advanced pfc
TRANSCRIPT
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Advanced Techniques in Power FactorCorrection (PFC)
Prof. Dr. Javier Sebastin
Grupo de Electrnica Industrial
Universidad de Oviedo (Spain)
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Outline
Introduction
Using a simple resistor to comply with the IEC 61000-3-2 in Class A
Using an inductor to comply with the IEC 61000-3-2 in Class A and in
Class D
Exploring the use of isolated Resistor Emulators as the only
conversion stage for medium-speed response applications
High-efficiency post regulators used to improve the transient
response of Resistors Emulators
Very simple single-stage PFCs
Very simple current shaping techniques for very low-cost applications
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Outline
Introduction
Using a simple resistor to comply with the IEC 61000-3-2 in Class A
Using an inductor to comply with the IEC 61000-3-2 in Class A and in
Class D
Exploring the use of isolated Resistor Emulators as the only
conversion stage for medium-speed response applications
High-efficiency post regulators used to improve the transient
response of Resistors Emulators
Very simple single-stage PFCs
Very simple current shaping techniques for very low-cost applications
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Cheap & reliable Input current with astrong harmonic content
Current
Focusing the problemIntroduction (I)
Electronic
circuitry
Power supply
DC/DC
converter
Line
Electronic load
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Load
Electronicload
Line impedance
Line
Load
Load
Introduction (II)
Current
Input
voltage
Distorted
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Power Factor (PF)
PF=Input power
Input voltage, rms X Input current, rms
THD=(Input current, rms)2 - (Its 1ST harmonic, rms)2
Its 1ST harmonic, rms
Introduction (III)Quantifying the problem
Total Harmonic Distortion (THD)
Each individual harmonic
Europeanregulations
Word used to describethe problem
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Introduction (IV)
Power Companies will:
High PF
No harmonics
Electronic equipment
manufacturers will:
Low cost
Reliability
Conflict of interest
Regulations aboutharmonics in the line
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Electronicload
Line impedance
Line
Electronicload
Electronicload
ActiveFilter
Introduction (V)Starting solving the problem (I)
Using active filters
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Introduction (VI)Starting solving the problem (II)
Modifying the electronic load Power Factor Correctors
Inputcurrent
Either
or
Electronic
circuitry
Power supply
DC/DC
converter
Electronic load
NewdevicesLine
Power Factor Corrector
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Introduction (VII)
However:the value of the Power Factor is notimportant.
According to the European Regulations, only the value
ofeach individual harmonic is important.
We should use words such as Low-Frequency
Harmonic Reduction and Low-Frequency
Harmonic Reducer instead of Power Factor
Correctionand PowerFactorCorrector.
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Focusing the courseIntroduction (VIII)
Line
Single-Phase
Three-Phase Conversion
AC/DC
AC/AC
Power
High power
Low-medium power(230V,
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What is the right choice in PFC?
It strongly depends on the application. There is notmagic solutions.
It depends on:
The regulations that must be applied The type of equipment
The output power
The input voltage range
The output voltage
The dynamic response needed
The main objective in the design
Introduction (IX)
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ClassA
YesBalanced3 equipment?
Portabletool?
No
Lightingequipment?
No
PC or TV &P
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Harmonic Class A [A] Class D [mA/W]
3 2.3 3.4
5 1.14 1.9
7 0.77 1.0
9 0.40 0.511 0.33 0.35
13 0.21 0.296
15 n 39 2.25/n 3.85/n
Introduction (XI)
Harmonic limits for Class A and Class D
Very Important!!
Limits in Class A are absolute values [A]
Limits in Class D are relative values [mA/W]
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Harmonic Limits in
Class A [mA]
Limits in
Class D [mA]
3 2300 340
5 1140 190
7 770 1009 400 50
11 330 35
13 210 29.6
15 n 39 2250/n 385/n
Introduction (XII)
Example #1: a 100 W (low-power) converter
Limits in Class A are less strict for low-power applications
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Harmonic Limits in
Class A [mA]
Limits in
Class D [mA]
3 2300 1700
5 1140 950
7 770 5009 400 250
11 330 175
13 210 148
15 n 39 2250/n 1925/n
Introduction (XIII)
Example #2: a 500 W (medium-power) converter
Limits in Class A and in Class D become more
similar for medium-power applications
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Introduction (XIV)
Battery
Example #1: a 100 W (low-power) batterycharger (Class A)
Linecurrent
Linevoltage
This waveform complies
with the regulations!!!
PF = 0.46 and
THD = 193.1%
Very cheap systems for low-frequency harmonic attenuation
can be used to obtain this type of waveform
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Introduction (XV)
Example #1: a 100 W (low-power) TV set (Class D)
Linecurrent
Linevoltage
A slightly more complex system
must be used (it is still very simple)
Linevoltage Line
current
PF = 0.748 and
THD = 88.8%
It does not comply with
the regulations
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Introduction (XVI)
Example #2: two 500 W (low-power) pieces ofequipment
The advantages of being Class A vanish at 500 W
Linevoltage Line
current
PF = 0.748 and
THD = 88.8%
Linevoltage
Linecurrent
PF = 0.705 andTHD = 100.5%
Class AClass D
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Introduction (XVII)
The complexity of the systems for low-frequency
harmonic attenuation increases with the power
Linevoltage
Linecurrent
PF = 0.705 and
THD = 100.5% PF = 0.963 andTHD = 28.1%
Linecurrent
Linevoltage
Example #3: same Class, different power
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Introduction (XVIII)
Influence of the input voltage range (I)
European range: 190 Vac 265 Vac
American range: 85 Vac 130 Vac
Universal range: 85 Vac 265 Vac
Two ranges (American and European), but a
mechanical switch permitted for changing therange
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Introduction (XIX)Influence of the input voltage range (II)
Electronic
circuitryLine
Power supply
DC/DC
converter
Electronic load
PFC
Single range (either European or American) and simple system for
low-frequency harmonic attenuation (PFC)
Moderate change in the input voltage of the DC/DC converter
Slight penalty in efficiency
Simple PFC with
single range
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Introduction (XX)
Electronic
circuitryLine
Power supply
DC/DC
converter
Electronic load
PFC
Universal range and simple PFC
Large change in the input voltage of the DC/DC converter
Significant penalty in efficiency
Complex PFCs which guaranty constant input voltage are interesting
Complex PFC with
universal range
Influence of the input voltage range (III)
I d i (XXI)
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Introduction (XXI)
Is it compatible with the
use of simple PFC?
Electronic
circuitry
Power supply
DC/DC
converter
Electronic load
Electronic
circuitry
Power supply
DC/DCconverter
Electronic load
230V
110V
Power supply for single
range without PFC
Power supply for double
range without PFC
Two ranges selected by a switch
Influence of the input voltage range (IV)
I d i (XXII)
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Introduction (XXII)
Simple PFC placed
on the DC side 110V
230V
Power supply
DC/DC
converter
SimplePFC
SimplePFC
110V
230V
Power supply
DC/DCconverter
Simple
PFCSimple PFC placedon the AC side
Influence of the input voltage range (V)
Two ranges selected by a switch and PFC
I t d ti (XXIII)
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Current
Introduction (XXIII)
Electronic
circuitryLine
Power supply
DC/DC
converter
Electronic load
Electronic
circuitryLine
Power supply
DC/DCconverter
as
Electronic load
ResistorEmulator
Current
Changing the place of the DC/DCconverter Resistor Emulator concept
I t d ti (XXIV)
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Introduction (XXIV) Using only a Resistor Emulator (I)
Line
Power supply
DC/DCconverteras
ResistorEmulator
Output
Current
Line
Power supply
DC/DC
converterOutput
Current
Energy stored at high voltage
(325 V DC) small size
Energy stored at the outputvoltage the size depends onthe voltage
It is not a good solution for low-voltage
(
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Introduction (XXV) Using only a Resistor Emulator (II)
Power
Line
Power supply
DC/DC
converterOutput
Current
Voltage
Line
Power supply
DC/DCconverter
as
Resistor
Emulator
Output
Current
VoltagePower
No devices to store
energy at 100 Hz
Energy stored here
The converter is in charge of
cancelling the output rippleLittle (or no) power processed at
specific moments the outputripple depends on the capacitor
It is not a good solution when
low output-ripple is needed
I t d ti (XXVI)
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Introduction (XXVI) Using only a Resistor Emulator (III)
Power
Line
Power supply
DC/DCconverter
as
ResistorEmulator
Output
Current
Voltage
Line
Power supply
DC/DC
converterOutput
Current
Voltage
Power
No devices to store energy at 100 HzEnergy stored here
The converter can get energy from the
capacitor to maintain the output voltage
when the output current changes
Little (or no) power processed at
specific moments no energyavailable to maintain the output
voltage when the output current
changesIt is not a good solution when fast
transient response is needed
Introduction (XXVII) I th f f t t i t
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Introduction (XXVII) In the case of fast transient responseneeded:
Power supply
Electronic
circuitryLine
Simple orcomplex
PFC
DC/DCconverter
Two separate
stages
Electronic
circuitryLine
Power supply
Simple
PFCsection
DC/DC
converter
section
One integrated
stage
A DC/DC converter
(or section) is needed
DC/DCconverter
DC/DC
converter
Introduction (XXVIII)
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Cost
Size
Weight
Efficiency
Only comply with the regulations
High Power Factor and low Total Harmonic
Distortion (for marketing reasons)
Introduction (XXVIII)
What are the design priorities?
