1. Zero Voltage Switching Quasi-resonant Converters
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2. Introduction Zero-current switching is not true soft-switching
It does have loss in the junction capacitor
For 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 high
An alternative method to cure this problem is to use zero-voltage switching.
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3.     Resonant switches
Zero –current switch Zero-voltage switch
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4. Configurations of the zero-voltage switch
general half-wave full-wave
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5. 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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6. 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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7. Equivalent circuit
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8. Capacitor charging stages
Switch SW is turned off at t0. Input current iLr rises linearly and is governed by the state equations
Solution:
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9. Half-wave
Uni-directional of resonant voltage on Vc
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10. Full-wave
Bi-directional voltage on Vc
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11. 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 / Io
Boundary condition: vCr =Vin
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12. Resonant state: Lr and Cr resonate and DF is on The solution is:
Resonant frequency
Impedance
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13. Duration and Boundary
Finish when the resonant capacitor voltage vCr reaches zero.
The duration of this state Td2 is
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14. 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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15. Voltage during the end of stage
At t2, iLr = Iocos
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16. Inductor recovering stage:
Resonance stops, Lr begins to be charged by the input voltage Vin
The solution:
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17. Finish of this stage
Finish 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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18. 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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19. 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):
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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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21. Why full-wave is no good
The 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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22. DC voltage conversion
The output voltage Vo can be solved by equating the input energy Ein and output energy Eo
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23. 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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24. 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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25. 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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26. Equivalent circuit
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27. 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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28. Half-wave Uni-directional of VCr
Similar to ZCS but the shapes of iLr and Vcr are swapped.
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29. Full-wave Bi-directional of VCr
Similar to ZCS but the shapes of iLr and Vcr are swapped.
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30. Capacitor charging stage
Resonant capacitor voltage vCr rises linearly due to the input current and is governed by the state equations:
Solution:
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31. Termination of stage 1
When 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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32. Resonant state Lr and Cr resonate and DF is on The solution is:
The definition of o and Z is the same
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33. Termination of this stage
This 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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34. Boundary The duration of this state Td2 is Boundary condition: vCr =0
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35. The current left in Lr
At t2, iLr = Iin (1-cos)
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36. Inductor recovering stage
 Inductor recovering stage (Fig. 8c): Resonance stops and Lr begins to be discharged through the output voltage.
The solution is:
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37. Boundary of Inductor recovering stage
Duration:
Boundary condition: iLr=0
Td3 = Lr Iin (1-cos ) / Vo
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38. 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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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
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40. Voltage conversion ratio (half-wave)
Depends on Load
Minimum is 1
Maximum approaches to infinity
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41. Voltage conversion ratio (full-wave)
Relatively independent of Load
Looks like the classical Boost converter characteristics.
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42. Other converter (Buck-Boost)
Can be formed similarly as the other two converters
Also inverted output voltage
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43. Equivalent circuit
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44. Voltage conversion ratio
Half-wave Full-wave
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45. Correct Full-wave for Buck-Boost
The previous is correct because:
Does not cover Vo/Vin =0 when fs/fo is close to one
When fs/fo is 0.2, the Vo/Vin is too large
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46. Duration of each stage in ZVS
Td1
Td2
Td3
Td4
Ts-Td1-Td2-Td3
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47. Comparison of Vo/Vin of the classical and QRC
Classical
ZCS
ZVS
Buck D g
1-g
Boost
Buck-Boost
Cuk
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48. 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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49. Practical implementation of ZVS quasi-resonant converters
Buck
Boost
Buck-Boost
Cuk
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50. Tutorial
A ZVS Quasi-resonant Buck converter is operated under the following condition:
Vin=50V
Vo=25V
fs=1MHz
Io=5A
Using 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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51. Answer
Full-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
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52. Answer (Cont)
After calculated Cr, the actual required Cr is to be deduced by the junction capacitance of the transistor.
=>
Actual Cr=8.05nF-100pF
=7.95nF
=>
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