I'm trying to aggregate information on config settings and current limit calculations for all the popular "StepStick" style motor drivers. However, I've noticed that while the older type (A4988 & DRV8825) just have a Vref = I / 2 style calculation, the new Trinamic TMC2xxx line of drivers all have calculations for both RMS and Peak current.

So, this leaves me with two main questions:

  • If my stepper motor data sheet says "Rated Current: 1.4A", do I need to calculate for the TMC drivers the Vref for RMS or Peak?
  • For the older drivers (A4988 & DRV8825) is it limiting the RMS or Peak current?
  • \$\begingroup\$ The current output has a DC shape, so peak = RMS. \$\endgroup\$ Aug 21, 2019 at 12:48
  • \$\begingroup\$ @MarkoBuršič current output of those drivers specifically? Or all stepper drivers? \$\endgroup\$
    – Adam Haile
    Aug 21, 2019 at 12:59
  • \$\begingroup\$ RMS fall behind peak current for micro stepping drive mechanism \$\endgroup\$
    – User323693
    Aug 21, 2019 at 13:33
  • \$\begingroup\$ Can you maybe explain more or provide a link? That doesn't really mean anything to me. \$\endgroup\$
    – Adam Haile
    Aug 21, 2019 at 14:26

1 Answer 1


If you look at DRV8825 datasheet you have detailed specification. At full step, the currents in both wingdings are 71% of the current setpoint value, meanwhile at 1/32 micro stepping the current is 100%, but it is sinewave like - AC.

For sine AC waveform it is true that \$I_{RMS}=0.707\cdot I_{peak}\$.

The total max. current is the sum of both phases. For full step it is equal to 1.41*I_set. For microstepping it is also 1.41*I_set.

The motor data is referring at full step current if not noted otherwise, therefore total of 2.8A, which turns into Iset=2A irrespective which mode you will use. The total flux will be unchanged, since it is the vector sum of both phases. The RMS current sum will be little lower when using microstepping.

  • full step : \$I_a=I_b=0.707\cdot 2A=1.41 A\$ Square pulses \$I_{RMS}=2\cdot 1.41=2.82A\$

    • microstepping \$I_a=2A\cdot sin\omega t\$ , \$I_b=2A\cdot cos\omega t\$ \$I_{RMS}=\sqrt{I_a^2+I_b^2}=\sqrt{2A^2\cdot (cos^2\omega t+sin^2\omega t)}= \sqrt{2A^2\cdot (1)}=2A\$

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