They also determine the right choice
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Outline
Introduction
Using a simple resistor to comply with the IEC 61000-3-2 in
Class A
Using an inductor to comply with the IEC 61000-3-2 in Class A and in
Class D
Exploring the use of isolated Resistor Emulators as the only
conversion stage for medium-speed response applications
High-efficiency post regulators used to improve the transient
response of Resistors Emulators
Very simple single-stage PFCs
Very simple current shaping techniques for very low-cost applications
Using a resistor (I)
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Using a resistor (I)
Looking for the simplest solution (I)
Line
Power supply
DC/DC
converter
(120 W)200 mF
4 X 1N4007
Capacitor voltage
Input current Class D
Using a resistor (II)
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Using a resistor (II) Looking for the simplest solution (II)
Input currentOrder Measured [A] Limits Class D [A]
1 0.542 -
3 0.527 0.4085 0.498 0.228
7 0.457 0.12
9 0.407 0.06
11 0.351 0.042
13 0.294 0.036
15 0.239 0.031
17 0.192 0.027
19 0.155 0.024
21 0.132 0.022
23 0.121 0.02
25 0.117 0.018
27 0.115 0.017
29 0.112 0.016
31 0.105 0.015
33 0.097 0.014
35 0.087 0.013
37 0.079 0.012
39 0.073 0.012The compliance is very far
0 5 10 15 20 25 30 35 400
0.2
0.4
0.6
Harmonic order
Input current [A]
Limits in Class D
Simulated
Using a resistor (III)
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Using a resistor (III) Looking for the simplest solution (III)
What about a Class A piece of equipment?
Line
Battery Charger
DC/DC
converter
(120 W)200 mF
4 X 1N4007 Battery
Capacitor voltage
Input current Class A
Using a resistor (IV)
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Using a resistor (IV) Looking for the simplest solution (IV)
Input current
It does not comply, but it
is very near to comply
Order Measured [A] Limits Class A [A]
1 0.542 -
3 0.527 2.35 0.498 1.14
7 0.457 0.77
9 0.407 0.4
11 0.351 0.33
13 0.294 0.21
15 0.239 0.15
17 0.192 0.13219 0.155 0.118
21 0.132 0.107
23 0.121 0.098
25 0.117 0.09
27 0.115 0.083
29 0.112 0.078
31 0.105 0.073
33 0.097 0.068
35 0.087 0.064
37 0.079 0.061
39 0.073 0.058
0 5 10 15 20 25 30 35 400
0.5
1
1.5
2
2.5
Harmonic order
Input current [A]
Limits in Class A
Simulated
Using a resistor (V) L ki f h i l l i (V)
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Using a resistor (V) Looking for the simplest solution (V)
Order Measured [A] Limits Class A [A]
1 0.528 -
3 0.5 2.3
5 0.448 1.14
7 0.378 0.77
9 0.3 0.4
11 0.225 0.33
13 0.164 0.21
15 0.128 0.1517 0.115 0.132
19 0.113 0.118
21 0.109 0.107
23 0.1 0.098
25 0.087 0.09
27 0.076 0.083
29 0.07 0.07831 0.067 0.073
33 0.066 0.068
35 0.063 0.064
37 0.058 0.061
39 0.053 0.058
Line
Battery Charger
DC/DC
converter
(120 W)100 F
4 X 1N4007
Let us change the value of the bulk capacitor
Capacitor voltage
Input current
Almost compliance
with 100 mF
Using a resistor (VI) L ki f h i l l i (VI)
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Using a resistor (VI) Looking for the simplest solution (VI)
However, the value of the bulk capacitor cannot be
freely chosen because:
Hold-up time requirements
Input voltage range of the DC/DC converter
Another solution must be found
Using a resistor (VII)
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Using a resistor (VII)
The simplest solution: to add a resistor
Electronic
circuitry
Class A
Line
Power supply
DC/DC
converter
DC side
Electronic
circuitryClass A
Line
Power supply
DC/DC
converter
AC sideR
R
Using a resistor (VIII)
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@ 230V ac, R =1.5 Wiinput peak = 4.12 A
Presistor= 1.85 W
Using a resistor (VIII)
Order Measured [A]with R=0
Measured [A]with R=1
Measured [A]with R=1.5
LimitsClass A [A]
1 0.542 0.539 0.538 -
3 0.527 0.52 0.516 2.3
5 0.498 0.484 0.474 1.14
7 0.457 0.433 0.416 0.77
9 0.407 0.372 0.347 0.4
11 0.351 0.304 0.273 0.33
13 0.294 0.237 0.2 0.21
15 0.239 0.173 0.135 0.15
17 0.192 0.12 0.084 0.132
19 0.155 0.084 0.056 0.118
21 0.132 0.067 0.053 0.107
23 0.121 0.066 0.057 0.098
25 0.117 0.067 0.058 0.09
27 0.115 0.065 0.052 0.083
29 0.112 0.058 0.041 0.078
31 0.105 0.047 0.029 0.073
33 0.097 0.036 0.021 0.068
35 0.087 0.028 0.02 0.064
37 0.079 0.025 0.022 0.061
39 0.073 0.026 0.024 0.058
Cbulk = 200 mFPconverter= 120 W
Using a resistor (IX)I f i h i
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@ 230V ac, R =1.5 Wiinput peak = 4.12 APresistor= 1.85 W
Using a resistor (IX)
Cbulk = 200 mFP
converter= 120 W
Input-current waveform with a resistor
Capacitor voltage
Input current
Capacitor voltage
Input current
@ 230V ac, R =0 Wiinput peak = 6.37 A
Using a resistor (X)
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Using a resistor (X)Design procedure
Obtain the resistor
(from graphs)
Use the simplestmethod
Othermethod must
be used
Choose bulk
capacitor
Input power
Calculate losses @ full
power, 190 Vac
Acceptablelosses?
