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TPS92520-Q1: Question about low-side sinking current limit threshold

Part Number: TPS92520-Q1

The Low-side sinking current limit threshold in TPS92520 is typical 1.5A. When ILS is over 1.5A, fault will be triggered.
As Buck converter, the low-side switch conduct the current when high-side switch turn off. If 1.5A(average) is required for the output current, that means the low-side switch will also conduct 1.5A, and this will trigger the threshold.
Are my understanding correct?
If yes, it is not possible to output 1.6A. But why datasheet mentions that the output current capability could go up to 1.6A?
Thanks.

  • Hello Zhang,

    Lets start by defining what constitutes a positive or negative current in each FET of the synchronous buck regulator.  Here is a diagram where we define the positive current in both the HS FET and the LS FET.

    We next have to understand that how the FET currents create the inductor current. 

    We see from the diagram above when the HS FET is ON it ramps up the current in the inductor.  When the LS FET is ON it allows the inductor current to ramp down.  Then we see that the inductor current is the sum of those currents.

         

    The TPS92520 can output an "average" current of 1.6A.  That isn't he peak current current.  There is a ripple current  that sets the min and max current seen by both the HS FET and the LS FET (in other words the inductor current IL ).  Depending on the average output current and the peak-peak ripple current we see peak current seen by the HS FET can change.  Same is true for LS FET.  The IHSx(ILIM) is the max positive current that the HS FET can see before it trips the over current condition. For the TPS92520 that is 2.7A typical.  The ILSx(ILIM) threshold is for sinking, which is the max negative current through the LS FET before it trips the over current condition.   

    You simply assumed the ILSx(ILIM) was positive, that isn't not true.  It is labeled as sinking and it means negative.

    -fhoude