[doanwannapickle]

I'm setting up a project with a G540.
My power supply is ~50v @ 8 amps.
These are the steppers I would like to use: Nema 17 2.1A Stepper Motor High Torque 92oz.in
There inductance is: 3mH±20%(1KHz)
When you run the numbers, the voltage come out to 55.4v which exceeds the capacity of my power supply and the 50v limit of the G540.

What would be the effect if I used them anyway?

[john-100]

If the 5.4V makes a noticeable difference
you have either chosen the wrong motor or stepper driver

(on a good day 3mH less 20% = 2.4mH which is ideal with your 50V power supply )

how precise are you going to set the motor current ?
2100 ohm resistors are available from the E96 range

mouser.com
Mouser No:
660-MF1/4DCT52R2981D


if buying the resistors locally restricts you to the E12 range
1800 ohm & 390 ohm resistors in series will give you 2190 ohms
(with resistors in series you only have one wire to soldered to each of the 9 pin plug terminals)

John

[doanwannapickle]

If the 5.4V makes a noticeable difference
you have either chosen the wrong motor or stepper driver

Yup. That's what I'm hoping to avoid.

(on a good day 3mH less 20% = 2.4mH which is ideal with your 50V power supply )

Definitely what I'm hoping to get.

how precise are you going to set the motor current ?

I thought I'd just series a couple of 1k's or stack a couple of 3k9's. But I'm open to suggestions. I have lots of resistors.

[john-100]

Gecko drive recommend 5% resistors are used to set the motor current

if you have a large number of 1K 5% resistors you may be able to find some close to 1K05
that two in series will give you 2K1

Attachment 394590

John

[doanwannapickle]

Sorry, I had in my head that the steppers were 2.0 amp. Oops.

[doanwannapickle]

I guess my original question should have been 'what is the effect of running a motor at less than the recommended Gecko voltage'?
I assume there will be a loss of holding power. Will it be significant? Will there be other effects?

[awerby]

If you give the motors less voltage, they won't run as fast. But 5.4v isn't too significant. You'd only lose torque if you gave them less current (than 2.1 amps). What are you trying to build, anyway?

[john-100]

all other things being equal

the stepper drivers supply voltage affects the maximum speed you can step the motor

if your motor has a 1.42 ohm coil resistance then applying 3V DC will result in a current of 2.1A
which is OK when the motor is still

using a motor supply voltage several times the voltage on the motors data sheet
has the advantage of forcing the current to build up to the desired current in a shorter time
enabling you to step the motor faster but the stepper driver needs to limit the current to 2.1A

on the other hand
the motor current determines the motor torque

so using a 50 V supply will slightly reduce the maximum speed you can step the motor
and setting the current to 2A with two 1 K resistors will slightly reduce the torque

John

[doanwannapickle]

That's what I wanted to hear. Reducing the voltage just slows things down proportionally. You still get rated power as long as your power supply has got the amps.

It's going to be a mini-lathe based on a pair of Parker slides. I've built a couple of other machines using them and I absolutely love 'em.

Thanks gents.

[ger21]

I guess my original question should have been 'what is the effect of running a motor at less than the recommended Gecko voltage'?
I assume there will be a loss of holding power. Will it be significant? Will there be other effects?

That recommendation is the Maximum voltage. At a lower voltage, you'll have less top speed, but it should be a negligible difference.

[automationtechinc]

I'm setting up a project with a G540.
My power supply is ~50v @ 8 amps.
These are the steppers I would like to use: Nema 17 2.1A Stepper Motor High Torque 92oz.in
There inductance is: 3mH±20%(1KHz)
When you run the numbers, the voltage come out to 55.4v which exceeds the capacity of my power supply and the 50v limit of the G540.

What would be the effect if I used them anyway?

if you order this NEMA 17 KL17H272-20-4A, it should be ok

you need to set the current to 2A

Applied Voltage: 32 x sqrt (4) = 64VDC


Current Per Phase:2.0A
Holding Torque: 9 Kg.cm (124 oz-in)
Rated Voltage: 3.2V
NO.of Phase: 2
Step Angle: 1.8° ± 5%
Resistance Per Phase: 1.6?± 10%
Inductance Per Phase: 4.0mH± 20%
Insulation Class: Class B
Dielectric Strength: 100Mohm

https://www.automationtechnologiesin...l17h272-20-4a/

[A_Camera]

If the 5.4V makes a noticeable difference
you have either chosen the wrong motor or stepper driver

(on a good day 3mH less 20% = 2.4mH which is ideal with your 50V power supply )

how precise are you going to set the motor current ?
2100 ohm resistors are available from the E96 range

mouser.com
Mouser No:
660-MF1/4DCT52R2981D


if buying the resistors locally restricts you to the E12 range
1800 ohm & 390 ohm resistors in series will give you 2190 ohms
(with resistors in series you only have one wire to soldered to each of the 9 pin plug terminals)

John

If he wants to be perfect and uses E12 series then it is better to use 2.2k in parallel with a 47k. That will give 2.10K, which is as good as he can get. There is no point in doing what you propose since it will result in only 10 Ohms below the standard value of 2.2k.

