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DRV8837

david sun77114
Expert 2151 points
Other Parts Discussed in Thread: DRV8837

Hello,

I have a question about DRV8837, I am using MAX4684 as a H-bridge as shown in following schematic, I am going to change to DRV8837. I removed JP49 and R124, R125

and put a 0ohm resistor at R107, 108, 109, 104, 123, and Jp66. though the motor runs but at much less strength. I don't understand why. 

the only difference is the MAX4684 runs at brake mode, and DRV8837 runs at Coast mode. I don't think Coast mode should reduce the strength of the motor.

the power for VCC and Vm are 3.8V. the PWM of the input is 30K, with 50% duty cycle. how to analysis the problem and how to trouble shoot it? thanks for all suggestions.

David Sun

  • 0
    Expert 2151 points

    blue is the input to PWM_PHA, PWM_PHB is always low here.

    pink is the output of PM_PHA, and green is output of PM_PHB. 

  • 0 in reply to david sun77114
    TI__Guru 55565 points

    David,

    The simple fix would be to make PWM_PHB = HIGH (always high instead of your existing always LOW condition).  This will put the device in a slow decay mode (recirculating current through the low side FETs) instead of the HIZ condition you have now when the PWM_PHA input cycles LOW. 

    This should fix your problem and provide more torque to the motor by recirculating the current during the "OFF" states instead of simply HIZ'ing the bridge.

     

  • 0 in reply to Ryan Kehr
    Expert 2151 points

    Hi Ryan,

    thank you for answer my question, I think I misunderstand "break mode" and "coast mode" by the literal meaning. If I want to know more detail why slow decay mode can provide more torque to motor, could you please send me link(name of) to reference book or application note? the reason I am struggling in this detail is that we try to reduce the noise of the motor when driven by PWM.

    Best Regards,

    David Sun

  • 0 in reply to david sun77114
    TI__Expert 7455 points

    Hi David,

    The reason why you see more shaft energy with slow decay than with fast decay has to do with average current. Since on fast decay current decays toward zero faster than on slow decay, the ripple current is larger on fast decay than on slow decay. At the same time this translates into a smaller average current which must imply a smaller torque. If you have less torque, then your ability to hit a faster speed diminishes.

    For DC motor  control you will want to use slow decay most of the time. The only reason why I would use fast decay is if coasting is a requirement.

    Hope the info helps. Best regards,

    Jose Quinones

  • 0 in reply to Ryan Kehr
    Intellectual 330 points
    Hi Ryan

    I come back to this project, I am reading Slva321, it mention Slow Decay, fast Decay. I am kind of clear about these terminology, I am not full understand what is real meaning.

    for example, which mode will provide best torque performance, which mode will consume less power. is there any application note analysis the mode and power and torque relation. so I will know how to optimize the performance over the power consumption.
    Sincerely,

    David Sun
  • 0 in reply to david sun
    TI__Mastermind 21305 points

    Hi David,

    We do not have an application note with this sort of analysis.

    It is a fairly linear relationship between torque and power. The more current that is driven to the motor, the more torque the motor can deliver. The more current that is driven to the motor, the more power your system is using.

    http://en.wikipedia.org/wiki/Brushed_DC_electric_motor#Torque_and_speed_of_a_DC_motor

    Fast and slow decay just deal with the way current is regulated in the motor winding. Fast decay has a higher ripple (lower average current) since current is directed back into the power supply during OFF periods and slow decay has a smaller ripple (higher average current) since current is recirculated in the H-bridge during OFF periods.

    Can you clarrify what you mean by "optimize performance over the power consumption"?

     

  • 0 in reply to Nicholas Oborny
    Intellectual 330 points
    Hi Nick,

    thank you very much for your prompt response, to make my question simple, let assume we are driving a DC motor with TI 8837 H bridge.
    if we drive with a defined voltage, let's say V = 3.0Volt. and the PWM is 10KHz with 50% duty cycle.
    is there any difference between fast decay and slow decay. we notice that in out design slow decay(short mos FET) has much higher torque than fast decay( Diode ). in other terminology is, in brake mode has higher torque than coast mode. Is this make sense. I thought coast mode will save some power for same torque. seems I am wrong. I hope I can have some application note analysis the difference between the performance of the torque and power of fast decay and slow decay.
  • 0 in reply to david sun
    TI__Mastermind 21305 points

    David,

    With similar voltage duty cycles slow decay will deliver more average current than fast decay. This is related to current ripple caused by the decay method. Since torque is linearly related to RMS current through the motor this means slow decay will deliver more torque than fast decay for a similar applied voltage.

    http://www.ti.com/lit/an/slva321/slva321.pdf

    Decay methods deal with how the H-bridge responds during the OFF period. Since current will continue to conduct through the motor (inductor) it must be directed somewhere. Please refer to Figures 3 and 5 in SLVA321 to understand how the current is directed in each decay mode. In summary, current will decay faster in fast decay. This causes increased current ripple, reducing the RMS current value.

    In general (as Jose mentioned), slow decay is the option used for brushed motors. Fast decay is generally used in stepper motors where rapid current response is required.