I am using Oriental's PK-243-03A stepper motor in bipolar series configuration. The rated current of the motor is 0.31A/phase according to its datasheet. My question is if I am supplying it with 12V, then what is the maximum current I can run my motor on? For both the phases to be on i.e. in full stepping my motor should draw 0.62A, isn't it? Is it okay if I supply my motor 2.5A? I have seen people saying that its okay to drive stepper on higher currents, like this guy here on this forum


But I have also read some totally contradicting views regarding high currents like in this article here http://smoothieware.org/cnc-mill-guide, it says

Each stepper motor model has a precise current rating. You can drive your stepper motor at a lower current, which will make it more silent, but also less powerful. But you can not drive the motor at a higher current than it is rated at, this would cause overheating, and possibly skipped steps

What is the correct way? higher currents or no higher currents? Really confused.

  • 1
    \$\begingroup\$ At 38 ohms per winding, how are you going to pass 2.5 amps through it? You'd need almost 100V. If yau're asking, is it safe to use a 12V supply rated at 2.5A, then yes it is. Because the motor will only take 12V/38 ohms, about 0.3A per winding. \$\endgroup\$
    – user16324
    Jan 9, 2015 at 11:42
  • \$\begingroup\$ Are you sure about the fact that the motor will take 12V, 0.3A/phase? Because I was using a supply for this one motor which was above its rated current, and my motor got hot. I searched around forums and it was advised not to use a supply above the rated current for the motor. \$\endgroup\$
    – alexhilton
    Jan 9, 2015 at 12:08
  • \$\begingroup\$ Easy to check. Measure the resistance of a winding. Compare it with the value in the datasheet you linked. If it's the same (within 10%) you're OK. If not, that isn't the motor in the datasheet and all bets are off. And steppers do get hot. 50-60C is about normal for full power on both windings. If you mean sizzle-when-you-spit hot, that's another matter. \$\endgroup\$
    – user16324
    Jan 9, 2015 at 12:13

1 Answer 1


Based on what can be gleaned from the website you linked you ought not to exceed the phase current specified. This is how I see but if you can find proper data sheets, they might tell a different story.

Firstly look at the torque-speed curve (linked on the page): -

enter image description here

If you do the math, you'd calculate that the maximum mechanical output power is about 1 watt (100 rpm and 0.1 N.m = 1.047 watts). This is about the same at 200 rpm ( 200rpm and 0.06 N.m = 1.26 watts). At 400 rpm output power is 0.837 watts hence, you can see the max output power is 1.26 watts.

With a coil current of 0.31 amps per phase and a coil resistance of 38.5 ohms (as stated), the power (heat generated not mechanical power) is 0.31^2 x 38.5 watts = 3.7 watts and this means your stepper motor can be running quite hot.

Having said that, this "apparent" inefficiency (some simplification and assumptions made here) of about 25% will not be at optimum mechanical output. At optimum power output ( I reckon about 100rpm), the power in will be about 2.1 watts assuming a peak efficiency of about 60%. This is about "normal" for steppers of this type.

So, if you are always going to be running about 100 rpm, the current into the coils will be lower than 0.31 amps BUT, the details in the link are really unclear about this so caution should be taken.

Conclusion - I don't think you can dare run the coils at more that 0.31 amps based on what the specification says. I recommend finding out more about the device. Try looking at one from a regular supplier i.e. one that has a proper pdf data sheet and deciding what information that data sheet provides that this one doesn't.

  • \$\begingroup\$ Hey that was a pretty cool explanation, makes sense. But if you notice the comments above, according to Brian regardless of what supply I use the motor will draw only 0.31 A because it has a winding resistance of 38 ohms. But you have brought in the speed factor as well, what do you think about that? \$\endgroup\$
    – alexhilton
    Jan 9, 2015 at 12:28
  • \$\begingroup\$ No, what Brian saying is that from a 12V supply, with a resistance of 38.5 ohms, the current, due to ohms law will be 0.312 amps. If you doubled the voltage you'd get twice the amps. However, if you are pulse-stepping the motor (as normal), the inductance of the coil slows the rate of change of current and you might find that at a high stepping rate the current can be 0.31 amps at 15 volts or thereabouts. Does the spec detail give winding inductance? I can't remember. \$\endgroup\$
    – Andy aka
    Jan 9, 2015 at 17:16
  • \$\begingroup\$ No there is no detail about winding inductance.Will it be appropriate to give it more current? what if I go for microstepping too. Will it require more current than 0.31A? please elaborate in a simplified way. \$\endgroup\$
    – alexhilton
    Jan 11, 2015 at 15:23
  • \$\begingroup\$ I can't suggest any more current than 0.31 amps. Without knowing the inductance you can't predict how to apply larger voltages than 12v (but for shorter periods). \$\endgroup\$
    – Andy aka
    Jan 12, 2015 at 9:13

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