
(dBΩ)
,
15.0
25.0
0
-80.0
-40.0
0
(°)
,in CLZ f
-160
-120 0inZ f
180inZ f Simulation
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1 10 100 1k 10k 100kresults
Why Does Zin Become Positive? The below expression only holds if Pout is truly constant
2in
inV
RP
Infinite input voltage rejection
It is true at dc, where loop gain is very high201 s
inoutP
p g j
60
40
, 2
0 0
1
1
load zSC OL
load L
RA s D
R r s sQ
,SC CLA f
RejectionAudiosusceptibility
100
80
,SC OLA sA
dB
Strong rejection
weakenssusceptibility
1 10 100 1 103 1 104 1 105 1 106 1 107120
,,C 1
SC OLSC LA s
T s
Loop gain quickly drops as f increases
g j
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Loop gain quickly drops as f increasesASC, audio susceptibility
Determining the Closed-Loop Impedance
Update the large-signal model with linear sourcesVZin
L2100u
R110m voutvin
B2Voltage
V(a,p)*V(d)/D Vcc
R81u 5.00V
L11T
5
R250m
a
B1CurrentI(Vc)*V(d)
16
B3CurrentI(Vc)*D
17
B4VoltageV(16,p)*D
p
c ca
20.0V 5.10V
10
VinVin
I1AC = 1
7
C1470u
R4Rload
R31u
p20.0V
5.00V
B2 needs to be linearized as d and Vin now include ac
0 0ˆ ˆˆ ˆin in in inV v D d dV v D
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0 0in in in inAnalytical analysis
Simplified Circuit Helps Analysis
VZin
2 5
L2100u
R110m voutvin Vc
Final closed-loop schematic to perform analysis
inV s c
L11T
2 5
R250m
14
CI s RLCZ s
10
VinVin
I1AC = 1
7
C1470u
R4Rload
B1CurrentIout*V(d)
B3CurrentI(Vc)*D
B4VoltageV(vin)*D+V(d)*Vin
H CI s 0D
D s
pH s
G s errV s
PWMG Sanity checkis mandatory!
outV s(-)
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ZRLC(s) is the filter input impedance
Derive Equations for Individual Variables
The duty ratio is linked to Vout by the error amplifier G PWM outD s G s G V s
From the previous slide, we have
0 i i
H sV s D V s D s V
Control-to-outputfunction of the buck 0out in in
in
V s D V s D s VV
0 PWM inD G G s H s V sD s
V G V G H
Divide by Vin to getthe transmittance Hp
in PWM inV G V G s H s
Current in terminal c is B4 applied to the RLC network
00
PWM inin in
in PWM inC
D G G s H s V sV s D V
V G V G s H sI s
Z s
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RLCZ s
Substitute Expressions and Rearrange The input current depends on the two input sources:
0in out CI s I D s I s D Buck dc output current
Static dutyratio
Substitute D(s) and IC(s) previously obtained Substitute D(s) and IC(s) previously obtained
0
00
PWM inin
PWM in in PWM in
D G G s H s V sD V
D G G s H s V s V G V G s H sI s I D
0in out
in PWM in RLC
I s I DV G V G s H s Z s
Factor 21 I G G H G G HD I D
T s
Rearrange
20 01 1
1 1in PWM PWMout
in in in PWM RLC PWM
I s G G s H s G G s H sD I DZ s V s V G G s H s Z s G G s H s
outi
VV
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0inV
D
Final Expression for Zin in Closed Loop Final expression involves two contributors
2 20 01 1
1 1T sD D
Z s R T s Z s T s
1 1in load RLCZ s R T s Z s T s
In dc or low frequency, T(s) is >> 1
2 2 R
2 20 01
1in load load
T sD DZ s R T s R
As s exceeds f and increases T(s) is << 1
20
loadin
RZ s
D
0s
As s exceeds fc and increases, T(s) is << 1
2 2
0 01 11
D DZ s Z s T s Z s
2RLC
in
Z sZ s
D
1in RLC RLCZ s Z s T s Z s 1
2
0 0
1
1RLC L load
s sQ
Z s r RR C
No gain means open-loop operationSame as 0D s
0D
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11RLC L loadC loads r R C Same as 0D s
Plot the Closed-Loop Input Impedance
100100
Magnitude becomes open-loop plot in high-frequency
0
60
80
,Oin LZ f Z f
Positive
10040
,in CLZ f ,in CLZ f
,Oin LZ f
Neg area
1 10 100 1 103´ 1 104´ 1 105´ 1 106´ 1 107´1 10 100 1 103´ 1 104´ 1 105´ 1 106´ 1 107´
20
A l l i hi h f
180
Neg. area
Argument meets open-loop plot in high-frequency Negative argument occurs only at low-frequency
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Open-Loop Output Impedance What is the buck converter output impedance?
outZ s1L
DinV
1
2C loadR
Voltage mode
Consider parasitic elements for L and CVoltage-mode
outout
V sZ s
I s
response
excitation
outI s2C1L
loadR outV s outI s excitation
Lr Cr
load out
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Buck Output Impedance
V
Let's find the term R0 in dc: open caps, short inductors
0 ||TL load
T
VR r RI
Lr loadRTITV
0sL r 1 1 0Lr s
The zeros cancel the responseLr
0?out zV s 1 0LsL r
1 0CrC
1 0LL
r sr
21 0Csr C
1sL2
1sC
loadR
out zI s2
2
1z
Cr C
11
z L
2CsC
Lr Cr
load 2C
121 1 C
L
LN s s sr Cr
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L
Low-Frequency Time Constants All elements are in their dc state Look at R driving L then R driving C
?R1 2
?R
r r
loadR
r r
loadR1L 2C 1L 2C
Lr Cr
R r R
Lr Cr
||R r R r L loadR r R ||L load CR r R r
11 2 ||L load C
L load
Lb C r R rr R
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L load
High-Frequency Time Constants
12
21
Set L1 in high frequency state and look at R driving C2
loadR loadR
?R ?R
1L 2C 1L 2C
Lr Cr Lr Cr
1 2 1 2
12 2 c loadC r R 2 1
1 ||L load C
Lr R r
L L 1 12 2 2 ||
||c load L load CL load L load C
L Lb C r R C r R rr R r R r
1b 2b redundancy
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2 1 2b 2 2 1b redundancy
Final Expression for ZOUT
We have our denominator!
