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I am using a MC33926 motor controller. Unfortunately, I am not extremely familiar with motor controllers and have a question. The circuit below shows the circuit that I am working with. The drivers datasheet is here. So I simply am pulling IN1 low and IN2 high and a PWM is driving D2.

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This image is the driving PWM from an MCU.

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This is my problem. The below signal is the signal on the output of the motor driver. When driving the motor control, if I drive it slow enough then the rise time is great but as in the picture below if I drive it faster(1.3kHz) then the rise time keeps getting worse. I have read through the data sheet and to my understanding I am driving it correctly and it says it can be driven at 20kHz. So if someone is familiar with this driver or same problems, am I driving the driver incorrectly?

enter image description here

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  • \$\begingroup\$ Are both images the driving PWM from the MCU? Check out page 14/30 on that datasheet - since D1 is just pulled to ground in your circuit, it may not be exactly how the driver was intended to be used \$\endgroup\$
    – Drewster
    Nov 30, 2018 at 16:26
  • \$\begingroup\$ D1 and D2 put the driver into tristate or OFF mode thus slow R/L slew rate. Use IN1 and IN2 only for direction and velocity (PWM d) and torque control. Slew rate control=off is used for PWM >11kHz \$\endgroup\$ Nov 30, 2018 at 16:42
  • \$\begingroup\$ What are you driving the input with? First picture at 5V/ shows ~5V wave, 2nd pic shows ~10V wave, is it a function generator? BTW, using the D1/2, IN1/2 controls as you are defeats the protection system on the chip (see 11.3 ff) as it constantly resets the fault. You may want to take the extra effort to rewrite the code to use the IN1/IN2 instead (I know the datasheet mentions using the D1/D2 for PWM, but IMO defeating/resetting overload circuitry usually leads to bad things happening. \$\endgroup\$
    – user201365
    Nov 30, 2018 at 16:44
  • \$\begingroup\$ I see slew rate should be set to off now and therefore removed R14, but now the signal is even worse. I will try now using IN1 and IN2 to control it. \$\endgroup\$ Nov 30, 2018 at 16:53

1 Answer 1

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I think I may be too late... but here it goes:

With this chip, the PWM to the IN2 and IN1 inputs. The D1 will set the output to high-impedance.

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