# quick stepper question

Discussion in 'General Electronics Chat' started by kutalinelucas, Feb 2, 2012.

1. ### kutalinelucas Thread Starter Active Member

Nov 20, 2007
98
0
Hey guys, I just have a quick question about steppers if thats ok

Say you have a bi-polar stepper running off 4.6v; quoted as '0.6a/phase (1.2a peak)'...considering that rotation is achieved through applying sequential pulses in sets of 2, when holding, should the motor be drawing 2.3v, 0.3a through each of the 2 energised wires?

So does the '1.2 peak' refer to the maximum amount the device could draw, if each coil was energised, even through the shaft wouldn't turn?

Cheers

2. ### JDT Well-Known Member

Feb 12, 2009
658
85
Bi-polar drive is where you have 2 coils (sometimes 4) and 4 wires (sometimes 8)? Unipolar drive is where you have 2 center-tapped coils.

I would say that 0.6A is the maximum continuous current in a single winding. So it will be OK to apply 0.6A per winding continuously when the motor is stopped and locked. The 1.2A probably applies to the two windings in parallel - if it's an 8-wire motor?

So it would be OK to apply 0.6A (1.2A) to both A and B windings at the same time. Making a possible total of 1.2A for a 4-wire motor or 2.4A for an 8-wire.

It will be OK to apply a current greater than the continuous maximum for a short time (milliseconds).

The maximum it could draw will depend on the voltage you apply!

I would treat any voltage specification with caution. Best to measure the DC resistance of the windings. Due to the inductance of the windings, when the motor is turning the current will be much lower. Faster it goes, the lower the current. Because of this, the motor torque will reduce with speed.

At the end of the day - if the motor is not getting too hot it's probably OK!