2 r RL 21
2 1 21 || C loadL load C
L load L load
r RLD s s C r R r s L Cr R r R
The complete transfer function is now:
L
12
,21
1 1||
1 ||
CL
out OL L loadC load
l d
Ls sr Cr
Z s r Rr RLs C r R r s L C
2 1 21 ||L load CL load L load
s C r R r s L Cr R r R
This is the open-loop output impedance
Open loop
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A Closed-Loop System The converter fights output perturbations
+
,out out OLI s Z s-
H(s)+
G(s)+
outV s refV s
Because the system is linear, superposition applies T 1 1
OLout ref
OL
T sV s V s
T s
2 ,out out out OL out OLV s I s Z s V s T s
out OLOL Z sT s
,
1 2 1 1out OLOL
out out out ref outOL OL
sT sV s V s V s V s I s
T s T s
,out CLZ The loop gain affects the final expression
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The loop gain affects the final expressionOpen loop
Loop Gain Impact on the Impedance
0
Plots superimpose as loop gain approaches 0
20
ZoutCL i 2 fk
,out OLZ f
0fCr
60
4020 logoutCL k
1
20 logZoutOL i 2 fk
1
LrC
Peaking is gonei l d l
80
,out CLZ f
in closed-loop
10 100 1 103 1 104 1 105100
fk
Having gain at f is important to damp the filterPublic Information
Christophe Basso –Input Filter Interactions79 3/7/2017
Having gain at f0 is important to damp the filter
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
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A Design Example with a Buck Assume the following 5-V/50-W buck converter
L222u
R110m vout
X2PSW1
vout> 1 2
R220
8 vout>ini
3
20m
C1
R4Rload
7Vin20 D1
mbr1645
C1680u+ -
4 5
X1COMPAR
VrampB1
V(err)>1.9 ?1.9 :V( ) 100 kHFV(err)
Specifications are less than 15 mA peak of input ripple
100 kHzswF
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Specifications are less than 15 mA peak of input ripple
What is the Necessary Attenuation? Use SPICE to analyze the input current signature
14.0 ini t .FOUR 100kHz I(V4), 5.4 Ain rmsI
10.0
11.13 A
in
4 94 A peakI
,
6.00
(A) 1 4.94 A peakI
2 00
2.00
10 µs
15 mA peakinI
10.054m 10.063m 10.072m 10.082m 10.091m
-2.00 10 µs
Attenuate by 3mor 50 dB2 2 2 25 4 2 85 4 6 AI I I
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or 50 dB, 5.4 2.85 4.6 Aac in rms dcI I I Current in the capacitor
Where do you Place the Double Pole? Insert a LC filter to attenuate the pulsating current
outI
20inI I
LC
outI inI
Position f to provide a 50 dB attenuation at 100 kHz
0 3 100 17 kHzfilter SWf A f m k
Position f0 to provide a 50-dB attenuation at 100 kHz
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Select the Filter Elements The resonant frequency lets us chose L and C
10 22 2
1 8.34 10 s4
LCf
2 204 f
Component selection depends on volume, cost etc. the capacitor sees the buck ac input current the capacitor sees the buck ac input current the inductor ripple current is small, consider dc only
Kemet F series 22 µHL 3 AinI
4 x 10 µFItot rms > 40 A 7447714220 Würth
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Itot,rms 40 A 7447714220 Würth
Always Consider Component ESR We can show that the complete attenuation is
I2
21
Cr Cr
C
Lr
L
outI
inI
21
2 2112
21
C
in
outC L
CILI r r L
CC
CL 11
CC
Check if attenuation is ok with selected components Check if attenuation is ok with selected components
5 mΩ 1.3 mΩ4Cr
IX 4
4
50 mΩLr
50.8 dB at 100 kHzin
out
II
Public InformationChristophe Basso –Input Filter Interactions85 3/7/2017
Control-to-Output Transfer Function The addition of the filter affects the converters
H s
ˆoutv
Lr Cr 1 s
inv
L C 2
1
1
zinH s V
s sQ
G sd
0 0Q Buck control-to-output
TF without filter.
The small-signal model needs an update
Public InformationChristophe Basso –Input Filter Interactions86 3/7/2017
A More Complex Architecture th
L222u
R110mvin
vout
The converter becomes a 4th-order system
V(c)9 5
R220m
Vc
R3
thZ sV(c)
7
20m
C1680u
R4Rload
d
B1CurrentIout*V(d)
B3CurrentI(Vc)*D
11
B4VoltageV(vin)*D+V(d)*Vin2
R350m
L122uH
15
R81.5m
C440u 680ud 22uH 40u
VmodAC = 1 pH s RLCZ s
Simplify the circuit before solving the function
Public InformationChristophe Basso –Input Filter Interactions87 3/7/2017
A Simplified Circuit
I
You can reuse previously-determined transfer functions
inV s outV s CI s
thZ s pH s V s D d s V
0CI s D outD s I
0in inV s D d s V
Small-signal model
Apply KVL and KCL to obtain the new expression
Small signal model
Public InformationChristophe Basso –Input Filter Interactions88 3/7/2017
Determine and Substitute Variables Determine the input voltage expression Vin(s)
0in th out CV s Z s I D s I s D
Terminal c current is the voltage V(c) divided by ZRLC(s)
0in inV s D D s V 0in inV s D D s V
0in in
CRLC
I sZ s
00
in inin th out
RLC
V s Z s I D s DZ s
0RLC th out in thD s Z s Z s I D V D s Z s
Th t t lt i l th RLC t itt H
02
0
RLC th out in thin
th RLC
V sZ s D Z s
The output voltage involves the RLC transmittance Hp
0out in in pV s V s D D s V H s
Public InformationChristophe Basso –Input Filter Interactions89 3/7/2017
Final Expression is Complicated Rearrange the final transfer function
2
011 thDs Z s
V
2 210
2
111
thout load in loadz
L loadload Lth
load L
V s R V RsC r RD s R r Ds s Z s
R rQ s s
0 0
0 0
1load LQ s s
Q
Classical buck, no filter Filter effect
L
2
1 1
1
fCf f
Lfth Lf
f Cf Lf f f
Lsr C s
rZ s r
sC r r s L C
1ff
f Cf Lf
LQ
C r r
01
ff fL C
1 Lfr 1 f Cf Lf f fsC r r s L C
Filter output impedance impacts the transfer function
1
1
f
zC fr C
2
Lfz
fL
Public InformationChristophe Basso –Input Filter Interactions90 3/7/2017
Plots Show a Distorted Transfer Function
Mathcad and SPICE match each other quite well
10020
40
outV f
outV fD f
100
0
20
0
out fD f
10 100 1 103 1 104 1 105
100
f
10 100 1 103 1 104 1 105
40
f fkfk
The notch occurs because Vin drops at resonanceControl-to-output transfer function
Public InformationChristophe Basso –Input Filter Interactions91 3/7/2017
The Extra Element Theorem at Work What is the EET principle? Identify an element whose presence complicates the analysis Calculate the transfer function with that element removed Calculate the transfer function with that element removed Apply a correction factor k to the transfer function: voilà!
