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EE4211 1
Zero Voltage Switching Quasi-resonant
Converters
EE4211 2
IntroductionZero-current switching is not true soft-switchingIt does have loss in the junction capacitorFor example, a switching device is operating under device voltage Vsw 300V DC and the switching frequency is 1MHz. The junction capacitance Cj is 500pF. The loss is then:
The loss is very highAn alternative method to cure this problem is to use zero-voltage switching.
W5.22W10300105002
1
2
1Loss 62122
sw sswsw fVC
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Resonant switches
Zero –current switch Zero-voltage switch
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Configurations of the zero-voltage switch
general half-wave full-wave
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Zero voltage switching Quasi-resonant (ZVS-QR)
Buck converter LF and CF, are large The analysis can be simplified by assuming the current through the bulk inductor is a current sink.
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4 stages of operation
Capacitor charging stages:(Fig. 4a)Resonant state: (Fig. 4b)Inductor recovering stage (Fig.4c)Free-wheeling stage: (Fig 4d)
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Equivalent circuit
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Capacitor charging stages
Switch SW is turned off at t0. Input current iLr rises linearly and is governed by the state equations
Solution:o
Cr I
dt
dVC
r
oCr C
ttIv
)( 0
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Half-waveUni-directional of resonant voltage on Vc
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Full-waveBi-directional voltage on Vc
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Boundary condition and Duration
The stage will finish when vCr increases to Vin, the voltage across DF becomes positive and it is in forward bias. The duration of this state Td1 =Cr
Vin / IoBoundary condition: vCr =Vin
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Resonant state:
Lrc
r
CrinrL
r
idt
dvC
vVdt
diL
)(cos
)(sin
1
1
ttIi
ttZIVv
ooLr
ooinCr
rro
CL
1
r
r
C
LZ ImpedanceResonant
frequency
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Duration and Boundary
Finish when the resonant capacitor voltage vCr reaches zero.
The duration of this state Td2 is
o
in-
o ZIV-
= 11 sinsinwhere
fullwave
halfwave
2<<2
3
2
3<<
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Termination of resonant stage
The termination of this state is similar to the zero-current switching.
For the half-wave mode, when resonant voltage vCr reaches zero, it cannot be reversed because the anti-parallel diode D of the switch conducts. The transistor T of the switch SW can be turned on after that to achieve zero-voltage switching.
For full-wave mode, vCr can reverse into negative because there is a series diode D with the transistor T. During vCr is negative, the transistor T should be turned off. When vCr resonates back from negative to zero, since T is already off, the resonance stops.
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Voltage during the end of stage
At t2, iLr = Iocos
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Inductor recovering stage: Resonance stops, Lr begins to be charged by the input voltage Vin
The solution:
V=dtid
L inLr
r
r
inoLr L
ttVIi
)(cos 2
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Finish of this stageFinish when iLr reaches the value of output current IO. DF no longer conducts because its current is now all conducted by Lr.
Duration: Td3 = Lr Io cos / Vin
Boundary condition: iLr=Io
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Free-wheeling stage: (Fig 4d)
Output current freewheels through Lr and switch SW. This stage finishes when the transistor turns off again at t4. t4 is the same as t0 in next cycle.
Duration: Td4= Ts-Td1-Td2-Td3
where Ts is the period of the switching
cycle.
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Zero-voltage switching condition
The AC component of the resonant voltage is required to be greater than its DC level in order to achieve the ZVS. In this case, the condition is:
This condition is the same as (12):
ino VZI
o
in-
o ZIV-
= 11 sinsinwhere
EE4211 20
Zero-voltage switching When to turn on and off
The instant to turn on the switch is also important. For half-wave mode, the transistor must be turned on after vCr reaches zero at t2 and before isw increases back to positive.
For full-wave mode, the transistor must be turned on when vCr is negative. When T is turned on, the series diode D is still in reverse bias and the voltage across T is virtually zero.
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Why full-wave is no goodThe charge stored in the junction capacitor of T cannot be conducted to outside, therefore when T is turned on this energy is dissipated internally. Therefore the switching loss is not zero. The current iSW through either T or D during t2-t4 is annotated in Fig.5.
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DC voltage conversion The output voltage Vo can be solved by equating the input energy Ein and output energy Eo
dtLrdtLrdtLr+dtLrV=E iiii
t
t
t
t
t
t
t
tinin
4
3
3
2
2
1
1
0
TIV=E sooo
)cos1(1
1
-++2f2
f=
V
V
o
s
in
oM
R
V
VZ
R
ZI
V n
in
oo
in
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DC voltage conversion(Half-wave)
Still varies between 0 and 1 which is a generally characteristic of Buck converters.The ratio now decreases with fs/fo increasing Depends on the load resistance R
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DC voltage conversion(Half-wave)
Still varies between 0 and 1 which is a generally characteristic of Buck converters.The ratio now decreases with fs/fo increasing Independent of the load resistance R
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Zero voltage switching Quasi-resonant Boost
converter A steady-state equivalent circuit using a current source Iin which represents the input current.
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Equivalent circuit
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4 stages of operation
Capacitor charging stages:(Fig. 8a)Resonant state: (Fig. 8b)Inductor recovering stage (Fig.8c)Free-wheeling stage: (Fig 8d)
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Half-waveUni-directional of VCrSimilar to ZCS but the shapes of iLr and Vcr are swapped.
