# Can a DC brushless motor produce the same torque at different power levels?

I am looking these Neo Brushless motors for a project. Here are the links to the data sheet and also to stall testing from the company:

So in the stall torque testing, on the last test with 80A limiting, they were able to achieve a torque of about 2.3 N-m.

You can see the current is about 25A and the voltage is about 10V. This gives a power of 250W at stall.

Now, looking at the data sheet, they list a stall torque of 2.6 N-m. They also give a voltage of 12V and current of 105A. This equates to power draw of 1260W. How is this same motor producing the same torque at wildly different power values? I assume this isn't gearing related since these are motor tests to figure out the limits of the motor itself. Does it have to do with current limiting?

• The graph doesn't show motor current or voltage. It shows input current and voltage. The motor controller can increase the current when it is using PWM. – Kevin White Sep 20 '19 at 19:14
• When stalled, motor outputs exactly 0 watts. $P=T\cdot\Omega$ Mechanical output power is not equal to electrical input power, which is the sum of mech.+power loss. – Marko Buršič Sep 20 '19 at 20:15
• Further, they produced a stall torque of 2.6Nm at 105A, and you 2.3Nm at 80A. If you compute kt you get almost the same value, so what's so different? – Marko Buršič Sep 20 '19 at 20:21

The motor controller acts as a buck type switching regulator with the motor windings acting as the inductor. Since the voltage across the windings when stalled is only the resistive drop from the current passing through them, there's no back EMF to overcome from mechanical work being done, the controller will be switching at a (fairly low) duty cycle that maintains the 80A limit through the windings, and this translates to only a smaller mean input current to the controller - the 25A you mention

As the motor warms up, and the resistance of the windings increases, the controller will increase the duty cycle to maintain that limit, and so the input current will rise for a steady phase current.

This is spelled out in the OP's linked page describing the testing -

Please take the following into consideration when interpreting the data below:

Average motor phase current (or winding current) is different than the average input current to the motor controller.

Average Input Current = Average Phase Current x Duty Cycle

Motor torque is proportional to phase current, not the input current. Therefore, it is important to control the phase current and not the input current.

2.3Nm at 80A and 2.6Nm at 105A are not too far out of line with each other.

• +1 for deciphering what really begs the OP. – Marko Buršič Sep 20 '19 at 21:11

Torque is proportional to current. Power out = torque x rpm.

At stall (locked rotor) there is no rpm, and therefore no power out. But torque is still proportional to current. Current is determined by the applied voltage and the motor's internal resistance, with the value determined by Ohm's Law (I = V/R).

If the motor is running at some rpm >0 then it also acts as a voltage generator, subtracting from the applied voltage and reducing the voltage across its internal resistance. Therefore to get the same current the input voltage must be higher, and the power input is also higher. This extra input power provides the mechanical output power that is now being produced. Some extra current is also required to overcome 'iron' losses inside the motor (which are proportional to rpm). However it is usually quite low - in this case 1.8A at 5676rpm, which reduces torque by less than 2%.

This is how the same motor can produce the same torque at 'wildly different power values'.

However,

You can see the current is about 25A and the voltage is about 10V. This gives a power of 250W at stall.

Now, looking at the data sheet, they list a stall torque of 2.6 N-m. They also give a voltage of 12V and current of 105A. This equates to power draw of 1260W.

When a controller is used to limit current the motor current may be higher than the power supply current.

The controller uses PWM to lower the effective motor voltage, relying on the motor's winding inductance to smooth out the current flow. Since the controller recirculates current through the motor during PWM 'off' periods, motor current is increased by the same proportion as voltage is reduced. This recirculated current is not seen by the power supply, so the power supply current may appear to be lower than expected.

This explains why the 80A torque test shows an input current of about 25A. The motor current is 3.2 times higher than the power supply current, which also implies that the motor voltage is ~3.2 times lower than the power supply voltage, ie. about 3V.

But what about the '2.6 N-m requires 12V at 105A = 1260W' anomaly? I suspect this is due to an invalid assumption that the empirical stall current is at the 'nominal voltage' of 12V. The motor probably has a lower voltage on it for this spec, just like in the 80A test. If the controller had to apply 33% PWM at 12V to get 105A motor current then the power would be 4V * 105A = 420W.

• This also equally answered my question. – Ryan C Sep 21 '19 at 2:28

A stall, the motor speed is 0 and so the power. This is the point that you are mentioning. The datasheet further mentions, the max power which is the middle speed position. The nominal output is somewhere near max efficiency output. You may also see that at stall the efficiency is zero, meaning all the electrical input power is converter into a loss- heat.

EDIT:

I see, what's is bugging you. The graph presents 12V DC link voltage, well that's not a motor voltage. It's the input voltage of the driver, which outputs the setpoint current, the voltage simply is $$\V=I\cdot R\$$ and not the DC link voltage, the driver does PWM.

How is this same motor producing the same torque at wildly different power values?

By generating mechanical power as well as dissipating power as heat.

If the motor is turning while exerting torque, it will be generating mechanical power (which is being absorbed by some load).