H s 1 s
ˆoutv
Lr Cr
20
0 0
1
1th
zinZ
H s Vs sQ
ˆ 0inv
L C
0 0Q
G sd 0thZ s
0thZH s H s k
Correction factor
Public InformationChristophe Basso –Input Filter Interactions92 3/7/2017
Correction factor
Determining the Correction Factor The correction factor requires two terms:1. The converter input impedance obtained in open-loop2 The converter input impedance obtained for V (s)=02. The converter input impedance obtained for Vout(s)=0
2CLr 1LTIExcitation
CrloadR
TV 1 0D
Response 0dp
The open-loop impedance has already been derived
0d s
2
0 02
0
1
1load L
D
s sQR rZ s sD
Public InformationChristophe Basso –Input Filter Interactions93 3/7/2017
p
The Null Propagates in the Circuit Despite excitation with IT, there is no ac response This is the principle behind nulling the response
CrLr1L
TI 0outI s
0 V
0CI s 0
0 Vexcitation inV s
Cr
2C
loadRTV
0CI s D outD s I
0C s
0in inV s D D s Vresponse
2C 0outV s
0 A in the load and 0 V at r -r implies 0 V across L 0 A in the load and 0 V at rL-rC implies 0 V across L1
0 0in inV s D D s V 0
inV sD s D
V
Public InformationChristophe Basso –Input Filter Interactions94 3/7/2017
inV
Expression for ZN Comes Easily Substitute/rearrange test and voltage expressions
0T out CI s I D s I s D I s I D
out
load
VR
0
in
in
V sD s D
V
0T out
T in
I s I DV s V
0
outVD
The input impedance ZN for a nulled response is:
2load
NR
Z sD
Nulled response implies infinite input rejectionIt is the incremental input resistance seen before
0D It is the incremental input resistance seen before
bulk 2V
We already had this with the SPICE simulation
Vin100
B1CurrentPout/V(bulk)
parameters
Pout=50W
20
0 2 20
out
in load
out out
l d
VDV R
RP V D
R
200 ΩRFor a buck
Public InformationChristophe Basso –Input Filter Interactions95 3/7/2017
loadR0 200 ΩR
Final Expression with EET The final expression includes the filter impact
2
11
th
load
Z sRs
V s R D
2
02
2
0 0
111
out load zin
thload L
V s R DVZ sD s R r s s
Q s sQR r
0 02
0 1load L
p
QR rsD
14 13
Comparison between EET and full formula
k
0
2 10 14
4 10 14
13
4 10 13
6 10 13
8 10 13
Argumentdifference
14
4 10 14
2 10 14
13
2 10 13
0
2 10 13
Magnitudedifference
Public InformationChristophe Basso –Input Filter Interactions96 3/7/2017
10 100 1 103 1 104 1 1056 10 14
10 100 1 103 1 104 1 1054 10 13
What is the Impact on Stability? When adding the filter, the loop gain is awfully ugly!
200
20
40
100
20
0
100
0(dB) (°)
40
10 100 10 10 103 4 5
T f
10 100 103 104 105
200
T f
Filter damping is necessary: th NZ s Z s
th DZ s Z s
Public InformationChristophe Basso –Input Filter Interactions97 3/7/2017
The Output Impedance is Affected
0
Closed-loop output impedance peaks as filter resonates
40
20
20 logZoutCLF i 2 fk
60
40g1
20 logZoutCL i 2 fk
1
1 thZ sZ s
Withoutfilter
100
80
0
,
1th
eout out Z s
th
in CL
Z sZ s Z s
Z sZ s
10 100 1 103 1 104 1 105
Filter damping is necessary: th eZ s Z s
Public InformationChristophe Basso –Input Filter Interactions98 3/7/2017
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions99 3/7/2017
Check Filter Output Impedance
60
Plot shows that design inequalities are not respected
(dBΩ)Open-loopimpedance th NZ s Z s No!
40
20 logZth i 2 fk
1
( )
NZ f
DZ fNulled-outputInput impedance
th DZ s Z sNo!
0
2020 log
ZinOL i 2 fk 1
20 logZN s( )
1
N f
20 thZ f
14.4fQ
10 100 1 103 1 104 1 105
fk
Gain and phase distortion of k must be minimized
Public InformationChristophe Basso –Input Filter Interactions100 3/7/2017
Filter Peaking Also Affects Zout
(dBΩ) 14.4fQ Input impedance
Inequality for the closed-loop Zout is not respected
20
20 logZth i 2 f k
( ) 14.4fQ
eZ f
Input impedancewith shorted output
0
g1
20 logZe i 2 f k
1
40
20
thZ f
10 100 1 103 1 104 1 10540
f k
th eZ s Z s Filter damping must ensure
Public InformationChristophe Basso –Input Filter Interactions101 3/7/2017
g
Look at the Correction Factor k S Q f th l t h d g i d i ti Sweep Qf for the least phase and gain deviation
1 th
load
Z sR
0
0.9fQ
20
21
1
th
Dk sZ s
s sQ
1
0.5
(dB)
fQ
1 3 dB0 02
0 1load L
p
QR rsD
1.5
10
10 100 103 104 105
Ph di i
Magnitude distortion k f-1.3 dB
1 2
0 2
1 1
1
z zth
s s
Z s Rs sQ
0
5
(°)
Phase distortion
-4.9°
0 0fQ
10
5
3 4 5
k f0.9fQ -6°1ff
f Cf Lf
LQ
C r r
Public InformationChristophe Basso –Input Filter Interactions102 3/7/2017
10 100 103 104 105f f f
Determine Maximum Filter Peaking Determine maximum peaking with selected Qf
0 2 2 2 20 0
fh
R QZ 1 2
1 2
0 0th z zMAXz z
Z
22 µH50 mΩ 14.4fQ
1.3 mΩV
10.7 Ωth MAXZ
Damp
40 µF
inV
0.9fQ
0 7 ΩZ Th t t0.7 Ωth MAXZ
What damping elements will reduce to 0.7 Ω?th MAXZ
The target
Public InformationChristophe Basso –Input Filter Interactions103 3/7/2017
What damping elements will reduce to 0.7 Ω?th MAX
Available Damping Techniques Damping means increasing the loss per cycle Increase power dissipation
dampR
Lfr fL Lfr fL
inV inVCfr
dampRCfr
fC fCParallel Series
rC-Cf parallel damping: Rdamp dissipates power rL-Lf parallel damping: Rdamp adds a zero and alters filter
Public InformationChristophe Basso –Input Filter Interactions104 3/7/2017
Adding a Blocking Capacitor The series capacitor blocks the dc component Literature recommends Cdamp = 10 x Cf
r LLfr fL
r R40 µFfC
inVCfr
dampRx10
fCdampC 400 µFdampC
Cdamp can be extremely bulky In ac Rdamp dissipation can be an issue
Optimum combination?
Public InformationChristophe Basso –Input Filter Interactions105 3/7/2017
damp
Analyzing the Filter Output Impedance
outZ s1L Lr
3C dampC out
1
Cr dampR0 V
23 p
Three storage elements with independent state variables: 3rd order system.
2 31 2 31D s b s a s a s
Public InformationChristophe Basso –Input Filter Interactions106 3/7/2017
1 2 3
Start with Dc Analysis s = 0
L Lr
outZ s
response
3C dampC
1L L response
stimulus3 damp
0 V IT
Cr dampR infR
0 inf||LR r R
Public InformationChristophe Basso –Input Filter Interactions107 3/7/2017
Rinf is the current source output resistance and ensures a dc path when IT = 0
Determine Time Constants Turn excitation off, current source is set to 0 A
?RLr
C
L1
?R
r R
0 V C3Cdamp
infR
Cr dampR
3 3 inf||C LC r r R 2 inf||damp damp LC R r R
Public InformationChristophe Basso –Input Filter Interactions108 3/7/2017
p p
Assemble Time Constants for b1
?RLr
L1
?R0 V 1L
C3 Cdamp
infR
Cr dampR1inf LR r
1 1 2 2 1 12 31D s s s s
11 1 2 3 inf 3 inf
i f
|| ||damp damp L C LL
Lb C R r R C r r R
R r
1 2 3 1 2 1 3 2 3 1 2 31D s s s s
Public InformationChristophe Basso –Input Filter Interactions109 3/7/2017
inf LR rinfR
1 1 2 2 1 12 3
Second-Order Time Constants 1 1 2 2 1 12 3
1 2 3 1 2 1 3 2 3 1 2 31D s s s s
t 1 t 1? ?r
LrL1 L1
tau1
tau3 tau2
tau1
tau3 tau2
?R ?R
Cr dampR
Lr
Cr dampR
C3 Cdamp C3 Cdamp0 V 0 V
12 infdamp dampC R R 1
3 3 infCC r R LrL1
tau1
tau3 tau2
?R
Cr dampR
C3 Cdamp 12
Reactance 1 is inits high-frequency state
What resistance
0 V
23 3 inf|| ||C damp LC r R r R
damp 2?R
What resistancedrives reactance 2?
Public InformationChristophe Basso –Input Filter Interactions110 3/7/2017
3 3 inf|| ||C damp LRinf is not added for clarity purposes
Assemble Time Constants for b21 1 2
2 1 2 1 3 2 3b
12 inf
inf
1
damp dampL
Lb C R R
R rL
C R
13 inf
inf
inf 3 inf|| || ||
CL
damp damp L C damp L
C r RR r
C R r R C r R R r
infR 2 1 3 3 ||damp damp damp L C L dampb L C C C R r C r r R
LrL1 123 1 2 3b
tau1 ?R
Lr
C3 Cdamp
L1
HF state
tau3 tau2
t 12t 3
Cr dampR 3 inf2
31 ||C dampC r R R
What resistancedrives reactance 3?
Public InformationChristophe Basso –Input Filter Interactions111 3/7/2017
tau12tau3drives reactance 3?
Build the 3rd-Order Denominator
2 31 2 31D s b s b s b s
Gather time constants and rearrange to form D(s)
||b L C C C R r C r r R
1 3damp damp L C Lb C R r C r r
2 1 3 3 ||damp damp damp L C L dampb L C C C R r C r r R
13 inf 3 1 3
infdamp damp C damp damp C damp
Lb C R R C r R L C C r R
R
3
2
1
||
damp damp L C LD s s C R r C r r
L C C C R C R
21 3 3
31 3
||damp damp damp L C L damp
damp C damp
s L C C C R r C r r R
s L C C r R
Public InformationChristophe Basso –Input Filter Interactions112 3/7/2017
p p
Determine N(s) Swiftly with Inspection
C C1L 1 0Z s 2 0Z s
Responseis nulled3C
dampC is nulled0TV
Cr dampRLr
0Z s
Three zeros when 1 0Z s 2 0Z s
3 0Z s
3 0Z s Three zeros when 11 0zZ s 22 0zZ s
131 1 1C damp damp
LN s s sr C sR C
r
33 0zZ s
Public InformationChristophe Basso –Input Filter Interactions113 3/7/2017
Lr
N
Run a Sanity Check to Verify Results
The complete low-entropy transfer function is thus 0out
N sZ s R
D s
131 1 1C d d
Lsr C sR C s
3
0 2 33 1 3 3 1 3
1 1 1
1 ||
C damp dampL
outdamp damp L C L damp damp damp L C L damp damp C damp
sr C sR C sr
Z s Rs C R r C r r s L C C C R r C r r R s L C C r R
The raw transfer function is || ||Z s Z s Z s Z sThe raw transfer function is 1 2 3|| ||outZ s Z s Z s Z s
40
60
50
0
2020 logHref i 2 fk
20 logH1 i 2 fk
0
arg H1 i 2 fk 180
arg Href i 2 fk 180
10 100 1 103 1 104 1 105 1 10640
20
fk10 100 1 103 1 104 1 105 1 106
50
fk
Public InformationChristophe Basso –Input Filter Interactions114 3/7/2017
Raw and full TF plots are superimposed: good to go!fk
Trying to Rewrite the Transfer FunctionIn this 3rd-order system, it is difficult to find a canonical form such as
N sZ R
0 2
0 0
1 1out
p
Z s Rs s s
Q
0 0p Q
If rL is larger than rC and CdampRdamp larger than C3rL
2 31 3 1 31 damp damp damp damp dampD s sC R s L C C s L C C R
Unfortunately, does not dominate at low freq. 1 damp dampsC R
Cannot factor the 3rd-order denominator
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S f f f 0
Neglect Parasitic Terms rL and rC Simplify the transfer function considering 0 rL and rC
13
2 3
1 1 1
1 ||
CL
L
damp damp
out
rr
rr r
Ls C sR C sZ s
s r rC R C s L C C C R C R s Lr Rr C C r
3 1 3 3 1 31 ||L C L Ldamp damp damp damp damp damp damp daC C mL pr rs r rC R C s L C C C R C R s Lr Rr C C r
Factor the L1/rL term 1 1L
damp dampsR C r
11
2 33 1 3 3 1 31 ||
LL L
damp damp
outdamp damp damp damp damp damp damp dampL L L
CsLL
Z s ss C R C s L C C
rr r r r rC R C R s L C C R
Have r go to zero Have rL go to zero
1 2 31 3 1 3
1
1damp damp
outdamp damp damp damp damp
sR CZ s sL
sC R s L C C s L C C R
Consider Cdamp = nC3
31 2 3 2
1
1 1damp
out
sR nCZ s sL
sR nC s L C n s L nC R
Public InformationChristophe Basso –Input Filter Interactions116 3/7/2017
3 1 3 1 31 1damp dampsR nC s L C n s L nC R
Determine the magnitude of this transfer function
Rearrange Expressions Determine the magnitude of this transfer function
2
1 3 12 2 2
3 1 3 1 3 3 11 1damp
outdamp
L nR C jLZ j
C L C L n j C R n C L
2 221 3 1
2 222 2 23 1 3 1 3 3 11 1
dampout
damp
L nR C LZ
C L C L n C R n C L
mag1
Check with Middlebrook’s definitions
10
LR dampR
Q sp s 01
03
RC 0
QR
0
p s 0
1 3L C
2 3
1
1 1out
p s nQp sZ s
nQp s n p s nQp s
2 2 2
0 222 2
1
1 1 1out
x n Q xZ R
n x xnQ x
0
x
mag2
Public InformationChristophe Basso –Input Filter Interactions117 3/7/2017
Design Techniques for Preventing Input-Filter Oscillations in Switched-Mode Regulators, R.D Middlebrook, Powercon, May 1978
Always Run a Sanity Check Check if expressions are ok versus Mathcad® calculations
60
40
20 logMag1 2 fk
20 logHref i 2 fk
0
2020 log
20 logMag2 2 fk
2020 logRload
1 10 100 1 103 1 104 1 10540
fk
Compare analytical results versus raw magnitude expression
Public InformationChristophe Basso –Input Filter Interactions118 3/7/2017
Compare analytical results versus raw magnitude expression
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions119 3/7/2017
The damping resistor value changes the resonant frequency
The Damping Resistor Affects Q and 0 The damping resistor value changes the resonant frequency
dampR 1
dampR1L No loss
Q 0
1 3
1L C
dampC
3C0dampR
No loss 0
1 No loss
Q 0
1 3 dampL C C
There must be a RdampCdamp couple optimizing Q What is the minimum in the maximum (minmax) peaking of |Zout|? For what optimum Q value does it occur and at what frequency?
