isolated bidirectional converter

8/13/2019 Isolated Bidirectional Converter

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Unified Analysis of IsolatedBidirectional Converters

for Battery Charging

Simone Buso, Giorgio Spiazzi

University of Padova - DEI


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S. Buso, G. Spiazzi - University of Padova - DEI 2/66

Outline

Introduction

Review of Considered Converter TopologiesSingle-Phase Dual Active Bridge (DAB)

Single-Phase Series Resonant DAB (SR-DAB)

Three-Phase Dual Active Bridge (DAB)Three-Phase Series Resonant DAB (SR-DAB)

Interleaved Boost with Coupled Inductor (IBCI)

Unified AnalysisSoft-switching conditions

Transferred Power Calculation


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Introduction

+

+ + + +

- - - -

V IN

Z IN

V 1 V 2 V OUT V BATT

Z OUT I OUT

ISOLATED DC-DC(STAGE #1)

CURRENT SOURCE(STAGE #2)

DC-LINK

Focus on the isolated stage

(V 1 = high voltage port, V 2 = low voltage port)


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Isolated Bidirectional Topologies

topologies with reduced switch count

Flyback

Cuk

Sepic

.

topologies with dual bridge (or half-bridge or push-pull)

configurationDual Active Bridge (DAB)

Interleaved Boost with Coupled Inductors (IBCI)

.

topologies with dual bridge configuration and highfrequency resonant networks

Series Resonant DAB (SR-DAB)

.


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ug

+

uo

+

-S 2S 1

L1

i1 i2

Ro

+u1

Co

L2C1

u2

C2+

Lm

Ld

1:n1:n

Reduced Switch Count Topologies

High voltage and/or current stress on active components

Limited soft-switching operation

Transformer leakage inductance requires suitable snubbercircuits

Example: bidirectional Cuk converter

Switch overvoltage!


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Dual Active Bridge (DAB)

Simple phase-shift modulation

Extended soft-switching operationExploitation of transformer leakage inductance

Optimum design for constant port voltages V 1 and V 2

Single-phase:

V2V1

i2

n:1

S 1aL S 1bL

Li1S 1aH S 1bH

S 2aL S 2bL

S2aH

SS2bHiL


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Dual Active Bridge (DAB)


Extended soft-switching operationExploitation of transformer leakage inductance

Optimum design for constant port voltages V 1 and V 2

Reduced input and output current ripples

Three-phase:

V2V1

i2

n:1S 1aL S 1bL

i1iLa

S 1cL

iLb

iLc

va1

vb1

vc1

va2

vb2

vc2

S 1aH S 1bH S 1cH

S 2aL S 2bL S 2cL

S 2aH S 2bH S 2cH

o1 o 2

L

L

L


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Series Resonant Dual Active Bridge(SR-DAB)

Same characteristics as single-phase DAB

Higher degree of freedom (two parameters: L and C)Inherent protection against transformer saturation (withC split between primary and secondary)

Single-phase:

vC V2V1

i2

n:1

S 1aL S 1bL

Li1

CS

1aH S 1bH iL

S 2aL S 2bL

S 2aH S 2bH


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Series Resonant Dual Active Bridge(SR-DAB)

Same characteristics as three-phase DAB

Higher degree of freedom (two parameters: L and C)Inherent protection against transformer saturation (withC split between primary and secondary)

Three-phase:

V2V1

i2

n:1S 1aL S 1bL

i1LiLa

S 1cL

iLb

iLc

va1

vb1

vc1

va2

vb2

vc2

S 1aH S 1bH S 1cH

S 2aL S 2bL S 2cL

S 2aH S 2bH S 2cH

o1 o 2

L

L

C

C

C


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Interleaved Boost with CoupledInductors (IBCI)


Extended soft-switching operation

Exploitation of mutual inductor leakage inductance

Duty-cycle control of port 2 switches for variable portvoltages V 1 and V 2Reduced port 2 current ripple (low-voltage high-current

port)

S 1aH

L

S 2aL

V1

S 2bL

V2

iL

ib

i1

S 1aL

np

npiaS 2aH S 2bH

Lm ima

Lm imb

nsnsi2

VCL

CCL C CLVCL

S 1bH

S 1bL


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Unified Analysis

All the aforementioned topologies control the powertransfer between the two ports by modulating the voltageapplied to a current shaping impedance

ZiL

vA vB

For DAB and IBCI topologies, Z is a simple inductor whilefor SR-DAB topologies Z is a series resonant L-C tank


