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UCC28600-Q1: larger resistor between VIN and VDD

Sherry Liang
Genius 10255 points
Part Number: UCC28600-Q1
Other Parts Discussed in Thread: UCC28600

Dear team,

Currently the resistor between VIN and VDD is 5*1.3Mohm, but the quiescent current of the system is too large, so we plan to add 5*3Mohm resistors between VIN and VDD. Is there any bad impact to do this?

Thanks & Best Regards,

Sherry

  • 0
    TI__Mastermind 40120 points

    Hello Sherry,

    Thank you for your interest in the UCC28600 flyback controller.

    It is certain that increasing the start-up resistance will result in longer start-up time.  How much longer depends on the input voltage.
    There is a possibility that the system will not start at all, depending on the lowest required input voltage for start-up.

    3M/1.3M is a 2.3X increase in resistance, which results in a 2.3X decrease in charging current to VDD cap C40.
    But it will take longer than 2.3X the previous start-up time because the UCC28600 pre-start current (Istartup) does not change.
    So cap C40 sees the difference of the lower charging current available and the fixed IC bias current before it starts-up.

    For C40 (22uF) to reach the 13V start-up threshold on VDD within 1 second, it needs 22uF*13V/1s = 352uA of current. 

    Accounting for Istartup(max) = 25uA of the IC, the minimum HV+ start-up voltage to achieve a 1-second start time = (352uA+25uA)*(5x3M) = ~5655V!
    If your minimum required start voltage is lower than this, start-up time will be much longer.  

    If your minimum required start-up voltage is lower than 375V (= 25uA*15M), typical parts with 12uA bias current may start after a long time, but worst case parts at 25uA may never start up.  
    Please consider the tradeoffs between required start voltage and start time, and also assess whether C40 can be made smaller without affecting other required performance.

    Regards,
    Ulrich