Was about to get a toy- then I did some research...

[NeoMorph]

If they use the same input connections for DC as AC as I assume they do then the rectifier and capacitors must still be in the circuit.

One other thing to consider: If you add an additional rectifier and capacitors to make a DC supply for an AC driver you will also increase the inrush current to the transformer and increase the risk of trips. Adding more capacitance for the sake of it to any DC supply (making Neale's audio amp PSU instead of a motor driver one) will have the same effect.

Yup. My brain was in full “doped up on morphine mode”... which is why I did a facepalm. It’s the reason I had to quit work. Not asking for sympathy, just stating the way things are. I seriously do not know this tech, BUT I WANT TO LEARN. It’s better to ask questions and act a fool than profess to be an expert and tell people incorrect info because “Well it works for me so must be okay”. A lot of my expertise is 20 years out of date and worked with super delicate security systems that required strong, stable power sources to detect intrusions. These days you could build the same thing for lest than 50p with tech from Asia.

But I love making things which is why I got into CNC in the first place. Anyhoo... I’m going to shut up until I can reduce my meds come spring.

[Kitwn]

Don't fret mate!
My radio/TV background doesn't allow for much ripple on a DC power supply so I was expecting to see huge quantities of expensive smoothing capacitors on a linear supply of the sizes discussed here. For this application you don't need that much.

I'm not brainy enough to use Faraday's Law to work out the inrush for a given transformer, but I suspect you need a pretty chunky lump of iron for that to be the only significant contributor to the overall inrush.

[NeoMorph]

Don't fret mate!
My radio/TV background doesn't allow for much ripple on a DC power supply so I was expecting to see huge quantities of expensive smoothing capacitors on a linear supply of the sizes discussed here. For this application you don't need that much.

I'm not brainy enough to use Faraday's Law to work out the inrush for a given transformer, but I suspect you need a pretty chunky lump of iron for that to be the only significant contributor to the overall inrush.

Hah... try doing math when your head is on the equivalent of 10 pints of lager. I read some of the posts here and go, “Errr... what the heck? The drives can take AC direct from the transformer... that ripple is going to be awful!”, without thinking it through that the drive itself will do the cleaning.

This is the problem with someone who knows a little bit of knowledge and thinking they know what to do (that’s me btw). Unfortunately, microelectronics which is VERY pick about power stability and a CNC drive that is going to take 60v from an AC transformer is two very different things.

Unfortunately I won’t be able to get my kit for at least a month apparently... nobody appears to be working at the factory/distributors in China. They are talking 2 weeks to even start back up again and maybe longer.

[Juranovich]

Do not rely on the OV protection of these drives, they will not save you from more than just a few volts. So if you get a spike then they will fry them.
They are good drives provided you leave a good safety margin.

You need to leave at least a 10% safety margin on the voltage and I wouldn't run much above 60Vac with 70Vac Max.

Regards the Transformer I use 750Va with these drives without any issues. I wouldn't go above 750Va because you will get troubles from inrush.

Do NOT use a regulated PSU like as been suggested by one member as it could cause you troubles with voltage clamping.

Now this gets fun! So, given that I want steppers around 4Nm. They usually (based on my google searches) come rated at 4-5A and 2-3mH inductance meaning total drawn amps are in the range of (Atot_= 2/3*pcs*Arated) 10.5-13.5Atot and Vmax (=32*√mH) 45-56V.. Now based on what I've picked up in this discussion I want to give my drives as much voltage as possible (minding 10% safety margin) and restrict the PSU to 750VA due to inrush. Say I supply 60V to 70Vmax drives, that means the 750VA PSU supplies 12.5A of current i.e. steppers should be rated at 4.5A (~12Atot) or lower in order to get max performance out of them (naturally I could have higher rated steppers, but that would be wasted monies, correct?). Am I correct in this logic?

Moreover, is the performance of the steppers linearly related to the current supplied or is there some leverage at play, i.e. say I supply a stepper with 4.5A, now will a 5A (arbitrarily chosen) rated stepper perform equally well as a 4.5A rated stepper would?

[JAZZCNC]

Say I supply 60V to 70Vmax drives, that means the 750VA PSU supplies 12.5A of current i.e. steppers should be rated at 4.5A (~12Atot) or lower in order to get max performance out of them (naturally I could have higher rated steppers, but that would be wasted monies, correct?). Am I correct in this logic?

Well, it depends on how many drives you intend to use now and in the future. Maybe you might add a 4th axis.? In this case, the 750Va allows a little overhead. However, what I didn't mention so as not to confuse the issue is that I've also used 625Va without any issues. But I knew this machine wouldn't need 4th axis because it was fitted with a 4th Axis that was powered from a separate smaller PSU.

Moreover, is the performance of the steppers linearly related to the current supplied or is there some leverage at play, i.e. say I supply a stepper with 4.5A, now will a 5A (arbitrarily chosen) rated stepper perform equally well as a 4.5A rated stepper would?

