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I want to be able to test a motor (permanent magnet motor + gearbox) continuously in steady state mode at different operating points (torque, RPM,) to measure the efficiency of the motor.

Instead of a dynamometer where all the losses go to heat (this is in a low pressure environment with low heat transfer,) the idea is to load the motor with an identical version of itself.

Current idea for setup: DC power supply (w/ a DIY power analyzer for V, I measurement) -> test engine -> torque transducer -> load engine -> power generated goes somewhere.

I have been told that the test motor is capable of demanding a certain level of torque based on "regeneration mode."

I have no experience controlling motors, and have tried researching this problem, but have no idea where to start. Currently, all I know is that we want to hook up the regen power to the "same DC link."

Is this setup reasonable/possible? If so, and the regen mode control is non-trivial, how do I go about figuring out how to make it work?

Secondly, if we have DIY power analyzer on the output end somehow, is the torque transducer redundant?

Remember, we need either Pout/Pin knowing the operating point, or T × ω/Pin, for efficiency.

Where would the power analyzer be measuring exactly? I know on the front end there's voltage outputs on the power supply to measure current and voltage (and filter current).

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  • \$\begingroup\$ What does this mean: the "same DC link" \$\endgroup\$
    – Andy aka
    Jan 20, 2021 at 17:19
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    \$\begingroup\$ 2 gearboxes in series reversed will be pretty lossy. You didn't mention the ratio but I hope it's pretty low. \$\endgroup\$
    – user16324
    Jan 20, 2021 at 20:49
  • \$\begingroup\$ Brushed or brushless? Motor specs? Gearbox ratio and type? "this is in a low pressure environment with low heat transfer" - exactly what is this 'environment'? Why do you need to measure efficiency, and how accurate does it need to be? \$\endgroup\$ Jan 21, 2021 at 5:46
  • \$\begingroup\$ @BruceAbbott Brushless PM synchronous motor. About 2kW operating point. 5:1 planetary gearbox. The environment is an artificial vacuum chamber operating at 40 Torr (so not much convection). I need to measure efficiency to judge the performance of the system...needs to be pretty accurate as it is the whole point of the test. I cannot give an exact resolution as I don't know what's possible. \$\endgroup\$ Jan 21, 2021 at 19:18

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Interesting idea. You may find that it's more trouble than it's worth compared to simpler dissipative methods.

An ideal permanent-magnet electrical machine generates a torque that's exactly proportional to the current through the armature, regardless of whether it's acting as a motor or a generator. So in a perfect world you'd just connect two motors back to back, and you'd load the motor under test by putting a current on the test motor and you'd be done.

This is aided by the fact that if you're driving a motor with a simple H-bridge drive, when you backdrive the motor you get power out of the drive. However, this regeneration is a problem in a lot of practical applications, which is dealt with by dumping the excess power into a resistor instead of stuffing it backwards out of the drive. This means that if you're choosing a motor drive for this you need to find one that comes set up to back-drive its supply, or that can be configured to do so.

Real motors have all sorts of losses that reduce the amount of torque for a given amount of current. Some of these are speed dependent (mostly bearings and wind resistance), some are electromagnetic and current dependent (armatures magnetically saturate), some are electromagnetic and dependent on voltage and speed (armatures have eddy currents). Moreover, gearboxes are lossy, too.

Up to a point, with two good motors and no gearboxes, you could probably still just connect the motors back to back. If your test motor drive is connected to the same electrical power rail as your motor under test's drive, then power will flow between them, and the electrical power consumption will be minimized. For real precision you'll still want to measure, and control for, torque.

With a lot of gear reduction, the losses in the driven gearbox will be so high that you just won't be able to control torque directly. In this case I think the safest approach would be to make a spring coupler between the test motor and the motor under test. Measure the torque, regulate the speed of both motors in an inner loop, and control the speed of the test motor to regulate the torque.

Note that this just won't work for (or at least will be very difficult for) a typical worm-gear drive, because you can't back-drive those. If the gearbox efficiency is 50% or less in the forward direction, then in general it cannot be backdriven at all, and because of the dynamics of friction, it'll be practically impossible to control the speed of the load motor when it's being driven.