NO YES
Using a resistor (XI)
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50 100 150 200 250 300
1
2
3
4
R [ ]
Output power [W]
Using a resistor (XI)Value of the resistor needed to complywith the IEC 61000-3-2 in Class A as afunction of the input power (bulk capacitor
inmF per watt as parameter)
0.5 F/W1 F/W
2 F/W
Using a resistor (XII)Ab l t l t f ll
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Using a resistor (XII)Absolute power losses at fullload and minimum line voltage(maximum line current)
Output power [W]
Power losses [W]
50 100 150 200 250 300
5
10
15
20
25
1 F/W
2 F/W
0.5 F/W
Using a resistor (XIII)Relative power losses (P /P at
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Using a resistor (XIII)Relative power losses (PR/Poutput) atfull load and minimum line voltage(maximum line current)
Output Power [W]
Relative losses [%]
50 100 150 200 250 300
2
4
6
8
10
0.5 F/W
1 F/W
2 F/W
Using a resistor (XIV)D i l
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g ( )Design example:Poutput=150W, C=150mF (1mF/W)
50 100 150 200 250 300
1
2
3
4
R [ ]
Output power [W]
1 F/W
2.5 W
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Using a resistor (XVI)Power limits for this solution
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g ( )Power limits for this solution
Output Power [W]
Relative losses [%]
50 100 150 200 250 300
2
4
6
8
10
1 F/W
Veryinteresting Not sointeresting
Using a resistor (XVII)Using this solution for Universal line
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Power lossesstrongly increase at
low line voltage
g ( )
Pconverter= 120 WCbulk = 200 mFR=1.5 W
Using this solution for Universal linevoltage range
Line
Quantity
@ 230V @ 110V
iinput peak 4.12 A 5.09A
iinput RMS 1.11 A 1.853 A
Plosses resistor 1.85 W 5.15 W
Line
Power supply
DC/DCconverter
PconverterCbulk
4 X 1N4007
R
Using a resistor (XVIII) Adaptation for operation in two ranges (I)
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g ( ) Adaptation for operation in two ranges (I)
DC/DCconverter
Electronic
circuitry
Class A
Line
Power supply
R/2
R/2
DC side
230V
110V
AC sideElectronic
circuitry
Class A
DC/DC
converter
Power supply
Line
R
230V
110V
Different operation (AC side & DC side)
Using a resistor (XIX) Adaptation for operation in two ranges (II)
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g ( ) Adaptation for operation in two ranges (II)
Electronic
circuitry
Class A
DC/DC
converter
Power supply
Line
R
230V
110V
iinput 230V
AC sideElectronic
circuitry
Class A
DC/DC
converter
Power supply
Line
R
230V
110V
iinput 110V
Both iinput 110V and iinput 230V passing through R
Using a resistor (XX) Adaptation for operation in two ranges (III)
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Adaptation for operation in two ranges (III)
DC side
DC/DC
converter
Electronic
circuitry
Class ALine
Power supply
R/2
R/2
230V
110V
iinput 110V
DC/DC
converter
Electronic
circuitry
Class ALine
Power supply
R/2
R/2
230V
110V
iinput 230V
iinput 110V passing through R/2and iinput 230V passing through R (better)
Using a resistor (XXI)Adaptation for operation in two ranges (IV)
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Adaptation for operation in two ranges (IV)
DC/DC
converter
Electronic
circuitry
Class A
Line
Power supply
R/2
R/2
230V
110V
110V
Electronic
circuitry
Class A
DC/DC
converter
Power supply
Line
R230V
110V
110V
Example:
Pconverter = 120 W
Cbulk = 2 X 400 F (series)R=1.5
Plosses resistors = 3.15 W
(total)
Plosses resistor= 5.27 W
Using a resistor (XXII)Adaptation for operation in two ranges (V)
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Impractical due to the fact that the power losses strongly
increase at low line voltage
Power R C losses
@ 230V
losses
@ 190V
losses
@ 110V
losses
@ 85V
100 W 1.6 W 2x220 mF 1.3 W 1.6 W 3.8 W 5 W
200 W 3.6 W 2x440 mF 8.5 W 11.5 W 29 W 50 W
Adaptation for operation in two ranges (V)
Electronic
circuitry
Class A
DC/DC
converter
Power supply
Line
R230V
110V
C
C
Using a resistor (XXIII)Adaptation for operation in two ranges (VI)
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Better results with the resistor split into two resistors
Adaptation for operation in two ranges (VI)
DC/DC
converter
Electronic
circuitryClass A
Line
Power supply
R/2
R/2
230V
110V
C
C
Power R C losses
@ 230V
losses
@ 190V
losses
@ 110V
losses
@ 85V
100 W 1.6 W 2x220 mF 1.3 W 1.6 W 2.1 W 3.1 W
200 W 3.6 W 2x440 mF 8.5 W 11.5 W 16 W 25 W
Using a resistor (XXIV)Experimental results (I)
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Line
R/2
R/2
230V
110V
C
C
1 A/div
@ 230V, 100W
2 A/div
@ 110V, 100W
@ 230V, 100W,2x0.82 , 2W
0
0.5
1
1.5
2
2.5
11 19 317 15 23 35273
Harmonic Order
Input current [A]
Limits inClass A
Measured
Experimental results (I)
Pconverter = 100 W
C = 2 X 100 F (series)
R = 2x0.82
Using a resistor (XXV)Experimental results (II)
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Line
R/2
R/2
230V
110V
C
C
2 A/div
@ 230V, 200W
2 A/div
@ 110V, 200W
Experimental results (II)
Pconverter = 200 W
C = 2 X 200 F (series)
R= 2x1.8
@ 230V, 200W,2x1.8 , 10W
0
0.5
1
1.5
2
2.5
11 19 317 15 23 35273
Harmonic Order
Input current [A]
Limits inClass A
Measured
Using a resistor (XXVI) Conclusions of the use of a resistor to
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comply with the IEC 61000-3-2regulations in Class A
This is the simplest possible solution Low-cost and low-size solution
Very interesting for low-power (P
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Outline
Introduction
Using a simple resistor to comply with the IEC 61000-3-2 in Class A
Using an inductor to comply with the IEC 61000-3-2 in Class A
and in Class D
Exploring the use of isolated Resistor Emulators as the onlyconversion stage for medium-speed response applications
High-efficiency post regulators used to improve the transient
response of Resistors Emulators
Very simple single-stage PFCs
Very simple current shaping techniques for very low-cost applications
Using an inductor (I)Another very simple solution: to add an inductor
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Another very simple solution: to add an inductor
Electronic
circuitry
Class A
Line
Power supply
DC/DC
converter
DC side
Electronic
circuitry
Class A
Line
Power supply
DC/DC
converter
AC side
L
L
Using an inductor (II)
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@ 230V ac, L = 2 mHiinput peak = 3.84 A
Order Measured [A]with L=0 mH
Measured [A]with L=1 mH
Measured [A]with L=2 mH
LimitsClass A [A]
1 0.542 0.552 0.545 -
3 0.527 0.531 0.515 2.35 0.498 0.493 0.459 1.14
7 0.457 0.438 0.384 0.77
9 0.407 0.374 0.299 0.4
11 0.351 0.303 0.214 0.33
13 0.294 0.232 0.138 0.21
15 0.239 0.167 0.079 0.15
17 0.192 0.11 0.046 0.132
19 0.155 0.067 0.039 0.118
21 0.132 0.042 0.04 0.107
23 0.121 0.036 0.036 0.098
25 0.117 0.037 0.028 0.09
27 0.115 0.037 0.02 0.083
29 0.112 0.032 0.017 0.078
31 0.105 0.025 0.016 0.073
33 0.097 0.019 0.016 0.068
35 0.087 0.016 0.014 0.064
37 0.079 0.015 0.011 0.061
39 0.073 0.015 0.009 0.058
Cbulk = 200 mFPconverter= 120 W
Using an inductor (III)Input-current waveform and harmonic
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Input current waveform and harmoniccontent with an inductor
Capacitor voltage
Input current
Example:
Cbulk = 200 mFPconverter= 120 W
L = 2 mH
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
Input current [A]
Limits in Class D
Simulated
Harmonic order
@ 230V, 120 W@ 230V, 120 W
Harmonic order
0 5 10 15 20 25 30 35 400
0.5
1
1.5
2
2.5
Input current [A]
Limits in Class A
Simulated
It complies
It does not
comply
Comparing input-current waveform withUsing an inductor (IV)