If using the Same Voltage then the 5A motor will have lower overall performance than the 4.5A motor. It will have a little more torque lower down the range but the RPM will be lower. This is mostly due to inductance because the higher current motor will have more inductance. It will also create more heat which robs performance.
In this case, half an Amp is neither here or there so wouldn't be a big difference and you'd hardly notice it. Thou any gains would be offset unless something else changed ie: Voltage. Everything comes at a cost.!

In a nutshell, Higher Amp's which often = higher inductance means higher voltage to reach the same speed. This is why often a 4Nm Nema23 will outperform a 4Nm Nema34 motor if using the same volts.
It's also why Large Nm Nema 34, 42 size motors require Very high or better still mains level voltages to allow any reasonable RPM's. All the machines I build that use above 8Nm motors use Mains voltage drives.

People often mistakenly think increasing the current will give more torque, which it does up to a point, but it also increases heat which affects the motor's saturation point which then creates resonance etc so stalls at lower RPM.

It's a complicated formula with several twists depending on motor spec, wiring, etc also with the advent of Digital drives allowing much better performance then Old Vmax (=32*√mH) Formula is even less relevant because the motors can be pushed harder and heat is controlled better along with resonance.

All I can tell you is that if you run the motors at the Rated current with a Voltage 10% lower than the drives Max V you'll be getting the best performance. If you need a little more Torque then increasing the current will provide a little extra but will cost in terms of heat and RPM.

[Juranovich]

All I can tell you is that if you run the motors at the Rated current with a Voltage 10% lower than the drives Max V you'll be getting the best performance. If you need a little more Torque then increasing the current will provide a little extra but will cost in terms of heat and RPM.

After much googling and talking to my uncle who's an electrical engineer I think I've got my head around the basics of this. One thing i still find confusing however is that logically you'd want to run the motors at they're rated current but I keep reading that the drives only draw 2/3 of that assuming parallel wiring (hence psu should be sized 2/3*Atot). Now, are the drives able to supply the motors the full rated current even if only 2/3 of that is supplied by the psu (through some magic I don't understand)? Or is this 2/3-rule applied simply due to the fact that the motors are seldom simultaneously drawing all of their rated current?

[AndyUK]

Or is this 2/3-rule applied simply due to the fact that the motors are seldom simultaneously drawing all of their rated current?

I think its that you'd never power all phases of the same motor to 100% of their current at the same time? But I'm sure one of the electronics experts around here will come up with a much more detailed answer! The 2/3rds rule is for parallel wiring, and for series wiring you can get away with 1/3rd.

I'd wager the answer is in here: http://homepage.divms.uiowa.edu/~jones/step/index.html (I just don't have the time to read it all!)

[JAZZCNC]

After much googling and talking to my uncle who's an electrical engineer I think I've got my head around the basics of this. One thing i still find confusing however is that logically you'd want to run the motors at they're rated current but I keep reading that the drives only draw 2/3 of that assuming parallel wiring (hence psu should be sized 2/3*Atot).

Be careful here when talking with electrical engineers ie: Domestic electricians or maintenance electricians because while they know how electricity and how circuits work etc I find they don't always understand or realize how different a Stepper driven machine differs to say typical AC motor system.
A stepper drive uses a chopping system to control the current/voltage that steppers require so it's not straight forward in terms of power draw etc like it is with say an AC motor connected straight to mains voltage.
The drives use a chopping system which uses PWM which only draws current 50% of the cycle on time. This power is taken from the capacitors in the DC system (AC drives just rectify inside the drives to DC) so during the Off cycle time the capacitors are recharging so only drawing power 50% of the time.
This is one of the reasons why the PSU can be sized lower than total Motor ratings. The other reasons being Not all Motors will draw full current all of the time and if they do then it's for very short periods and the Capacitors and drives will deal with any shortfall.

Now, are the drives able to supply the motors the full rated current even if only 2/3 of that is supplied by the psu (through some magic I don't understand)? Or is this 2/3-rule applied simply due to the fact that the motors are seldom simultaneously drawing all of their rated current?

Above should explain this hope fully.! . . . . Don't try to overthink this, I understand the need to understand how it works but if you want to build a good machine then what's been suggested will work great. You could spend weeks or months learning how it all works and you'll still end up back at what's been suggested.
Go with what's proven to work and you won't go wrong.

[Kitwn]

The rated current of a stepper motor is the maximum permitted continuous current in each winding. This will flow in the windings when supplied at the rated fixed voltage, something you will never do in practice for the reasons Dean has outlined above.

[Juranovich]

Above should explain this hope fully.! . . . . Don't try to overthink this, I understand the need to understand how it works but if you want to build a good machine then what's been suggested will work great. You could spend weeks or months learning how it all works and you'll still end up back at what's been suggested.
Go with what's proven to work and you won't go wrong.

This! Even though I'll probably end up building my machine according to the various suggestions I've got, I still cannot bring myself to do something without having built an understanding of my own of the matter. Hence, the occasional silly questions :)