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Phase-Shift Modulation

LiL

vA vB

2

2f sw t

2f sw t

2f sw t

vA

vB

iL

2

2

iL(0)

iL() iL() = - i L(0)

V A

-V A

V B

-V B

Single-phase DAB

VA > V B

Example: power transfer from v A to v B


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Plus Phase-Shift Modulation

LiL

vA vB

(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

S 2aL S 2aH

S 2bH S 2bL

S1aH

= S1bL

S 1aL = S 1bH

1 2 30

2(1-D)

4 5 6 2

IBCI converter

(D-1/2) < < /2

Case B:

0 < < (D-1/2)

Case A:

D > 0.5


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(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

1 2 30

2(1-D)

4 5 6 2

IBCI converter

0 1

L

S 2aL

V1

V2

iL

ib

i1

S 1aL

np

npia

Lm ima

Lm imb

ns n si2

S 1bH

S 2bL

0v A =

1B Vv =


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(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

1 2 30

2(1-D)

4 5 6 2

IBCI converter

1 2

L

S2aL

V1

V2

iL

ib

i1

S 1aL

np

npiaS 2bH

Lm ima

Lm imb

ns n si2

CCLVCL

S 1bH

( )

ACLp

s

CLggp

sA

VVn

n

VVVnnv

==

+=

1B Vv =


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(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

1 2 30

2(1-D)

4 5 6 2

IBCI converter

2 3

S 1aH

L

S 2aL

V1

V2

iL

ib

i1

S 1aL

np

npiaS 2bH

Lm ima

Lm imb

ns n si2

CCLVCL

S 1bH

S 1bL

( )

ACLp

s

CLggp

sA

VVn

n

VVVnnv

==

+=

1B Vv =


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(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

1 2 30

2(1-D)

4 5 6 2

IBCI converter

3

S 1aH

L

S 2aL

V1

S 2bL

V2

iL

ib

i1

S 1aL

np

npia

Lm ima

Lm imb

ns n si2

S 1bH

S 1bL

0v A =

1B Vv =


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v A is a three-level voltage of amplitude V A while v B is asquare wave voltage of amplitude V B

The duty-cycle of port 2 switches is controlled so as toobtain the condition V A=V BThe power transfer between ports 1 and 2 is controlledthrough the phase-shift angle

Inductor current has a piecewise linear behavior

0 /2 v A v BP

- /2 0 v A v BP

D1V

nn

Vnn

V 2p

sCL

p

sA

== 1B VV =

( )( )D1

VVD1VVDV gCLgCLg

==


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LiL

vA vB

(D-1/2)-

(D-1/2)+

VA

VB

I1

I2

1

2

3

0

2(1-D)

I4 = -I1

I0

4

5

6

2

vA()

vB()

iL()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

0 < < (D-1/2)

(VA

= VB)

IBCI converterD > 0.5

Case A:

Power flow


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(D-1/2) < < /2

1 2 30

2(1-D)

I2

I1

4 5 6

I0 I4 = -I1

I5 = -I2

(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

iL()

2

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

(VA

= VB)

D > 0.5

Case B:

LiL

vA vB

IBCI converter

Power flow


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Phase-Shift Modulation in Three-Phase Converters

V1 /3

2V1 /3

nV2 /32nV 2 /3

vao1

= vA

nvao2 = v B

va1

vb1

vc1

vo1 2V 1 /3V1 /3

/3-

0 /3 2/3

V1

< /3 < /3 < /3 < /3

Phase a voltages(Hp: symmetry)

V1

S 1aL S 1bL

i1iLa

S 1cL

iLb

iLc

va1

vb1

vc1

S 1aH S 1bH S 1cH

o1

L

L

L


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Phase-Shift Modulation in Three-Phase ConvertersDecoupled phase behavior (perfect symmetry)

v A and v B are six-step voltage waveforms

The power transfer between ports 1 and 2 is controlled

through the phase-shift angle

Two different situations (power from port 1 to port 2):

0 /3

/3 /2

Inductor current has a piecewise linear behavior (DAB)

0 /2 v A v BP

- /2 0 v A v BP


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Systematic Steady-State Analysis

The analytical determination of the current waveformsrequires a systematic method for complex topologies (e.g.three-phase resonant DAB).

The outcome is the mathematical expression of phasecurrents as a function of phase-shift and other designparameters.

In addition, soft switching conditions can be analyzed in

detail by extracting current values at the moment ofswitch commutations.

