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DRV8840, unexpected behavior with inputs I0 and I4, hardware bug?

Stefan Zimmer
Prodigy 30 points

while designing a pcb using the Motor Driver IC DRV8840PWP I have found an unexpected behavior.

 In operating mode the current set inputs I0 .. I4 work as expected. They sink a low current only as specified  (pull down resistor of approx.. 100k).

When the IC is set into the sleep mode by driving the nSleep input to GND the input current on I0 and I4 rises extremely. Driven with 3.3V and a series resistor of 220Ohms the input current is approx. 1.5mA per input in this state.

The behavior of I1, I2 and I3 is independent to the state of the nSleep input.

For my PCB I have a solution (drive nSleep and I0, ... with the same signal), but I am interested if I've found a hardware issue or if I have overseen something.

 

  • 0
    TI__Guru 55615 points

    Stefan,

    This is very interesting.  I am investigating.

    Alternatively, you could tie V3P3OUT to the Ix pins as V3P3OUT collapses when going to SLEEP state. 

     

  • 0 in reply to Ryan Kehr
    Prodigy 30 points

    Dear Ryan,

    thank you for your answer and for the suggestion to use the V3P3OUT.

    With my first prototype I wondered why the idle current of the PCB was higher than expected. The prototype has I0, I1 and I4 tied together with one 220 Ohms resistor to VCC (5V). I know that this is not the best idea, because it would be a load of 150µA in sleep mode with 100K pull down at each input.

    With the evaluation board, that has separate resisitors for each input I found out that I0 and I4 seem to have the described behaviour.

     

    I am looking forward to the result of your investigation.

     

  • 0 in reply to Stefan Zimmer
    TI__Guru 55615 points

    Stefan,

    It turns out we have some test circuits tied internally to I0 and I4 that repurpose these pins during our final testing.  The simplified internal circuit (looking into our device) is a 1k resistor and series diode connected to V3P3OUT.  When V3P3OUT collapses in SLEEP state, then you are looking into a diode + resistor to 0V. 

    I = V/R = (3.3V - 0.7V) / (1 + 0.22) = 2mA.  The 3.3V is what you applied externally and the 220 ohm in the equation is your series resistor. 

    Hope that answers your question.  First time seeing this question as most customers just connect pins to V3P3OUT and never see this.  Good find! 

  • 0 in reply to Ryan Kehr
    Prodigy 30 points

    Dear Ryan,

     

    this answers my question of course!

    Thank you for this detailled explanation.

    (Sorry for the delayed answer, but I was not in the office last week and I didn't receive an automatied notification in my mail account as last time.)