If steppers are run in series, the voltage per stepper drops off - supposedly to half.
But can that reduction be made up partially by increasing current for series steppers? (Assume the driver can deliver the current and that supply is 24V vs 12V)

Applying Ohms Law - likely way too simplistically - to the following example:

In parallel: 1A stepper, 2.5V, 2.5ohms (easy numbers to work with) Operating: 1A x 2.5V = 2.5W. So, 2.5W of heating power, the max the motor can take without overheating.

In series: Same steppers. Operating: (current the same in series circuit) 1A x1.25V (per motor with V divided across two motors) = 1.25W. So only half the heating power. The motor should be able to take a higher current

Can the stepper take up to almost 2A when in series? 2A x 1.25V = 2.5W, the same power load as it would see in the parallel example.

Edited 1 time(s). Last edit at 04/10/2019 07:20PM by DanL.

The drivers used in 3D printers are constant current drivers. As long as the supply voltage is high enough, the motor will get the current that you set. If you connect the motors in parallel, they share the current.

In your example, you have steppers that are rated at 1A and have a phase resistance of 2.5 ohms. You want to run them at 1A peak current (in practice you would normally run them at a lower current).

- If you connect them in parallel, set the driver to 2A. Each motor phase will get 0 to 1A (depending on the microstep position) and the voltage drop across the parallel pair of windings at standstill will be 0 to 2.5V.

- If you connect them in series, set the driver to 1A. Each motor phase will get 0 to 1A (depending on the microstep position) and the voltage drop across the series pair of windings at standstill will be 0 to 5V.

The main differences between the two configurations is:

- In the parallel configuration, the driver will run much hotter because it will be dissipating more than 4x the power, because of the higher current

- In the series configuration, the speed at which torque starts to drop will be 50% lower than for the parallel configuration. This is because the higher the speed, the more voltage is needed to force the current through the windings. Eventually the voltage needed becomes higher than the driver supply voltage, so the full current can no longer be maintained. There is a motor EMF calculator at [reprapfirmware.org] to help you estimate this speed.

Edited 6 time(s). Last edit at 04/12/2019 02:47AM by dc42.

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Large delta printer [miscsolutions.wordpress.com], E3D tool changer, Robotdigg SCARA printer, Crane Quad and Ormerod

Disclosure: I design Duet electronics and work on RepRapFirmware, [duet3d.com].

Thanks DC42. I see my error. I halved the V for each motor. But if current taken to 2 A, with 2.5ohms per motor, V drop per motor is 5V....above the motor V rating.
Your explanation is clear.