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Short circuit RMS current for DRV8301 ( TPS54160 )

Derek King1
Intellectual 365 points
Other Parts Discussed in Thread: DRV8301, TPS54160, TPS54060

I'm trying to properly size the output inductor for the buck circuit on the DRV8301 to handle an short circuit for an extended amount of time. 

There is another post the specifies that the DRV8301 uses the TPS54160 as the buck converter http://e2e.ti.com/support/applications/motor_drivers/f/38/p/223933/791680.aspx#791680

The TPS54160 datasheet mentions the typical current limit threshold is 2.7Amps.  To determine RMS inductor current, would this be a good starting point?:

current_rms = 2.7 - inductor_rippple_pk_pk/2

For my design, the nominal current is pretty low (~250mA).  To prevent pulse skipping the current ripple is  250mA pk-pk, and the inductance is (33uH).  

The RMS current for a dead short would be about 2.6Amps, which would require a rather large inductor.  Is there a good way to limit output current to allow a smaller inductor?  

Normally, I would use an IC with a lower current limit like the TPS54060, but I don't have that option because I want to use the built-in DRV8301 DC/DC. 

  • 0
    TI__Mastermind 35165 points

    Hi Derek,

    The equation you give is a good estimation of the rms current. The ripple current typically has a very small contribution to the rms current as is the case with your ripple.

    For completeness and to be more precise you can use Equation 30 on page 29 of the datasheet. The current estimated with the equation you give is the average current.

    Unfortunately there are no quick and easy ways to limit output current. One possibility would be to add an external current sense resistor and amplifier to send a signal to the TPS54160 if the current is too high.

    Best Regards,
    Anthony

  • 0
    TI__Guru**** 191445 points

    Its a little more complicated than that.  The RMS current will be given by eq. 30 in the TPS54160 datasheet.  The TPS54160 uses frequency foldback so under a hard short condition, the output voltage will be around 0 V.  In any case, the contribution to the RMS current by the ripple current is small, maybe a few 10's of mA.  So you can work out the equation or just estimate the Irms = Iss.  The typical current limit is 2.7 A but the maximum is around 3.6 A (unspecified).  So you will want to size your RMS inductor current rating to whichever you feel comfortable with, typical or max.  You will end up with a physically large inductor if you need to protect from long term short circuit.

  • 0 in reply to JohnTucker
    Intellectual 365 points

    From page 20 of the datasheet : "During overcurrent conditions that pull the output voltage low, the error amplifier responds by driving the COMP pin high, increasing the switch current. The error amplifier output is clamped internally, which functions as a switch current limit."

    I thought I could clamp the COMP voltage to a lower value with a zener, but I'm not sure how it would effect the compensation circuit.   I was hoping somebody had tried this already?

  • 0 in reply to Derek King1
    TI__Guru**** 191445 points

    The operating voltage range of the COMP pin is pretty small.  Also the source and sink current is only +/-7 uA.  It is not designed to drive anything other than the compensation network.  Ideally if long term operation into a short circuit is a major concern, you would want to look at a converter with hiccup mode current limit rather than pulse by pulse. You could externally sense the current (or output voltage under voltage detect) and then use that to shut down and restart via the EN, effectively providing for hiccup mode current limit, but that may be more circuitry than you care to add.