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DRV8871-Q1: Driving a light inductive load with a 7KHz square wave

Art Mecina
Genius 16505 points
Part Number: DRV8871-Q1

Hello Guys,

Our customer is driving a toroidal inductance of about 500uH (not a motor). As he understands it, in order to "change direction" there are inductive effects to related to changing the direction of the current need to be overcome and wait for it to settle down; and there is the amount of delay time it takes for the circuits in the chip to turn off one high-side and one low-side transistor and to turn on the other transistors in the opposite driver.

Customer is interested to know how long this time would be - how long to turn off one half-H drive and turn on the other - say for a purely resistivity load. There is no timing info related to this on the datasheet that they could find.

Bottom line is whether the part can "change direction" (To describe it using motor terminology) at 7KHz?

Thanks!

Art

  • 0
    TI__Guru 56650 points

    Art,

    "how long to turn off one half-H drive and turn on the other"

    It can be immediately to turn off one half-H drive and turn on the other. With an inductor load, it would take some time to let the current reverse the direction.

    For example: if Vin=12V; L=500uH and cycle time is 143us (7kHz), di=V*dt/L= 12V*143us/500uH=3.4A. So, if the input is 12V, it is capable to change 3.4A with 500uH and 7kHz. Please check the load current level is less than 3.4A/2=1.7A 

  • 0 in reply to Wang5577
    TI__Genius 16505 points

    Thanks for looking into this Wang!

    How much time it would take for the internal delays in the chip to turn off one side of the H-drive and turn on the other? Will it be a good proportion of 143us?
    Said a different way, what is the propagation delay in the chip to turn on/off the half H-drives that you'd add to the amount of time it takes to reverse the current in the inductor?

    Thanks!

    Art

  • 0 in reply to Art Mecina
    TI__Guru 56650 points

    Art,

    The propagation delay should not be a concern because the amount of time to reverse the current in the inductor is much longer than the propagation delay.

    The amount of time to reverse the current in the inductor is depended on Vin, output inductance; current level: dt=di*L/V. It is an inductor and follows the inductor's physics law.

  • 0 in reply to Wang5577
    TI__Genius 16505 points

    Thanks Wang!

    Do you think 143uS is a good proportion of that propagation delay? I apologize for customer is asking approximate number.

    Thanks!

    Art

  • 0 in reply to Art Mecina
    TI__Guru 56650 points

    Art,

    The propagation delay tpd is 0.7us in the datasheet spec table. 143us (7kHz cycle time) is picked by the customer. They are different and not related.

  • 0 in reply to Wang5577
    TI__Genius 16505 points

    Thanks Wang!

    Art