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Cbulk = 200 mFPconverter= 120 W
Comparing input current waveform withan inductor and a resistor for Class Aequipment
@ 230V ac, R =1.5 Wiinput peak = 4.12 A
Presistor= 1.85 W
Capacitor voltage
Input current
@ 230V ac, L = 2 mHiinput peak = 3.84 A
Capacitor voltage
Input current
Comparing input-current waveforms withUsing an inductor (V)
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0
5
10
10 ms0
C = 200 F
C = 800 F
Comparing input current waveforms withdifferent bulk capacitor values
Slightly influence of
the capacitor value
340 V
288 V
8.38 A
C = 200 F10 ms 20 ms0
Capacitor
voltage
Input current
0
312 V
300 V
7.55 A
C = 800 F
10 ms 20 ms0
Capacitor
voltage
Input current
L = 3.3 mH
Pconverter
= 400 W
Using an inductor (VI)Looking for the most restrictive harmonics (I)
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3 5 7 9 11 13 15 17 19
0.5
1
1.5
22.5
3
3.5
Harmonic Order
@ 230V, 100 W,
1.7 mH & 47 mF
Input current [A]
Limits inClass A
Simulated
20 ms-5
0
5
3.85 A
10 ms0
Input current [A]
Time
g ( )
Example: 100 W,1.7 mH & 47 F
Harmonics 13th-17thare the most restrictive
at low power
Using an inductor (VII)Looking for the most restrictive harmonics (II)
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Example: 600 W,7.8 mH & 330 F
Harmonics 3rd-5th are
the most restrictive at
high power
g ( )
0.5
1
1.5
2
2.5
3
3.5
3 5 7 9 11 13 15 17 19
Harmonic Order
@ 230V, 600 W,
7.8 mH & 330 mF
Input current [A]
Limits inClass A
Simulated
-10
0
108.68 A
10 ms 20 ms0
Input current [A]
Time
Using an inductor (VIII)Value of the minimum inductor needed to comply
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100 200 300 400 500 600
2
4
6
8
L [mH]
Output power [W]
0.5 F/W
2 F/W
p ywith the IEC 61000-3-2 in Class A as a functionof the input power (bulk capacitor inmF per wattas parameter)
Using an inductor (IX)Comparing the influence of the bulk capacitor for
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Comparing the influence of the bulk capacitor forthe case of the inductor and the resistor
0.5 F/W
2 F/W
100 200 300 400 500 600
2
4
6
8
L [mH]
Output power [W]
50 100 150 200 250 300
1
2
3
4 R [ ]
Output power [W]
0.5 F/W
1 F/W2 F/W
Lower inductor values with
high bulk capacitor values
Erratic influence of the
value of the bulk capacitor
Design procedureUsing an inductor (X)
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Design procedurefor Class A
Obtain the inductor
(from graphs)
Use this method
Other
method mustbe used
Choose bulk
capacitor
Input power
Calculate the inductor
size
Acceptablesize?
NOYES
Using an inductor (XI)Design example:
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100 200 300 400 500 600
2
4
6
8
L [mH]
Output power [W]
0.5 F/W
2 F/W2.7 mH
g pPoutput=200 W, C=100mF (0.5mF/W)
Using an inductor (XII)What about the inductor size?
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We must know the maximum peak value of the input current (at full
load and minimum line voltage) determine the gap and number ofturns
We must know the maximum RMS value of the input current (at full
load and minimum line voltage) determine the wire size (diameter)and losses
Input power[W]
L [mH] Ipeak [A] IRMS [A]Equivalent
ferritecore size
Power losses(%)
200 2.7 5.33 @ 230V6.07 @ 190V
1.6 @ 230V1.88 @ 190V
E30/15/7 0.8
Using an inductor (XIII)Inductor size and losses for different
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power levels
Input power[W]
L [mH]Equivalent
ferritecore size
Power losses(%)
100 2 E20/10/5 0.53
200 2.7 E30/15/7 0.8
300 3.4 E42/21/15 0.3
400 4.4 E42/21/15 0.66
500 6.8 E42/21/20 0.57
600 7.8 E42/21/20 1.66
Using an inductor (XIV)Magnetic materials for the inductor (I)
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10 100 1000 10000 100000
0.5
1
1.5
2
B [T]
H [A/m]
B [mT]10 100 1000 10000
0.01
0.1
1
10
100
Plosses [kw/m3]
g ( )
Silicon steel laminationcore (instead of ferrite)
Example: RG11
High induction levels(1.4 T) are possible
Magnetic materials for the inductor (II)Using an inductor (XV)
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DC-side or AC-side inductor?Using an inductor (XVI)
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Line current
with DC-side
inductor
Capacitor voltage
with DC-side inductor
Time
with AC-side
inductor
with AC-side
inductor
DC-side inductor
AC-side
inductor
Exactly the same result if the converter is
working in strong DCM
Example: Cbulk = 200 mF, L = 2 mH, Pconverter= 120 W
Using an inductor (XVII)What about complying with the IEC
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0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6Input current [A]
Limits in Class D
Simulated
Harmonic order
@ 230V, 120 W
Example: Cbulk = 200 mF, L = 2 mH, Pconverter= 120 W
Low-frequency harmonics are the most significant ones
A considerable increase in the inductance value is needed
61000-3-2 regulations in Class D?
Looking for the minimum value of L toUsing an inductor (XVIII)
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3 5 7 9 11 13 15 17 19
0.1
0.2
0.3
0.4
0.5
Harmonic Order
@ 230V, 100W, 41mH & 200mF
Input current [A]
Limits in
Class D@ 100W
Simulated
-2
-1
0
1
21.42 A
0 10 ms 20 ms
Input current [A]
Time
Looking for the minimum value of L tocomply with the regulations in Class D (I)
Example: Cbulk
= 200 mF, L = 41 mH, Pconverter
= 100 W
An inductor of 41 mH is needed for 100 W
Looking for the minimum value of L tocomply with the regulations in Class D (II)
Using an inductor (XIX)
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comply with the regulations in Class D (II)
Example: Cbulk = 1200 mF, L = 7 mH, Pconverter= 600 W
An inductor of 7 mH is needed for 600 W
If we increase the power, the limits will also increase a similarinput-current waveform is enough to comply with the regulations
-10
0
108.12 A
0 10 ms 20 ms
Input current [A]
Time3 5 7 9 11 13 15 17 19
1
1.5
2
2.5
3
Harmonic Order
0.5
@ 230V, 600W, 7mH & 1200mF
Input current [A]
Limits inClass D
@ 600W
Simulated
0
Using an inductor (XX)Value of the minimum inductor needed to comply
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with the IEC 61000-3-2 in Class D as a functionof the input power (bulk capacitor inmF per wattas parameter)
0.5 F/W
2 F/W
L [mH]
Output power [W]
100 200 300 400 500 600
10
20
30
40
50
The value of the inductors
inductance decreases
when the power increases,
but the size increases
(because it depends on the
square value of the peak
current)
Using an inductor (XXI)Inductor size and losses for different
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power levels
Input power[W]
L [mH]Equivalent
ferritecore size
Power losses(%)
100 41 E42/21/15 1
200 21 E42/21/15 2
300 14 E42/21/20 1.1
400 10 E42/21/20 1.25
500 8.7 E42/21/20 1.8
600 6.9 E42/21/20 2.18
Using an inductor (XXII)Comparing the value of the minimum inductor
d d l i h h IEC 61000 3 2 i
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needed to comply with the IEC 61000-3-2 inClass A and in Class D
0.5 F/W
2 F/W
L [mH]
Output power [W]
100 200 300 400 500 600
10
20
30
40
50Minimum inductor tocomply in Class D
100 200 300 400 500 600
2
4
6
8L [mH]
Output power [W]
0.5 F/W
2 F/W
Minimum inductor tocomply in Class A
Lower L values at low power Similar L values at high power
Using an inductor (XXIII) Inductor size and losses for different power levels
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Similar L sizes at high powerLower L sizes at low
power in Class A
Input power[W]
L [mH] inClass A
Equivalentcore size in
Class A
Powerlosses in
Class A (%)
L [mH] inClass D
Equivalentcore size in
Class D
Powerlosses in
Class D (%)
100 2 E20/10/5 0.53 41 E42/21/15 1
200 2.7 E30/15/7 0.8 21 E42/21/15 2
