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DRV8256EEVM: Drive waveform output exception

Cherry Zhou
Mastermind 22235 points
Part Number: DRV8256EEVM

Hi team,

Here's some issues may need your help:

1) Regarding the drive waveform of 8256E, the description in spec is "The EN input drives the waveform PWM, and the PH pin inputs the direction signal so that out1 and out2 output the corresponding waveform". 

The following two figures correspond to 100% and 50% of the EN pin input (blue input and yellow output), and the output waveforms are 50% and 25%, respectively,  which is different from what the spec shows:

2. When the motor is hooked up, it sees no load, so the resistive process is removed, that is, the chip can still input the PWM waveform without output, and it is found that the load is still not hooked (no output from the chip). But the MCU side is still feeding the chip input waveform (possibly chip protection), after a while the chip is burned out.

When the chip is already protected, will it still burn out if it is still being fed into the signal?

Tested with no load after removing the resistive process, no problem, and theoretically, load of the motor with a full load of around 3 A does not trigger protection (that is, the drive chip appears to be auto-protected, but normally it is not protected in this case. If the load is halved, it can operate normally, but at a slower speed, it appears as full load).

Could you help check this case? Thanks.

Best Regards,

Cherry

  • 0
    TI__Guru 55605 points

    Cherry,

    It is a US holiday today.  Expect a response tomorrow.

    Regards,

    Ryan

  • 0 in reply to Ryan Kehr
    TI__Guru 55605 points

    Cherry,

    For the first question, what is the state of the PH pin for scope captures shown?

    For the second question, the chip will protect itself even if input is still toggling.  If chip is burned, something else is going on.  Possibly the VM voltage is climbing too high and damaging the chip?

    Regards,

    Ryan

  • 0 in reply to Ryan Kehr
    Expert 1910 points

    1. When the load is too high, the PH pin will pull down.

    2. We used 8256E before, now we use 8256P, and add A TVS tube, it will not burn the chip.  

    3.  What are the current protection conditions of this chip?  Instantaneous current or average current over time?  

    thanks!

  • 0 in reply to Reed
    TI__Guru 55605 points

    Hello!

    For current protection, Iocp = 8A and Tocp = 2us.  If current exceeds 8A for a time period of 2us, an over-current event will occur.  Please see section 7.3.7.3 in the datasheet.

    For current limiting (regulation), please see section 7.3.2.  This is an average current limitation.  

    Regards,

    Ryan

  • 0 in reply to Ryan Kehr
    Expert 1910 points

    Ryan，

    It's currently a smart curtain device.  The power supply part can monitor the current. The protection state will not be triggered when the starting current becomes larger when going up, current<8A. The protection will be triggered when the starting current is small when going down, which may be caused by the reverse current.  But it is not known how to measure and circumvent this reverse current.  

    Any good suggestions?

    Thanks！

  • 0 in reply to Reed
    TI__Guru 55605 points

    Hello,

    When you say the "protection" is triggered when starting current is small (curtain going down), which protection are you referring to?  DRV device or other protection on power supply?  It must not be OCP if level is below trip point. 

    Please help to explain more and maybe some scope captures of the current in the load (using current probe) and voltage on the supply to the DRV would help make it more clear for me.  

    Regards,

    Ryan