Public InformationChristophe Basso –Input Filter Interactions120 3/7/2017
For what optimum Q value does it occur and at what frequency?
Pl i d i Q f C 10C
Changes Induced by Damping Resistor
20 logMag3 x 0( )
Plot magnitudes versus various Q factors, Cdamp = 10C3
100Q 0dampR ,outZ f Q
60
20 logMag3 x 0.01( )
20 logMag3 x 0.1( )
100Q p
Rdampdecreases
4020 log
Mag3 x 1( )
20 logMag3 x 10( )
Mag3 x 100( )
optQ
2020 log
Mag3 x 100( )
20 logMag3 x Qopt
opt0.1 1 100
x
This point isindependent of Q
10n
It is the point at whichpeaking is the lowest!
opt0
Public InformationChristophe Basso –Input Filter Interactions121 3/7/2017
independent of Q peaking is the lowest!
Optimum Z magnitudes versus different values of n
Optimum Peak Depends on Cdamp Optimum Zout magnitudes versus different values of n
1n ,outZ f n 2
20 inVL
max
opt
30
4020 log
Mag5 x 1( )
20 logMag5 x 5( )
5n
20 in
out
LogP
opt
20
30
20 logMag5 x 10( )
20 logMag5 x 100( )
10n opt
opt
10
20 logRload
100n
Q Q0.1 1 100
x
There is an optimum R and n to match a given Z
optQ Q0
Public InformationChristophe Basso –Input Filter Interactions122 3/7/2017
There is an optimum Rdamp and n to match a given Zout
The frequency at which the minmax occurs is immune to Q
Determine the Optimum Frequency The frequency at which the minmax occurs is immune to Q Calculate the sensitivity of Zout to Q and cancel itWork on Zout² to get rid of square roots
22 2 2 22 2
0 222 2
1
1 1 1out
x n Q xd dZ Q RdQ dQ n x xnQ x
3 6 2 22 2 2 2
0out
Qn x nx xd Z QdQ D Q
2 22 2 0nx x
2 2 2x n The point at which 2 2
22
xn
x
The point at which Qopt occurs is: 0
1 32 2opt n n L C
Public InformationChristophe Basso –Input Filter Interactions123 3/7/2017
0
Update the magnitude definition to have |Z | at
Calculate the Magnitude at opt Update the magnitude definition to have |Zout| at opt
2 2 2
0 222 2
1
1 1 1out
x n Q xZ R
Q
0
22
optxn
2 21 1 1n x xnQ x
10
2 2 2 2out opt
n n LZ R
C
This is the valueof |Z | at
3p n n C
We want to minimize |Zout| at opt
of |Zout| at opt
Differentiate Zout² with respect to x², find the optimum Q Replace and2A x A x
2
2 2 2
220
10
1 1 1out
A n Q AZ Ad ddA R dA n A n AQ A
Public InformationChristophe Basso –Input Filter Interactions124 3/7/2017
Apply brute force differentiation with Mathcad ®
Determine the Optimum Value of Q Apply brute-force differentiation with Mathcad ®
2 4 3 2 2 2 4 4 2 2 3 2 2 2 2 2 22 2 2 2 10outZ A AQn A Q n A Q n A Q n A n A n A AQ nd
dA R D
0dA R D
4 3 2 2 2 4 4 2 2 3 2 2 2 2 2 22 2 2 2 1 0AQ A Q A Q A Q A A A AQ
Extract Q
4 3 2 2 2 4 4 2 2 3 2 2 2 2 2 22 2 2 2 1 0AQn A Q n A Q n A Q n A n A n A AQ n
3 5 3 2 3 2 2 3 3 22 2 2 2 2A n A n A A A An An A n A n
2A
2 4 4 4
2 2 2 2 2opt
A n A n A A A An An A n A nQ
A n A n
Result from2
An
2
2 2
4 3 23 10 82 4 2 4opt
n nn nQn n n n
Result fromDr Middlebrook
Public InformationChristophe Basso –Input Filter Interactions125 3/7/2017
Apply the Technique to the Buck The target is to reduce the filter impedance peak to 0.7 Ω
fLR 22
0f
RC
2 2out mmZ n
20 0 0
2
43.5
out mm
out mm
R R R Zn
Z
2
0
mm
R n
4 3 2n n 0 0.487damp optR R Q
target
2
4 3 20.65
2 4opt
n nQ
n n
141µFdamp fC nC
This is a rather large capacitance value An electrolytic capacitor and its ESR can do the job Watch for temperature effects as ESR increases at low temp!
