I'm using STK 672-330 to drive a unipolar 2 phase stepping motor. The situation can be seen roughly in the circuit diagram below. example circuit diagram

I set Vref so that the peak phase current becomes 1.2 A. When I measured the current in the phase A (blue arrow), i confirmed that it correctly peaked at 1.2A . But then I measured the current from 24V (red arrow) it looks more like DC 400 mA ish.

I did a few research and I found out that according to this source (around last page) supply current is calculated using this formula:

Isupply = IM · ( VM ⁄ Vsupply )

where IM is motor winding phase current, VM is motor winding voltage.

In my situation: Vsupply= 24V, IM = 1.2A, VM = 3.6V (winding resistance is around 3 Ohms) so it gives me Isupply = 180 mA for each phase. Which gives a total 0f 360 mA. This result more or less agrees with my measurement.

But why does this equation holds? I find it strange we need to divide with Vsupply. If the current flowing in the phase peaks at 1.2A, why there is no such peak appear in the 24 V line? Where does the current come from and go to?

I have a hunch this has something to do with the PWM chopping by the FETs inside the driver IC. But I can't find satisfactory explanation for the continuity of current.


Yes, you are right, this is 'something to do with the chopping'.

Look up how a Buck DC to DC step down converter works. The inductance of the stepper motor acts as the inductor in a buck converter, while its resistance acts as the load resistance.

With the driver acting as a loss-less step-down buck converter, the power supply only needs to supply enough power to meet the power losses in the load. As the voltage is high, the current can be low.

  • \$\begingroup\$ Thank you for your response Neil. I looked up Buck Converters and found this [diagram] (radio-electronics.com/info/power-management/…). It shows that the input current enters the circuit as pulses. I see now that total power of the input pulses equals the DC output of the circuit. But how come I can't see the input pulse waveform? From this explanation I'm sure it peaks at 1.2 A also. \$\endgroup\$ – iinhebat Jul 28 '16 at 9:17
  • \$\begingroup\$ The 1.2A is delivered from time to time by the 100uF cap, which charges steadily at 400mA from the supply. \$\endgroup\$ – Neil_UK Jul 28 '16 at 9:26
  • \$\begingroup\$ Ah, I see. Sorry I missed the capacitor existence. Thanks a lot for the clue. It helped a lot. \$\endgroup\$ – iinhebat Jul 28 '16 at 10:54

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