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Zero Voltage Switching Quasi-resonant

Converters

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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

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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

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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(

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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

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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

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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

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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

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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

=>

=>