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Question about DRV8813 behavior

Go
Genius 4190 points
Other Parts Discussed in Thread: DRV8813, DRV8818, DRV8811

Hi,

I got some question about DRV8813 behavior,

1. My customer tested two different stepping motor with DRV8813 as following settings,

   VM=12V , Decay=Low(slow) Vref=0.32V , RISENSE=0.47Ω (so, Ichop=0.32/(5*0.47)=0.136A)

   As a result, on of them seemed to work correctly (Ichop=0.134A)

  but the Ichop of the another motor exceeded the setting value (Ichop=0.257A)

  and when the Decay set to High(fast), both of them seemed to work correctly (Ichop was about 0.12A)

 What is the possible cause of this behavior?

 2. Regarding the attached waveform,

     Is the Aout1 output voltage waveform correct?

     Why does the Aout1 output current increase when Aout1 output voltage is low?

3. what is the advantage and disadvantage of these two Decay mode (fast and slow)?

Thanks

Go

 

  • 0
    TI__Guru** 114905 points

    Hi Go,

    Can you provide more details on the motor parameters? What is the resistance, inductance and motor current per phase of each motor? This may explain the reason that one motor appears to operate properly and not the other.

    1) The description implies that  the DRV8813 is losing current regulation. This can occur in slow decay when the motor is turning slowly or being held. The blanking time injects more current than the slow decay mode can remove during the remainder of the pwm cycle.

    Fast decay does not have this problem because more current is pulled out of the winding than injected.

    2) The waveforms appear to be correct. The current will increase depending on the quadrant and the direction of the current probe. In this case, the current increasing when AOUT2 is being driven high and AOUT1 is being driven low. Please capture both AOUT2 and AOUT1 along with the current. This will allow confirmation that the signals are being driven properly.

    3) When properly regulating current, the advantages of slow decay are more torque and less current ripple. The disadvantage is loss of regulation under certain circumstances.

    Fast decay has less torque and more current ripple, but does not lose current regulation.

    Thank you.

     

  • 0 in reply to Rick Duncan
    Genius 4190 points

    Hi, Rick

    Thank your for your advice,

    Here is the specs of 2 motors as follows.

    (Motor1) KCL42SCK560E(this one doesn't work correctly in slow decay mode)

    resistance : 5.8Ω　+/-10％

    inductance 　：　6.3mH　@1KHz

    motor current ：　500mA

    (Motor2) KFG20A120Z　(this one works correctly in slow decay mode)

    resistance : 34.5Ω　+/-10％

    inductance 　：　20.6mH　@1KHz

    motor current ：　200mA

    and attached file is Aout1 and 2 waveform for each motor (slow decay mode)

    as see in the attached file , themoMotor2( KFG20A120Z) seems not to work correctly when Vref=0.6 V.

    What is the behavior?

    Thanks

    go

    Aout1・2_waveform for each motor(Decay=L).pdf
  • 0 in reply to Go
    Genius 4190 points

    here is the Aout1 and 2 waveform for each motor (fast decay mode).

    Aout1・2_waveform for each motor(Decay=H).pdf
  • 0 in reply to Go
    TI__Guru** 114905 points

    Hi Go,

    Can you describe what is not working correctly at VREF=0.6V?

    From the waveform, it looks like Channel 1 and 4 should go together and Channel 2 and 3 for the KFG20A120Z motor with decay=Low.

    Is the motor spinning or holding when these plots were taken?

    Thanks.

  • 0 in reply to Rick Duncan
    Genius 4190 points

    Hi, Rick

    Thank you for your comment.

    Is it no problem that the pulse width of Aout1/2 output waveforms with KFG20A120Z are not stable?

    and could you tell us How we can figure out from motor specs if we can drive a motor in slowdecay mode or not?

    Thanks

    Go 

  • 0 in reply to Go
    Genius 4190 points

    Hi, Rick

    and as for your question as follows,

    >Is the motor spinning or holding when these plots were taken?

    The motor was holding.

