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The above is the specifications of a wheelchair DC motor I'm trying to use as my hobby project.

My question is how can a motor of 320 Watt and 24 V DC be of 3.0 A max rating when the current should be around 13 A?

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  • \$\begingroup\$ Parv, sorry, but i think this should be on electronics... \$\endgroup\$ – Martynas Nov 24 '14 at 12:16
  • \$\begingroup\$ You can messure the resistance of terminals, insert it in OHM's law and you will get stall current. This is probably nominal current. \$\endgroup\$ – Martynas Nov 24 '14 at 12:36
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Sure, they can print anything they want on the sticker. As you noticed, \$24\:\mathrm V \cdot 3\:\mathrm A = 72\:\mathrm W\$. To get 320W in any sense (mechanical, electrical, or thermal power) would require more voltage, more current, or both, or a violation of the law of conservation of energy.

One then wonders, how do they get 320W?

My guess (since the sticker is too brief to say): "320W" is the maximum theoretical mechanical power this motor could produce.

A DC motor driven by a fixed voltage has a maximum no-load speed, and a maximum torque which is developed at zero speed. That maximum no-load speed is limited by the battery voltage (here, 24V), and the maximum torque is limited by the stall current, which isn't specified but which is probably (a lot) more than 3A. You can get a rough idea by measuring the winding resistance with an ohmmeter, then calculating what the current would be into that resistance at 24V with Ohm's law. Sometimes you have to jiggle the rotor to get the commutator to line up.

A linear function then describes the torque vs. speed relationship for a given motor, at a given voltage. From Understanding D.C. Motor Characteristics:

torque vs rotational speed

Mechanical power is the product of torque and speed, and thus, the motor develops maximal mechanical power at the midpoint of this line, where the motor is running at half its maximum speed and delivering half its maximum torque.

"320W" likely refers to this maximum theoretical mechanical power. Of course, the motor will overheat if you actually run it under those conditions for very long. Briefly however, it will be fine, as long as you don't generate more heat than would be generated by 3A continuously, or generate so much torque as to damage the motor mechanically.

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  • \$\begingroup\$ Setting the current limit to exactly the rated current of the motor is generally not what you want to do. You may need/want the extra current for acceleration or for heavier loads during intermittent operation. \$\endgroup\$ – Eric Nov 25 '14 at 6:51
  • \$\begingroup\$ @Brad sure, you can run any component out of its specified operating parameters for a brief time and usually things will be fine. Usually. \$\endgroup\$ – Phil Frost Nov 25 '14 at 11:52
  • \$\begingroup\$ Phil, the current on the nameplate is the continuous rated current (assuming the manufacturer follows NEMA or IEC standards and doesn't specify otherwise). Running a motor at higher currents for a brief time is NOT outside of the specified operating parameters since the rating on the name plate is a continuous rating. In my experience, the reason people might set their current limit at or below the rating is to limit torque to protect the load, not to limit current to protect the motor. \$\endgroup\$ – Eric Nov 25 '14 at 13:20
  • \$\begingroup\$ @Brad You aren't wrong, but I don't want to get to deep into what "intermittent" means, because the question isn't really about that. I also don't want to give ambiguous advice to a beginner. Does the answer look better with edits? \$\endgroup\$ – Phil Frost Nov 25 '14 at 14:21
  • \$\begingroup\$ Good point. And yes, the edits look good. \$\endgroup\$ – Eric Nov 25 '14 at 14:30
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In my opinion this is the nominal specifications. And the power is written as peak :) This is like on cheap stereo written for example 3000W, but the RMS is always hidden and much lower:))

Anyway, i think this specification of power shows peak power or stall current.

UPDATE

If you want to use it for wheelchair this is not the motor you need. Because it is C.C.W - counter cock wise, that means this motor isn't recommended for clockwise turning. It is limited due to angled brushes. H-class is insulation class. H stand for 180C.

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  • \$\begingroup\$ I doubt that - the motor is a wheelchair motor - 70W wouldn't be nearly enough for a wheelchair. \$\endgroup\$ – RJR Nov 25 '14 at 6:12
  • \$\begingroup\$ So what do you want to say that 320W stands for? It says you the max current. \$\endgroup\$ – Martynas Nov 25 '14 at 6:15
  • \$\begingroup\$ I expect the power is the maximum continuous power the motor can dissipate without damage. The mechanical power delivered would be the electrical power times the efficiency of the motor. Generally, this would be in the datasheet - if there is one. \$\endgroup\$ – RJR Nov 25 '14 at 6:41
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    \$\begingroup\$ @PhilFrost It can't. Neither can it draw 320W at the given voltage and current. As such, and per my answer below, we know that we are missing some specification. As long as there isn't a datasheet that explain what the values on the nameplate means, the best way to find out is to measure the motor's characteristics. \$\endgroup\$ – RJR Nov 25 '14 at 23:17
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    \$\begingroup\$ @Martynas I think the OP is saying the motor came from a wheelchair - not that he wants to use it as such. \$\endgroup\$ – RJR Nov 25 '14 at 23:24
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We used the same one in our wheelchair. Basically 320watt is safety power operating rating for the motor at load condition. Our wheelchair takes continuous 30-40amp with good grades. This doesn't mean that motor will burn @40 or even 100amp, because of the PWM Controller. It follows this equation:

\$ V_{out}=Dutycycle*V_{in} \quad \textrm{and} \quad P=Vout*I_{Load} \$

So with a heavier load the controller reduces the duty, reduces speed and the average voltage. It may be possible get 50amp continuous, for times more than a minute.

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Experts, correct me if I'm wrong, but often the maximum current specification on DC motors is the maximum current the motor will draw at the specified voltage under no load (maximum speed). Stall current (under load) will be much higher than that.
See for example these datasheets, which specify a maximum current but then have a power/torque graph that allows currents much higher than the specified maximum.

I suggest you hook up the motor to a (current limited) 24V power supply and measure the current under various conditions. I expect you'll see it draw about 2A under no load, and this current will increase when you put a load on the motor. It is well possible that if you stall the motor, it will draw more than 13A (320W) and as such, your driving circuitry needs to be able to shut down or limit the current to prevent burning out the motor.

Basically maximum current can mean a number of things:

The conclusion is that the information on the nameplate is insufficient to determine either. Assuming all values are somehow correct and it isn't a 70W motor or the current rating is actually 13A, measuring the motor's behaviour is your best bet.
Also, good info here.

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  • \$\begingroup\$ Typically, the current found on a motor nameplate/datasheet is the rated current. That is the current the motor can continuously run at without overheating. \$\endgroup\$ – Eric Nov 25 '14 at 6:44
  • \$\begingroup\$ @Brad I agree there's no way to be sure just from the nameplate. That's why I suggest measuring the motor's behavior. Ideally, a datasheet would be available and specify continuous, freewheel and stall currents. Having said that, the datasheet I linked to in my answer doesn't specify the definition of their current rating either... \$\endgroup\$ – RJR Nov 25 '14 at 6:50
  • \$\begingroup\$ A motor that draws 2A at no load would be surprising. Just because there's a graph in the datasheet with currents over the rated current does not mean the motor can actually operate at that load for very long without overheating. Graphed != allowed. \$\endgroup\$ – Phil Frost Nov 25 '14 at 11:51
  • \$\begingroup\$ @PhilFrost I'm not saying either. All I'm saying is that there is insufficient information to answer the OPs question. As such, measuring the motor's characteristics is the best way to determine its specifications. \$\endgroup\$ – RJR Nov 25 '14 at 23:18

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