I am trying to determine if it is safe to apply 24V across a 12V rated motor. I am working on a robot that weighs approximately 150lb and has a friction intensive drive system. I believe the motors are drawing >60A as my motor controller is giving an error.

If I switch to 24V would the current being drawn be cut in half?

Here is the spec sheet of the motor: http://files.andymark.com/CIM-motor-curve.pdf

Thank you for your help

EDIT: Also would it be possible to use two relays per motor instead of a motor controller to allow for the motor to draw higher currents?

  • \$\begingroup\$ "@MAX POWER: 67.9A" I wouldn't be surprised if they were. \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 25 '14 at 3:23
  • 2
    \$\begingroup\$ Surely this is first of all a physics problem, the question being does the required power (torque) to move the robot exceed what the motors are rated to provide? Perhaps simply gearing the system differently, to reduce speed but increase torque, would improve things? \$\endgroup\$ – John U Apr 25 '14 at 7:48

Ok, a couple of things:

  1. It's pretty much never a good idea to surpass recommended brushed DC motor voltage ratings, as doing so will definitely damage/overheat your motor leading to safety issues.
  2. If you need to increase your power/torque, the way to do that is to add another motor controller/motor and in your code set them to be equal/inverse.
  3. I've worked with that motor before, and it definitely should not be drawing >60 amps, even 40 amps is alot. Try using a multi-meter to test current, or put in a 40amp fuse in series with your motor wiring.
  4. What motor controller are you using? Considering you're using a CIM I assume it's FIRST (Talon, Victor, Jaguar, etc). If you motor was drawing over the maximum current you would not get an "error," rather it would simply stop working. I recommend revisiting your wiring diagram. Additionally, these motors draw a lot of current, make sure your power supply can handle them.

I would love to hear more about your project so I could better understand your issue.

  • \$\begingroup\$ I am using a Sabertooth dual 60A motor controller. The project is not for FIRST but for NASA RMC. Each motor has a 1:64 gear box and spins a 40lb steel screw from the inside. It looks like this: auto123.com/ArtImages/129370/Screw-Propelled-Vehicles-i001.jpg Also the motor controller has a red error light that goes on when the controller is overcurrent \$\endgroup\$ – brad123664 Apr 25 '14 at 3:47
  • \$\begingroup\$ Ok, try using a multi-meter or putting in a fuse to make sure you're really drawing that much current. Are the motors being stalled? I've done FIRST and for me it was rare when they took too much current. One possible explanation is that your input voltage is higher than the reccomended 12v, probably another thing to measure. \$\endgroup\$ – nanogru Apr 25 '14 at 4:01
  • \$\begingroup\$ The input voltage is 12.8V. The motors barely turn. They are close to being stalled or stalled some of the time. \$\endgroup\$ – brad123664 Apr 25 '14 at 4:03
  • \$\begingroup\$ Can you post a diagram or picture or your wiring? Something's definitely off here. Take my advice with the multimeter, if you don't have one try reading current off the sabertooth, although I highly reccomend purchasing one. \$\endgroup\$ – nanogru Apr 25 '14 at 4:08
  • \$\begingroup\$ @nanogru, doubling the voltage doesn't imply that the current will double. All it implies is that the no-load speed will double. Assuming the same torque, the current draw would stay (roughly) the same. \$\endgroup\$ – Eric Apr 25 '14 at 14:58

If you re-wind the coils for the higher voltage (twice the turns), then yes, you would see lower current draw for the same load.

But if you simply apply higher voltage to the existing motor, it will simply force more current through it, not less.


In general, you shouldn't apply more than the rated voltage to a DC motor. First, it will not cut the current in half. But it won't, as the other answers said, double the current either. In a DC motor, current is proportional to torque and speed is proportional to voltage. So doubling the voltage will double the speed. If your load is something like a fan or a pump where torque increases with speed, then your current will increase. If the load has constant torque as the speed increases, then your current will stay the same. If your load is constant power as the speed increases, then your current will actually decrease as the speed goes up. So without more information, it is difficult to tell you how increasing voltage (and therefore speed) will affect current.

Now, all that is assuming you don't have a control. Assuming your control is rated for at least 24 V AND assuming that you dial the speed back down to your original speed using the control, then the current draw of the control (that is, the input current to the control) will decrease. But the current that the control outputs to the motor will stay the same (for the same reason I mentioned above - current is proportional to torque). This will not fix your problem.

Your problem seems to be simply that your motor can't provide enough torque for your load. This could mean that you need to add a gearbox to get more torque (assuming you are okay with less speed) or that you need to find a bigger motor (assuming you can't sacrifice speed).


You might find these two blogs to be helpful: Controlling Motors With The Talon SRX and FIRST Robotics (FRC) Motor Modeling

They show you how to model/simulate a CIM motor and the Talon SRX controller. You can see how the voltage, current, and mechanical loads all interact. You can see the impact of gearing and multiple motors.

enter image description here


Motors may be damaged if any of the following occur (may not be a completely exhaustive list)

  1. Any part of the motor reaches a higher temperature than it can tolerate.

  2. The temperature differential between different parts of the motor causes parts to expand or contract out of relative tolerance.

  3. Any part of the motor is subjected to mechanical forces beyond what it can handle.

  4. The magnets in the motor are subjected to stronger magnetic fields than they can handle.

  5. Insulating parts of the motor have higher voltage across them than they can tolerate.

In general, DC motor ratings are based upon whichever of the above would cause failure first under "normal" operating conditions. If for a particular motor one knows which of the above factors was responsible for limiting the specified operating voltage, and knows that particular factor won't apply for some reason (e.g. the limiting factor was #1 but the motor will be operating at an ambient temperature of 4C, and will be operated for short enough intervals that it won't overheat) it may be possible to use the motor at a higher voltage without damage.

It's important to note, however, that even if one could ensure that the factor which had been limiting the voltage wouldn't be a problem, that would only raise the allowable voltage to the point where the next factor would come into play. A motor which is designed for 12V is unlikely to have insulation which is rated for 12,000V even if one could otherwise drive the motor with very short 12,000V pulses without exceeding any of its other ratings.

If the motor's insulation can handle it, driving the motor at 24V half the time and shorting it half the time, switching between the two operations fast enough that the inductance of the motor keeps the current fairly stable, would draw about half as much "average" current from the supply was would driving the motor at 12 volts, and the stress to parts of the motor other than the insulation would be similar to those resulting from 12-volt operation. Some motors would work just fine under such conditions, but others may not. A motor manufacturer should be able to offer guidance as to whether a higher-voltage pulse drive should be usable on any particular device.


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