    Thanks

    Go

  • 0 in reply to Go
    TI__Guru** 114905 points

    Hi Go,

    It is no problem. The current appears to be regulated during this time. If Channel 1/4 and Channel 2/3 go together for the KFG20A120Z motor with decay=Low, the outputs are PWM'ing at approximately 50kHz. At each rising each current is injected until the trip point. Then slow decay is started for the remainder of the PWM cycle. The slow decay time can be as long as the PWM cycle minus the blanking time or as short as a few hundred ns.

    It is difficult to determine if you can drive a motor in slow decay. Factors such as VM, motor parameters, step speed, blanking time, and PWM frequency, and use conditions all have an impact. Many customers tune the motor in system to obtain the best operation, paying close attention to the current waveforms when running and holding.

    A more detailed description of decay modes is available at http://www.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=slva321&fileType=pdf

  • 0 in reply to Rick Duncan
    Genius 4190 points

    Hi, Rick

    Thank you for your support. I really appreciate it.

    I got a strong request from customer that they need the method to determine if they can drive a motor in slow decay.

    Do you have any good idea? 

    (the method to determine a rough indication would be OK.)

    and as for the request, I have some question as follows,

    1. I think that VM, Motor winding Inductance, blanking time are more influential factors to determine

        a motor can use in slow decay or not, is ti correct?

    2. I think that the DRV8813 wil never lose current regulation in fast decay mode, is it correct?

    3. Is there any way to set the PWM frequency lower than 50kHz?

    (When the DRV8813 loses a current regulation in slow decay mode, 

     I thnik we can use this motor with slow decay mode if we can set the PWM frequency to lower value.)

    4.　As for the document about decay mode as follows,

       http://www.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=slva321&fileType=pdf

    In slow decay mode, Current will decay as slow as the LR time constant,

    where L is the motor winding inductance and R is two times the low side switch RDSon.

    Is a motor winding resistor not included in LR time constant?

    Thanks,

    Go

  • 0 in reply to Go
    TI__Guru** 114905 points

    Hi Go,

    Do you have any good idea? 

    >> We generally tell customers to tune the motor in system. This is because there are many variables such as motor inductance, speed, voltage, etc that can affect this. In general slow decay does not work well when current is being removed from the motor. Mixed decay is a compromise using fast decay for 75% of the period then switching to slow decay for the remainder.

    >> Slow decay also can have problems when the motor is holding. If the blanking time injects more current than the remaining time of the period can remove using slow decay, the motor current will continue to increase. Motor 1 would increase to 2A while motor 2 would increase to 350mA.

    1. I think that VM, Motor winding Inductance, blanking time are more influential factors to determine

        a motor can use in slow decay or not, is ti correct?

    >> Yes those factors are very influential.

    2. I think that the DRV8813 wil never lose current regulation in fast decay mode, is it correct?

    >> Yes that is correct.

    3. Is there any way to set the PWM frequency lower than 50kHz?

    >> No, this is a fixed 50kHz PWM frequency. The DRV8811/DRV8818 provides the ability to control the PWM frequency but does not have the voltage range of the DRV8813.

    (When the DRV8813 loses a current regulation in slow decay mode, 

     I thnik we can use this motor with slow decay mode if we can set the PWM frequency to lower value.)

    4.　As for the document about decay mode as follows,

       http://www.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=slva321&fileType=pdf

    In slow decay mode, Current will decay as slow as the LR time constant,

    where L is the motor winding inductance and R is two times the low side switch RDSon.

    Is a motor winding resistor not included in LR time constant?

    >> Motor winding resistance can be included, but the time constant will only get worse.

  • 0 in reply to Rick Duncan
    Genius 4190 points

    Hi, Rick

    Thank you for your great help.

    I got a question about the behavior when lose current regulation in slow decay mode,

    Why does the current become stable at a certain current like I attached before?

    Thanks

    Go

  • 0 in reply to Go
    TI__Guru** 114905 points

    Hi Go,

    This may not answer your question directly. When you zoom out during a motor hold condition, what is the maximum current seen in the winding(s)? The current should become stable once the current in the inductor cannot increase any more.

    The best way to observe loss of regulation is to enable the device in slow decay during a motor hold condition. When enabled, observe the current chopping. If the current regulation is lost, the current will increase each PWM cycle until the maximum current through the winding is observed. For the second motor, this could be up to 2A @12V.