N

BA

N

Z

Vvv

Vv

==A

B

VV

k =

General case:Z

iL

vA vB


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The half switching period is divided into m subintervals. For eachsubinterval ( i = 1 m ), the values of the current shapingimpedance state variables x i at the end of the interval arecalculated, in normalized form, as a function of their value x i-1 at

the beginning, i.e.:


i1 += iiii NxMx

FxMNNMxMx ii +=+

+=

=+ 01,mmm

1m

1ii1,m01,mm

i j

j

ik k i, j = = MM

[ ]Tm21 = L

We can iterate, obtaining

where

and is a column vector containing m i elements, i.e.


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from which the initial state variable values are found:

001,mm xFxMx

=+=

( ) FMIx 11,m0 =

Exploiting the waveform symmetry, we can write:

that can be used to derive the current waveform expressionsand to discuss soft-switching conditions


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Example: IBCI

Base voltage: V N = V A

Base impedance: Z N = sw LBase current: I N = V N /Z NBase power: P N = V N 2 / Z N

Base variables:

The half switching period is subdivided into 4subintervals ( m = 4 ).

Two situations has to be considered:Case A : 0 < < (D-1/2)

Case B : (D-1/2) < < /2


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Example: IBCI

1 2 3 4

A k -k

21

D 1-k ( )D12 -k

21

D

B k

21

D 1+k

21

D 1-k

D23

-k

21

D

i=

(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

S 1aH = S 1bL

S 1aL = S 1bH

1 2 30

2(1-D)

4 5 6 2

Case B


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Example: IBCI

Defining j( ) = i( )/I N as the normalized inductorcurrent:

m,,1iforJJ ii1ii K=+=

i1 += iiii NxMxComparing with:

m,,1iforN

1M

ii

iK=

==


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Example: IBCI

From:

the normalized initial inductor current value is:

For both cases A and B we have:

=

=4

1iii0 2

1J

( )

+=

2kD1J 0

( ) FMIx 11,m0 =

For plus phase-shift modulation k = 1 :

=21

DJ 0


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Example: SR-DAB

Base voltage: V N = V A

Base impedance: Z N = Z rBase current: I N = V N /Z NBase power: P N = V N 2 / Z N

Base frequency: N = r

Base variables:

C

LZr =

LC

1r =

ZiL

vA vB


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Example: SR-DAB

Current shaping impedance state variables:

( )

( )

+

+

=

+

=

n

0L

n

0C

n

C

n0L

n0C

nL

f

sinJ

f

cosU

f

cos1u

fcosJ

fsinU

fsin j

=C

L

u j

xNormalized state variable vector:

C LiLuC


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Example: SR-DAB

10 2

VA

VB

vA()

vB()

2

The half switching period is subdivided into 2subintervals ( m = 2 ).

i = 1 i = 2 1+k 1-k


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Example: SR-DAB

Current shaping impedance state variables:

{

Matrix M

i

n

i

n

i

1Ci

1Li

n

i

n

i

n

i

n

i

Ci

Li

fcos1

fsinUJ

fcos

fsin

fsinfcosUJ

+

=

for i = 1,2

{

Matrix N

==

nn

nn1,21,m

fcos

fsin

fsinfcosMM


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Example: SR-DAB

k1

fcos

fcos2

fsin2

fsin

fcos1

fsin

nn

nn

n

n

+

=F

Fx

1

nn

nn0

fcos1

fsin

fsinfcos1

=

Matrix F:

Initial conditions:


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Example: SR-DAB

+

+

+

= k

fcos

fcos

fcos1

fsinfsin

0f

sin

fcos1

1

nnn

nnn

n

0x

Initial conditions:

( )

+

+

=

n

nnn0L

f

cos1

fsinfsinkfsinJ


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Example: Three-Phase DAB

V1 /3

2V1 /3

nV2 /32nV 2 /3

vao1 = v A

nvao2 = v B

/3-

0 /3 2/3

The half switching period is subdivided into 6subintervals ( m = 6 ).Two situations has to be considered:

Case A : 0 < < /3


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Example: Three Phase DAB


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Example: Three-Phase DABand SR-DAB

1 2 3 4 5 6

A3

k 1+ 3

k 1

3

3k 2

( )

3k 12

3

3

k 21

3

k 1

3

B 3k 21+

3

3

k 1+

3

2

3

k 2 +

3

3

k 2

3

2

3

k 1 3

3

k 21

3

2

i=

V1 /3

2V1 /3

nV2 /32nV 2 /3

vao1 = v A

nvao2 = v B

/3

0 /3 2/3

Case B


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Soft-switching conditions

Single phase DAB: port 1

1k1

2

( ) 0k12

k)0( jL

=

Power flow

2

2f sw

2f sw t

2f sw t

vA

vB

iL

2

2

iL(0)

iL() iL() = - i L(0)

