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Original question:

DRV8833: About measuremen value t of IVMQ

DRV8833: About Motor current control by current detection resistor

Y.Ottey
Mastermind 6145 points
Part Number: DRV8833

Dear all,

Could you give me some advice about the following content.

Our customer measured motor current and both-end voltage of R2 in following circuit diagram.

※Our customer's situation

VM=6.0V

VIN1,VIN2=3.3V

C4：ceramic capacitor

As the results of measurement, the following waveform data is measured.


As shown in the above waveform, when the motor is driven, the motor current is suppressed to about 1A by the function of the current detection resistor.

However, a 1.5 V peak pulse was observed at the both-end voltage of R2 just before the transition to the slow decay mode.

Although it is equivalent to about 7.5A in current conversion, such fluctuation is not seen in the motor current waveform.


We believe that DRV8833's dead time caused this peak.
When we confirmed the waveform data, the width of this peak seems to be about 500 ns.
According to datasheet, tDEAD(DEAD TIME) is 450ns(typ,VM=5V) and this value is close value compared the value from waveform data.

Our customer would like to know the cause and solution of this phenomenon .

It would be great if you could let me know your thoughts this cause.

 

Y.Ottey

 



  • 0
    Mastermind 6145 points
    Dear all,

    I would like to ask more questions below.

    1. Is the flow to change from Forward drive(Reverse drive) to slow decay in the following form?
    Forward drive(Reverse drive)→Dead time →Fast decay(about 500ns)→Slow decay

    2. Although the absolute maximum rating of the Isense pin is 0.5V, should it be considered that 1.5V(both-end Voltage of R2)@500ns will damage the device?

    3. if 1.5V(both-end Voltage of R2)@500ns is condition damaged the device, Is there any recommended circuit to clamp?

    Y.Ottey
  • 0 in reply to Y.Ottey
    TI__Guru 56650 points

    Ottey-san,

    1. From Forward drive to slow decay:  keep IN1=1 and flipping IN2 as PWM. There are two states:

    a. xIN1=1, xIN2=0; xOUT1= high and xOUT2=low; Forward mode

    b. xIN1=1, xIN2=1; xOUT1= low and xOUT2=low; slow decay

    xIN2 from low to high, xOUT1 from high to low, the internal dead time prevents any risk of cross-conduction (shoot-through). Customer side doesn't need to insert a special mode (such as fast decay...) to create dead time.  

    2. The internal dead time also prevents any risk of cross-conduction (shoot-through) between the two bridges due to timing differences between the two bridges. The schematic shows paralleling operation. That means both AIN2 and BIN2 has an same input. If the propagation delay is same between "A" bridge and "B" bridge, we should not see any shoot-through. Would you zoom in xIN2 and xIN1 signal to see the rising edge or falling edge sharp enough?

    3. Let us understand that 1.5V spike first. And then, discuss about damage or protection. So, please only change xIN2 to do forward to slow decay and monitor xIN2 transient with 1.5V current signal.

  • 0 in reply to Wang5577
    Mastermind 6145 points
    Hi,Wang.

    Thank you for your reply.
    The xIN1,xIN2 signal data that you mentioned in the previous post will be checked with our customer.
    I would like to ask more questions.

    According to Customer information, the flow to change from Forward drive to slow decay is in the following form.

    Forward drive→Itrip(or Vtrip )→Slow decay

    We assume that there will be dead time to prevent short circuit(shoot through) while it becomes slow decay after Itrip(Vtrip).
    With the dead time, all internal FET's turn off for 500ns.
    Therefore, current flows in the following form like fast decay.

    GND → XOUT1 →motor→XOUT2→VM

    At this time, isn't 1.5V rising like the signal data of out customer?

    Best Regards,

    Y.Ottey
  • 0 in reply to Y.Ottey
    TI__Guru 56650 points
    Ottey-san,

    I agree with:
    a. Forward drive→Itrip(or Vtrip )→Slow decay
    b. With the dead time, all internal FET's turn off for 500ns and the current flow likes the fast decay.

    To debug this, can they separate PWM output: (AOUT1 and BOUT1) or (AOUT2 and BOUT2) and check does AOUTx have the same Itrip point or timing as the BOUTx? If AOUTx Itrip and BOUTx Itrip timing difference than 450ns, we may see a high current spike there.
  • 0 in reply to Wang5577
    Mastermind 6145 points
    Hi,Wang

    Thank you for your reply.

    According to previous post, you say that we may see a high current spike there if AOUTx Itrip and BOUTx Itrip timinig difference than 450ns.
    If this is the cause of 1.5V peak's phenomena, is there any solution?

    Please give me some advice.

    Best Regards,

    Y.Ottey
  • 0 in reply to Y.Ottey
    TI__Guru 56650 points
    Ottey,

    Let us find the root cause first. And then, think about a solution.
  • 0 in reply to Wang5577
    TI__Guru 56650 points
    Ottey,

    Still waiting here. Let me temporary close this thread. You can re-open it by sharing the debugging result.
  • 0 in reply to Wang5577
    Mastermind 6145 points
    Hi,Wang

    Thank you for your reply.
    I received the board layout and signal data from Our customer.
    However, there are included confidential business information.
    So, I would like to exchange e-mails about this subject.
    Could you please tell me your e-mail address?
    I look forward to your reply.

    Best Regards,

    Y.Ottey
  • 0 in reply to Y.Ottey
    TI__Guru 56650 points

    Ottey,

    You can send me an e2e private message to discuss it by putting cursor on my name and selecting private message.