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UCC28950: ZVS unbalance on all MOSFETS

VictorTagayun
Intellectual 655 points
Part Number: UCC28950

I am using an application with Vout = 150V Pout = 800W, Vin = 380V.

I see that Qa and Qb need higher output current to reach ZVS, around 2.0-3.8A output but Qc and Qd would reach ZVS at lower output current of 0.1 - 0.4A (100uH).

Qa = 3.8A

Qb = 2.0A

Qc = 0.4A

Qd = 0.1A

I also used 50uH.

Qa = 6.9A

Qb = 3.6A

Qc = 0.1A

Qd = 0.1A

Do you experience this and how to solve this?

  • 0
    TI__Mastermind 25570 points

    Hello Victor

    This behaviour where the CD switches maintain ZVS down to lower currents than the AB switches, is inherent to the PSFB topology. The basic reason is that there is more energy available to drive the CD switches transition than the AB switches transition. The energy available to drive the CD switches transition is that contained in the resonant (leakage + shim), magnetizing and output inductors.

    However during the AB transition the secondary is short circuited by the rectifiers (this is true for both SR and Diode rectification) and of course this short circuit is reflected into the transformer primary too. This means that the energy available to drive the the AB switches transition is that stored in the resonant (leakage) inductance only. ie 1/2 Lr (i_mag_pk + I_Lout_min *Ns/Np)^2

    The ZVS range can be extended by reducing the magnetizing inductance so that the magnetizing current, which flows in the resonant (leakage) inductor of course, is increased or by increasing the resonant inductance. You have seen this already when you changed Lr from 100uH to 50uH.

    I'm not quite sure why the QA transitions need more current than the QB transitions, the circuit should be symmetrical in that the capacitance at the switched node is the same for both devices - are the transformer currents the same for both transitions.

    The UCC28950 offers a burst mode of operation. The entry point for burst mode operation may be set at the point where the power train is just losing ZVS.

    Hope this helps

  • 0 in reply to Colin Gillmor
    Intellectual 655 points

    Hello,

    Our application is 150V 800W (5.3A, nominal/ave) out with 380Vin.

    The design is required to have a peak current of 15A (2,250W) for 20mS.

    Initially I design the shim inductor around 100uH so that one of the worst Fet would be able to reach ZVS around 50% of the full load.

    So I retest again some waveforms will be shown.

    ==================================== QB 100uH =============================================

    Qb = 4.5A ZVS , CH2 (blue) = output current, CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    Qb = 3.9A before ZVS , CH2 (blue) = output current, CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    Qb = 3.4A before ZVS , CH2 (blue) = output current, CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    ==================================== QA 100uH =============================================

    Qa = 1.8A ZVS , CH2 (blue) = output current, CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    Qa = 1.7A before ZVS , CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    ==================================== QC 100uH =============================================

    Qc = 0.4A ZVS , CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    Qc = 0.3A before ZVS , CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    ==================================== QD 100uH =============================================

    Qd = 0.3A ZVS , CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    Qd = 0.2A before ZVS , CH3 (pink) = Vds, CH4 (green) = Vgs via pulse transformer.

    =======================

    At 5.3A output load, these are the waveforms.
    Ch1 = OutA
    Ch2 = Pri Inductor current
    Ch3 = Pri Transfomer voltage
    Ch4 = OutD

    Another waveform showing pri inductor voltage
    Ch1 = OutA
    Ch2 = Pri Inductor voltage
    Ch3 = Pri Transfomer voltage
    Ch4 = OutD


    As you can observe that even though my controler is already at max duty, pri transfomer is not. Some voltage is dropped on the primary inductor.
    When I try to load to 7A, output would drop to 120V due to transformer turns ratio not enough.


    Then I changed primary inductor to 50uH this is the waveform I get when loaded to 7A and the output is still within regulation.

    regards,

    Victor

  • 0 in reply to VictorTagayun
    Intellectual 655 points

    Here are other info about the configuration of the controller.

    Fsw = around 84khz

    Rab = 56k

    Rdc = 56k

    Ref = 68k (not using sync rect due to high voltage output and low current)

    Rtmin = 15k

    Rt = 70k

    Rsum = 62k (to GND)

    Rdcmhi = 33k

    Rdcm = 1k

    Radel = 0 ohm

    Radelef = 0 ohm

    Xformer info:

    Np = 26

    Ns = 9 + 9 (18)

    Inductance = 2.3mH

    Sec inductor = 760uH

  • 0 in reply to VictorTagayun
    TI__Mastermind 25570 points

    Hello Victor

    It looks like you are making good progress. There is always a certain 'loss of duty cycle' due to the voltage across the primary leakage+shim inductor as you correctly point out. This duty cycle loss is greater with greater inductances. You will need to alter the turns ratio if you need the higher inductance value.

    There is a certain amount of ringing on the top of the primary current waveform due to interwinding capacitances - especially those between primary and secondary. If these appear on the waveform applied to the CS pin then you may have instability at certain duty cycles where the dv/dt rate of the signal at the CS pin is very slow. You may be able to filter the signal or else design the windings in the transformer to minimise these capacitances.

    Regards

    Colin