VA

-V A

V B

-V B

V1

S 1aL S 1bL

Li1S 1aH S 1bH iL


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( )k12

( ) 0k12

)( jL

=

Power flow

2

2f sw

2f sw t

2f sw t

vA

vB

iL

2

2

iL(0)

iL() iL() = - i L(0)

VA

-V A

V B

-V B

V2

i2

n:1

iLs

S 2aL S 2bL

S 2aH S 2bH

Single phase DAB: port 2


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Single phase DAB

For a power flow from port 1 to port 2 the phase-shiftinterval is 0 /2. Thus , if k 1 the soft switchingcondition is satisfied for any value and for bothbridge switches.

1k12

The same consideration holds for a power flow fromport 2 to port 1 where voltages v A and v B are sweptand k = 1/k. Now, if k 1 ( k 1 ) the soft switchingcondition is satisfied for any value.

Port 1:

( )k12

Port 2:


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Single phase DAB

V1

S 1aL S 1bL

Li1S 1aH S 1bH iL

For a bidirectional power flow, if k = 1

the soft switching condition is satisfiedfor any value between /2 and /2.

1k12

Port 1:

( )k12

Port 2:

f h d


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Single phase SR-DAB converter: port 1

vC V2V1

i2

n:1

S1aL

S1bL

Li1 CS 1aH S 1bH iL

S2aL

S2bL

S 2aH S 2bH

0 0.333 0.667 14

3.2

2.4

1.6

0.8

0

0.8

1.6

2.4

3.2

4

NORMALIZED W LH

J L0()

J L1()

( ) 0

fcos1

fsinfsinkfsinJ

n

nnn0L <

+

+

=

Power flow

S f i hi di i


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Single phase SR-DAB converter: port 2

vC V2V1

i2

n:1

S1aL

S1bL

Li1 CS 1aH S 1bH iL

S2aL

S2bL

S 2aH S 2bH

0 0.333 0.667 14

3.2

2.4

1.6

0.8

0

0.8

1.6

2.4

3.2

4

NORMALIZED W LH

J L0()

J L1()

( ) 0k

fcos1

fsin

fcos1

fsin

fsin

J

n

n

n

nn1L >

+

+

+

=

Power flow

S f i hi di i


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Single phase SR-DAB converter

vC V2V1

i2

n:1

S1aL

S1bL

Li1 CS 1aH S 1bH iL

S2aL

S2bL

S 2aH S 2bH

( ) ( ) ( )1k

f cos1

f sin0J0J

n

n1L0L

+

==

0 0.333 0.667 14

3.2

2.4

1.6

0.8

0

0.8

1.6

2.4

3.2

4

NORMALIZED W LH

J L0()

J L1()

Worst case: = 0

K = 1

Power flow

S ft it hi diti


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Interleaved Boost with Coupled Inductors

S 1aH

L

S 2aL

V1

S 2bL

V2

iL

ib

i1

S 1aL

np

npiaS 2aH S 2bHL

m ima

Lm imb

n snsi2

VCL

CCL C CLVCL

S 1bH

S 1bL

Power flow

Lets analyze the port 1 switch commutations first

Soft switching conditions


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S 1aH

LV1

iL

i1

S 1aL

n s n s

S 1bH

S 1bL

D > 0.5

(D-1/2)-

(D-1/2)+

VA

VB

I1

I2

1 2 30

2(1-D)

I4 = -I1

I0

4 5 6

2

vA()

vB()

iL()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

( )D12k

( ) 02

kD1

kJJ 01

+=

+=

Case A : 0 < < (D-1/2)



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S 1aH

LV1

iL

i1

S 1aL

n s n s

S 1bH

S 1bL

1 2 30

2(1-D)

I2

I1

4 5 6

I0 I4 = -I1

I5 = -I2

(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

iL()

2

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

D > 0.5

( )

( ) 0k12

21Dk1JJ 12

=

++=

2

1k

Case B : (D-1/2) < < /2



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S 1aH

LV1

iL

i1

S 1aL

n s n s

S 1bH

S 1bL

21kCase B : (D-1/2) < < /2

( )D12k Case A : 0 < < (D-1/2)

Case B is included in case A!