Public InformationChristophe Basso –Input Filter Interactions126 3/7/2017
p p
Check Margins with ZD and ZN
The overlap is gone for both impedances60
(dBΩ)
40
20 logZth i 2 fk
1
(dBΩ)
Z f
DZ f
0
20
20 logZinOL i 2 fk
1
20 logZN s( )
1
NZ f
20
01
thZ f
10 100 1 103 1 104 1 105
fk
Gain and phase distortion of k are minimized
Public InformationChristophe Basso –Input Filter Interactions127 3/7/2017
p
Check Margin with Ze
60
The output impedance should not be affected by the EMI filter
damped
40
Zth i 2 fk
pfilter(dBΩ)
0
2020 logth k
1
20 logZe i 2 fk
Z f20 dB
20
1
Z f
eZ f
input impedance
10 100 1 103 1 104 1 10540
fk
thZ fwith 0-Ω Rload
Public InformationChristophe Basso –Input Filter Interactions128 3/7/2017
k
Verify Results after Damping - Magnitude
40
Check the resulting control-to-output transfer function
damped
20
40
filter Dampedfilter
0
20 log Tfilter i 2 fk
20 log TfilterM i 2 fk Undamped
40
20
undamped T f
10 100 1 103 1 104 1 105
fk
Magnitude distorsion has disappeared after damping
Public InformationChristophe Basso –Input Filter Interactions129 3/7/2017
Magnitude distorsion has disappeared after damping
Verify Results after Damping - Phase Original phase margin is unaffected after damping
200damped Damped
100
180
dampedfilter
Dampedfilter
0
arg Tfilter i 2 fk 180
arg TfilterM i 2 fk 180
Undamped
100
undamped T f
Undamped
10 100 1 103 1 104 1 105200
fk
f
Public InformationChristophe Basso –Input Filter Interactions130 3/7/2017
Closed-Loop Output Impedance
0
Peaking effects of the EMI filter are now gone
damped Damped
20
ZoutCLFD i 2 fk
dampedfilter ,out CLZ f Damped
filter
Undamped(dBΩ)
60
4020 logZoutCLFD i 2 fk
1
20 logZoutCLF i 2 fk
1
80
1
undamped
10 100 1 103 1 104 1 105100
fk
Public InformationChristophe Basso –Input Filter Interactions131 3/7/2017
Opting for a Different Strategy What if you only try to get rid of the overlap, ignoring ZD and ZN? Plot the closed-loop input impedance and build margin (10 dB)
60
40
(dBΩ)
2020 log
Zth i 2 fk 1
20 lZinCL i 2 fk
,in CLZ f
10 dBΩ10 dB
0
20 log1
10 dBΩ
3.3 Ω
10 100 1 103 1 104 1 105
20
f
thZ f
Public InformationChristophe Basso –Input Filter Interactions132 3/7/2017
fk
Calculate the New Damping Elements Different damping elements are now required
4 3 2n n 220 0 0 4 out mm
R R R Z
2
4 3 22.4
2 4opt
n nQ
n n
2 0.5
mm
out mm
nZ
3.3Ω
0 1.8damp optR R Q 20 µFdamp fC nC 10
10
0
20 logZth i 2 fk
1
(dBΩ)
30
20 thZ fPlot includesall ohmic losses
Filter outputimpedance
Public InformationChristophe Basso –Input Filter Interactions133 3/7/2017
10 100 1 103 1 104 1 105
Verify Results after Damping - Magnitude
The new filter effect can be observed when it resonates
40
20
40
20 l T i 2 f
damped
20
0
20 log Tfilter i 2 fk
20 log TfilterM i 2 fk
40
20
T fundamped
10 100 1 103 1 104 1 105
fk
Gain distortion is noticeable before crossover
Public InformationChristophe Basso –Input Filter Interactions134 3/7/2017
Gain distortion is noticeable before crossover
Verify Results after Damping - Phase
200
Phase distorsion appears but do not jeopardize phase margin
100
arg Tfil i 2 fk 180
damped
0
arg Tfilter i 2 fk
arg TfilterM i 2 fk 180
100
T fundamped
10 100 1 103 1 104 1 105200
fk
Original phase margin is unaffected in this case
Public InformationChristophe Basso –Input Filter Interactions135 3/7/2017
Original phase margin is unaffected in this case
Verify Results after Damping - Zout
0
Peaking can be observed but it remains limited
20
ZoutCLFD i 2 fk
,out CLZ fdamped
60
4020 logZoutCLFD i 2 fk
1
20 logZoutCLF i 2 fk
1
80
1
undamped
10 100 1 103 1 104 1 105100
fk
Public InformationChristophe Basso –Input Filter Interactions136 3/7/2017
Average Model Simulation Template Both damping strategies can be tested with SPICE
0.487
141 FdampR
C
or
1.8
20 FdampR
C
141µFdampC 20 µFdampC
1 2
L222u
R110m vout
vouta c48
R350m
9
L122uH
R220m
dPWM sw itch VM p5
X1PWMVM
R81.5m
Vi 11
X3PSW1
Rdamp1.8
3
C1680u
PWMVML = 22uFs = 100kG
AIN
7
XPWMGAINK = 1/Vp
V(err)>1.9 ?
10C440u
Vin20
11
V6unknown12
Cdamp22u
Type 3
B1Voltage
V(err) 1.9 ?1.9 :V(err)
The load is stepped from 50 to 100% in 1 µs
G s
Public InformationChristophe Basso –Input Filter Interactions137 3/7/2017
The load is stepped from 50 to 100% in 1 µs
Average Response Shows Oscillations Oscillatory but stable response to a load step
5.16
5.321.8
20 µFdamp
damp
R
C
5.00(V)
4.840.487
141µFdamp
damp
R
C
10 5m 11 5m 12 5m 13 5m 14 5m
4.68 outv t
Public InformationChristophe Basso –Input Filter Interactions138 3/7/2017
10.5m 11.5m 12.5m 13.5m 14.5m
Switching Model Simulation Template
L222u
R110m vout
X2PSW1R3
50mL122uH
This is a simple model compensated by the type III structure
1 2
22u 10m
R220
8 vout9
IV4
10
50m
11
22uH
X3PSW1
3
20m
C1680u+
7
X1COMPAR
D1mbr164512
R81.5m
C440u
Vin20
13
V6unknow n14
Rdamp1.8
Cdamp22u
680u+ -4 5
COMPAR
VrampB1
V(err)>1.9 ?1.9 :V(err)
G s
Type 3
You can verify the step response but also the filtered signature
Public InformationChristophe Basso –Input Filter Interactions139 3/7/2017
