Hi, I am about to try to set up two FIT0278 on a BTT SKR MINI E3 V2 and the data says these are 1.7AMPS/PHASE but I am needing to know the peak current in RMS to safely set up the motor on the board. Not having much luck verifying that information from the web. Has anyone got insight into the peak current in RMS for the FIT0278 at all please?
I’m feeling I want to use 1.2A based upon 1.7A/sqrt(2)?
Cheers,
Ray
Can you explain why you need to know this and how you intend to use it? Stepper motors are digital devices (the coils are either on or off) and RMS values are not usually relevant.
Note that the current rating for the motor is a per phase value, so the required current capacity of the driver will depend on how you are controlling the stepper, because that determines how many phases are active simultaneously. Is that where the figure of 1.2A comes from?
Steppers are not digital per se as a series of motor pulses probably means RMS is actually relevant (since your dealing really with more of an AC not DC current), not to mention the discussion on multiple sites where over heating of steppers is dealt with by ensuring your setting the amps in the stepper controller config to RMS rather than peak etc. I think you might be thinking of DC (but not AC) solenoids? So the problem is the amperage cited is not clear if its RMS or peak etc. and I am assuming its peak at the moment, since there is no clarity on which in the brochure.
Steppers are driven by a signal that energises the coils in a particular sequence. The frequency of the signals is related to the speed of rotation and the sequence in which the coils are energised.
Just because the signal turns on and off (like any digital signal) does not mean that you can consider it as AC. Even if you were to consider the signal as AC then RMS would not be relevant because it is a calculation that is applied to a voltage that varies as a sine wave, and for an on/off signal it is a very poor approximation to the power in a digital pulse train. If for some reason you want to consider the power in a digital signal you simply take the voltage and the on time.
You might see RMS mentioned in respect of stepper motors in the driver implementation of microstepping, where the current is ramped up and down for each pulse. But that calculation only applies to each pulse. The driver specifications will indicate the percentage of the maximum current that corresponds to the load for each type of microstepping and the ramping option selected. That’s why I said “…so the required current capacity of the driver will depend on how you are controlling the stepper”. So if you are using a driver with microstepping then you can set you maximum current according to the microstep that you have selected and the corresponding table entry. Generally, this allows you to set a higher current limit for fewer microsteps. That table might be calculated using a formula similar to the formula for RMS, but it does not mean that the manufacturer is considering the controller as an AC device. For your driver the table of current for each step of a microstepping pulse is listed at P64. The data behind this table is at P67 - you will notice that it is nothing like a sine wave. This rating is expressed in terms of percentage of maximum current - that is, it allows you to derate (uprate?) the motor specs according to the mode in which it is driven. But the base point for the calculation (100%) is the rated current for the motor.
The formula you have used might be a reinterpretation of the common advice to run the driver at about 70% of the motor rated current if you are using double stepping implemented in software. And that, in turn, might have come from an assumption that the use of RMS in the description of pulse ramping had something to do with a sine wave.
Hi Ray
AC voltages swing negative WRT ground. AC= Alternating Current. That is the current in a circuit reverses direction every half cycle. The requirement here is a DC pulse switching to different coils to achieve rotation. When you look at a pulse train with an oscilloscope the result might look like AC but you should see it never goes negative.
You can make it go negative by passing through a capacitor which will remove the DC component but once this happens you require a split power supply for any further processing.
That does not happen here. The pulse train remains on and off DC.
Cheers Bob
Actually I could be wrong here. I think the DC is switched to the coils to control rotation and a pulse train is used to drive the logic that does this switching.
The polarity flips so it is ‘alternating’ but the sequence in which the pulsing occurs and the pulse ramping (or not, depending on the driver mode) means that there is no realistic comparison with an AC sine wave. Even when there is ramping it is done as current-controlled PWM.
The only respect in which this variation can be relevant for the driver maximum current setting is where the driver is pulsing more than one coil at a time, which occurs with microstepping, but an intelligent driver automatically allows for this. However, if this overlap is done by software (“Double” mode) then the driver maximum should be set somewhat below the motor maximum, which is possibly where the question comes from.
Hi Ray,
Thanks for your post, for calculating run current the Voron documentation has a great guide on it, but put simply it is as below:
Amp rating x 0.707 = Run Current
For the FIT0278 it would be
1.7 x 0.707 = 1.2A
Don’t worry about setting your hold current in Klipper, its not needed.
Dan, thanks so what I missed was the story about the hold current, as opposed to peak v RMS. That’s helpful cheers.
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