I have a hub motor: https://www.uumotor.com/4-5-inch-350w-hub-servo-motor.html

I know that the motor requires 12A but I can only supply 2.5A using my motor controller. The problem I have is when I want to turn in place using these motors, I need 2.5A at low RPM but for some reason, I cannot supply more than 1A and can only supply above 1A if I increase the speed.

Why is that? Why can't I supply a maximum current (using PID) at low speed?

  • \$\begingroup\$ What is the voltage doing when you are failing to get more than 1 A out of it? \$\endgroup\$ Jun 11 '19 at 17:26
  • \$\begingroup\$ Do you have a torque load on said motor? \$\endgroup\$
    – Voltage Spike
    Jun 11 '19 at 17:33
  • \$\begingroup\$ @evildemonic I am supplying 24V, do you mean the voltage going to the motion of the motor? \$\endgroup\$
    – aikhs
    Jun 11 '19 at 17:52
  • \$\begingroup\$ @laptop2d I guess \$\endgroup\$
    – aikhs
    Jun 11 '19 at 17:53
  • \$\begingroup\$ Are you saying that you can "turn" at high speeds but not at low speed? \$\endgroup\$
    – Andy aka
    Jun 11 '19 at 17:54

I cannot supply more than 1A and can only supply above 1A if I increase the speed.

Why is that?

Motor speed is proportional to voltage. To run at lower speed you have to lower the voltage. But you only have a fixed supply voltage of 24V, so you are reducing rpm by lowering the PWM ratio in the controller, which lowers the average (effective) voltage at the motor. Why does this matter?

During the 'on' part of the PWM cycle the controller feeds supply current to the motor, but during the 'off' part it disconnects the supply and recirculates current through the motor. As PWM ratio (throttle level) is lowered the ratio of motor current to supply current increases in inverse proportion. At 50% PWM ratio the motor draws double the (average) supply current, at 25% PWM it draws 4 times as much. Thus at low throttle the current you measure at the supply is not the motor current!

If your controller limits motor current then at low throttle the supply current will be less than the current limit. The peak supply current is also equal to the motor current, so your supply might shut down even though the average current (measured with a meter) is below the limit.

Understanding why the motor phase current is different to the power supply currents

enter image description here


You have inadequate power and there is no better solution than a 24V battery and a 15A ESC.

Motor speed with no load is proportional to voltage.

Torque is proportional to current which starts at Vm/DCR where your current will be ~ 10x rated load current due to this winding resistance , DCR (Ohms) .

If you can pulse at > 20kHz the motor (V-) with a transistor at 20% duty cycle from your 24V to motor V+, with a low-side switch rated for >10A and similar >10A power diode in reverse across across switch to gnd. Then you have a “Buck Regulated” step down voltage at 20% of input voltage with current boosted . This will give more torque but 20% of rated RPM. This may be the lowest starting speed above stall speed.

Adding a large cap (>470uF) to 24V with low ESR will boost the current if the supply does not have one.

Once the motor speeds up the inductance and back EMF reduces the AC current rectified by motor commutation to DC from source. Then you can adjust duty cycle with a pot using a Schmitt trigger Astable CMOS Logic clock by pulling the input high and suitable R ratios and C for frequency and then be able to regulate the voltage say from 0 to 30% duty cycle.

You seem to need a 5V supply just to get started not a 24V supply with only 24W on. 150W Motor. I need your target RPM, and load. I can only “guestimate”. But this motor is rated for 150W~300W power and 10x surge current with 24V.

  • \$\begingroup\$ I can’t search now , but I have designs on this site for variable PWM clocks \$\endgroup\$ Jun 11 '19 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.