I'm trying to build a dyno for electric motors. The dyno consists of two motors: a motor under test (MUT) and a load motor. The two motors are connected together through a shaft. On the shaft there's a torque sensor mounted.

For a load motor, I'm using a Revolt 160 Pro (12kW) with a SEVCON GEN4 controller. The idea is to run the motor from a battery or a power supply (connected to the SEVCON controller). When the load motor is acting as a generator, the three phase terminals of the load motor are gonna be switched to a circuit with a dump load to dissipate all the generated power.

My question is: let's say that I have an 8Nm clockwise input on the MUT and a 4Nm clockwise input on the load motor (since I want to load the MUT with 4Nm). The net result on the shaft it's going to be 4 Nm (clockwise referred to the MUT). Therefore, the load motor is going to act as a generator. When it's acting as a generator, the SEVCON controller and power supply are no longer part of circuit, the load motor is just connected to a circuit with a dump load. How can I control the load motor (now acting as a generator) so now it loads the MUT with exactly 4Nm? Do I need to have a variable and controllable dump load to achieve this application?

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EDIT: This is a rough schematic of the circuit of the dynamometer on the load side. There's two circuits, one used for the motor when it acts as a generator and another one used when the motor acts as a motor (and therefore it's controlled by a SEVCON controller). My question was how to control the motor when it acts as a generator, i.e. how to control the generator's load on the shaft.

enter image description here

  • 1
    \$\begingroup\$ Just to be clear. You plan on an input power equivalent to 8Nm and the MUT is 50% efficient? Because, if you drive a motor in torque mode without an equivalent counter torque, it will continuously accelerate. To test a motor at a constant velocity you must apply the same counter torque as the output of the MUT. \$\endgroup\$
    – lakeweb
    Sep 20, 2017 at 15:57
  • \$\begingroup\$ Those values are just an example to explain my question. The idea of the whole dynamometer is to test motors that will be fitted into an electric bike. Therefore, my idea is to be able to input road profiles, with uphills and downhills, to the load motor and see how the MUT behaves. Reading your answer as well as Charles Cowie's, I think I need to use torque control on the load motor and speed control on the MUT side. Still not clear how can I implement torque control on the load motor when it is acting as a generator. \$\endgroup\$
    – Uwe
    Sep 21, 2017 at 8:09
  • \$\begingroup\$ Ah, OK. It makes a lot more sense knowing the application. My first thought would be to buy an old power supply load tester. They have a bank of transistors/heatsinks. Then you can hack into the drivers to create a widely variable load. Because, of course, the voltage from the load motor will be proportional to the MUT speed, so all over the place. You will probably want to use a microprocessor to control the load. \$\endgroup\$
    – lakeweb
    Sep 21, 2017 at 15:42
  • \$\begingroup\$ PWMing a resistive load to vary effective resistance is easy and effective. (Long (long ...) ago I built exercise bike electronic load controllers). In my case we had a 3 phase alternator as part of the exercise bike and I rectified that to DC and PWMd a load resistor. Depending on how rapidly the load changes you can probably just monitor torque and RPM and control load to suit. \$\endgroup\$
    – Russell McMahon
    Dec 26, 2019 at 9:52

1 Answer 1

  1. A dynamometer must provide negative torque, torque that opposes the motion of the shaft driven by the MUT. Dyno torque plus motor torque = zero.

  2. The same parameter, torque or speed, can not be controlled by both the motor controller and the dyno controller. In this case, it is probably to control MUT speed and dyno torque.

  3. The dyno controller is useless unless it can control negative torque and return energy to the source.

  4. The easiest way to control the dyno is probably to not control the dyno motor at all. Just operate it as an uncontrolled generator. Rectify the generator output and use a chopper to control the load current. The current can be set manually or the current reference can be derived from a torque control error signal.

  • \$\begingroup\$ The reason to choose a motor as a load for my dynamometer, instead of just using a hysteresis brake, is that I also want to provide positive torque to the MUT. That is, I want to be able to provide both negative and positive torques to the MUT since the MUT will be part of a motorbike and I want to be able to simulate uphill and downhill profiles on the MUT. Thanks for the answer, I'm going to look into implementing a chopper to control the generator load. \$\endgroup\$
    – Uwe
    Sep 21, 2017 at 7:54

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