300 3.4 E42/21/15 0.3 14 E42/21/20 1.1
400 4.4 E42/21/15 0.66 10 E42/21/20 1.25
500 6.8 E42/21/20 0.57 8.7 E42/21/20 1.8
600 7.8 E42/21/20 1.66 6.9 E42/21/20 2.18
Using an inductor (XXIV)Adaptation for operation in two ranges (I)
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Different operation (AC side & DC side)
AC sideElectronic
circuitry
Class A
DC/DC
converter
Power supply
Line230V
110V
L
DC/DCconverter
Electronic
circuitry
Class A
Line
Power supply
L/2
L/2
DC side
230V
110V
Adaptation for operation in two ranges (II)Using an inductor (XXV)
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AC side
Both iinput 110V and iinput 230V passing through L
Electronic
circuitry
Class A
DC/DC
converter
Power supply
Line
230V
110V
L
Electronic
circuitry
Class A
DC/DC
converter
Power supply
Line
230V
110V
Liinput 110V
iinput 230V
Adaptation for operation in two ranges (II)Using an inductor (XXVI)
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DC side
L/2
L/2
DC/DC
converter
Electronic
circuitry
Class ALine
Power supply
230V
110V
iinput 110V passing through L/2and iinput 230V passing through L
DC/DC
converter
Electronic
circuitryClass ALine
Power supply
L/2
230V
110V
L/2iinput 110V
iinput 230V
Experimental results (I)Using an inductor (XXVII)
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Line
L
C
0.5 A/div
@ 230V, 100W
Class D
Pconverter = 100 W
C = 47 F
L = 41 mH
5 9 13 21 25 29 33
0.1
0.2
0.3
3717
@ 230V, 100W,41 mH
Harmonic order
Input current [A]
Limits inClass D
Measured
Experimental results (II)Using an inductor (XXVIII)
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Line
L
C
1 A/div
@ 230V, 100W
Class A
Pconverter = 100 W
C = 47 F
L = 1.7 mH
@ 230V, 100W,1.7 mH
Harmonic order
Input current [A]
Limits inClass A
Measured
3 15 19 23 27 31 35
1
2
119
Conclusions of the use of an inductorto comply with the IEC 61000 3 2
Using an inductor (XXVIII)
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to comply with the IEC 61000-3-2regulations in Class A and Class D
This is a very simple solution Low-cost and high-efficiency (low-losses) solution Very interesting for low-power (P
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Outline
Introduction
Using a simple resistor to comply with the IEC 61000-3-2 in Class A
Using an inductor to comply with the IEC 61000-3-2 in Class A and in
Class D
Exploring the use of isolated Resistor Emulators as the onlyconversion stage for medium-speed response applications
High-efficiency post regulators used to improve the transient
response of Resistors Emulators
Very simple single-stage PFCs
Very simple current shaping techniques for very low-cost applications
Using only a RE (I) Passive (L or R) versus active systems toreduce the harmonic content
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Line
Power supply
DC/DC
converter Outpu
t
R or L
Current
Line
Power supply
DC/DCconverter
as
ResistorEmulator O
utput
Current
Low-cost
Either low-losses or low-size Non-sinusoidal waveform solutions for low power
Unregulated voltage across thecapacitor solutions for limited linevoltage range (many times, voltage
doubler needed)
Sinusoidal waveform solutions for anypower
Regulated voltage across the capacitorsolutions for universal line voltage range
A good solution if only the
Resistor Emulator were enough
Using only a RE (II) Is only a Resistor Emulator enough toimplement the overall power supply?
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p p pp y
Power supply
DC/DC
converteras
ResistorEmulator
Output
Energy stored at the output
voltage the size depends onthe voltage
From the point of view of the capacitor size, it is not a bad
solution for medium and high voltage applications (>12 V DC)
Power supply
DC/DC
converter Output
R or L
Energy stored at high voltage
(325 V DC) small size
Using only a RE (III) And, what about the dynamics?
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DC/DCconverter
Example of Resistor Emulator control:control based on an analog multiplier
Lowpassfilter
Why a lowpass filter
here?
Using only a RE (IV)The lowpass filter influence (I)
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DC/DCconverter
LowpassfilterVeaVea
Input voltage
Current Reference=VeaSinus
Vea
Input voltage
Current Reference=VeaSinus
Filter with very-
low cut-offfrequency
Filter with
high cut-offfrequency
A filter with low cut-off frequencyis needed if a perfect sinusoidal
is required
Using only a RE (V)The lowpass filter influence (II)
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DC/DCconverter
LowpassfilterVea
Filter with very low cut-off frequency:
Perfect sinusoidal line current
Very poor dynamic response
If yes, the use of only a Resistor
Emulator as overall power
supply becomes very attractive
And, what about the
dynamic response?
Filter with high cut-off frequency:
Non-perfect sinusoidal line currentBut, can we achieve compliance withthe IEC 61000-3-2 and reasonable
dynamic response?
Using only a RE (VI) Line current waveform as a function ofthe voltage regulator pole frequency fp
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p
fp: 10 Hz
fp: 100 Hzfp: 1000 Hz
fp: 500 Hz
Voltage
regulator
fp
timef [Hz]
-135
-90
-45
0
45
1 10 100 1000 10000
AR []
AR [dB]
-40
-20
020
40
60
fp fp fpfp = 1kHz is a practical limit (no
significant phase shift at 100Hz)
Using only a RE (VII) Line current waveform as a function ofthe voltage regulator DC gain AR
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AR = 100AR = 50
fp: 10 Hz
fp: 100 Hz
fp: 1000 Hz
fp: 500 Hz AR = 100
is a practical limitdue to the voltage
levels in the
controller
Using only a RE (VIII)Looking for the worst case
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AR 100fp1000 Hz
Line current
3 11 21 31 390
1
2
3
2.3 A
Harmonic Order
Input current [A]
Limits inClass D@ 100W
Simulated
Theoretical harmonic content: Onlythe third harmonic is present
Using only a RE (IX) Why is the third harmonic the only onepresent in the line current? (I)
V i tFor 0 wt
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DC/DC
converter
LowpassfilterVea
Viref
V1sinwt
Rs
Viref
= V1
sinwt (Veao
+ Vea
sin2wt)
Veao+ Veasin2wt
V1sinwt
Viref( t) = VeaoV1sinwt +0.5V1Veacoswt - 0.5V1Veacos3wt
For 0 wtiline DC
iline DC ( t) = (VeaoV1sinwt +0.5V1Veacoswt - 0.5V1Veacos3wt)/Rs
Therefore, for 0 wtiline AC ( t) = iline DC ( t) = (VeaoV1sinwt +0.5V1Veacoswt - 0.5V1Veacos3wt)/Rs
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Using only a RE (XI)Looking for the maximum power compatiblewith complying with the IEC 61000 3 2
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IEC 61000-3-2 regulations in Class A can be
complied up to very high power levels
with complying with the IEC 61000-3-2regulations in Class A (I)
AR 50-100fp1000 Hz
Line current
AR Output ripple=1 % Output ripple=2 %
50 3680 W 3400 W
100 3400 W 1700 W
Using only a RE (XII) Looking for the maximum powercompatible with complying with the IEC
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compatible with complying with the IEC61000-3-2 regulations in Class A (II)
Theoretical
Theoretical
Line current obtainedby simulation
AR = 100,fC: 1 kHz
0
1
2
3
AR = 100,fC: 500 Hz
0
1
2
3
Simulated
Simulated
The theoretical and the simulatedwaveforms are slightly different
The cause is the output voltageripple.
Due to this, the actual ripple is
not exactly sinusoidal
AR Output ripple=1 % Output ripple=2 %
50 3600 W 2500 W
100 2600 W 1300 W
Compliance up to very high
power levels is achieved
Using only a RE (XIII) Can we get a very fast transientresponse if we have a very fast
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response if we have a very fastoutput voltage feedback loop?
