I've been searching around the internet on and off now for over a month and now, and I fear my question is so basic that I can't find an answer. Or I've seen the answer and just recognize it. All I am trying to do (at this time) is power a small dc motor that I bought from Radio Shack from a "wall wart" ac/dc converter. I have an on/off switch. And while I know enough to wire this all together, I don't know how to power the motor with out burning it up.

The motor is the Radio Shack Hi-Speed 9-18Vdc Motor, #2730256. Radio Shack does not provide specs with its components, but I was able to find this online: "The 12-18V within the 1.98A it requires. The motor is designed to run at 1.98 AMPS MAX!!"

The first power supply that I tried was the wall wart from an old laptop that was rated at 19V 3.4A. This worked great when the motor did not have any load applied. It pulled around 0.25 amperes. When I put a load on it, however, the motor then pulled over 4 amperes.

"No problem," I though, I'll just purchase a more appropriate power supply. On eBay if found an 18V 2A power supply. Perfect! When that arrived, I hooked it up, switched it on and the motor then drew 4.5 amperes!

So now I'm lost. My naive assumption was that a 2 ampere power supply would not supply more that 2 amperes. Is this not true, or is the power supply I purchased junk?

Or, do I really need to be looking for some sort of current limiting circuit that limits the current to just under 2 amperes? This would allow me to return the supply I purchased and just use the one I had. It would also protect the motor from current spikes. I've tried to read up on this, but the more I read, the more confused I get.

The application I have in mind will require lots of torque, so I'm trying to keep the current supply as close to its max as is safe. My next step will be to control the speed using PWM with either a 555 timer or an Arduino board. But for now, I'd be happy just to switch it on and have it run for 6 hours and not burn out the motor. How do I do this? How do I keep the dc motor from drawing more current than is healthy for it?

  • 2
    \$\begingroup\$ How did you measure the 'load' you place on the motor? The specs specify between 150 g.cm (9V) to 290g.cm for 18V with a corresponding 1.1A (9V) to 2.4A (18V) loaded or did you just stall the motor? Too much load = high 'stall' current. \$\endgroup\$ Oct 6, 2013 at 17:13
  • \$\begingroup\$ I didn't, nor would I know how to, measure the load. I simply attached the disk I need this motor to rotate and threw the switch. Does all this suggest I'm asking too much of this motor? To spin this massive a disk at full throttle (well, no throttle) is unreasonable? \$\endgroup\$ Oct 7, 2013 at 3:47
  • \$\begingroup\$ A large 'heavy' disk has a lot of rotational inertia - so expecting a small motor to spin it up to (high) speed from stationary could be a bit too much for it. 290g.cm is not a lot of torque. \$\endgroup\$ Oct 7, 2013 at 15:55
  • \$\begingroup\$ If you are spinning a disc up to speed and then keeping it there, don't worry if it draws (say) twice its continuous rating while getting up to speed, as long as it draws less than rated current at speed. If you are accelerating it frequently, that's a different matter... \$\endgroup\$
    – user16324
    Oct 8, 2013 at 11:46

4 Answers 4


What's important is to not overheat the motor or any of its parts (windings, brushes, bearings, etc.)

The heat in the motor comes from current-squared times resistance.

If the motor is rated for 1.95 amps max, then you have three options to avoid overheating it:

  1. Make sure the load is never so high that the motor stalls or overloads.
  2. Make sure that the voltage is so low that the current through the windings will never be higher than rated.
  3. Use a current controller to drive the motor that can limit the current at the given maximum. (Typically, average current is limited through PWM.)

The reason your "2A power supply" didn't do the limiting is that it wasn't built with "continuous current limiting" as a feature. There exists power supplies that have this feature, but they are generally more expensive as it's usually harder to build that feature than either an unlimited power supply (that you can destroy by overloading it) or a intermittent-limiting power supply (that turns off entirely on overload or overheat.)

The amp rating for a power supply is generally how many amps it can safely deliver -- NOT a limit or exact number, like the voltage rating. The reason is that you don't "push" amps into a load; the load "draws" amps based on the voltage you supply and its internal construction (resistance, impedance.)

My recommendation for you is to get a 12V/2A power supply and see how much current is drawn if you stall the motor. If < 2A, great! If > 2A, reduce voltage even more, until the current drawn when stalled is acceptable.

If that's not good enough, then get a power supply that lets you establish a current limit, where the response of that power supply is to reduce voltage until current is under that limit. You can buy these as "components" from places like Jameco or Digi-Key, or you can buy a cheap 18V/3A benctop power supply from Amazon (which will also come with handy digital read-outs.)

  • \$\begingroup\$ The next step in my plan was to use some PWM to control the motor. Is it possible that using PWM to throttle back the duty cycle will keep the motor "happy and safe"? I guess I don't understand what is critical for the motor's well being - never letting the current exceed the specs, in which case PWM won't help. Or, keeping the current in line over time, in which case PWM would be just the thing. You mentioned "average current", is this the key concept I'm missing? \$\endgroup\$ Oct 7, 2013 at 3:33
  • \$\begingroup\$ Never let the average current exceed the specs, where the average is calculated over some time window that may be measured in seconds or minutes, depending on the motor and application. Also, the higher the peak current is, the shorter the window. PWM is just fine! The inductance in the motor coils will also serve to average out the PWM signal, in addition to the thermal mass of the windings/motor. Also, some parts (brushes that make/break contacts) are more sensitive than others to instantaneous current (bearings are only sensitive to average heat, for example.) \$\endgroup\$
    – Jon Watte
    Oct 8, 2013 at 15:20

Motors are fairly robust and you need not worry about the difference between - for example - 1.98 and 2A. However over-current can shorten the life of a motor in two distinct ways :

1) In the long term, the motor can overheat. It is not unusual for a motor rated at 2A continuous to have a higher short term current rating - maybe 4A intermittent, for max. 10 minutes in any half hour. The key is to keep the temperature of the motor down - either by limiting the current to 2A, or allowing it to cool down, or by improving cooling e.g. with a fan.

