Looking at the datasheet, while it doesn't lie, it is definitely on the edge of misleading.

Note that the "max power" stated occurs at almost exactly half the unloaded RPM and half the stall current. This is indeed the "max power" point for such a motor, but the datasheet fails to mention that it is also the nominal 50% efficiency point, thus dissipating 12V*68A-337W = 479W in that tiny motor - destroying it, probably in minutes.

(Ideally, exactly half the power would be delivered, about 400W shaft and 400W heat, but the motor isn't ideal).

The motor is probably suitable for 100-150W continuous output and 200-250W short term.

So practically you must operate the motor at the upper end of the speed range, and if the speed falls below (say) 70% of the unloaded speed (or the current rises to 30% of the stall current) then - unless this is strictly temporary, like starting a heavy load or hitting a chilled spot while machining a cast iron surface, you need to cut the current and protect the motor.

Then the question of which side of the torque speed curve doesn't apply - unless the protection has tripped, you should be on the high speed side.

You can get circuit breakers that will allow short-term overcurrent. These are "motor rated" or Class C breakers for the AC motors used in most machine tools. I don't know of anything suitable for 12V DC though. I'd be looking for a 12V DC supply that can be set to trip if its output exceeds 40A for more than a couple of seconds. And as Olin says, if you want to monitor it yourself, measuring the current is definitely the way to go.

On the Faulhaber website there is available for download very extensive information about the motor including definitions of all of the constants, equations of the relationships among the constants and data items etc. Complete analysis of your set-up and the data obtained will probably explain everything. Each applied motor voltage creates a different speed vs. torque characteristic. Each speed creates a different generator voltage. Each generator voltage creates a different generator load. Each generator load creates a different motor torque. All of that can be calculated.

The Faulhaber motor current doesn't need to be controlled. To characterize the motor that is being used as a generator, it is sufficient to control the driving speed and the load resistance.

Note that the Faulhaber motor motor efficiency is only 37.3% at rated speed and load.

Rated speed = 4130 RPM

Rated torque = 2.9 mNm

Mechanical output power = 4130 X 2.9 / 9549 = 1.25 Watts

Rated voltage = 12 V

Rated current = 0.28 A

Electrical input power = 12 X 0.28 = 3.36 Watts

Total losses = 3.36 – 1.25 = 2.11 Watts

Motor resistance = 19.8 ohms

Losses at rated speed and torque:

Copper loss = 0.28^2 X 19.8 = 1.55 Watts

Friction torque = 0.08 mNm

Friction loss = 0.08 X 4130 / 9549 = 0.035 Watts

2.11 – 1.55 – 0.04 = 0.52 Watts windage and other losses

## Best Answer

$$\Omega=2\pi f= \dfrac{2\pi N}{60} = \dfrac{\pi N}{30} $$ $$\Omega=\dfrac{3.14\cdot 730}{30}= 76.4 [rad/s]$$