The dynamics depends on
the capacitor
The capacitor is recharged
each 10ms (100 Hz) the
faster response is 10 ms
Line
Power supply
DC/DCconverter
asResistorEmulator
Out
put
No devices to storeenergy at 100 Hz
PowerVoltage
Little (or no) power processed at
specific moments no energyavailable to maintain the output
voltage when the output current
changes, except the energystored in the capacitor
Current
Energystored here
Using only a RE (XIV)Simulating the dynamic response
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The output voltage takes 90 ms in
recovering the steady state
fC = 10 HzOutput voltage
20 40 60 80 100 120 140 160
360
380
400
420
40 ms
90 ms
Time (ms)
4300 W 1700 W
fC = 1kHz
20 40 60 80 100 120
Time (ms)
360
370
380
390
400
410
2600 W 400 W
10 ms
Output voltage
The output voltage takes 10 ms in
recovering the steady state
Using only a RE (XV) Resistor Emulator topologies: low power
Voltage
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Line Power supply Load(Electronic
circuitry)
Voltage
Current
Voltage
Current
Line Power supplyLoad(Electronic
circuitry)
Flyback
based
SEPICbased
Using only a RE (XVI) Resistor Emulator topologies: medium power
Current-fed Push-Pull based
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Line
Power supply
Load(Electronic
circuitry)
Voltage
Current
Line
Power supply
Load(Electronic
circuitry)
Voltage
Current
Cu e t ed us u based
Using only a RE (XVII) Resistor Emulator topologies: high power
Current fed Full bridge based
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Line
Power supply
Load(Electronic
circuitry)
Voltage
Current
Current-fed Full-bridge based
Line
Power supply
Voltage
Current
Load(Electronic
circuitry)
Using only a RE (XVIII) Example of application: a power supplyfor a 300 + 300 W audio amplifier (I)
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p ( )
Flyback based
Universal line voltage
Flyback with 2 Cool-MOS in parallel
10 ms dynamic response is good enough for this application
Line
Power supply
300 W audioamplifier
(Channel Right)
300 W audioamplifier
(Channel Left)
+70 V
-70 V
GND85-250 V
Using only a RE (XIX) Example of application: a power supplyfor a 300 + 300 W audio amplifier (II)
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p ( )
300 + 300 W
audio amplifier
Power supply
For Behringer
Developed at the
University of Oviedo
(GEI group)
Using only a RE (XX) Experimental results: line waveforms
R i t E l t b d 300 W b t t
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0.5 A/div 5ms/div
AR = 10
fC = 1 kHz
AR = 25 fC = 1 kHz
0.5 A/div 5ms/div
AR = 40 fC = 1 kHz
0.5 A/div 5ms/div
0.5 A/div 5ms/div
AR = 10
fC = 10 Hz
Simulated ResultSimulated Result
Simulated Result Simulated Result
Resistor Emulator based on a 300 W boost converter
Using only a RE (XXI) Experimental results: transient response
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1/3 Full load Full load
60 ms
fC = 10 Hz
10 ms
fC = 1kHz
1/3 Full load Full load
Using only a RE (XXII) Conclusions of the use of isolated ResistorEmulators as the only conversion stage formedium-speed response applications (I)
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medium speed response applications (I)
Many applications do not need fast dynamic response. Inthese cases conventional Resistor Emulators (like flyback)
can be used directly as power supply with no second stage
and with several advantages:
Low cost and size (no second stage)
Very low harmonic content
Can be used in high and low power applications.
Can be used with universal line voltage
Using only a RE (XXIII) Conclusions of the use of isolated ResistorEmulators as the only conversion stage formedium-speed response applications (II)
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medium speed response applications (II)
The limitations in the transient response are:
The 100-120 Hz output voltage ripple only depends on thecapacitor value
This ripple ripple cannot be reduced by increasing thecorner frequency of the output-voltage feedback loop
The maximum effective corner frequency is about 1kHz(10 times the ripple frequency)
The minimum response time is 10-8.3 ms (one 100-120 Hzcycle)
Using only a RE (XXIV) Conclusions of the use of isolated ResistorEmulators as the only conversion stage formedium-speed response applications (III)
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ed u speed espo se app cat o s ( )
This solution should not be used if the output voltage is
relatively low (lower than 12 V) due to the fact that the bulk
capacitor is placed just at the output, which means:
Energy stored at low voltage Large value of thecapacitor size
High current levels passing through the capacitorLarge capacitor losses due to the ESR
Outline
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Introduction
Using a simple resistor to comply with the IEC 61000-3-2 in Class A
Using an inductor to comply with the IEC 61000-3-2 in Class A and in
Class D
Exploring the use of isolated Resistor Emulators as the only
conversion stage for medium-speed response applications
High-efficiency post regulators used to improve the transient
response of Resistors Emulators
Very simple single-stage PFCs
Very simple current shaping techniques for very low-cost applications
High-efficiency post-regulators (I)
fp: 1000 Hz
Can we improve the dynamic responseof a Resistor Emulator with a lowpenalty in the converter efficiency?
Line curent
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DC/DCconverter
Lowpassfilter
p 000
fp: 10 Hz
90 ms
10 ms
AR
Output voltage
20 40 60 80 100 120
Time (ms)
f [Hz]
-135
-90
-45
0
45
1 10 100 1000 10000
-40
-20
0
20
40
60
AR [dB]
AR []
Time (ms) 10
fp fp
The minimum response time is
10-8.3 ms (one 100-120 Hz cycle)
Another stage can be connected to
improve the transient response
High-efficiency post-regulators (II) Characteristic of the high-efficiencypost regulators
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DC/DCconverter
Lowpassfilter
Line
+
-
V1+
-
VO
Output
High-efficiency
post-regulators
Common characteristics of all high-
efficiency post-regulators:
Low additional cost and size
Only a fraction of the total powerundergoes a power switching processing
Very high efficiency: 96-98%
No short-circuit protection in the post-regulator
V1 and VO are voltages of similar values
High-efficiency post-regulators (III) Use of the high-efficiency post regulatorsin multiple-output applications
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Lowpassfilter
Line +
-
V1+
-
VO Output
High-efficiency
post-regulators
10 ms
fp: 1000 Hz
90 ms
fp: 10 Hz
Some slow or medium-speed outputs andsome fast response outputs
fp: 10 Hz
fp: 1000 Hz
High-efficiency post-regulators (IV) Operation principle of the high-efficiency post regulators
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DC/DCconverter
Lowpassfilter
Line
+
-
V1+
-
VO Output
How can we implement thevoltage source?
+-VS
Time
v1
vO
vS
High-efficiency
post-regulators
High-efficiency post-regulators (V) Implementing thevoltage source VS(I)
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?DC/DC
converter
Lowpassfilter
Line
+
-
V1+
-
VO Output+- VS
High-efficiency
post-regulators
Small
DC/DC
converter
Where should weconnect the input port
of this converter?
High-efficiency post-regulators (VI) Implementing thevoltage source VS(II)
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DC/DC
converter
Lowpassfilter
Line
+
-
V1
Option #1: connect the input port to anadditional Resistor Emulator output
+
-
VO Output
High-efficiencypost-regulators
+- VS
Small
DC/DC
converter
One additionaloutput
High-efficiency post-regulators (VII) Implementing the voltagesource VS(III)
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DC/DC
converter
Lowpassfilter
Line
+
-
V1
Option #2: connect the input port to theResistor Emulator output
+
-
VO Output
High-efficiencypost-regulators
+- VS
SmallDC/DC
converter
High-efficiency post-regulators (VIII) Implementing the voltagesource VS(IV)
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+
-
V1
+
-
VO
High-efficiency post-regulators
+- VS
Small
DC/DC
converter
+
-
V1
+
-
VO
High-efficiencypost-regulators
+- VS
Small
DC/DCconverterV2
Option #1: connect the input
port to an additional output of
the Resistor Emulator
Option #2: connect the input
port to the Resistor
Emulator output
Two-Input Buck (TIBuck) Series-Switching post-Regulator (SSPR)
High-efficiency post-regulators (IX) Why is the efficiency of these post-regulators very high?
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Time
v1 vO
vS
V1, VO >> VS
P1, PO >> PS
+
-
VO
DC/DCconverter
Lowpass
filter
+
-
V1
High-efficiencypost-regulators
+- VS
Small
DC/DCconverterIO
The Small DC/DC converter is processing
only a small part of the output powerLow losses in the post-regulator
High efficiencypost-regulator
High-efficiency post-regulators (X) Why is not possible to implement aover-load or short-circuit protectionin these post-regulators?
