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DRV8256PEVM: Load current does not follow IN1 and IN2 PWM signals

Rob Helvestine1
Intellectual 295 points
Part Number: DRV8256PEVM
Other Parts Discussed in Thread: DRV8256

Tool/software:

Hello. I'm using the DRV8256PEVM, hooked up to an MCU that is driving the IN1 and IN2 inputs. My motor is actually a 2.8 ohm resistor, which is what our application needs (bipolar drive across a 2.8 ohm resistive load). 

Right now, for the most basic test, IN1 is pulsed at 20% duty cycle (25kHz frequency) and IN2 is low, as shown below.

I would expect the motor voltage to follow the truth table of Table 7-3, being in Brake for 80% of the period, and Forward for 20% of the time. 

However, that is not the case. Resistive load voltage is shown below using a diff probe:

Where VM=5V. So +5V for ~10% duty cycle and -5V for ~10% duty cycle. Why is this the case? It seems to violate the Table 7-3 Truth Table. 

Am I missing something regarding the operation of this motor driver IC? Could it be related to Decay Mode?

Thanks,

Rob

  • 0
    TI__Mastermind 45205 points

    Hi Rob,

    Thank you for posting in this forum.

    Rob Helvestine1 said:
    I'm using the DRV8256PEVM, hooked up to an MCU that is driving the IN1 and IN2 inputs. My motor is actually a 2.8 ohm resistor, which is what our application needs (bipolar drive across a 2.8 ohm resistive load). 

    The chopper current regulation scheme requires some level of load inductance to slow down the rise time and fall time of current to have effective tON and tOFF cycles. For example a BDC motor or a solenoid load. With a pure resistive load the inductance is negligible. The current rises too fast as well as falls instantly. After tON has lapsed depending on the DECAY mode set in the design the off time tOFF current direction would be in the opposite drive during fast decay time. If there were inductance of a few mH in the load this reversal would cause rapid ramp down of the current but not a negative current. 

    With a resistive load this is expected current waveform with fast decay component in the current decay during tOFF. The DRV8256 current regulation is designed to work only with inductive load. 

    The original Figure 7-6 in the datasheet shows the load current (saw teeth) with L-R type of load such as a BDC motor. I added in red what would be the current flow with pure resistive load. This would be true for most chopper current regulation schemes.

    If you don't need current regulation, and avoid current chopping you can set VREF to MAX 3.3 or 5 V  and make sure VM / Rload <  5 A by either reducing VM or increasing Rload. With this condition you can see the true PWM ON / OFF with the load current with the input PWM. Thank you.

    Regards, Murugavel 

  • 0 in reply to Murugavel4637
    Intellectual 295 points

    Thank you very much Murugavel! You are correct. Making VREF = 3.3V solved the issue. Great thorough response.