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  • \$\begingroup\$ It's going to take a while for me to digest all of your answer due to my background, but in the meantime, I want to clarify that part of the test is test the whole engine efficiency (including the gearbox effect). We expect a 98% gearbox, and we have confirmed it can be driven in reverse. We will eventually decouple and test the motor alone. \$\endgroup\$ Jan 20, 2021 at 21:19
  • \$\begingroup\$ If the gearbox efficiency is 98%, you may just be able to couple the motor under test and the test load with a rigid coupling, and regulate torque by regulating the test load current. If possible, I'd test that with just a resistor on the test load, and maybe spinning its input shaft by hand, if that's sensible. \$\endgroup\$
    – TimWescott
    Jan 20, 2021 at 22:11
  • \$\begingroup\$ I was told that instead of supplying power back to the source, a simpler way would be to dissipate the power. Fair enough. Now, downstream of the regen motor and controller, I was told that I would need a 3-phase rectifier followed by a DC programmable load. I'm a bit confused, because I was under the impression that the regen motor controller controlled the current requirement, so what would be the need for a DC programmable load here? Couldn't I just hook up a resistor beyond the rectifier and have the current be controlled by the controller? \$\endgroup\$ Jan 27, 2021 at 20:44
  • \$\begingroup\$ "Have been told" is pretty vague. You need to control the current in the load machine one way or another -- you can do this with a 3-phase motor controller running in regen mode (complex), or you can do this with a bridge rectifier and a controlled load (less complex) or you can do this with a bridge rectifier and a resistor (less complex and less control). But it's not even remotely sensible to do more than one at a time. \$\endgroup\$
    – TimWescott
    Jan 27, 2021 at 21:34
  • \$\begingroup\$ Sorry, "was told" by an experienced engineer who I got 10 mins with. Thanks, I think that answers my question. Basically, the programmable load and the regen controller are redundant. And I don't need a programmable load if I have a regen controller \$\endgroup\$ Jan 27, 2021 at 21:54
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The described setup is possible. It is reasonable to the extent that the gearbox can be reasonably driven from either side. With a commutator DC motor, the motor armatures can be connected together such that the power generated by the load machine can help to power the machine under test. Electrical power would be measured at the input to the machine under test. Mechanical power would be calculated from the torque transducer output and a speed measurement at the same point.

A practical implementation, would probably require a means of regulating the speed of the machine under test and regulating the load torque presented by the load machine. That might mean an output voltage controlled DC to DC converter for the machine under teat and a current controlled DC to DC converter transferring current to the DC link from the load machine. A third converter will be required to supply system losses to the DC link.

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  • \$\begingroup\$ So can this be powered by a power supply less than the actual input power? It's a 2kW engine at full operation, but let's say the output power is 1.8kW, can we power it using a power supply less than 2kW? \$\endgroup\$ Jan 21, 2021 at 19:20
  • \$\begingroup\$ The outside power is equivalent to the losses in all of the components. If the gear-motor combination has 90% efficiency, you might assume that the control units for each motor might each also have 90% efficiency. That would make the overall operating efficiency .9 x .9 x .9 x .9 = .66, so you would need something like .34 x 2000 = 688 watts to supply the power to the test setup. If the power supply that does that is also 90% efficient, you would need 764 watts from the outside. \$\endgroup\$
    – user80875
    Jan 21, 2021 at 22:30
  • \$\begingroup\$ Yes, that is what I meant. Assuming the TOTAL system losses are 200W, or let's say 764 W using your more realistic scenario, would we only need a DC power supply capable for > 764 W? I'm just wondering if there are other reasons for which you would still need the whole 2kW on the input supply side? \$\endgroup\$ Jan 21, 2021 at 23:13
  • \$\begingroup\$ The only reason that you might need to supply more than the losses, whatever they are, would be the need to build up stored energy. From what I have see so far, I don't think that would be significant. \$\endgroup\$
    – user80875
    Jan 21, 2021 at 23:24
  • \$\begingroup\$ What if instead of supplying power back to the source, we simply tried to dissipate the power. Now, downstream of the regen motor and controller, I was told that I would need a 3-phase rectifier followed by a DC programmable load. I'm a bit confused, because I was under the impression that the regen motor controller controlled the current requirement, so what would be the need for a DC programmable load here? Couldn't I just hook up a resistor beyond the rectifier and have the current be controlled by the controller? I'm confused because it feels like the load motor IS the programmable load. \$\endgroup\$ Jan 27, 2021 at 20:44
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It can be done of course. The output of the system under test needs to go to a low-friction transmission with output speed higher than the input speed (eg. a multi-v belt stage or two). The whole point here is to make it cheap and easy – using whatever fancy transmission you use in the DUT will be likely a waste. A couple bearing blocks and shafts mounted to a frame is easy, heck, the frame can be probably made out of lumber if you don’t want to weld.

