Running small DC motor at a higher voltage with PWM and feedback

[daqq]

Hello,

I have a 2.4V DC motor and a 5V power supply. I would like to run it directly from this, like chopper drives for stepper motors do - measure current/voltage on the motor and adjust the PWM so that there's a 2.4V average voltage.

Would this work? Anything to watch out for?

Thanks,

David

[brucehoult]

First!  And least qualified ahahaha.

Yeah, should be no problems, as long as the PWM frequency is well above any possible mechanical or electrical (LC) resonances.

Using 100V or something would be asking for trouble with breaking down insulation / air gaps etc, but 5V no problems if you don't go (much) over 50% duty cycle.

[Benta]

No problem.
Just feed it a 50% duty cycle, the extra 0.1 V will likely be lost in the power switch.
Don't go over 5 kHz, the iron losses will start giving trouble otherwise (and live with the whine).

[IanB]

You can do this maybe at lower voltages, but it could be a problem at higher voltages.

For instance, I once tried running a 12 V motor at 24 V with a 50% duty cycle, and the motor rapidly overheated. It didn't like it at all.

I didn't have the technical knowledge to know why at the time, but I did observe vastly more arcing at the commutator, which might have had a lot to do with it. Or it might be that the core saturated with the higher pulsed currents.

[Benta]

For instance, I once tried running a 12 V motor at 24 V with a 50% duty cycle, and the motor rapidly overheated. It didn't like it at all.

Too high PWM frequency? (see above).

[DavidAlfa]

50%? I'd say it's 25% for 200% voltage!

Let's say it takes 1A with 2.5V = 2.5W
With 5V it'll be 2A, so 10W, 50% PWM would do 5W.

[Benta]

50%? I'd say it's 25% for 200% voltage!

Let's say it takes 1A with 2.5V = 2.5W
With 5V it'll be 2A, so 10W, 50% PWM would do 5W.

Except it'll also be 2 A only 50% of the time.

[DavidAlfa]

Yes, 2A, doing 400% the work during that 50% of the time.
You're thinking in amps, not in power.
The average would be 1A at 5V right? That's 5W, twice the original power.

Back to math 101? (You deleted that later ) Maybe, or maybe not... we'll see!

[Benta]

Yes, 2A, doing 400% the work during that 50% of the time.
You're thinking in amps, not in power.
The average would be 1A at 5V right? That's 5W, twice the original power.

Back to math 101? (You deleted that later ) Maybe, or maybe not... we'll see!

No.

[DavidAlfa]

This is power, not a RC filter or something, which would effectively take the average voltage of 2.5V, what you need is the RMS voltage.

RMS=V*SQRT(DUTY).
5*SQRT(0.5) = 3.53V, 5W with 2.5 ohms.

To keep the original power rating, 25%..

I won't tell I knew the formula, I've never used it as normally you have dc or sinewave ac which is plain simple ( *sin(45) or /SQRT(2) ).

[Zero999]

This is power, not a RC filter or something, which would effectively take the average voltage of 2.5V, what you need is the RMS voltage.

RMS=V*SQRT(DUTY).
5*SQRT(0.5) = 3.53V, 5W with 2.5 ohms.

To keep the original power rating, 25%..

I won't tell I knew the formula, I've never used it as normally you have dc or sinewave ac which is plain simple ( *sin(45) or /SQRT(2) ).

That is true for purely resistive loads. Double the voltage at 50% duty cycle will indeed result in double the power dissipation in a resistor, but a motor is not a resistive load. It behaves like a very large capactior, with a resistor in parallel and another resistor in series with a small inductor.

I would measure the speed at the rated voltage, then connect it to 5V with the PWM duty cycle set very low and gradually increase the duty until the speed reaches that when run at the rated voltage.

[Siwastaja]

Yes, 2A, doing 400% the work during that 50% of the time.
You're thinking in amps, not in power.
The average would be 1A at 5V right? That's 5W, twice the original power.

Back to math 101? (You deleted that later ) Maybe, or maybe not... we'll see!

A motor is not a resistive load, as explained to you by others. This is not the first time you are being corrected from the exact same misinformation.

RMS voltage is meaningless for a motor. Root mean SQUARING models those physical phenomenon where squaring is the correct mathematical operation to do, which is for voltage->power relationship for RESISTORS, but not every other component or phenomenon. Specifically for a motor, increasing voltage increases RPM, which linearly increases back-EMF voltage. Current driven to the motor is not relative to input voltage, but (input voltage minus back emf voltage), therefore the quadratic relationship only applying during edge cases and transients (e.g. stall or start-up).

But you really want active current limiting if the motor is capable of large stall current.

