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TPS629210-Q1: IBB current limit calculation

Part Number: TPS629210-Q1

Hi support team,
This is a question regarding equation (7) of SLVAFC5.

https://www.ti.com/lit/an/slvafc5/slvafc5.pdf

The "Low-side FET current limit" of 1.3A is applied to IL(max).
Isn't the "High-side FET current limit" of 1.5A applied to the inductor peak current?

Regards,
Dice-K

  • Hi Dice,

    Good question. It should be low-side FET current limit.

    If the IL(max) exceeds the Low-side FET current limit but not exceeds the high-side FET current limit, the device would not protect itself immediately. However, when the high-side FET off and low-side FET on, the high-side FET would not be turned on until the inductor current decreases below the low-side FET current limit which would influence the duty and output voltage by this protection behavior. Here using the low-side FET current limit is just to ensure the duty and output voltage would not be influenced by any protection of the device under IBB application.

  • Hi Athos,
    Thank you for your great support.

    Please let me know about the calculation of "Low-side FET current limit".
    According to the calculation method, is it determined based on the peak current rather than the valley current?

    Best regards,
    Dice-K

  • Hi Dice,

    The "Low-side FET current limit" is triggered when inductor current is high than the limit value. The low-side FET would keep on to decrease the inductor current until it reach the limit value as below waveform FYR. So it's indeed valley current. When this is triggered, the output voltage usually decrease due to more off-time.

    However, the purpose of this IBB calculation is to ensure that the peak inductor current would not exceed the "Low-side FET current limit" value so that no protection would be triggered and the device could performance well with correct output voltage.

  • Hi Athos,
    Thank you for your reply.

    I have understood it as below.

    The current limit is detected by the inductor valley current.
    However, in order to provide some margin in the calculation, the inductor peak current is used.

    Best regards,
    Dice-K

  • Hi Dice,

    Yeah, you can understand it like this.