Same condition holds for reversed power flow

k = 1 is used to minimize the inductorcurrent crest factor



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Interleaved Boost with Coupled Inductors

S 1aH

L

S 2aL

V1

S 2bL

V2

iL

ib

i1

S 1aL

np

npiaS 2aH S 2bHLm ima

Lm imb

n snsi2

VCL

CCL C CLVCL

S 1bH

S 1bL

Power flow

For port 2 switch commutations we have to analyze thecurrents i a and i b which depend also on duty-cycle:

Lp

smaa in

nii += L

p

smbb in

nii =



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IBCI

( ) 0Inn

Ii 2p

smpk2b =

D > 0.5

(D-1/2)-

(D-1/2)+

VA

VB

I1

I2

1 2 30

2(1-D)

I4 = -I1

I0

4 5 6

2

vA()

vB()

iL()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

Case A: 0 < < (D-1/2)

S 2aL S 2bL

V2 ibnp

np

iaS 2bH

Lm ima

Lm imb

i2

CCLVCL



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IBCI

( ) 0Inn

Ii 3p

smvl3b =

D > 0.5

(D-1/2)-

(D-1/2)+

VA

VB

I1

I2

1 2 30

2(1-D)

I4 = -I1

I0

4 5 6

2

vA(

)

vB()

iL()

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

Case A: 0 < < (D-1/2)

S 2aL S 2bL

V2 ibnp

np

iaS 2bH

Lm ima

Lm imb

i2

CCLVCL



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IBCI

( ) 0Inn

Ii 1p

smpk1b =

D > 0.5

1 2 30

2(1-D)

I2

I1

4 5 6

I0 I4 = -I1

I5 = -I2

(D-1/2)

(3/2-D)- (D-1/2)

VA

VB

vA()

vB()

iL()

2

S 2aL S 2aH

S 2bH S 2bL

S 1aH = S 1bL

S 1aL = S 1bH

Case B: (D-1/2) < < /2

S 2aL S 2bL

V2 ibnp

npiaS 2bH

Lm ima

Lm imb

i2

CCLVCL


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The active clamp operation requires the clamp current tohave zero average value. This means that the upperswitch current must reverse polarity during their

conduction interval ( help soft-switching )A non negligible magnetizing inductor ripple helps tosatisfy the soft-switching conditions especially at lowpower levels

Considerations:

S 1aH

L

S2aL

V1

S2bL

V2

iL

ib

i1

S 1aL

np

npia

S 2aH S 2bHLm ima

Lm imb

nsnsi2

VCL

CCL C CLVCL

S 1bH

S 1bL

Normalized Transferred Power


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( ) ( )

( )

=

=

0L

N

d j1P

P

LiL

vA vB

Single-phase DAB

2

2f sw

2f sw t

2f sw t

vA

vB

iL

2

2

iL(0)

iL() iL() = - i L(0)

VA

-V A

V B

-V B Power flow



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( )

= 1k

0 0.25 0.5 0.75 10

0.2

0.4

0.6

0.8

NORMALIZED TRANSFERRED POWER

NORMALIZED PHASE-SHIFT ANGLE

1

0

( )

10

A

B

VV

k =

Single-phase DAB

LiL

vA vB

Power flow



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4k2max

=

=

0 0.25 0.5 0.75 10

0.2

0.4

0.6

0.8



1

0

( )

10

Single-phase DAB

LiL

vA vB

max

Power flow



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Single-phase SR-DAB LiL

vA vB

C

vC

0 0.125 0.25 0.375 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1



( )

=

=

1

f2cos

f2 2coskf2

P)(P)(

n

nn

N

Power flow



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0 0.125 0.25 0.375 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1



( )

=

=

1

f2cos

1kf2

2

n

n

max

max

Single-phase SR-DAB LiL

vA vB

C

vC

Power flow


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6



Dtest,( )

( ) ( )

=

221

-D for 21

Dk1k

21

-D0 for D1k22

LiL

vA vB

IBCI

Power flow



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LiL

vA vB

Three-phase DAB

( )

=

2318

323

2

2

0 0.125 0.25 0.375 0.50

0.2

0.4

0.6

0.8NORMALIZED TRANSFERRED POWER


( )

Power flow



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Three-phase SR-DAB

0 0.083 0.167 0.25 0.333 0.417 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8


1 ( )

No closed formexpression for ()

LiL

vA vB

C

vC

Power flow

Conclusions


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Different isolated bidirectional topologies, belonging tothe family of dual active bridge structures, have beenconsidered

A unified analysis has been carried out to calculate thesteady-state current waveform responsible for the powertransfer

Soft-switching conditions have been investigated for eachconverter topology

The transferred power and its relation with the phase-shift angle has been calculated

isolated bidirectional converter

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