Oscillations in the Switched Model O
5 16
5.32
Oscillations are present but not too disturbing
1.8
20 µFdampR
C
4.84
5.00
5.16
20 µFdampC
4.68 outv t
5 00
5.16
5.32 0.487
141µFdamp
damp
R
C
4.68
4.84
5.00
outv t
Public InformationChristophe Basso –Input Filter Interactions140 3/7/2017
10.6m 11.6m 12.5m 13.5m 14.5m
Input Current Signature is Good ff
-2.7501.8
20 FdampR
C
No difference in the signature – amplitude is within specs
, 14.8 mAin peakI
-2.760
20 µFdampC , p
-2.770(A)
-2.780
10 005m 10 016m 10 026m 10 037m 10 047m
-2.790 ini t0.487
141µFdamp
damp
R
C
Public InformationChristophe Basso –Input Filter Interactions141 3/7/2017
10.005m 10.016m 10.026m 10.037m 10.047m
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions142 3/7/2017
A Practical Case with NCL30186
R2R
C4Cvinavg 707mV Input impedance Zin
This LED driver featuring PFC requires the insertion of a filter
4
R1R
C1C
3 8
R C
B2Voltage
VFB6
X3
GA
X4GAIN
D2unknown
2
R5
vinavg
R73Meg
1.80V 1.80V
707mV
707mV
707mV
p p in
PFCSV
X2
ton vca
tch
BCM
ton
X1PWMBCMCM2
X3AMPSIMP
AIN GAIN
K = 0.2528
V65
VCvinavg
8.2k C2470p Vin
90.0V449mV
9.95V
5.00V
QR model
14
X2XFMRRATIO = -N
18
ESRESR
13
Fswvout
c
PW
M s
wip
Fsw
(kH
z)
V221
7 21
L31m
12
R6600m
C71p
C8220nF
1
R13100m
Vout
Iout
46
L41GV11
90.0V
90.0V-130V 42.8V
42.8V28.4fV
59.4V
0V 0V 0V
0V 0V
90.0V
90.0V
Input filter Zth
CoutCoutIC = IC
LpLp
B3Voltage
V(VC)/(Ri)
Ipk5
R424
44
R1123m
45
R1232m
VinVin
V100 48
unknownAC = 1
89.9VB12CurrentI(V11)
1.47V 21.8V0V 0V 90.0V
Public InformationChristophe Basso –Input Filter Interactions143 3/7/2017
Average model by S. Cannenterre
Plot the Impedance Ratio Zout/Zin
° dBΩ
Analysis reveals a negative phase margin: stability issue
60.0
120
180
360
1th
in
Z fZ f
th
in
Z fZ f
00
-60.0-180
th
in
Z fZ f
10m 100m 1 10 100 1k 10k 100k 1Meg
-120-360
233th
in
Z fZ f
Public InformationChristophe Basso –Input Filter Interactions144 3/7/2017
g
A Simulation Shows Oscillations
(V)
t
The negative phase margins brings a diverging filter voltage
150
200 inv t
100
150
50
28.9m 86.7m 144m 202m 260m
0120 V
90 V C1 = 0 pF
Public InformationChristophe Basso –Input Filter Interactions145 3/7/2017
Act on the PFC Input Voltage
inV sR
Reduce the phase stress by inserting a zero in the PFC chain
1R 1C
20
1 2
RH
R R
Addedzero
SV s
1 11 1 1C R 2 1 2 1 2||R R C C
11
s SV s
R C
10 0
2
11 1
zsH s H H
ss
2R
2C p
1
1z
2
1p
Public InformationChristophe Basso –Input Filter Interactions146 3/7/2017
Insert the Zero to Reduce Phase Stress
360 120° dBΩ
Adding a zero in the PFC control input brings phase margin
180
360
60.0
120
th
in
Z fZ f
1th
in
Z fZ f
0 0
-180 -60.0
th
in
Z fZ f
135thZ fZ f
10m 100m 1 10 100 1k 10k 100k 1Meg
-360 -120
inZ f
Phase margin is 45°
Public InformationChristophe Basso –Input Filter Interactions147 3/7/2017
“Interaction between EMI filter and PFC with average current control”, G. Piazzi, J. Pomilio, IEEE Transactions on Power Electronics, 1999
The Zero Insertion Cancels Oscillations
(V)
t
Adding a zero builds phase margin and tames oscillations
150
200 inv t
100
150
50.0
28.9m 86.7m 144m 202m 260m
0C1 = 22 pF90 V
120 V
Public InformationChristophe Basso –Input Filter Interactions148 3/7/2017
Practical Experiments Placing a zero at 2.5 kHz (22-pF cap.) stops oscillations
However, PFC operations are affected: need to damp the filter!
Public InformationChristophe Basso –Input Filter Interactions149 3/7/2017
However, PFC operations are affected: need to damp the filter!
2
Explore Another Option and Damp Filter
10
3
LR
C
2
0
2 2out mmZ n
R n
220 0 0
2
4 out mm
out mm
R R R Zn
Z
2
4 3 22 4opt
n nQ
n n
mm
0damp optR R Q
out mmZ This is the minimum of Zout peaking to avoid overlap
2Vinout mm
out
VZ k
P For a 40-W output and a 108-V input:
291Ωout mmZ Assume 10% margin, k = 0.9 262 Ωout mm
Z
0.584n
2.18optQ
220 nF 0.128 µFdampC n
0 147 Ωdamp optR R Q
Public InformationChristophe Basso –Input Filter Interactions150 3/7/2017
optQ 0damp opt
Ch k b f l i h M h d ®
Plots with L1 = 1 mH and C3 = 220 nF
60Minimum resistance at Vin = 108 V rms
Check absence of overlap with Mathcad ®
4020 log
H1 i 2 fk
10
0
2020 log
Href i 2 fk
10
R
No overlap
20
20 logRload
10 100 1 103 1 104 1 105 1 10640
fk
Rdamp = 146 Ω and Cdamp = 0.128 µF
Public InformationChristophe Basso –Input Filter Interactions151 3/7/2017
fk
Si l i l V 108 V P 40 W l
Check Results on the Simulation Template Simulation results Vin = 108 V rms Pout = 40 W: no overlap
54040 0
dBΩ °Rdamp = 180 Ω and Cdamp = 0.1 µF
360
540
20.0
40.0
1800
Z f
5-dB margin
0-20.0
th
in
Z fZ f
th
in
Z fZ f
10m 100m 1 10 100 1k 10k 100k 1Meg
-180-40.0
Public InformationChristophe Basso –Input Filter Interactions152 3/7/2017
10m 100m 1 10 100 1k 10k 100k 1Meg
Final Q is affected by several other factors:
Practical Implementation in the Driver Final Q is affected by several other factors: PCB traces, dielectric losses in the caps but also iron losses in the inductor. True peaking is often lower and gives design margin.
