Current for a Printer with 5 Nema17 Steppers

I'm working on a 3D Printer that works with 5 Nema17 Steppers (1,7A/Phase). I don't need a heated bed or a heater for the extuder.
The maximum current the Printer could consume should be 17A which would never happen. Just from what i've read (this is my first printer) i should be fine with 7A - 10A if i am correct.
But yesterday i measured the current when moving the different axes (just moved them through Pronterface, one axis at a time) i measeured a maximum current of 100mA. I mean i have to made a mistake when measuring right? Or does the power comsumption go up significantly when you print a real model and move more then one steppers at a time?

On another note: I made my U_ref = 0.67V on the drivers. Do i need double of that for the z-Axis if i have two steppers?

Ok maybe i just figured it out myself. Because my multimeter is very sluggish i changed the axis Speed to 5mm/min. Thats probably why the current was so low.

So.. i should be fine with 7A - 10A right?

Edited 1 time(s). Last edit at 09/13/2016 08:32AM by MilloMille.

You don't drive stepper motors with dc, so they don't draw 1.7A/Phase (its another miss leading stepper motor number for worsts case condition, ie DC)

I cant recall the maths of it, but what I can say is that a standard ramps all 5 steppers and the hot end are on a circuit with a 5 amps fuse.

The hot end is typically 2.5-3 amps

So steppers are not exceeding 2-2.5 amps.

ok thanks a lot that makes sense

1.7A per stepper happens at a much lower coil voltage than the common 12V and the driver output works in PWM mode. So the average ( DC-equivalent ) current asked from the PSU is 5-6 times lower than the coil voltage.

The 1.7A label rating of the motor is a kind of maximum rating, and NOT the value it should run at, not even something you should aim for. Most likely should make a driver setting at 0.7-1.0 amp or around those values. And values 10x of thess, like 7A-10A are ... offcharts and impossible.

Also you can deduct the missing variable, for example if the motor label is 1.7A and 5V, then you can deduct from ohms law that the 3rd variable, the coil resistance will be 2.9 ohms. And to add to the confusion, you can note that even if the motor label says 5v, that is NOT the voltage the motor is meant to run at, either.

So, when you make a driver setting which according to driver datasheet, lets say equates to a 1.0A current, that is (read datasheet) I(peak). Note the word "peak". The driver lets the current rise in the coil, and when the current reaches a certain preset point Ipeak, driver shuts curent off. Also "peak' is NOT "average", they mean very different things. The average of current drawn by motor is who knows what, for starters its rise time shape/formula also depends on the L inductance, and fall time depends also on decay mode, but its beyond the point to make the exact average anyway, it would be a lot of work for some useless result. The point of the potentiometer setting is that when you need less, or more power you move it accordingly, thats all (the coil energy named flux depends only on current I, not v).

I wanted to point out at a few of what i think are usual misconceptions around the issue. Good luck.

Quote
MilloMille
Ok maybe i just figured it out myself. Because my multimeter is very sluggish i changed the axis Speed to 5mm/min. Thats probably why the current was so low.

So.. i should be fine with 7A - 10A right?

You haven't said what power supply voltage you are using. The higher the power supply voltage to the drivers, the less current they will draw. If you intend to use a 12V supply, then one rated at 4A should be adequate for the motors.

---

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].