I have a Astrosyn Miniangle Stepper motor that I just got and I need to know how much voltage and current I need to drive it. I am having trouble finding the specs online so I am trying to understand what the stuff written on the motor means. It says:

2.3 V/Phase 2.3 A/Phase 1 Ohm/Phase

Does this mean that It needs 2.3 volts and 2.3 Amps, and that the coils are giving 1 ohm of resistance?

  • \$\begingroup\$ Here are all of the datasheets: astrosyn.com/page.php?xPage=technical You haven't said which one you have though. \$\endgroup\$
    – Kellenjb
    Dec 28, 2011 at 21:09
  • \$\begingroup\$ The Type is 23LM-C355-28, Which I didn't see there, and the part No. is T9810-02 \$\endgroup\$ Dec 28, 2011 at 21:12
  • \$\begingroup\$ It's possible you might have a custom variant part number built on a more standard chassis, so you may be able to get some hints of general things such as thermal limits, comparison of single vs. two coil current limits, etc from the manufacturer's other similar sized models. Of course the model numbers could also change as the technology of things like the permanent magnets evolves. \$\endgroup\$ Dec 28, 2011 at 21:37

2 Answers 2


A datasheet would help, but what you have there sounds good enough to design the drive circuitry. Yes, it looks like each winding has 1Ω DC resistance, and that in steady state can handle 2.3A, which means 2.3V. A drive that never exceeds 2.3A into any coil will be safe.

Because the stepper motor windings look heavily inductive to the driving circuit, it is allowable to apply more than the steady state voltage when switching. The real spec is the current because the limiting factor is heating of the winding, and that is the current squared times the winding DC resistance.

The simplest safest approach is probably a fixed voltage drive. That means the current will ramp up as a voltage step is applied, approaching the steady state value of 2.3A with a exponential decay. If you want to get more fancy and need the fastest possible stepping speed, then current control is better. The circuit will attempt to regulate the current to 2.3A, which will automatically start at the maximum voltage, then decay down towards 2.3V in steady state.


Yes, it is designed for 2.3 volts and would draw 2.3 amps, with two major caveats.

First, if you use a step mode which includes a time where both coils are energized, you may end up with higher than 2.3 amp current, so you might want a more capable supply. However, the overall current limit may not be the winding - it may be the risk of damage to the magnets if the temperature or magnetic field strength becomes too high, and those effects are somewhat additive, so you may not want to supply full current to each winding. That's where a full data sheet would be helpful.

Second, the coil has only 1 ohm of DC resistance. Once you start hitting it with pulsed step currents, the inductive reactance of the coil comes into play, and it will have a higher effective total impedance. With only 2.3v supply, you will see less and less current as you increase the step rate. To counteract this, high performance drivers use a supply voltage up to many times the rated voltage, and use a pulse width (chopping) current regulator to limit the current to the desired value - usually by measuring the voltage drop across a fractional-ohm power resistor in the return supply to the drive transistors/FETs. Even many of the small IC drivers have this mode.


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