3. Six-lead unipolar/bipolar stepper motor connections

If you have a six-lead unipolar stepper motor, you can connect it to the driver as a bipolar stepper motor in two different ways.

Half-coil

You can connect half of each coil to the driver; for example, you could connect stepper leads A and A′ to one pair of driver outputs and stepper leads B and B′ to the other pair of driver outputs, leaving C and D disconnected. Since 6- and 8-lead stepper motors generally have their ratings specified for unipolar operation (with the rated current through half the winding producing the rated torque), driving the motor in this way will cause it to perform as rated.

Full-coil series

You can connect both full coils to the driver with the half-coils in series by connecting stepper leads A and C to one pair of driver outputs and stepper leads B and D to the other pair of driver outputs, leaving A′ and B′ disconnected. This changes the characteristics of the motor: it will have twice the rated resistance and four times the rated inductance. Because this makes the current in the motor coils ramp up more slowly, wiring the motor in this way decreases the maximum step rate that the motor can achieve for a given input power (compared to half-coil operation).

The main advantage of full-coil series wiring is that at low speeds, you only need half the rated current to produce the rated torque of the motor (since torque is proportional to the current times the number of turns in the coil, and you are effectively doubling the number of turns by connecting the half-coils in series). You might be able to push the motor’s torque even higher by using more than half the rated current; however, you might encounter diminishing returns and a loss of microstepping accuracy as the motor becomes magnetically saturated, and you risk overheating the motor if the power dissipation is higher than what it is designed for. More precisely, you should not use more than about 70% of the motor’s rated current with full-coil series wiring to avoid exceeding the power that the motor would normally draw with half-coil operation at the full rated current. (``P = I^2R``: power is equal to current squared times resistance, so if you double the resistance, you must divide the current by ``sqrt(2)`` to keep the power the same.)