Peak or RMS

[m.marino]

Does anyone know if the Amperage on the nema23(4) 3Nm stepper motors is listed in Peak or RMS?

The international standard is to list it in RMS but we are dealing with China here so that is not a for sure in any way. I am just wondering as it greatly affects how you set up the drivers. Which set up properly is not a problem and should have smooth running steppers. Step up the wrong way and you are either running hot and going to cause their useful life to shorten a good bit or you will be under powering them massively and there by have step loss issues and such. Just asking as It would be useful to know.

Michael

[Jonathan]

Assuming the motors are wired in bipolar parallel then set them to the nearest to 4.2A peak. Keep an eye on the motor temperatures to start with - up to 80°C on the case is OK.

[m.marino]

That is assuming that the current listed is Peak and not RMS/phase amps. Which is what I really want to know is which is it. Not the assumption of use it as Peak and therefore add to the safety margin of the motors operating life span. IF it is phase/RMS then using 4.2 as peak reduces the current that motor can use to .707 of what RMS would be.

Let me write it this way:

Which is the 4.2A on said motor? Peak or Phase? Why?

Peak is 1.414 * RMS (or put list the inverse RMS = Peak * .707)

IF it is Peak then fine set accordingly and everything should run fine.

IF RMS then either set as RMS (IF Driver lists RMS Amps) or do the math to arrive at what peak current you can use for said RMS.

IF set as Peak and IS RMS/Phase then you have a problem RMS/phase being .707 of Peak means that you take 4.2 * .707 = 2.9694A is your operating Amperage (RMS/Phase) IF that was peak; this section we are assuming RMS so you have reduced you usable power by 30%.

Now for folks coming into this discussion and not having a clue why it is important:

RMS/Phase is the current you stepper motors run at under normal full current load (no micro-stepping no stand by mode).

Peak is the max that the motor can handle without damage to the motor from use under these conditions (normally for sustained periods).

While you can use RMS as peak it decreases the effectiveness of the motor as well as it's torque.

This is why I am chasing this information down to find out which it is so I can set up the best option on my machine for power usage.

Michael

[m.marino]

After a cal to Zapp (Thanks for your time Gary), according to him set them as Peak on M752 type series drivers. Other drivers or full stepping may require different settings.

Again good things to know for the best signal to the motors.

Michael

[Jonathan]

Just to confuse matters further

Which is the 4.2A on said motor? Peak or Phase? Why?

Peak is 1.414 * RMS (or put list the inverse RMS = Peak * .707)

...
RMS/Phase is the current you stepper motors run at under normal full current load (no micro-stepping no stand by mode).

It's rather odd that in the stepper driver datasheets they state the RMS current and the current multiplied by 1.414 for peak current since the current waveform varies so much from sinusoidal. It depends on the driver and the speed the motor is running at - on some drivers the current is just chopped, so without microstepping it's closest to a square wave. With microstepping the waveform approximates a sinusoid (but quite poorly at the low microstepping settings we use) and for better drivers at medium-high speed they do transform to a sinusoidal waveform, but still an approximation. The only waveform for which the peak value is 1.414*RMS is a sinusoid, so this formula generally doesn't apply...yet they still confuse the ratings by including it.

Anyway..

The current rating on the stepper motor is the maximum constant current you can apply to the coil/phase due to the thermal limit (i.e. in steady state). Since in the 3Nm motors this rating is 2.1A, they're 4.2A in parallel so you know the current cannot be greater than 4.2A so it's safe to use 4.2A peak (4.09A) is nearest on PM752.

4.2 * .707 = 2.9694A is your operating Amperage (RMS/Phase) IF that was peak; this section we are assuming RMS so you have reduced you usable power by 30%.

Power is proportional to the square of the voltage, (P=I^2*R etc) so reducing the current by 1/2^0.5 reduces the power by the square of that, so (1/2^0.5)^2=0.5. Therefore you have reduced the available power by 50%. However the motor torque is (roughly) proportional to phase current, so you would be correct in saying the torque is reduced by 30% in this hypothetical situation.

[blackburn mark]

can you run your eye over this for me jonathan :)
my motor spec

new drivers


i assumed id be setting peak for 2.84 (closest?)
am i going to smoke anything?

[Jonathan]

Setting to 2.84A as you can't get 2.8A is only 3% more heat, which is negligible so you'll be fine. It's always a good idea to keep an eye on the motor temperature to start with just to be sure.

[blackburn mark]

cheers man x

It's always a good idea to keep an eye on the motor temperature to start with just to be sure.

im a nervous bugger when it comes to electric.... ill be like a gun-slinger on the E-stop :)

[blackburn mark]

another conundrum
my bob has EL817 opto isolators all over it and my drivers state "Opto-isolated signal I/O"
the drivers came with this bob thrown in?
i thought it was unwise to have optos on both?
any thought anybody?

[Jonathan]

i thought it was unwise to have optos on both?
any thought anybody?

Yes, double isolation is an additional pointless delay, but if the breakout board is well designed the difference should be negligible. Try it and if it doesn't work just bypass the breakout board isolation on just the outputs.