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then VO=V1 0
+
-VO
DC/DCconverter
Lowpassfilter
+
-V1
High-efficiencypost-regulators
+- VS
SmallDC/DC
converterIO
The over-load or short-circuitprotection must be implemented
in the Resistor Emulator
0
A over-load occurs
If VS = 0,
High-efficiency post-regulators (XI) Introducing the Two-Input Buck(TIBuck)
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This is a Buck converter withtwo inputs instead of one
+
-
V1
+
-
VO
High-efficiencypost-regulators
+- VS
Small
DC/DC
converterV2
+-
V1
V2
+
-
VO
DC/DC converter
High-efficiencypost-regulators
VS+
-
High-efficiency post-regulators (XII) Single-output Resistor Emulatorbased on a Flyback + a TIBuckpost-regulator
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+
-VO+
-
V1
V2
Standard
controller
ResistorEmulatorcontroller
p g
TIBuckpost-regulators
High-efficiency post-regulators (XIII) Multiple-output Resistor Emulatorbased on a Flyback + a TIBuckpost-regulator
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p g
+
-VO+
- V1
V2
Standard
controller
ResistorEmulatorcontroller
+
-V3
-
+V
4
TIBuckpost-regulators
V V+ V -
High-efficiency post-regulators (XIV) Comparing Buck and TIBuckconverters
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V1 > VO
VQMAX = V1
VDMAX = V1
VO = V1d(d is the duty cycle)
V2 > VO > V1
VQMAX= V2-V1
VDMAX= V2-V1
VO= V2d + V1(1-d)
(from volts-second balance)
Buck
V1
VO
+ VQ -
+VD
-
TIBuck
V2
V1
VO
+ VQ -
+VD
-
High-efficiency post-regulators (XV) DC equivalent circuitfor the TIBuck
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VO= V2d + V1(1-d) =
Poorlyregulated
Controlled
(V2-V1)d + V1
(V2-V1)d
V1
VO
+PWM -
Regulated
High-efficiency post-regulators (XVI) Relationship between inputand output voltages (I)
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V2 range
Voltages
Time
VO
V2
V1
+VD
-
VO
+ VQ -
PWM +-
V1 range
ALWAYS
V2 > VO > V1
High-efficiency post-regulators (XVII) Relationship between inputand output voltages (II)
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+
-VO+
-V1
V2
VO
Voltages
Time
v2
v1
Transient responseSteady state
ALWAYS V2 > VO > V1, taking into account theworse case of transient response and ripple
Case of being used as post-regulator of a Resistor Emulator
High-efficiency post-regulators (XVIII) Comparing filter inductance forBuck and TIBuck converters (I)
LB
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132
Buck
V1
VO+VFilter
-
LB
TIBuck
V2
V1
VO
LTB
VFilter
-
+
Time
VFilter
VOV2
V1
VFilter
Time
VOV1
Lower value in the case ofthe TIBuck converter (inpractice, 3 times lower)
High-efficiency post-regulators (XIX) Comparing filter inductance forBuck and TIBuck converters (II)
Boundary between continuous and discontinuous conduction modes
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CCM: 2L/RT > KCRIT
DCM: 2L/RT < KCRIT
Boundary between continuous and discontinuous conduction modes
TIBuckV2 /V1=4
KCRIT
d(duty cycle)
Buck
1
00 1
1.25
2
3
)d1(CRITKBuck:
1)11V/2V(d
)11V/2V)(d1(d
CRITKTIBuck:
Lower value in the caseof the TIBuck converter
High-efficiency post-regulators (XX) Explaining the high efficiency ofthe TIBuck converter (I)
Realistic case for a Buck converter:
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d = 0.55PLosses = 10 W
= 90 / 100 = 90%
VOB = 50 VIO = 1.8 APOB = 90 W
Realistic case for a Buck converter:
VG = 100 VIG = 1 APG = 100 W
R= VOB/IO = 27.8
VG
VOB
Buck
R
d = 0 55
High-efficiency post-regulators (XXI) Explaining the high efficiency ofthe TIBuck converter (II)
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VG
VOB
Buck
VOB = 50 VIO = 1.8 APOB = 90 W
VG = 100 VIG = 1 APG = 100 W
d = 0.55PLosses = 10 W
IO = 1.8 A
IO = 1.8 A
V1 = 300 VP1 = 540 W
We are processing 540 W FREE !!
= (90 + 540) / (100 + 540) = 98.4 %
IO = 1.8 A R1 = V1/IO = 166.7
V1 = 300 V
P1 = 540 W
V1 IOIO
R1
High-efficiency post-regulators (XXII) Explaining the high efficiency ofthe TIBuck converter (III)
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VOB = 50 VIO = 1.8 A
POB = 90 W
IO = 1.8 AV1 = 300 VP1 = 540 W
IO = 1.8 AV1 = 300 VP1 = 540 W
VG = 100 VIG = 1 A
PG = 100 W 27.8
166.7
d = 0.5
I2
I1
d = 0.5
V1
V2
194.5IO = 1.8 AVO = VOB+ V1= 350 VP
O= P
OB+ P
1= 630 W
= 630 / 640 = 98.4%
I2 = IG = 1 AI1 = IO- IG = 0.8 AV2 = VG+ V1= 400 VP2 = 400 W
P1 = 240 WPi = P2 + P1 =640
High-efficiency post-regulators (XXIII) Explaining the high efficiency ofthe TIBuck converter (IV)
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The closer V2 and V1 (and,therefore VO) the higherthe efficiency
V2
V1
VO
V1
V2 V1
VO-V1
V1
V1
V2 V1
VO-V1
V1
V1
V2 V1
VO-V1
V1
High-efficiency post-regulators (XXIV) Explaining the high efficiency ofthe TIBuck converter (V)
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TBis the TIBuck efficiency
V1
V2 V1
VO-V
1
V1TB=1
B =50%
75%
90%
85%
0.4 0.6 0.8 1
100
80
60
TIbuck efficiency
V1/VO
O
1B
BTB
V
V)(1-1-
Bis the Buck-part efficiency
High efficiency TIbuckwith a limited efficiency
in the Buck part
TB=96.6%
High-efficiency post-regulators (XXV) Small-signal transfer functions ofthe TIBuck converter
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Similar to the case of a Buck converter, but faster dueto the lower values of the output filter components
V2
V1
VO
RCTB
LTB
Output filter
1sR
LsLC TB2TBTB
1OvD
V2-V1
1-D
++
+
2v
1v
d
The quantities with hats
are the perturbations
High-efficiency post-regulators (XXVI) Implementing the transistor driver
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Requirements:
Galvanic isolation
Wide duty cycle operation
High-efficiency post-regulators (XXVII) Experimental results ofTIBuck-based prototypes (I)
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TIBuck DC/DC post-regulators
V2 V1 VO IO LTB CTB fS
TIBuck 1 440-400 V 360-320 V 380 V 1-0.1 A 1 mH 250 nF 100 kHz
TIBuck 2 67-57 V 52-42 V 54.5 V 4-0.4 A 51.4 mH 4.7 mF 100 kHz
High-efficiency post-regulators (XXVIII) Experimental results ofTIBuck-based prototypes (II)
TIBuck 1: overall efficiency
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V2=400V
V1=320V
V2=440VV1=360V
V2=420VV1=340V
100 200 300Output power [W]
100
99
98
Efficiency [%]
High-efficiency post-regulators (XXIX) Experimental results ofTIBuck-based prototypes (III)
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100
95
90
85
Efficiency [%]
0 500 1000Output current [mA]
V2 = 80 VV1 = 0 VVO = 40 V
V2 = 180 VV1 = 100 V
VO = 140 V
280,200240
380,300,340
V1
V2VO
TIBuck 1 efficiency with V2 & V1 variable, V2-V1 =80 V, VO=(V1+V2)/2
Being V2-V1 a constant, the
closer V2 and V1 (and, thereforeVO) the higher the efficiency
High-efficiency post-regulators (XXX) Experimental results ofTIBuck-based prototypes (IV)
TIBuck 1 and Buck-part efficiencies
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The experimental results fit very well with the calculated ones
TIBuck(calculated)
TIBuck(measured)
Output current [mA)]
Efficiency [%]100
90
80
70200 400 600
Buck-part(measured)
TBis the TIBuck efficiencyV1
V2 V1
VO-V1
V1
=1
Bis the Buck-part efficiency
O1B
BTB
V
V)(1-1-
High-efficiency post-regulators (XXXI) Experimental results ofTIBuck-based prototypes (V)
TIBuck 2: overall efficiency and small-signal modelling
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100 1,000 10,000
0
-20
20
40
Gain [dB] , phase
Frequency [Hz]
-180
-90
0
Measured
0 100 200
100
96
92
Output power [W]
Efficiency [%]
y g g
Theoretical
Theoretical
51 4 H25CPF40
High-efficiency post-regulators (XXXII) Experimental results ofTIBuck-based prototypes (VI)
Resistor Emulator based on a
Flyback converter + TIBuck 2
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1.5mF
1.5mF
51.4 H
4.7 F
IRF7403
10T045 54V4A85-264V
IRFPC50
25CPF40
24 t
9 t
3 t
UC3854
UC3825
80 120 160 200Output power [W]
Overall efficiency [%]
84
86
88220V
110V0.01A/ s
2.1A 3.23A
IOV2 5V/div
VO 1V/div
Transient response
200 ms/div
High-efficiency post-regulators (XXXIII) Experimental results ofTIBuck-based prototypes (VII)
Voltage ripple cancellation in the case of the Resistor
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g pp
Emulator based on a Flyback converter + TIBuck 2
+
-
VOV1
V251.4 H
4.7 F
IRF7403
10T045
UC3825
V2 2V/div.
