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TIDA-010054: The mosfet turn off loss when ZVS is not achieved in DAB.

jin juil
Prodigy 10 points
Part Number: TIDA-010054

Tool/software:

Hello, I have a question about that application note.

I’m trying to calculate efficiency theoretically, and I’m particularly curious about the MOSFET turn-off loss when the current is negative.

In a dual-active-bridge (DAB) converter, if the power transfer is low, it might not achieve Zero Voltage Switching (ZVS).

When ZVS is not achieved, the turn-on current of the MOSFET is positive, but the turn-off current becomes negative.

Here’s my confusion: if the turn-off current of the MOSFET is negative, according to the power loss equation:

Psw​=1.2​⋅Vds​⋅Ids​⋅toff​⋅fsw
where toff​ is the MOSFET turn-off time, and fsw is the switching frequency, the resulting power loss would be negative.​

Until now, I thought that a negative value means zero loss. However, it suddenly occurred to me: is this correct?

I asked ChatGPT, and the answer was that the current value should be treated as absolute when calculating power loss.

This has left me confused, so I’d like to ask for clarification. What is the correct approach?

Thank you for your help in advance.

  • 0
    TI__Expert 4400 points

    Hi,

    Thanks for reaching out. 

    For turn on of a MOSFET a negative current means smaller losses, since in that case the body diode is conducting already, clamping the voltage which is turned on. This is called zero-voltage switching. 

    If you are turning of a MOSFET with negative current, you are just shifting the current from the channel to the body diode. This means neither the current nor the voltage are changing significantly, so you are not really switching the FET. This means there will be no real switching losses, but you will get the conduction losses of the body diode and you will have the switching losses of the other other FET in this leg since he needs to turn on this current and he will see this negative current as a positive current.

    Hope this helps.


    Best regards,
    Andreas

  • 0 in reply to Andreas Lechner
    Prodigy 10 points

    Thank you for helping me better understand the problem.

    However, I have one more question.

    You kindly explained, "This means there will be no real switching losses, but you will get the conduction losses of the body diode and you will have the switching losses of the other other FET in this leg since he needs to turn on this current and he will see this negative current as a positive current., and it will see this negative current as a positive current."

    But isn't this the same under ZVS conditions? What I mean is that, even in a DAB operating under ZVS conditions, the diode incurs conduction losses, and the other FET in the same leg also experiences turn-off losses, as it seems.

    The key point I want to ask is this: If it is possible to ensure that only one condition—either turning the MOSFET on or off—results in negative current, would the losses remain the same regardless of the choice from a loss perspective?

    As far as I know, MOSFET turn-on time is usually longer than turn-off time. However, with modern MOSFETs, turn-on and turn-off times are nearly the same, so the losses seem similar.

    Thank you in advance for your help.

    Sincerely,
    Juil Jin

  • 0 in reply to jin juil
    Prodigy 10 points

    I’ve added pictures: one shows ZVS achieved, and the other shows Non-ZVS achieved, with negative current flowing during the MOSFET’s turn-off time.
    When ZVS is achieved, the losses include S2 off loss and S1 diode conduction loss. When Non-ZVS is achieved, the losses include S1 on-loss and S2 diode conduction loss. The total losses in each condition seem similar. Does this mean that the losses appear to be similar in DAB regardless of whether ZVS is achieved or not?

    Thank you in advance for your help.
    Best regards,
    Juil Jin

  • 0 in reply to jin juil
    TI__Expert 4400 points

    Hi Juil,

    ZVS can only be done for turn on. There is no such thing as ZVS for turn off off a FET. The equivalent to ZVS for turn of is zero current switching ZCS.
    With standard modulation schemes only ZVS can be achieved for a DAB.
    SO for turn off there will be always some losses in a DAB. However for typical switches the turn off energy is lower than the turn on energy, which means ZVS can avoid the major part of switching losses.

    The losses will not be the same for turn off and turn on regardless of the sign of the current. For a detailed analysis each of these conditions needs to be analyzed separately.

    Best regards,
    Andreas