Stepper question: torque re: current

[Mariss Freimanis]

Easy; it really is. Only a couple of things to know.:-)

Tourqe = ampere-turns. Fancy way of saying multiply the current by the number of turns of wire it passes thru to know your torque. It's easy, a series connection has twice as many turns of wire as parallel (or unipolar for that matter). It takes half as much current in series to get the same torque as in parallel.

2 Questions then: 1) Why not keep the same current and get twice the torque? 2) Why ever would you want to run the motor in parallel?

1) If iron didn't saturate, you could. Iron is magnetic. You can think of it as being made of many little bar magnets called magnetic dipoles. These are randomly oriented when no current is passing thru a coil and the iron has no bulk magnetism. It just sits there.

When you pass some current thru a coil, some of the dipoles line up and the iron develops a magnetic field. Pass even more current thru a winding and even more dipoles line up. Sounds good so far.

Problem is, there is not an infinite number of dipoles. At some current, all will be lined up; past that point increasing the current won't increase the magnetic field (and torque) any higher. That is called magnetic saturation.

Motor iron is about 80% saturated at its rated phase current. That is why current must be halved in series. That last 20% is very elusive because it is non-linear, meaning it might take 10 times the rated current and a burning motor to get it.

2) Any winding has an electrical characteristic called inductance. Inductance increases as the square of the turns of wire; a series connection has twice the turns and therefore 4 times the inductance of a parallel connection.

Inductance limits motor power. Inductance has a property called inductive reactance and it is measured in Ohms. Unlike resistance, it increases proportionately with frequency. Double the speed (frequency) and you double the inductive reactance.

According to Ohm's law, doubling the reactance halves the current (I = V / R). Torque is proportional to current so doubling the speed cuts torque in half.

Power on the otherhand is speed times torque. Doubling the speed while cutting torque in half leaves power unchanged. The step motor has a perfectly flat power vs. RPM curve.

The only way to increase power is decrease inductance or to increase the power supply voltage. Look at Ohm's law: I = V / R.

Holding and low speed torque stays the same. The drive's job is to see to that by current limiting the winding to its rated value. What changes is the speed to which this torque stays constant. Doubling the voltage doubles this speed. Cutting inductance 4-fold also doubles the speed (and power) if the voltage is left unchanged.

I leave it as a puzzle for someone to solve why output power is proportional to the inverse of the square root of the inductance. It's not that hard.

Mariss

[bunalmis]

Originally Posted by bunalmis This motor have 8 coils. One coil inductance is L. 1) Serial Bipolar operations. One phase inductance Ls=2L and phase current = I = 2.8A We can find the storaged energy in the inductance. W = 0.5 * L * i^2 = 0.5 * 2*L * 2.8 * 2.8 = 7.84 * L 2) Parallel Bipolar operations. One phase inductance Lp=L/2 and phase current = 2*I = 5.6A We can find the storaged energy in the inductance. W = 0.5 * L * 0.5 * i^2 = 0.5 * 0.5 * L * 5.6 * 5.6 = 7.84 * L You can see we find same results.

I found an error in my text.

If one coil inductance is L. Serial inductance is not 2L.
Because two inductance magnetic coupled therefore Ls=4L

Lp=L (two coils use same magnetic core)

Ws=(1/2)*4L*i*i
Wp=(1/2)*L*2i*2i

Ws=Wp

[Novec]

All this parallell/series talk made me kinda confused, but I was already confused, so who cares The thing is I'm using five of the PowerMax II motors mentioned in the first post, but it has different specs:

Parallel: 3.50A 0.53ohms 2.0mH
Series: 1.75A 2.12ohms 8.0mH

I've been told to use 35V power supplies, but that isn't necessarily correct... The main problem is the amperage, which I honestly don't know how to figure out, and I'll probably end up making this PSU myself. First of all - do these motors draw 3.5/1,75A continously, regardless of speed, even at stand still? Considering it's a mill, only two of them will be active at once, sometimes three.

I'm using bipolar drivers from Quasar Electronics. They're rated at 50V/5A, but with some extra cooling, it shouldn't be a problem to go a bit further. There's also a chance I'll get a hold of some Pacific 5410 chopper drivers, but that depends on whether they drive more than one motor per driver.

Also, there's the issue of parallel vs. series, which I'm too much of a newbie to figure out from this thread. My mill is for metal working, so it won't be moving very fast, only a couple of revolutions pr. second. If the 140 oz-in for some reason isn't enough for driving the screws directly, I have 1:4 gears to help it. Does that mean series is the way to go, needing only half the amperage?

Bear with me guys, you were all newbies once