Zero Voltage Switching Quasi-resonant Converters EE4211

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. EE4211

    Resonant switches Zero –current switch Zero-voltage switch EE4211

Configurations of the zero-voltage switch general half-wave full-wave EE4211

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. EE4211

4 stages of operation Capacitor charging stages:(Fig. 4a) Resonant state: (Fig. 4b) Inductor recovering stage (Fig.4c) Free-wheeling stage: (Fig 4d) EE4211

Equivalent circuit EE4211

Capacitor charging stages Switch SW is turned off at t0. Input current iLr rises linearly and is governed by the state equations Solution: EE4211

Half-wave Uni-directional of resonant voltage on Vc EE4211

Full-wave Bi-directional voltage on Vc EE4211

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 EE4211

Resonant state: Lr and Cr resonate and DF is on The solution is: Resonant frequency Impedance EE4211

Duration and Boundary Finish when the resonant capacitor voltage vCr reaches zero. The duration of this state Td2 is EE4211

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. EE4211

Voltage during the end of stage At t2, iLr = Iocos EE4211

Inductor recovering stage: Resonance stops, Lr begins to be charged by the input voltage Vin The solution: EE4211

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   EE4211

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. EE4211

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): EE4211

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. EE4211

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. EE4211

DC voltage conversion The output voltage Vo can be solved by equating the input energy Ein and output energy Eo EE4211

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 EE4211

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 EE4211

Zero voltage switching Quasi-resonant Boost converter A steady-state equivalent circuit using a current source Iin which represents the input current. EE4211

Equivalent circuit EE4211

4 stages of operation Capacitor charging stages:(Fig. 8a) Resonant state: (Fig. 8b) Inductor recovering stage (Fig.8c) Free-wheeling stage: (Fig 8d) EE4211

Half-wave Uni-directional of VCr Similar to ZCS but the shapes of iLr and Vcr are swapped. EE4211

Full-wave Bi-directional of VCr Similar to ZCS but the shapes of iLr and Vcr are swapped. EE4211

Capacitor charging stage Resonant capacitor voltage vCr rises linearly due to the input current and is governed by the state equations: Solution: EE4211

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 EE4211

Resonant state Lr and Cr resonate and DF is on The solution is: The definition of o and Z is the same EE4211

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. EE4211

Boundary The duration of this state Td2 is Boundary condition: vCr =0 EE4211

The current left in Lr At t2, iLr = Iin (1-cos) EE4211

Inductor recovering stage  Inductor recovering stage (Fig. 8c): Resonance stops and Lr begins to be discharged through the output voltage. The solution is: EE4211

Boundary of Inductor recovering stage Duration: Boundary condition: iLr=0 Td3 = Lr Iin (1-cos ) / Vo EE4211

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

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 EE4211

Voltage conversion ratio (half-wave) Depends on Load Minimum is 1 Maximum approaches to infinity EE4211

Voltage conversion ratio (full-wave) Relatively independent of Load Looks like the classical Boost converter characteristics. EE4211

Other converter (Buck-Boost) Can be formed similarly as the other two converters Also inverted output voltage EE4211

Equivalent circuit EE4211

Voltage conversion ratio Half-wave Full-wave EE4211

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 EE4211

Duration of each stage in ZVS Td1 Td2 Td3 Td4 Ts-Td1-Td2-Td3 EE4211

Comparison of Vo/Vin of the classical and QRC   Classical ZCS ZVS Buck D g 1-g Boost Buck-Boost Cuk EE4211

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

Practical implementation of ZVS quasi-resonant converters Buck Boost Buck-Boost Cuk EE4211

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. EE4211

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 EE4211

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 => EE4211