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8/13/2019 Isolated Bidirectional Converter
1/66
Unified Analysis of IsolatedBidirectional Converters
for Battery Charging
Simone Buso, Giorgio Spiazzi
University of Padova - DEI
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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)
Simple phase-shift modulation
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)
Simple phase-shift modulation
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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Plus Phase-Shift Modulation
(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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Plus Phase-Shift Modulation
(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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Plus Phase-Shift Modulation
(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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Plus Phase-Shift Modulation
(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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Plus Phase-Shift Modulation
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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Plus Phase-Shift Modulation
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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Plus Phase-Shift Modulation
(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.:
Systematic Steady-State Analysis
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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Systematic Steady-State Analysis
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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Soft-switching conditions
( )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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Soft-switching conditions
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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Soft-switching conditions
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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Soft-switching conditions
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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Soft-switching conditions
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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Soft-switching conditions
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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Soft-switching conditions
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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Soft-switching conditions
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)
Soft switching conditions
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Soft-switching conditions
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
Soft switching conditions
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Soft-switching conditions
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
Soft switching conditions
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Soft-switching conditions
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 =
Soft-switching conditions
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Soft-switching conditions
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
Soft-switching conditions
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Soft switching conditions
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
Soft-switching conditions
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Soft switching conditions
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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Soft-switching conditions
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Soft switching conditions
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
Normalized Transferred Power
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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
Normalized Transferred Power
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4k2max
=
=
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
Single-phase DAB
LiL
vA vB
max
Power flow
Normalized Transferred Power
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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
NORMALIZED TRANSFERRED POWER
NORMALIZED PHASE-SHIFT ANGLE
( )
=
=
1
f2cos
f2 2coskf2
P)(P)(
n
nn
N
Power flow
Normalized Transferred Power
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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
NORMALIZED TRANSFERRED POWER
NORMALIZED PHASE-SHIFT ANGLE
( )
=
=
1
f2cos
1kf2
2
n
n
max
max
Single-phase SR-DAB LiL
vA vB
C
vC
Power flow
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Normalized Transferred Power
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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
NORMALIZED TRANSFERRED POWER
NORMALIZED PHASE-SHIFT ANGLE
Dtest,( )
( ) ( )
=
221
-D for 21
Dk1k
21
-D0 for D1k22
LiL
vA vB
IBCI
Power flow
Normalized Transferred Power
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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
NORMALIZED PHASE-SHIFT ANGLE
( )
Power flow
Normalized Transferred Power
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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
NORMALIZED TRANSFERRED POWER
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