ini t ini t
sv t sv t
No damping7.5 mH and 220 nF (no C2)after bridge. Pout > 40 W
Damping 100 nF + 180 Ω7.5 mH and 220 nF (no C2)after bridge. Pout > 40 W
Public InformationChristophe Basso –Input Filter Interactions153 3/7/2017
Vin = 120 V rms Vin = 120 V rms Curves and experiments by J. Turchi
Course Agenda
A Switching Regulator as a Load EMI Filter Impact EMI Filter Impact An Introduction to FACTs Buck Converter Input/Output Impedances Buck Converter Input/Output Impedances Filtering the Input Current Damping the Filter Damping the Filter Optimum Component Selection A Practical Case Study A Practical Case Study Cascading Converters
Public InformationChristophe Basso –Input Filter Interactions154 3/7/2017
Cascading ConvertersWhen cascading converters, impedances matter
inV s
thZ s thV s inZ s
in
Boost Buck outZ s inZ s
+
The system can be also be modeled by a gain TM
1V V thV s inV s-
th inZ Z
1in th
th
in
V s V sZ sZ s
T s
Public InformationChristophe Basso –Input Filter Interactions155 3/7/2017
MT s
Use SPICE Models First Si l h b i d Simulate the boost converter output impedance
12
L133u
R57m
R1018m3
Vin11
2
duty_cycle
Rload
voutvout
11
c
c
WM
sw
itch
CM
p
duty
-cyc
le
0 Z f
Closed-loop Zout
16
C5820u
11 8.3vca PWX3PWMCML = 33uFs = 200kRi = -300mSe = 82k
C3282p
I1AC = 1A
40 0
-20.0
outZ f
1
R111.66k
85
9
R250k
C210n
R41m
-60.0
-40.0
R31.35k
6
X2AMPSIMPVHIGH = 10VLOW = 10m
V22.5
Verr
1 10 100 1k 10k 100k 1Meg
-80.0 (dBΩ)
Public InformationChristophe Basso –Input Filter Interactions156 3/7/2017
11-15 V/24 V – 3 A
Use SPICE Models First Si l h b k i i d Simulate the buck converter input impedance
duty cycleZin
Vout
Resr14m1
V4
3 4
L18u
7
5
duty_cycle
R44m
R2
vout
vc
a c
PWM switch CM p
duty-cycle
6
L21kH
I1
5.00V
24.0V
5.05V 5.05V
210mV
24.0V
2.37V
Closed-loop Zin
19.8
2
Cout820u
24R2416m
PWMCMX2L = 8uFs = 200kRi = 176mSe = 30k
C31n
I1AC = 1 5.00V
19.0
19.4 inZ f
9
R11k
11
vout
8
R615k
C13.5n
R71m
2.50V
2.50V
2.37V
2.37V
18.6
R31k
10
V22.5
Verr
(dBΩ)10 100 1k 10k 100k 1Meg
18.2
Public InformationChristophe Basso –Input Filter Interactions157 3/7/2017
24 V/5 V – 12 A
Compare Magnitude Curves
39.0
19.0
iZ f N l 1outZ f
-1.00
(dBΩ) inZ f No overlap
1out
inZ f
-21.0
outZ f
10 100 1k 10k 100k 1M
-41.0
Public InformationChristophe Basso –Input Filter Interactions158 3/7/2017
10 100 1k 10k 100k 1Meg
Compare Phase Curves
270
180
iZ f
Neg. up to 1 kHz
90.0
inZ f(°)
0
Z f
1 10 100 1k 10k 100k 1M
-90.0
outZ f
Public InformationChristophe Basso –Input Filter Interactions159 3/7/2017
1 10 100 1k 10k 100k 1Meg
Check Minor Loop Gain Th i i Th i blZ Z
dB °
There is no gain, . The system is stableout inZ Z
-90 0
0
-50 0
-40.0
-180
90.0
-60.0
50.0
-270-70.0
Z f
outZ fZ f
-360-80.0
out
in
Z fZ f
inZ f
Public InformationChristophe Basso –Input Filter Interactions160 3/7/2017
1 10 100 1k 10k 100k 1Meg
Cascade Converters for Simulation
12
L133u
VoBoost
R57m
10
dc_buck
Iout
11.0V
11.0V209mV
4
R1018m
3
Vin11
2
dc_boost
11
vcac
PW
M s
witc
h C
Mp
duty
-cyc
le
X3
VoBuck
19
Resr14m
14 15
L28u
18
10 R44m v out
vc
a c
PWM switch CM p
duty-cycle
PWMCMI1unknown24 1V
24.1V
11.0V
545mV
11.0V
5.00V
5 00V
5.05V 5.05V
2.37V
16
C5820u
va PX3PWMCML = 33uFs = 200kRi = -300mSe = 82k
C3282p
13
Cout820u
X1L = 8uFs = 200kRi = 176mSe = 30k
C6
unknown24.1V 5.00V
1
R111.66k9
R250k
C210n
282p
21
R91k
v out
24
R615k
C13.5n
1n
R71
2.50V
2.50V
2.50V
2.37V
Boost
R31.35k
6
7
X2AMPSIMPVHIGH = 10VLOW = 10m
V22.5
VerrBoost
R111k
22
V32.5
23
VerrBuck
1m
2.50V
2.02V
2.50V
2.37V
Buck
Public InformationChristophe Basso –Input Filter Interactions161 3/7/2017
Check Transient Response T i i d f h b k d b i bl
5.10
Transient response is good for the buck and boost is stable
Buck
4.95
5.00
5.05
(V)
t4.90
outv tIout from 5 to 12 A in 1 µs
24.00
24.05
24.10
(V)
23.90
23.95
outv tBoost
Public InformationChristophe Basso –Input Filter Interactions162 3/7/2017
100u 300u 500u 700u 900u
Practical Measurements – Zin I i d i d di d i i Input impedance measurement requires a dedicated circuit
1QinV
SwitchingConverter1R
1C3R
22 Dcinput
dc load
2RinI
senseRNetworkanalyzer inV
sense
senseVacV
inin sense
sense
V sZ s R
V s
Public InformationChristophe Basso –Input Filter Interactions163 3/7/2017
“Practical Issues of Input/Output Impedance Measurements”, Y. Panov, M. Jovanović, IEEE Transactions on Power Electronics, 2005
Practical Measurements – Zout Y d i j h b d l d V You need to inject enough current to observe a modulated Vout
1C 2CoutV
ISwitchingConverter
D Q
1R3R
outI
Dcinput
1Q
RR
biasV22
dc load
V2RsenseR
4RThermalstabilizationoutV
acV
Networkanalyzer
senseV
out
out sensesense
V sZ s R
V s
Public InformationChristophe Basso –Input Filter Interactions164 3/7/2017
“Practical Issues of Input/Output Impedance Measurements”, Y. Panov, M. Jovanović, IEEE Transactions on Power Electronics, 2005
Literature“Fundamentals of Power Electronics”, R. Erickson, D. Maksimovic, Springer, 2001
“Designing Control Loops for Linear and Switching Power Supplies”, C. Basso, Artech House, 2012
“Practical Issues of Input/Output Impedance Measurements” Y Panov Practical Issues of Input/Output Impedance Measurements , Y. Panov,
M. Jovanović, IEEE Transactions on Power Electronics, 2005
“Physical Origins of Input Filter Oscillations in Current Programmed Converters”,
Y. Jang, R. Erickson, IEEE Transactions on Power Electronics, 1992
“Design Consideration for a Distributed Power System”, S. Schultz, B. Cho,
F Lee Power Electronics Specialists Conference June 1990 pp 611-617F. Lee, Power Electronics Specialists Conference, June 1990, pp. 611-617
“Input Filter Considerations in Design and Applications of Switching Regulators”,
R. D. Middlebrook, IAS 1976
“Design Techniques for Preventing Input-Filter Oscillations in Switched-Mode Regulators”
R.D. Middlebrook, Proceedings of Powercon, May 4-6 1978, San-Francisco
“Linear Circuit Transfer Functions” C Basso Wiley & Sons IEEE Press 2016New book!
Public InformationChristophe Basso –Input Filter Interactions165 3/7/2017
Linear Circuit Transfer Functions , C. Basso, Wiley & Sons, IEEE Press, 2016
Conclusion Incremental input resistance of switching converter is negative Inserting a LC filter impacts the switching converter performance Y h t d i t t i You have two design strategies you design the filter together with the converter you add a filter to an unknown content converter you add a filter to an unknown-content converter You must determine open- and closed-loop input impedance Cascading converters requires knowledge of Zout and Zing q g out in
Test in the lab. dynamic responses when the filter is added Explore how damping is maintained at temperature extremes
Merci !Thank you!
Xiè-xie!
Public InformationChristophe Basso –Input Filter Interactions166 3/7/2017