VO 0.5V/div.
V1 2V/div.
Voltage ripples
Can we improve theripple cancellation?
Voltage-Mode control
High-efficiency post-regulators (XXXIV) Experimental results ofTIBuck-based prototypes (VIII)
Other TIBuck control methods to improve the voltage ripple cancellation
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+
-
VOV1
V2 51.4 H
4.7 F
IRF7403
10T045
UC3825
Input voltage feedforward
Other TIBuck control methods to improve the voltage ripple cancellation
Feedforward
Input voltage feedforward
Current mode control (average current mode control)
+ V2+ V1
R2
R1 CextRext
+ Vdc
vC UC3825Feedforward
implementation
High-efficiency post-regulators (XXXV) Experimental results ofTIBuck-based prototypes (IX)
Average current mode control Voltage ripples
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Average current mode control
+
-
VOV1
V2 51.4 H
4.7 F
IRF7403
10T045
UC3825TL082
VO 2mV/div.
V2 2V/div.
V1 2V/div.
Voltage ripple attenuation 66dB (1900 times). Also,
excellent transient response
V2 (5 V/div)
VO (20 mV/div)
100 ms/div
Transient response
0.01A/ s
2.1A 3.23A
IO
High-efficiency post-regulators (XXXVI) Introducing the option #2: Series-Switching Post-Regulator (SSPR)
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+
-
V1
+
-
VO
High-efficiency post-regulators
+- VS
Small
DC/DC
converter
Option #2: connect the input
port to the Resistor
Emulator output
Series-Switching post-Regulator (SSPR)
+
-
V1
+
-
VO
+- VS
Small DC/DCconverter
+
-
V1 +
-
VOSmall DC/DC
converter
+
-V
S
Re-drawing
High-efficiency post-regulators (XXXVII) Introducing the SSPR basedon a Forward converter
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+
-
V1 +
-
VOSmall DC/DC
converter
+
-VS
Re-placing
the capacitor
+
-
V1+
-VO
Small DC/DC
converter
+
-
V1 +
-
VOSmall DC/DC
converter
+
-VS
Controller
The controlled outputvoltage is VO instead of VS
High-efficiency post-regulators (XXXVIII) Other SSPR implementations
Implementation based on a Flyback
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+
-V1 +
-
VOSmall DC/DC
converter
Controller
The implementation based on a Flyback
becomes a Boost converter if n1=n2
A Boost converter has a very high
efficiency if the input and output voltages
are very close
+
-
V1
+
-
VO
+- VS
Small DC/DCconverter
+
-
V1 +
-
VO
Controller
n1 n2
If n1=n2
High-efficiency post-regulators (XXXIX) Single-output ResistorEmulator based on a Flyback+ a Forward-type SSPR
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+
-V
O
+
-
V1
Standardcontroller
ResistorEmulatorcontroller
Forward-type SSPR
High-efficiency post-regulators (XL) Multiple-output ResistorEmulator based on a Flyback+ a Forward-type SSPR
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+
-VO
+
-
V1
Standardcontroller
ResistorEmulatorcontroller
+
-V2
-
+V3
Forward-type SSPR
High-efficiency post-regulators (XLI) Computing SSPRs efficiency (I)
VS +-I1 IOIO
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VO+
-
V1+
-
SSPRss
VO = V1 + VS
I1
= IiDC
+ IO
C=VSIO
V1IiDC
Being KS=VS/V1
Small DC/DCconverter
C
PWM +-
IiDC
SS=
VOIO
V1I1 =
1+KS
1+KS
C
High-efficiency post-regulators (XLII) Computing SSPRs efficiency (II)
Example:100
SS [%]
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C = 80%
KS = VS/V1 = 0.1 ss = 97.7%
The lower KS, the higher the efficiency
However, VS must reaches VSmaxand must be always positive
Voltages vO
Time
Steadystate
Transient
response
vSmax
v1
vS
KS=0.3
0.10.2
60 70 80 90 10080
85
90
95
KS=VS/V1
c [%]
11DQ10
High-efficiency post-regulators (XLIII) Experimental results of aForward-type SSPR (I)
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11DQ10E20
12CTQ045
47 H, E20
47 FSMP20N20
6,800 F28 : 28 : 15
fS = 100 kHz
V1 = 47V
+
-
VO= 54.5 VIO = 4 A
+
-
VS = 7.5V+
-
0 50 100 150 200 25095
96
97
98
Output power [W]
Efficiency [%]
High-efficiency post-regulators (XLIV) Experimental results of aForward-type SSPR (II)
Average current mode control
V (1V/di )
Voltage ripples
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V1
+
-VO+
-
UC3825TL082
V1 (1V/div)
VO (10mV/div)
10 ms/div
V1 (5V/div)
VO (20mV/div)
200 ms/div
Transient response
0.01A/ s1.83A
3.67A
IO
(Attenuation 50dB)
Conclusions of the use of High-efficiency post-regulators toimprove the transient responseof Resistors Emulators (I)
High-efficiency post-regulators (XLV)
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of Resistors Emulators (I)
Low additional cost and size
Low output voltage ripple and fast dynamic response
Very high post-regulator efficiency (96-98%)
Very low harmonic content
Can be used in high and low power applications
Can be used with universal line voltage
Very interesting for multiple-output applications withdifferent transient response specifications
Conclusions of the use of High-efficiency post-regulators toimprove the transient responseof Resistors Emulators (II)
High-efficiency post-regulators (XLVI)
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of Resistors Emulators (II)
V1 and VO are voltages of similar values
It is not a good solution for low output voltage applicationsbecause the energy is stored near the output voltage
No short-circuit and/or overload protection can beimplemented in the post-regulator (it must be implemented in
the Resistor Emulator)
However, short-circuit overcurrent from the bulk capacitorcan be diverted through an additional diode (see next slide)
High-efficiency post-regulators (XLVII) Additional diode to divert short-circuit overcurrent from the bulkcapacitor
Small
DC/DC
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+
-
VO+
-V1
V2
CB1
CB2
V1
+
-VO+
-
+
-
V1
+
-
VO
High-efficiencypost-regulators
+- VS
DC/DC
converter
overcurrent
The overcurrent i