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UC2844AQ: Faster Start-up

Hideki Hayashi1
Genius 10130 points

Part Number: UC2844AQ

Hi team,

Customer is evaluating UCC2844AQ, but the slow power startup time is an issue. In the figure shown in the datasheet below, C2 (Vcc cap) is 100µF, and the customer believes this capacitance is a place to improve. Could you please advise whether it's possible to reduce this capacitance, and if there would be any problems in doing so? 

REgards,

Hayashi

  • 0
    TI__Genius 12835 points

    Hi Hayashi, 

    The startup time is determined by the RC time constant, in this case R2 and C2+C3.

    You can calculate the startup time by using t = R*C*ln*(Vin/(Vin-Von)), where Von is the UVLO turn-on threshold of the controller and Vin is the input voltage. 

    Thanks,

    John

  • 0 in reply to John Gomez
    TI__Genius 10130 points

    Hi John,

    Other than the RC time constant, how can customer decide the value of R2 (Rh in figure 7-1), C2+C3(Vcc cap)?

    Regards,

    Hayashi

  • 0 in reply to Hideki Hayashi1
    TI__Genius 12835 points

    Hi Hayashi,

    The startup time completely depends on both the R and C value. Additionally, the R value should be small enough such that it is larger than the startup current requirement of the controller, which is .5mA maximum. 

    This means that Vin/R >= .5mA. 

    Thanks,

    John

  • 0 in reply to John Gomez
    TI__Genius 10130 points

    Hi John,

    There are no other factor to decide RC, right?

    Vin/R >= 0.5mA

     t = R*C*ln*(Vin/(Vin-Von))

    From Vin/R >= 0.5mA, R is limitted by the maximum value, I'm not sure about the tradeoff. Bigger resistance is better or smaller is better?

    The datasheet mentioned operating supply current is 11mA typical. It is current that Vcc pin sink, so capacitance of Vcc doesn't affect it, correct?

    Regards,

    Hayashi

  • 0 in reply to Hideki Hayashi1
    TI__Genius 10130 points

    Hi John,

    Is there any comment for questions about RC selection?

    Regards,

    Hayashi

  • 0 in reply to Hideki Hayashi1
    TI__Genius 12835 points

    Hi Hayashi, 

    Somehow the follow-up question didn't go through when E2E re-structured the organizations supporting these devices. 

    Hideki Hayashi1 said:
    Bigger resistance is better or smaller is better?

    A bigger resistance is typically 'better' due to lower power dissipation. The startup resistor will always dissipate power according to (Vin - Vcc)^2/R. For this reason, you should choose the resistor to only a bit smaller than the result of this equation: Vin/R >= 0.5mA.

    Hideki Hayashi1 said:
    The datasheet mentioned operating supply current is 11mA typical. It is current that Vcc pin sink, so capacitance of Vcc doesn't affect it, correct?

    The capacitance will affect the voltage ripple on the VCC pin. You can start off with a value of C  according to the equation -> t = R*C*ln*(Vin/(Vin-Von)). Then increase C if the voltage ripple is too large during steady state operation. If the ripple is too large, the controller could go below the UVLO turn-off threshold. 

    Thanks,

    John