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Full-waveBi-directional of VCr
Similar to ZCS but the shapes of iLr and Vcr are swapped.
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Capacitor charging stage Resonant capacitor voltage vCr rises linearly due to the input current and is governed by the state equations:
Solution:
inCr
r Idt
dVC
r
inCr C
ttIv
)( 0
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Termination of stage 1When vCr increases to the value of Vo, DF becomes forward biased.
The duration of this state: Td1 =Cr Vo / Iin
Boundary condition: vCr =Vo
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Resonant state Lr and Cr resonate and DF is on
The solution is:
The definition of o and Z is the same
LrinCr
r
oCrLr
r
iIdt
dvC
Vvdt
diL
))(cos1(
)(sin
1
1
ttIi
ttZIVv
oinLr
oinoCr
EE4211 33
Termination of this stageThis stage finishes when the resonant capacitor voltage reaches zero. For half-wave mode, the anti-parallel diode D of the switch SW conducts. The resonant voltage across SW cannot be reversed. For full-wave mode, the transistor T of the SW should be turned off when the resonant voltage across the switch is negative. At this time, the series diode D is in reverse bias and T can then be turned off under zero-voltage switching.
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Boundary
The duration of this state Td2 is
Boundary condition: vCr =0
IZV-
= in
o-
o
11 sinsinwhere
fullwave
halfwave
2<<2
3
2
3<<
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The current left in LrAt t2, iLr = Iin (1-cos)
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Inductor recovering stage Inductor recovering stage (Fig. 8c): Resonance stops and Lr begins to be discharged through the output voltage.
The solution is:
V-=dtid
L oLr
r
r
oinLr L
tVIi )cos1(
EE4211 37
Boundary of Inductor recovering stage
Duration:
Boundary condition: iLr=0
Td3 = Lr Iin (1-cos ) / Vo
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Free-wheeling stage Output current freewheels through switch SW. This stage finishes when the transistor turns off again at t4. The t4 is the same as t0 in the next cycle.Duration: Td4= Ts-Td1-Td2-Td3
EE4211 39
DC voltage conversion The output voltage Vo can be solved by equating the input energy Ein and output energy Eo.
The output current Io for the derivation of the Eo can be obtained from the current of the diode DF that is the same as the iLr
TIV=E
dti+dtiV=E
sininin
Lr
t
tLr
t
too
3
2
2
1
)cos1(1
1
-++2f2
f=
V
V
o
sin
o
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Voltage conversion ratio (half-wave)
Depends on LoadMinimum is 1Maximum approaches to infinity
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Voltage conversion ratio (full-wave)
Relatively independent of LoadLooks like the classical Boost converter characteristics.
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Other converter (Buck-Boost)
Can be formed similarly as the other two convertersAlso inverted output voltage
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Equivalent circuit
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Voltage conversion ratio
Half-wave Full-wave
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Correct Full-wave for Buck-Boost
The previous is correct because:Does not cover Vo/Vin =0 when fs/fo is close to oneWhen fs/fo is 0.2, the Vo/Vin is too large
0
1
2
3
4
5
0 0.2 0.4 0.6 0.8 1
fs / fo
Vo
/ V
in
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Duration of each stage in ZVS
o
o
ocos1
Stage Duration
Td1
Td2
Td3
Td4 Ts-Td1-Td2-Td3
EE4211 47
Comparison of Vo/Vin of the classical and
QRC
g1
1g
1D1
1
g
g
1
g
g1
D
D
1 g
g
1
g
g1
Classical ZCS ZVS
Buck D g 1-g
Boost
Buck-Boost
Cuk
D
D
1
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Summary of the quasi-resonant ZVS converters
Advantage Disadvantage
True soft-switching High voltage rating is needed
No power loss during switching
ZVS Depends on load condition
Not good for very high voltage input
EE4211 49
Practical implementation of ZVS quasi-resonant
converters
Buck Boost
Buck-Boost Cuk
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TutorialA ZVS Quasi-resonant Buck converter is
operated under the following condition:1) Vin=50V2) Vo=25V3) fs=1MHz4) Io=5AUsing the condition of the converter just
able to achieve ZVS, suggest the Lr and Cr that is needed. Assume the junction capacitance of the transistor is 100pF.
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AnswerFull-wave cannot be used because of the junction capacitor of the transistor, therefore Half-wave is used
Therefore: for just ZVS, Vin=ZIo => 50=Z*5 =>Z=10
Hence, you can obtain fo=1.976MHz
)cos1(1
1
-++2f2
f=
V
V
o
s
in
o
o
in-
o ZIV-
= 11 sinsinwhere
foe-++2f2
e=
o/6988.015.0))2/3cos(1(
1
12/3
1611
50
25
EE4211 52
Answer (Cont)
After calculated Cr, the actual required Cr is to be deduced by the junction capacitance of the transistor.
10r
r
C
LZ
orr
o fCL
21
Hff
ZL
nFeZf
=C
oor
or
805.0*2
10
2
05.810*6976.1**2
1
2
1
Actual Cr=8.05nF-100pF
=7.95nF
=>
=>
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