[Doctorandus_P]

You can even run the motor directly from the 5V power supply for short durations. It may be very handy to have some peak power and torque in some applications. But with such an over voltage the motor heats up quickly, and it releases it's magic smoke when it gets too hot.

I have had a motor designed for 2 NiCd cells run on a 16V power supply for about 10 years before it finally wore out. It wore out because the copper of the commutator was completely worn away. In those 10 years, it controlled some roller blinds and it ran for around 20seconds twice a day. It was controlled by PWM and PiD, and it also had some extra lines of software to detect fault conditions (overcurrent, stalling motor, too long runtime) The PWM output was also limited to around 12%. If it ever got over that, the motor controller was put in a fault condition.

It all worked quite well. The main problem was that the motor itself was underpowered from the beginning, but I only noticed that after I built the complete system. And when it was built, I thought, "well, I've built it now, let's see how long it will last". And that turned out to be (a bit over) 10 years.

And for motors, I prefer a low PWM frequency (400Hz) above a higher PWM frequency. At a low PWM frequency, the hum caused by the PWM mostly drowns in the other motor noises (such as from the gearbox) while at higher PWM frequency the whining of the PWM is annoying.

[Siwastaja]

You can even run the motor directly from the 5V power supply for short durations. It may be very handy to have some peak power and torque in some applications. But with such an over voltage the motor heats up quickly, and it releases it's magic smoke when it gets too hot.

Yeah. More accurately, when a DC motor is fed with higher-than-rated voltage directly, two things happen:

1) Current at low RPM (e.g. stall) increases quadratically, so if you spend a lot of time generating a lot of torque, thermal overload of windings or brushes is likely. Active current control (measure, terminate PWM cycles on overcurrent) solves this issue completely and offers nice torque limitation even if input voltage seriously exceeds motor nominal voltage,

2) Speed at no-load condition increases linearly, and while this power dissipation wise isn't a big problem, brushes can't handle unlimited speeds, they will seriously arc. On a PWM controller, this condition is easy to avoid by simply capping the duty cycle.

[Benta]

A motor is not a resistive load, as explained to you by others. This is not the first time you are being corrected from the exact same misinformation.

Thanks. I was starting to write the same myself, but dismissed it. Why fight windmills.

[DavidAlfa]

Benta that wasn't your point of view, so don't talk like it was.
You specifically said 50% pwm would be the same, which is not, regardless of inductive and/or bemf effects on motors

[This is not the first time you are being corrected from the exact same misinformation.

Can't tell, I don't remember talking about this matter.
My intention was just to show 50% pwm on a load is not the equivalent of half the voltage in dc.
I never said a motor was a purely resistive load?
However a small DC motor will probably behave pretty close.

[Benta]

 

[Siwastaja]

However a small DC motor will probably behave pretty close.

No, it won't; no matter how small it is, if rotor is not locked it will generate a back-EMF voltage which needs to be subtracted. Your power calculation is bogus, that's all. Life goes on.

[Kleinstein]

A small motor will indeed have more resistive part than a typical larger motor. Unless one goes really tiny and very low efficiency (e.g. those motors used for vibrations in a phone), the inductive / back EMF part are still be relevant. If really wanted one could add a little inductor in series to the motor to add external inductance.

The suitable PWM ratio may be in between the 50% for a inductive and 25% for a resistive load, maybe around 40%. Depending on the needed speed one can adjust it.

With PWM drive the losses in the motor are a bit higher than with DC power - so one has to accept a reduced power (less maximum torque). The maximum voltage for the small motors is anyway not a hard barrier. It is a compromise between speed and longlivety.

One should have a free wheeling diode in parallel to the motor too. This way current can flow also from the inductive kick back instead of causing more ringing and EMI problems.

[Zero999]

I never said a motor was a purely resistive load?

By posting that formula, you implied it's a purely resistive load.

This is power, not a RC filter or something, which would effectively take the average voltage of 2.5V, what you need is the RMS voltage.

RMS=V*SQRT(DUTY).
5*SQRT(0.5) = 3.53V, 5W with 2.5 ohms.

To keep the original power rating, 25%..

I won't tell I knew the formula, I've never used it as normally you have dc or sinewave ac which is plain simple ( *sin(45) or /SQRT(2) ).

You got it wrong, no big deal, suck it up and don't allow your ego to get in the away of you learning.

[Siwastaja]

Remember, a motor is neither resistive, inductive, or resistive+inductive: the equivalent circuit consists of resistor, inductor and a voltage source in series. For a very efficient motor, you can maybe ignore the resistance. You never can ignore the inductance and especially not the voltage source.

(Voltage source can also be replaced with a huge capacitor, maybe with variable parallel resistor which models external mechanical force.)