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UCC28881: Limit the peak current in a Flyback using UCC28881

Part Number: UCC28881
Other Parts Discussed in Thread: UCC28704, UCC28910

Hi Experts!

I designed a Flyback using UCC28881, 

Input power is 300mW,

input DC-rail is regulated at 300V

and I am using a hysteresis control scheme (manipulating the FB pin) to regulate the output voltage at 700V.

However, the efficiency is very low (in the range of 40 ~50%). I know that this IC powers-up from the high-voltage sinking somewhere around 100uA in HVIN pin.

I am trying to limit the peak current to reduce losses and increase the efficiency a bit. Is there any way to do this in this package?

Any other ways to increase the efficiency?

Thanks!

  • ALIREZA TAJFAR

    UCC28881 is a highly integrated device that is designed for low system cost.  What this means is that the ability to modify the design to improve efficiency is limited.  The first place you can optimize to improve efficiency is to adjust the feedback network.  Section 9.2.1.2.7 (page 20) of the UCC28881 datasheet goes into detail on what changes.

    If you find that the efficiency cannot meet your requirements, we have alternative device such as UCC28910 and UCC28704.  These devices have advanced features such as valley switching and AM/FM control law that enable them to have higher efficiency than UCC28881.

    Best Regards,

    Eric

  • Hi Eric,

    Thanks for your response. Another feature we need is integrated high-voltage current source for power up and since the input current charging the bulk cap is only 500uA, we need this current source to sink very low current. Does UCC2890 meet this requirement? Do you know how much current is sunk on DRAIN at startup and before the auxiliary winding takes over the power?

    Regards,

    -Alireza

  • Alireza

    For UCC28910 the startup current source is specified in the electrical characteristics on page 8 of the UCC28910 datasheet.  This energy comes can come from the bulk capacitor, so as long as you have a large enough bulk capacitor on the input voltage there should be sufficient charge that will allow the startup current source to operate properly.

    Best Regards,

    Eric