[john-100]

1800 + 270 would of been a better solution

for members who have little practice soldering
I still think its easier to have two resistors in series
easy to insulate with a plastic sleeve

Attachment 394646

Attachment 394648

the main thing to use the best solution for you

john

[A_Camera]

1800 + 270 would of been a better solution

Why? That's still giving the wrong value since 30 Ohms are missing.

2k2 parallel with 47k is 2102 Ohm (2101.6260162601626016260162601626 Ohms to be more precise) and that's DEFINITELY better than 2070 since the difference is only 2 Ohms. I have no idea how important it is, but really, why would 2070 Ohm be better than 2102 Ohm if you need 2100 Ohm? Connecting two resistors in parallel is just as easy as connecting two in series, so what is the benefit? Why would 1800 + 270 be better than a more exact value?

[john-100]

duplicate post

[john-100]

using resistors 5% tolerance resistors in series

1800 + 270 would of been a better solution in the sense that its about 1.5% lower than the desired 2100 ohm

instead of the 1800+390 posted in error in post 2 ( not sure what caused my error )
as an alternative 1800 +330 would be 1.4% high

from a practical point of view
I expect some will find it easier to wire the resistors in series and that's why I posted that solution
as I don't think you need to be accurate to the Nth degree setting the motor current

although I admit I would use a single 2100 ohm E96 resistor

John

[A_Camera]

from a practical point of view
I expect some will find it easier to wire the resistors in series and that's why I posted that solution

That's because people generally understand how to connect resistors in series AND calculate the new value but have no clue about how to calculate resistor values for parallel connections.

although I admit I would use a single 2100 ohm E96 resistor

John

I would probably look in my boxes and take what I have, if I don't have a 2k1 then I'd use my calculator to get the right value. If I don't have a single resistor with the right value I'd use parallel or serial connected resistors to get the right value.

...but like I said, I have no idea how critical this is for the Gecko because I don't have that driver... Probably it doesn't really matter if you use 2k1 or 2k2 resistor.

[doanwannapickle]

...but like I said, I have no idea how critical this is for the Gecko because I don't have that driver... Probably it doesn't really matter if you use 2k1 or 2k2 resistor.

Well, that was going to be my next question since the whole resistor thing is being wrung out. Just how close is close enough? Is it better to go high or low? What is the effect of each?
I don't have a problem putting them together series or parallel or a combination thereof as long I can cram them in the shell.

Backing up a bit, I'm now thinking of going with the motors suggested by Automationtech.
Inductance Per Phase: 4.0mH± 20%.: This calcs out to 64v using the Gecko formula but it has now been pointed out that this is a max voltage and a lower voltage is just going to produce less rpm.
Holding Torque: 9 Kg.cm (124 oz-in).: That's a lot of grunt in a small package.

My screws have a 5mm pitch and the total travel is just a bit over 4" so I don't think I need a lot of speed but torque is good.

[john-100]

with the Gecko G540 the motor current is 1A per 1000 ohms across pins 1 & 5 on the 9 pin plug
up to 3500 ohms to set the current to the G540's maximum of 3.5A

getting the resistors within 2 or 3% is not going to be a problem

various possible ways of setting the G540's motor current
Attachment 394718

one issue some users have is when they omit the 3K5 resistor
because the G540 default current without a resistor is 3.5A
the catch is without the resistor the G540 does not reduce the current when the drive is idle and not being stepped
the result is the motor runs hotter

one handy online calculator for 2 resistors in parallel can be found here :-

Resistor calculator - compute the best resistors combination to obtain a given value

with the calculator you can easily see the effect of using resistors with different tolerances
10% restricts you to 12 possible values per decade (10 ,12 , 15 , 18 , 22 , 27, 33 , 47 , 56 , 68 )
1% gives you 96 possible values per decade

the new motors calculated 64V supply verses the G540's 50V limit can be a problem
as the speed you can step the motor is going to be reduced by approximately 22%

john

PS
a motor with 2.8mH 3.5A coils would be closer to the G540s ideal limit

something like this

https://www.automationtechnologiesin...23h2100-35-4a/

then again bigger is not always better

so splitting the difference
a motor with coils having an inductance of 2.8 to 3 mH at 2.5A to 3A could be what you need to look for

its a pity we don't have feed back from any Gecko G540 users