I have a Faulhaber DC motor (model 1524T012SR) that I coupled to a cheap DC motor in order to characterize the cheap motor to use as a generator (and get RPM and torque from voltage and current). The Faulhaber motor uses a motion controller (MCDC 3006S) that was purchased with the motor years ago. I am using the motion controller to step through various motor speeds from 1000 to 9000 RPM.

I expected the current output from the Faulhaber motor to be linear for a linear change in speed but it is not. The current increases until about 2500 RPM and then starts to drop as the RPM increases, similar to a stepper motor response. The problem that I have is that the datasheet provided with the motor only gives a single torque constant (11.5 mN-m/A) so I have no idea how to get the actual torque supplied to the coupled motor.

Is this a problem caused by using the speed controller for the Faulhaber motor? What can I do to determine the actual torque output by the Faulhaber motor?

Here is my setup: enter image description here

Here is the current drawn by the driving motor for a linear increase in speed: (it says torque but it's just the current times Kt.) enter image description here

  • \$\begingroup\$ Maybe a torque gauge for a shaft? google.com/… \$\endgroup\$
    – Voltage Spike
    Commented Jun 26, 2017 at 16:16
  • \$\begingroup\$ You could also calibrate the faulhaber motor for known torques and speeds and only take data in the steady state condition. One problem you will face is the other motor is a variable load \$\endgroup\$
    – Voltage Spike
    Commented Jun 26, 2017 at 16:17
  • \$\begingroup\$ The idea behind doing it this way was to not spend $3k+ on a rotary torque sensor and just use existing equipment, understanding that there would be some inaccuracies involved. Since the Faulhaber motor already provides a torque constant, shouldn't it be more or less "calibrated"? (I understand this isn't exactly true but I'm trying to avoid all the cost associated with torque measreuments). \$\endgroup\$
    – Kyle Jones
    Commented Jun 26, 2017 at 16:32
  • \$\begingroup\$ So if you went the calibrated route, if you know the speed then you know the torque and the amount of power. Dynamics will be a problem but if you take measurements at steady state then you should be fine (the load needs to be constant also), otherwise you'll have to model the dynamics. \$\endgroup\$
    – Voltage Spike
    Commented Jun 26, 2017 at 16:50
  • \$\begingroup\$ @CharlesCowie The voltage for each of these points increases linearly, as I would expect for the current as well (which isn't the case). The controller limits the current to 0.28A but the peak here is only 0.22A. I tried increasing the current limits on the controller and it had no effect on the response. There is a 6 wire connector between the motor and controller but the motor has an added encoder on the back. \$\endgroup\$
    – Kyle Jones
    Commented Jun 26, 2017 at 17:01

3 Answers 3


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

  • \$\begingroup\$ Thanks Charles. I think the issue that I was having is that I had assumed the torque/speed response would be linear all the time, not only for a given voltage. I should be able to use this information you provided to get the motor characteristics for what I need. \$\endgroup\$
    – Kyle Jones
    Commented Jun 27, 2017 at 12:59

A brushed DC motor has a linear current vs torque curve. The upper boundary of the curve is the stall torque and current. The lower boundary is zero torque, zero current. The slope of this line is the proportionality of motor current to torque (A/Nm), and is called the current constant. The reciprocal of this slope is the torque constant of the motor (Nm/A).

The motor spec sheet lists the stall torque as 6.52 mNm and the torque constant as 11.5 mNm/A. From this we can compute that the stall current is 6.52 mNm/11.5 mNm/A or ~0.57 amps. You can confirm this experimentally by locking the rotor and measuring the current.

So in summary, simply measure the motor current, multiply by 11.5 mNm/A, and you will have the mNm of torque developed by the motor.

If you wish to calculate RPM, first measure the resistance (R) of the motor (take several readings, rotating the shaft each time, and take the most commonly read value). The spec sheet lists this as 19.8 ohms. The Back-EMF constant, Ke, for your motor is listed as 1.21 mV/min-1.

Now monitor motor current (I) and voltage (VO) and calculate RPM as:

RPM = ((VO-(I*R))*1000)/Ke

  • \$\begingroup\$ This is essentially what I am trying to do by I'm wondering why the torque output by my driving motor is increasing with RPM at lower speeds and then changes to decreasing with increasing RPM at higher speeds. \$\endgroup\$
    – Kyle Jones
    Commented Jun 26, 2017 at 21:06
  • \$\begingroup\$ If you use the two equations together you can then model the RPM to Torque relationship. Is this what you wish to do? \$\endgroup\$
    – Glenn W9IQ
    Commented Jun 26, 2017 at 21:16
  • \$\begingroup\$ Your original question was "What can I do to determine the actual torque output by the Faulhaber motor?" I answered that you measure your motor current in amps and multiply by 11.5 to get the torque in mNm. Do you need assistance with other parameters? \$\endgroup\$
    – Glenn W9IQ
    Commented Jun 27, 2017 at 11:00

The Current required by the driving motor is not necessarily a linear function of speed.

The current into the driving motor depends on the torque required of it as well as a other factors. The torque depends upon its own friction as well as the output torque to drive the load. There will also be current required to account for electrical losses. These other terms may not be linear with speed.

To determine the torque being absorbed by the generator first plot a curve of the motor current without the generator attached - this will give you the current required just to spin the motor. Then repeat the experiment when driving the generator. The difference in current should be a function of the torque to drive the generator.

You should perform the test with various loads on the generator at each speed to get a complete characteristic.

Subtract the plot with no-load from the ones with a load. That will represent the additional current required to drive the generator. Multiply that current by your torque constant to get the actual torque required by the generator.

  • \$\begingroup\$ I'm referring to the current input to the driving motor. I expected the current used by the driving motor should be linear with a linear change in voltage (speed). Is that not correct? \$\endgroup\$
    – Kyle Jones
    Commented Jun 26, 2017 at 16:09
  • \$\begingroup\$ No, not necessarily true. \$\endgroup\$ Commented Jun 26, 2017 at 16:14
  • \$\begingroup\$ Added in the diagrams for clarity. Where am I wrong in my thinking then? \$\endgroup\$
    – Kyle Jones
    Commented Jun 26, 2017 at 16:33
  • \$\begingroup\$ Your diagrams are not working. Your presumption that current is proportional to speed is incorrect. Back EMF is proportional to speed. Current is proportional to torque. Torque is not proportional to speed. \$\endgroup\$ Commented Jun 26, 2017 at 16:48
  • \$\begingroup\$ I'm not assuming current is proportional to speed. I just want to know how current relates to torque so that I can get a torque v speed curve to characterize the driven motor. \$\endgroup\$
    – Kyle Jones
    Commented Jun 26, 2017 at 17:05

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