I'm relatively new to the cnc concept and was shopping around for steppers. I kniw that holding torque is more or less the "strength" of a stepper and that the voltage and amperage are used in power supply consideration. But I'm a bit lost with inductance and what all these specs mean on different steppers. IE; what does a high vs low inductance rating do for steppers. I see steppers with huge torque ratings but only a fraction of the amp rating on lower torque steppers. Are these just inverted torque/RPM charts or fibbed Amp ratings?

Hey there, I'm in the same boat here, that's why I'll be tagging along and hoping somebody answers this. but here's a link somebody gave on another thread that may help. http://www.geckodrive.com/support.html here's a quote: "Inductance (L) has a property called inductive reactance, which for the purposes of this discussion may be thought of as a resistance proportional to frequency and therefore motor speed." maybe we can think of it as how fast can the motor go? if that's the case then we'll need higher inductance for CNC (especially if screw driven) hope someone can give a better answer.

From that description it sounds like the KV rating that rc model motors use. A relation of voltage and RPM, the higher the KV rating the more RPM per volt.

I just thought a bit more about it, Inductors naturally resist current spikes, so that probably means that higher inductance = less tolerance for higher frequencies which = Less speed. in these terms, less inductance is better as we can get higher speeds from the stepper? I'm not 100% sure about this. would be great if someone could confirm it.

Stepper motors have the highest holding torque at the lowest speed. That means that when they're not moving, but powered they're at their strongest. This is also when they use the most current. When they begin to rotate their torque drops. The higher the speed the lower the torque. This drop isn't linear though and it's given by manufacturers in the form of a torque chart. It is dependant on motor size, design, number of windings, size of wire, and how it's wound among other things. In simple terms, the inductance rating is the amount of resistance to current change. Lower the number then the faster the ability to charge the windings. This results in a more linear torque curve. A higher inductance motor will have its torque drop off with higher rpms sooner than a low inductance motor. The windings not being able to charge quick enough is one of the reasons. The trade off is higher current often equates to a higher torque at lower speeds. All of this is considering we are talking about similar sized motors.

So ideally you'd want a stepper optimized for torque in your expected RPM range? So the inductance number itself isnt important so much as the resulting torque curve?

Kinda yes. However, finding a happy medium of a machine that can zip from one cut entry location to another without losing steps in between and can cut a material confidently at speeds the spindle can handle is the goal. It's a tricky balance and usually there is some give and take. There are plenty of folks using high inductance (not really that high compared to what's avail. in industry. I'm just talking general hobby level) steppers without issues. Your best bet is to chug through the math for each axis and size your motors from there. It can become a real task depending on your motion system type. A handy formula for helping with the size motor you should consider is F=T/(d/2) F is output force in oz. T is Manufacturers torque rating in oz-in. d is diameter at which torque is being translated. For example. One has a 16mm ballscrew and 425 oz-in rated motor. The ballscrew root diameter is 15mm Looking at the formula we see that it is using inches and mm. So we'll have to do some conversion. 15mm/25.4= .590" is the root dia. in inches.(it can be argued that translation happens somewhere between root and pitch dia) Now apply the formula. 425/(.590/2)=1441oz. of force at the root of the ballscrew. iirc that's about 90lbs of force. Have to keep in mind that this is holding torque and there is friction to overcome in the ballscrew. A conservative place to sit with ballscrew efficiency is 90% 1441*.90=1297oz. force on that ball screw. 1297 is around 80lbs of force. Again, this is only standing still. There is also the static friction of the load bearings that need to be overcome. As far as I'm aware there is no rule of thumb to determine reduction in torque at a given rpm based on motor stats or holding torque. I know there are stats on the effects of step settings on overall torque (which should also be considered when reading this). EDIT:manufacturers motor torque chart can be handy! If you take the mass you intend to accelerate and figure out the cutting speeds you'll be working at (For velocity). You can narrow things down pretty well by working backwards with this formula. Hope I didn't bore yous. Joe

The inductance of the motor relates to how quickly the motor will develop full torque after voltage is supplied. This is why slower stepping rates result in greater torque since the motor has more time for voltage in the coil to reach it's maximum. In order to overcome the inductance people run them at higher voltages than the base ratings. A higher voltage reduces the amount of time it takes for the coil to reach maximum potential meaning the motor can be pulsed faster with less penalty on torque. The lower the voltage and inductance rating a motor has the more beneficial a fixed higher voltage will be. For instance a motor rated at 200 oz-in | 3V | 4A | 2mH will benefit more from a 36v power supply than a motor rated at 300 oz-in | 4V | 3A | 3mH. And by benefit I mean it will be able to step faster while still maintaining closer to it's rated torque. This is my understanding based on past reading. If I am off base please feel free to chime in. Regards, Mike