bipolar parallel or serial - volt, amere resistance and inductance?

[atair]

hi

i have a "Mean Well SP-240-30" 30V / 8A power supply

now the question:
with these motors (3-axis) - http://www.motioncontrolproducts.com...TH76-2808S.pdf - i still dont get the combination of the different values / units, to calculate if i can run the motors in parallel mode.

the controler is TB6560AHQ, which says it can run at 3A..
now the datasheet of the motors say - in parallel they need 4A - does this mean i cant set them in parallel or what other values (inductance/resistance) play a role in this?
is there some formula to calculate these relations?

thanks!

[KOC62]

Generally, more current = more torque.
Your controller can handle 3 amps and limits your maximum current output.

You can use the Unipolar connection, which needs 2.8 amps. This should work fine. There is no exact right answer due to too many variables.

[atair]

yes - but the power supply gives only 8 amps - and max. 3A * 3axis = 9amps?
isnt this a problem?
and if yes - having the current switch on my board down to 75% or 50% - would that resolve it? (less current = more amps available?)

[KOC62]

Not really. The motor amps are peak usually.

What do you think you will cut that requires all three motors, X, Y, Z, to go full bore for any length of time? Statistically you will use less current because some motor is hardly turning.

I'm not sure what you mean by current switch. If you are referring to the motor current setting on the controller then yes you can throttle down the motor current that way. But also understand that your motor power output also goes down (and torque). Then you might as well buy smaller motors.

Have you oversized your motor requirements?

If you have already got the parts, just try it in unipolar mode and see how it works.

[H500]

The 4amp rating means it can take up to 4 amps without overheating. If you put in 3 amps, you will get 3/4 of the rated holding torque. This is quite respectable.

I would recommend bipolar parallel with the current set to 3 amps. That will give you the best performance. Your 8 amp power supply is more than enough. I'm running two 4 amp per phase motors with a 2 amp supply. The duty cycle is not 100%, so you will never need 9 amps.

[mike_Kilroy]

any reason you dont just wire them in series so you get bipolar full torque and only need 2 amps then for total max of 6 on ur 8amp supply?

[H500]

any reason you dont just wire them in series so you get bipolar full torque and only need 2 amps then for total max of 6 on ur 8amp supply?

The high inductance would make the torque drop quickly as the speed increase.

[atair]

actually i have them in bipolar series now - as said i was unsure to put them in parallel as the amps would add up to 12 and the power supply only can deliver 8.
but after a day of tweaking with the settings i found out that the only way i get smooth running (no stuttering, loosing steps etc.) is when i put the current setting to 25% - everything else makes some "clicking" sound every couple of seconds - which i think are lost steps. (TB6560AHQ)

i still don't get the math, but i assume as the current goes down, the available amps increase - right? so i can put them now in parallel for sure? (with the goal of having higher top speed - i know i loose torque, but planning on a ball screw i need rather speed than torque)

[H500]

It's hard to explain the math, but if you understand how a welder is able to convert 15 amps into 300, then you will understand how a chopper drive like the tb6560 is able to do the same. If you like details, read up on switching power supplies.

As I've said, I'm temporarily supplying a calculated 16 amps to my motor from a 2 amp supply without issues. 8 amps more than adequate for your puny setup.

[KOC62]

atair,

It is not a math problem.
The model you use to explain reality is flawed.
Let's assume a motor specification of 3 amps per motor coil.
The motor current is set in a chopper by a sense resistor so that the peak current of 3 amps can't be exceeded. In a linear controller it will be a power resistor or a constant current source to limit the maximum current to 3 amps. Your simple math adds up all the individual currents of say three motors to get a total of 18 amps. This is correct as simple math goes. What you don't know is whether each motor coil actually draws 3 amps all the time(i.e. average). If it did then you would need 18 amps of current.

Hence, something is happening during the cutting process that the average current is not 3 amps all the time. This has to do with the actual usage of each motor during a cutting process based on the g-codes received.

If you really want to know the average power supply current used during a cutting job you will need to insert an ammeter in series with one DC power supply lead. Then, if you want to know what each motor contributed to this average power supply current, you will need six ammeters, one for each motor coil, and sum the average current of each motor coil. This math works. Now you can model motor currents based on actual current consumption, rather than assumptions about what is consumed.

H500 maintains that his operation works fine with 2 amps average supply current while using two motors with 4 amps per coil current - if I interpreted his statements correctly.

[mike_Kilroy]

atair,
It is not a math problem.
The model you use to explain reality is flawed.......

Actually H500 is right, you are not. And it is not hard to explain how you can provide significantly more amps into a motor than is pulled from the power supply....

It is a simple case of power in = power out. If you start with this 30v supply and run it thru a PWM drive to a stepper that needs 2v as this one does, that is a ratio of 30/2. so to provide 4.0 amps into a motor at stall, it will draw 4/15 or only 0.33 amps from the power supply.

The mat his simple if you want to look at it as power in = power out. 2v*4a=8watts out. so from the power supply 8w=30v*i or i=8/30= .3amps

Granted, as you go faster, it will require more speed so more voltage and the power out will go up. But what goes up must come down as Newton said, so it will not stay at high speed forever, so it is only short time.

So if one is not happy with the torque available at higher speeds with the windings in series, one should try putting them in parallel and see if the average current is not less than 8 amp total from the power supply.

[KOC62]

I'm sorry Mike if I gave you the wrong impression.
I used H500's power supply as the practical outcome of "average" current, to assure atair that he can use his power supply without worrying about it. Perhaps I left the sentence such that you thought I disagreed. My bad.

Regarding the statement you made, quote "And it is not hard to explain how you can provide significantly more amps into a motor than is pulled from the power supply...."
This statement I find problematic as stated since all motor current comes from the power supply. Kirchhoff's current law can't be violated. What I interpret you to mean is that peak motor current can be higher than average power supply current. This, I thought, I also explained by using "average" current (i.e. "area under the curve"), which you also implied with your PWM explanation.

Both our explanations, (and H500), show that atair can't simply add all the peak currents to arrive at an "average" power supply current.

I'm hoping not to increase atair's confusion with our techno-babble.