1
\$\begingroup\$

I am using a DC motor as a generator for my small toy/experiment. The problem is that the motor/generator can not provide enough counter EMF or torque or whatever the correct word is. So the rotor on the other side just keeps spinning at too high RPMs. I would like to control the rpm of the rotor to do so I will change the motor. My question is what is the right characteristic to check while buying the new one. Power supply (Watt), the speed constant (RPM/V) , the supply voltage (V) and torque constant (mNm/A)

Clarification:

I am using it with a small wind turbine model. It is connected to a circuit where I can play with the resistance. I change the resistance and read the voltage output by that I get the power extracted. It is being extracted, so it should be connected correctly I guess. But the resistance range of 10 to 10000 Ohm at a constant incoming wind speed, the rpm of the rotor only changes from 5000 to 5200. surely The RPM goes higher with an increasing incoming wind speed but the change of rpm with same resistance range is very limited. so somehow the counter EMF is not enough need to buy a new motor?

Right now the characteristics of the motor I am using:

  • Maxon RE 10 Ø 10 mm,
  • Precious Metal Brushes,
  • 0.75 Watt Values at nominal voltage
  • Supply voltage 12 V
  • No load speed 11500 RPM
  • No load current 5.37 mA
  • Nominal speed 2790 RPM
  • Nominal torque (max. continuous torque) 0.731 mNm
  • Nominal current (max. continuous current) 0.081 A
  • Stall torque 1.01 mNm
  • Starting current 0.106 A
  • Torque constant 9.55 mNm/A
  • Speed constant 1000 rpm/V
  • Speed / torque gradient 11900 rpm/mNm
\$\endgroup\$
  • 1
    \$\begingroup\$ could you explain what you are trying to achieve in more detail. I'm confused as to why you can't control speed. \$\endgroup\$ – Andy aka Dec 24 '13 at 10:58
  • \$\begingroup\$ Are you asking how you can use a DC motor as a brake? Or how you can use it to generate electricity? Or how you can power it and control its speed? Or what? \$\endgroup\$ – Brian Drummond Dec 24 '13 at 11:34
  • \$\begingroup\$ I worked on a project where we used a small dc motor as a brake for a scale-model hydraulic turbine. If that's what you are trying to do, please clarify and perhaps the question will be reopened. \$\endgroup\$ – Joe Hass Dec 24 '13 at 12:02
  • \$\begingroup\$ The driving motor high RPM's mean it's not seeing enough load. Questions; is the generator turning the right direction to produce power? Is there an electric load on the generator? \$\endgroup\$ – Optionparty Dec 24 '13 at 14:10
  • \$\begingroup\$ Is the stator terminals left unloaded? if that is the case then yes it will just accelerate off based upon the shaft torque and any additional frictional loads. Do some quick sums and put a power resistor (right R, right W) across the stator terminal and you should see the speed settle out. Then you need to change that resistor into an active powerload \$\endgroup\$ – JonRB Dec 24 '13 at 15:48
1
\$\begingroup\$

A typical DC motor may be reasonably accurately modeled as an ideal motor in series with a resistor and inductor. At any moment in time, the voltage on an ideal motor will be proportional to the rotational speed, and the current proportional to the magnetic torque. The proportionality constants for speed and torque will be such that one watt of electrical power will yield one watt of mechanical energy.

When trying to use a motor as a generator, if shorting its leads doesn't remove enough mechanical energy from the source, you'll have to add a gearbox or use a different motor. Trying to extract maximum energy from a given source of mechanical energy may be tricky, but if even a dead short doesn't result in enough mechanical loading there's no need to work things further.

\$\endgroup\$
  • \$\begingroup\$ Quote question above; My question is what is the right characteristic to check while buying the new motor. Power supply (Watt), the speed constant (rpm/V) , the supply voltage (V) and torque constant (mNm/A) \$\endgroup\$ – user2375049 Dec 25 '13 at 21:03
  • \$\begingroup\$ If your source RPM is limited to a particular value, the maximum current you're going to get out is going to be the speed constant times that speed, divided by the DC resistance. The maximum usable power at that RPM will be obtained at half that current, and will be half the earlier-computed voltage times half the earlier-computed current. \$\endgroup\$ – supercat Dec 25 '13 at 22:05
  • \$\begingroup\$ so the motor i am using has a speed constant of 1000rpm/V and my rpm varies with incoming wind speed but lets say for constant wind speed, 5000rpm => 5V when it is loaded with 10 Ohm. This corresponds to 0.5 Amps. Then the power would be 0.25*2.5=0.625W u are saying.Again the motor i am using is 0.75W. So the conclusion is I should try to find a more powerful motor (watt-wise) or i should be checking the speed constant? I understand what your answer is but cant connect to the optimum motor. here is photo for my problem. i1372.photobucket.com/albums/ag353/emre1371/… \$\endgroup\$ – user2375049 Dec 25 '13 at 22:43
  • \$\begingroup\$ You haven't mentioned the DC resistance of the motor; based upon the stall torque, torque constant, and nominal voltage, I would infer that to be about 113 ohms. Thus, when you apply a 10 ohm load at 5000rpm, it would see about 5V/(113+10)ohms = 40.7mA at about (40.07mA*10ohms) = 0.407V. Power would be about 16.5mW. If you used a 113-ohm load, the current would be reduced to about 22mA, but your voltage power would be increased to 2.5. Your power would be about 55mW. \$\endgroup\$ – supercat Dec 25 '13 at 22:58
  • \$\begingroup\$ I would like to refer you to my main question in the top of this, page first 2 paragraphs. There I explain everything so it would be wrong to redo it here now. But i have varied the electrical resistance from 10 to 10k Ohm at different rpms (or wind speeds) if you click this you see a sample graph of it, and if you read the question asked on top of the page you will see what I am seeking. i1372.photobucket.com/albums/ag353/emre1371/… \$\endgroup\$ – user2375049 Dec 25 '13 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.