Controlling a motor to get desired speed and torque

Discussion in 'Electronics Resources' started by CircuitZord, Apr 11, 2014.

  1. CircuitZord

    Thread Starter Member

    Oct 8, 2012
  2. Sensacell

    Senior Member

    Jun 19, 2012
    Nope, more torque only by using a mechanical gear reduction.

    You can drive it slower, but the stall torque will remain the same.
  3. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    How much torque do you need?

    Are you talking about vertically winching 50 kg loads?

    Are you trying to spin a pallet wrapper deck?

    Are you looking to run a 5m 30degree slope conveyer belt?

    With a proper belt drive link I could see that motor used for any of those with what I would consider reasonable speed.

    You can pulse it with a higher voltage jolt to get some extra starting torque.

    If you use PWM to try and increase torque then you would need to switch to a capacitor voltage boosted drive. Keep in mind that current is what is most likely to kill the motor. That is how you can drive it to a higher voltage with a PWM because you use PWM to limit the current.

    A 200w (1/4hp) worm drive is going to cover a lot of uses.
  4. Austin Clark

    Active Member

    Dec 28, 2011
    You can get a lower speed with more torque simply by applying a load.

    Electrically, the voltage at the motor terminals controls the RPMs and the current controls the torque. As you apply a load, the RPMs decrease, decreasing the reverse voltage generated by the motor, which increases the voltage across the coils, and ultimately increasing the current passing through them producing more torque.

    It's worth noting that it's physically impossible to apply more torque to, say, the rotor of a motor without increasing it's RPMs. By using gears, or by purchasing a motor with a lower KV rating, you increase the total available torque, but not the no-load torque. Hopefully that makes sense.
  5. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    Your post is well written but I would like to add something. :)

    In a lot of practical applications the DC motor torque is limited by the PSU's ability to supply current. For instance when using voltage adjustment to slow the motors speed.

    In a situation where the PSU can only supply 12v 1A the motor torque at any speed is limited by the max of 1A though the motor.

    But if driven with PWM and a freewheel diode across the motor coils you can get a much greater current than the PSU current. That is because the motor inductance and PWM duty acts like a SMPS and gives you a current transformation ratio.

    So with a 12v 1A limited PSU, driving a motor with 25% PWM, the motor voltage can be 3v (/4) and the motor current can be 4A (*4).

    That situation (which is common in modern DC PWM motor drives like cordless powertools) can give th4e overal machine greatly increased torque at lower RPMs because of the PWM drive. Maybe that is what the OP is asking about?
    Austin Clark likes this.
  6. Austin Clark

    Active Member

    Dec 28, 2011
    Wow, I never heard that before, but I suppose it makes sense. I'm not sure how strong the effect is practically, but 4X seems like a bit of a stretch at any duty cycle. I wouldn't expect to see so many gears otherwise. It probably depends on the PWM frequency and motor inductance among other things though, so it may be possible under certain conditions.
  7. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    *4 is nothing in the current transformation.

    I did a PWM motor driver running a DC motor to move a heavy robot at very low speed, and got good motor torque at low speed with only 70mA(!) drawn from the main DC supply. The current transformation on that one was probably in the region of *10.

    Basically the PWM, motor inductance and freewheel diode acts like a SMPS. The same basic efficiency factors apply, so FET Rds, motor winding resistance and diode voltage drop all reduce the efficiency as you expect but assuming you can easily achieve 80% overall efficiency the PWM can be 25% and the current transformation is close to *4 *0.8 = *3.2.

    On my robot motor example the PWM was less than 10% and the overall efficiency was closer to 90% because I used a modern hbridge driver chip with synchronous rectification, so the FET Rds losses were very low and the diode forward drop was replaced with a FET with very low drop. So I expect I was getting close to 700mA through the motor winding with only 70mA drawn from the 12v PSU. :)
    Austin Clark likes this.