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TPS61178: Questions on Ilimit

Elder Costa
Prodigy 141 points
Part Number: TPS61178


Tool/software:

Question 1: about ILIMIT calculation

Per data sheet:

I am having some difficulty to correlate the result of equation above with RLIMIT=80.6k (ILIMIT=9.24A) with graphs on Figures 18 and 19 and values on the specifications (section 7.5):

Could you provide some explanation for the apparent (to me) discrepancies between the formula, graphs and table? What am I missing?

Question 2: what value should I use?

According to my understanding of the diagrams of the data sheet, the limit should be set based on the peak inductor current. Per my math and Webench, maximum peak inductor current (min Vi, max output current) is ~4A during startup. Based on this and giving 50% tolerance to deal with component variation (guestimate based on the table excerpt above), I would set the current limit to 6A. However, Webench sets the RLIMIT to 86.6k, ~8.7A per the formula above, 8.9A per Webench panel, way above the 6A I would use.

What am I missing?

Thanks in advance.

  • 0
    TI__Mastermind 18530 points

    Hi,

    We will reply after weekend. Thanks.

  • 0
    TI__Genius 13085 points

    Hi Elder,

    There seems like some mistakes on the ILIM set, I think the formula should be a design target but the value in table is an actual measurement value.

    Seems like there -20%/+17.5% variation on the ILIM set for auto PFM based on the table, so if a min 4A current limit is required for you application, recommend to set the RLIMIT > 128.96k ohm.

    Also the current limit value varies with the duty cycle as shown in Fig.19, so also need to leave some margin for this.

    Let me know if you have more questions.

    Regards,

    Nathan

  • 0 in reply to Nathan Ding
    Prodigy 141 points

    Hello, Nathan.

    Thank's for your comment. Unfortunately it does not answer my questions.

    I am still not sure how data in table correlates with Figures 18 and 19. The table states those limits are with 80.6k, but at wich duty cycle? Same question about equation 1 (RLIMIT).

    Is the horizontal axis on Figure 19 correct? In my design (Vin max = 16V, Vout = 19V), Duty Cycle would be ~0.18 (D = 1-Vin*n/Vout) where n is the efficiency. Figure minimum  is 0.4.

    Could you elaborate how did you get to that value for RLIMIT? (BTW, I gess you meant RLIMIT < 128.96).

    Thank you.

  • +1 in reply to Elder Costa
    TI__Genius 13085 points

    Hi Elder,

    So sorry that I make a mistake before... pls ignore my response on the last thread.

    The data in the table is a test data, which is tested in test mode and should be at very large duty cycle condition, so the value may not make sense in other condition, but the variation does make sense (-20%/+17.5%). The formula should be design targeted on a specific condition also.

    The current limit varies with duty cycle is caused by slop compensation, for duty cycle < 0.4, there is no slop compensation so the current limit is same as it is in duty cycle=0.4.

    I think you can refer to Fig.19 to get the current limit with RLIMIT=80.6ohm and duty=0.4 (9.8A based on the fig), then calculate the RLIMIT for your condition like 6A (the product should be constant for a fixed duty cycle condition), then leave some margin considering the variation of current limit.

    Hope this can answer your question.

    Regards,

    Nathan