I'm working on a project using this stepper motor it is rated at 1.5 A/Phase current. Initially I planed of using this driver, so I checked if it can provide enough current for the motor. The website says:

can deliver up to 2 A per coil with sufficient additional cooling

however further down it says:

can deliver up to approximately 1 A per phase in full-step mode

I assume the difference if only significant in the case of 6-wire stepper (unipolar) and each coil is two phases?

The linked stepper has 4 wires (bipolar) so current per phase = current per coil? Can the stepper be used with this driver?


1 Answer 1


You've taken your quote out of context. A more complete quote is:

The Black Edition achieves its higher performance through its four-layer printed circuit board (PCB), which better draws heat out of the A4988 driver—while our original carrier can deliver up to approximately 1 A per phase in full-step mode without a heat sink or air flow, the Black Edition can deliver up to approximately 1.2 A under the same conditions.

This is a bipolar stepper drive where the amps-per-phase and amps-per-coil are the same thing. The reduced current described (1.2A for the Black Edition) is for use of the drive without a heat-sink or airflow. You can either add some sort of cooling (heat-sink &/or fan) or run your motor at a reduced current. If your motor only runs occasionally and you don't need holding torque, you could run at a reduced duty-cycle with the /enable pin held high which would allow the drive to cool.

Many drive manufacturers will also build in the capability of reducing the drive current to a lower holding current when the motor is not stepping allowing the motor and drive to cool. This is possible with the A4988 IC by reducing the Vref voltage when not stepping, although the Pololu board referenced does not have this capability.

  • \$\begingroup\$ So regardless of 4 or 6 wire per coil = per phase? \$\endgroup\$
    – Michal
    Commented Dec 4, 2014 at 20:42
  • \$\begingroup\$ For bipolar, per coil = per phase. \$\endgroup\$
    – Tut
    Commented Dec 4, 2014 at 20:45

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