2) DC motors usually have brushes - either carbon blocks or metal contacts rubbing against the commutator. The latter are especially prone to wear through melting, either from sparking or simply too high current. Carbon brushes are tougher, but they can also wear and overheat. On some motors they are easily replaceable, which is one way to increase life from a hard driven motor!

In your case, if you want 6 hours continuous run, you need to consider the long term current draw.

And that really means, limiting the torque on the motor.

Limiting the current would only stall the motor when the torque is exceeded - that can actually overheat the motor faster if it has an internal cooling fan!

Better to run the motor faster and gear it down until it can supply the torque you need without excessive current.

  • \$\begingroup\$ The specifics of how long you can overheat a motor before it takes damage varies by the motor. I've used servos based on Maxon motors (high quality!) that would overheat enough to take damage by being stalled at 5 seconds with max rated voltage applied. \$\endgroup\$
    – Jon Watte
    Oct 6, 2013 at 18:05
  • \$\begingroup\$ @Jon Watte Absolutely! One factor is the ratio of stall current to rated current; for a high performance servo this will be huge and the lifetime at stall, very short. Can you check the ratings at a moderate 2:1 overload (as in my example) rather than stall, for the Maxon? \$\endgroup\$
    – user16324
    Oct 6, 2013 at 18:40
  • \$\begingroup\$ If you are careful with limiting current you can drive DC motors quite hard, in the OP's case he needs to spin something heavy up to speed. So, you may stress the motor at startup by running it near its limit, but once it's going it's well below its spec. \$\endgroup\$
    – John U
    Oct 8, 2013 at 8:40

What you need to do is determine exactly how much torque you need to accelerate your disk to the speed you want it to spin at and then size your motor appropriately. This is a fairly simple physics problem and all you need to know is the moment of inertia of your load and how fast you want to accelerate it. For a fairly simple object like a large disk, you'll need more torque to get it up to speed than you will to keep it moving.

The formula for torque is \$\tau = I*\alpha\$, where \$\tau\$ is torque, \$I\$ is moment of inertia, and \$\alpha\$ is angular acceleration. The moment of inertia for a disk is the same as for a cylinder, \$I = \frac{1}{2} m*r^2 \$, where \$m\$ is the mass of your disc and \$r\$ is the radius.

So when you turn your motor on, it is going to draw as much current as it can to accelerate this disk. If you come up with some method to limit current to 2 amps (which, for a given motor, is equivalent to limiting the torque to a certain value) then that is equivalent to limiting the acceleration of your load (\$\alpha = \frac{\tau}{I}\$). Your other option would be to limit how quickly you accelerate the load, which would limit how much torque you need. This could be as simple as slowly applying more voltage over a period of time instead of hitting the motor with 18 V instantaneously. Since it sounds like you plan on controlling speed anyway, this might be something you'll want to try.

And this all assumes, of course, that your motor is sized correctly. I'd encourage you to run the calculations to figure out your moment of inertia and then use your acceleration requirement to determine the torque you'll need. It is very possible that the motor you have won't work at all for this (i.e., even when the load is up to speed it requires more torque to keep it moving than what your motor can safely give you for 6 hours at a time).

Also, here is a note about motor ratings. Most motor manufacturers rate their motors based on how its internal temperature changes with respect to ambient temperature and what kind of insulation it uses. So for a given type of insulation, the motor manufacturer will load the motor such that the motor windings rise (say) 85 \$^{\circ}\$C over the ambient air temperature over a long period of time. The torque, current, speed, etc. on the name plate the data from that test. So if a motor says it is rated for 2 Amps, that means if you run it at 2 amps for long periods of time (like you are at 6 hours) then you should expect the windings of the motor to get as hot as the insulation is rated for. It also means that you could possibly overload the motor for a short period of time as long as the motor windings don't get hotter than that temperature. The problem for you is that you don't know what kind of insulation is in your motor and you probably don't have the test equipment to properly determine how much you could overload the motor. Basically all I'm saying is that drawing 4.5 Amps for a short period of time (say, 1 minute) isn't going to necessarily burn out your motor but given the technical information you can get from Radio Shack, you probably aren't going to be able to determine this by yourself.


Why didn't anyone just tell him to use a resister? This is how blower fans in cars' AC systems are safely limited. You figure out which resistance value you need to pass the amount of current at a given voltage. This will prevent the motor from ever drawing more current than the resister would pass through.

  • 1
    \$\begingroup\$ This isn't a forum so you can't comment in the Your answer box. If you feel that a resistor solution is suitable then write an answer to that effect for the OP. You might point out that a high powered resistor is required. \$\endgroup\$
    – Transistor
    Jan 25, 2020 at 13:33

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