I want to make an order with this configuration

  • Arduino MEGA 2650 R3
  • Ramps 1.6 Plus
  • 2 TMC2130
  • 2 Stepper motors 17hs3401
  • 1 Fan


But I am not sure what voltage and current would be enough to make it work. The Ramps 1.6+ board has an input of 12V as you can see in the picture, but I have read that other people had to increase the voltage with a DC-DC converter.


So, would be 12V and 15A enough to make work that configuration?

  • $\begingroup$ Are you going to be powering a heated bed from the same PSU too? And perhaps a heated nozzle? $\endgroup$ Mar 24, 2020 at 20:24
  • $\begingroup$ @AndrewMorton Thanks for your insterest. Actually, I am not going to mount a 3D printer. I am going to use those printer parts for another purpose. I will connect just the few components I have listed in my question $\endgroup$
    – ChesuCR
    Mar 24, 2020 at 22:40
  • 1
    $\begingroup$ 15A of 12V is 300W... that's a LOT of power. $\endgroup$ Mar 27, 2020 at 20:42

2 Answers 2


The easiest way to know how powerful the PSU should be is to download from https://github.com/rcarlyle/StepperSim the Excel workbook which simulates the power absorbed by the stepper motor. Input the motor specifications, check in the graph the max speed at which you plan to run it, check the absorbed power, add 20% for the various losses. Once you know the absorbed power, you can multiply by the number of stepper motors and obtain the current required, add 20% not to stress the PSU too much.

In you case, this is the result

enter image description here

Your motor will never absorb more than 7 W, 14 W for the two motors, around 20 W considering losses in the motors and overcapacity of the PSU. This means that around 2 A PSU will be perfect for your setup.

Previous answer, which may not be completely correct but still provides a useful background knowledge:

The TMC drivers limit the current to the value you set, which is most of the time lower than the datasheet. In no case you will need more than 1.3 A * sqrt(2) * 2 motors = 3.6 A. In fact, you won't need this much either.

When there is no field in a coil, the driver applies full voltage, but the current is low (initially zero) so you don't hit the 1.3 A per coil.

The current increases (the higher the maximum voltage available to the driver, the faster it increases) and the driver (probably) keeps the full voltage until the preset current is reached. Just a moment before that, the current is almost there, but you still have full voltage from the power supply. This is the theoretical worst case, but it applies only for a very short amount of time.

As soon as the current reaches the preset, the driver "cuts the voltage" to keep 2.4 ohm * 1.3 A = 3.2 V (because V = R * I). This means that the power supply sees 3.2 A/12 V*1.3 A = 0.35 A.

When running, the motors almost never start from zero to max current: both coils are powered and when one increases, the other one decreases.

In fact, the microstepping makes the steppers act more or less like AC motors with two phases. This means that overall the current is the max current per phase multipled by sqrt(2). Also, when using microstepping one phase (coil) is not completely shut off, but two of them work at the same time (with different current levels). This means that in total one compensates the other, and the power supply only provides, more or less, 0.35 * 1.4 = 0.5 A per stepper. You have two, so it's 1 A total, therefore 2 A PSU considering the inefficiencies.

A very easy and complete explanation is provided here:

By controlling the duty cycle of the chopper, an average voltage and an average current equal to the nominal motor voltage and current are created.


As the current increases, a voltage develops across the sensing resistor, which is fed back to the comparator. At the predetermined level, defined by the voltage at the reference input, the comparator resets the flipflop, which turns off the output transistor. The current decreases until the clock oscillator triggers the flip-flops, which turns on the output transistor again, and the cycle is repeated

So you never have coming out of the PSU more than the preset current.

Supply current is not the same as the motor current in a copper drive. It is the motor current multiplied by the dutycycle, at standstill typically Isupply = IM · ( VM ⁄ Vsupply )


Depending on how the H-bridge is switched during the turn-off period, the current will either recirculate through one transistor and one diode (path 2), giving the slow current decay, or recirculate back through the power supply (path 3). The advantage of feeding the power back to the power supply is the fast current decay and the ability to quickly reduce to a lower current level.

  • 2
    $\begingroup$ You do not need anywhere near 6 A. The stepper motors do not draw their rated current from the supply because they operate at a (much) lower voltage than 12 V. From the phase resistance being 2.4 Ohms and the phase current being 1.2 A, you can deduce that the rated voltage is 1.85 V. So the steppers only draw around 560 mA from the power supply (2 steppers x 1.3 A x 1.85 V x sqrt(2) / 12 V). A 2 A supply would easily suffice. $\endgroup$ Mar 26, 2020 at 18:35
  • $\begingroup$ @TomvanderZanden, that information is entirely incorrect. (your answer actually crawled my blood). The phase resistance will result in around 5 amps if operated at 12 volts. V= IR $\endgroup$
    – user77232
    Mar 26, 2020 at 21:33
  • 1
    $\begingroup$ @user77232 It is incorrect to use V=IR and assume that the current will be 12V / 2.4 Ohms = 5A. The stepper motor does not only have resistance, it also has inductance. The inductance, combined with the chopping action of the stepper driver, limits the current. $\endgroup$ Mar 27, 2020 at 6:32
  • 2
    $\begingroup$ @FarO Of course the stepper sometimes sees a higher voltage applied across the coil, but that doesn't change the calculation. The 1.85 V figure is an average. Moreover, the energy put into the coil when increasing the current is recovered when decreasing the current. In theory the increase of current in one coil balances out with the decrease in the other, so no additional current is drawn from the power supply. Of course there are inefficiencies, losses and mechanical work (hence why I recommended a 2A supply) but the static current draw is a good approximation of the actual current draw. $\endgroup$ Mar 27, 2020 at 7:24
  • 1
    $\begingroup$ @FaO, are we confusing current mode stepper drivers with voltage mode ones? $\endgroup$
    – user77232
    Mar 27, 2020 at 18:36

Yes, A power supply that can deliver 12 volts and up to 15 Amps will work. Since that board says 12 V, that means it is designed to work at 12V. The components on the board might not survive 24V. The TMC stepper motor drivers specified can tolerate a max of 2.5 amps. That current draw is dependent on the per phase winding impedance of the motors that are used. @2.4 ohms phase resistance, that 12 volts could generate a max of 5 amps (more than the Trinamic can tolerate). Therefore, if you configure the stepper drivers to operate at more than 50% power, they will overheat and shutdown.

  • $\begingroup$ Are you saying that I cannot use the TMC2130 with the stepper motor in my question because it will overheat? $\endgroup$
    – ChesuCR
    Apr 1, 2020 at 11:59
  • 1
    $\begingroup$ @ChesuCR, I am saying that if when you are making your configurations in Marlin you set the power level for the TCM2130 to more than 50% then it will overheat. However, it may only do so when running at low speed. I doubt it will over heat if running at a high speed. $\endgroup$
    – user77232
    Apr 1, 2020 at 12:30
  • $\begingroup$ Thank you. Anyway, I still have some doubts. I have written another question $\endgroup$
    – ChesuCR
    Apr 2, 2020 at 11:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .