Please bear with me, if this is a stupid question.

I obtained a linear actuator of a family of linear actuators, here are the specs

| Bipolar linear actuator | 04     | 12       |
| Operating voltage       | 4 VDC  | 12 VDC   |
| Current / phase         | 0.2 A  | 0.07 A   |
| Resistance/ phase       | 20 Ohm | 180 Ohm  |
| Power consumption       | 1.6 W  | 1.6 W    |

I own the 04 model that is rated for 4 VDC and which is a bipolar linear actuator.

The first question for me is: When I am configuring a stepper motor voltage, the total current, that the 04 version of the actuator uses, would be $$0.2 A \cdot 2\space phases = 0.4 A $$ is this correct?

The fact that $$ 4V \cdot 0.2A \cdot 2 = 1.6W$$ also somehow supports my understanding :D

Now to the real question: When I use a stepper driver whose voltage would be 12VDC, what does it actually mean to the 4 VDC version of the actuator itself?

Higher voltage might saturate the coils quicker, as far as I understand, but the effect is probably not that big as the actuator for 4 VDC was designed to saturate the coils already in the most efficient way, increasing voltage also might just increase power consumption and little other effects?

When I apply 12 VDC to the actuator, I would have to limit the total current to the specs of the 12 VDC version, then or still go with the 4 VDC?

Would it be $$ 0.07 A \cdot 2 = 0.14A$$

or am I wrong and would have actually to drive it with the current / phase for the 4 VDC which would be 0.4 A? Which however would mean that the power consumption increases from 1.6 W to $$ 12 V \cdot 0.4 A = 4.8 W $$

affecting the resistance per phase also, increasing it from 20 Ohm to $$ \dfrac{12 V}{0.2 A} = 60 \Omega $$

So for what current should I adjust the stepper to use the 4 VDC actuator with 12 VDC (if at all).

Insights appreciated :)


3 Answers 3


This will depend on your specific motor driver a little, however if you can set a current limit on the driver, you can use the lower voltage device with a higher supply voltage, your driver will get a little hotter depending on the method it uses, but it also will give your actuator a higher maximum torque and higher top speed in general.

This is quite common in larger stepper motors, the motor itself would cook if fed with a constant 48V, but by having the controller limit the current once the inductance is overcome it runs fine. the perk being a higher voltage gives a higher torque,

When I played with this in the past I was dumb enough to use a linear current limit off a 72V supply for 12V steppers, they absolutely flew, and could almost stop on a dime, and while the motor stayed fairly cool, the current limiting transistors where practically glowing after a few minutes

In your case, It would apply 12V, the inductance of the phase would take some time to be overcome, so during this time the voltage would be fairly constant with the current rising, as it approaches the current limit, the voltage will fall down to the resistive voltage drop for the given current, and you would end up with about 4V * 0.2A on that phase, I would assume its a switching controller, so you would be drawing ~12V * 0.07A on the input (essentially a buck converter) at this point, and may fold back to a lower holding current, again based on what your controller is set up to do.

  • \$\begingroup\$ To what would you suggest I should adjust the current limit then to? 0.14 A on 12 V for both phases, or to 0.4A on 4V (*2 since I will use both phases)? It theoretically shouldn't really matter because at the end it will be run with 12 V anyway and the current limit potentiometer would be at the same position? \$\endgroup\$
    – Samuel
    Sep 1, 2019 at 12:19
  • 1
    \$\begingroup\$ as you have the 4V stepper, you would be using its single phase current for the current limit, as the current is measured on the stepper side. in this case 0.2A Now as for why only a single phase's current, with most microstepping controllers, you end up with 1 phase being a Sine, and one a Cosine, they add up to 1*however many amps, and stay fairly constant around this point, there is not really many times you need 2 phases 100% on. \$\endgroup\$
    – Reroute
    Sep 1, 2019 at 12:23
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    \$\begingroup\$ As you have not said what controller your using, I'm going to assume the current sense is between ground and the stepper phase, as such it doesn't matter the voltage, only the current, \$\endgroup\$
    – Reroute
    Sep 1, 2019 at 12:40
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    \$\begingroup\$ pololu.com/file/0J450/A4988.pdf It measures current per phase, take a look at page 14-16, this controller uses the sine/cosine. so yep, set it to 0.2A limit, and it will within margin of error not draw more than that. \$\endgroup\$
    – Reroute
    Sep 1, 2019 at 13:29
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    \$\begingroup\$ It is measuring each phase independently, so your device is not seeing "Phase1+Phase2>0.2A" but "Phase1>0.2A or Phase2>0.2" \$\endgroup\$
    – Reroute
    Sep 1, 2019 at 14:41

Usually modern stepper drivers are designed to control the current to the coils by rapidly switching on and off so that the current through the coil is at a predetermined value (which should be selected to be within the motor specifications). A sense resistor allows the current to be measured. By using a higher supply voltage than the DC rating of the coils the motor can turn fast with better torque because the higher voltage overcomes the coil inductance more quickly.

If your proposed driver circuit is not a chopper type you can use series resistors (40 ohms in your case, to drop 8V each), which is a simpler approach and has similar performance benefits, however the resistors will waste 4.8W - 1.6W = 3.2W.


..and little other effects

If you put 12V on the 4V actuator, one of the little effects could be that you let the magic smoke out.

Your 60Ohm calculation is correct which means you have to add another 40 Ohms per phase.

Your power consumption calculation is partly correct. You will use 4.8W but not all that will be used by the actuator. 3.2W will be used by your resistors.


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