The output of that has to be coupled into the motor that presents the load, and that motor’s output energy would be dumped into a regenerative resistor brake connected to the motor servo drive’s DC link. The resistor’s connection is controlled by the DC link voltage (all regen energy dumps work that way, but sometimes the switching element is within the drive itself, sometimes it’s with the resistor pack).

Then you’d set the load motor’s drive for torque control, and apply whatever load torque you want for testing.

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  • \$\begingroup\$ The load side motor's control seems very daunting and non-trivial to me. How do I go about "applying whatever load torque" I want, for example? Do I assume the drive has a regenerative mode that is easy to control and I just need to wait and read the manual (it's not been sent to me yet, I'm assuming until after I purchase)? Would I need a rectifier post-controller on the load side? And is the "DC link" a shared rail after the DC power supply that the output power also flows into? Obligatory note: no electronics background so sorry if these are dumb questions \$\endgroup\$ Jan 20, 2021 at 22:01
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the idea is to load the motor with an identical version of itself...

I have been told that the test motor is capable of demanding a certain level of torque based on "regeneration mode."

If you can control the braking torque of the load motor then you just need an rpm measurement to calculate power. Accuracy of this setup will depend on how accurately the load motor's torque is maintained. If you need higher accuracy then mount the motor on a base that can rotate and attach a load cell to measure torque directly.

The driven gearbox will have some loss that should be taken into account. A high quality 5:1 planetary gearbox should have very low loss so you could just estimate it. If that is not good enough then measure it separately with a dynamometer.

If ... the regen mode control is non-trivial, how do I go about figuring out how to make it work?

That will depend on the controller you are using. You could try reading the operating manual.

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  • \$\begingroup\$ If I choose to simply dissipate the generated power, would I need a programmable load? If the regen motor is in "current/torque mode", then wouldn't a rectifier downstream of the controller, followed by a dumb resistor, be sufficient? What would be the need for a programmable load. I guess I'm confused because it feels like the load motor IS the programmable load that is being controlled by its driver. \$\endgroup\$ Jan 27, 2021 at 20:47
  • \$\begingroup\$ The load motor itself is not the load. It generates power that has to go somewhere. Where it goes depends on the controller. It may use a resistor bank to dissipate the energy and prevent the supply voltage from rising due to current back-fed into it, or may use this effect to feed AC power back into the mains. Either way the generated power has to be properly routed to its destination. You can't just whack a resistor bank 'downstream of the controller' and expect the controller to work. OTOH you could do it with a bare motor, then measure current to get torque. \$\endgroup\$ Jan 27, 2021 at 22:57
  • \$\begingroup\$ "What would be the need for a programmable load" - If you use a fixed resistor bank then it will have a fixed rpm to torque ratio. To test a motor you need independently variable torque. This can be done with a variable or switched resistor bank, or a single resistor with PWM applied. \$\endgroup\$ Jan 27, 2021 at 23:05
  • \$\begingroup\$ If I'm understanding correctly...are these two equivalent? 1) bare motor w/ no controller + rectifier + DC programmable load; 2) motor w/ driver controlling current draw and a simple resistor that burns this power up. I think I'm a bit lost and my lack of EE experience is not helping \$\endgroup\$ Jan 27, 2021 at 23:38
  • \$\begingroup\$ Yes, they are equivalent. The controller effectively is a programmable load when it is braking. And torque is proportional to current, so if you know the torque constant you can measure the current to get torque, or even adjust the load to maintain a set current and then torque will be constant. \$\endgroup\$ Jan 28, 2021 at 3:00

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