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Lee Roberts
29-08-2009, 01:22 AM
Hi,

What I would like to know is, when selecting a driver unit for use with stepper motors, should we be matching the driver amp rating to the same amp rating of the motor or do we really only need to select a driver that is half the amps of the motor ?

So if I had a motor that was rated at 4a, would that be 4a total regardless of the number of phase’s or divided by the number of phase’s?

So in this case it is a two phase motor rated at 4a, two phase’s at 4a = 8a so I would need an 8a driver or am I right in thinking only half of the motor (one phase) is in use at any one time?

What happen's when the current is moving from one phase to the next are they both in use at the cross over stage?

Thanks,
:feedback:

irving2008
29-08-2009, 01:33 AM
Hi,

What I would like to know is, when selecting a driver unit for use with stepper motors, should we be matching the driver amp rating to the same amp rating of the motor or do we really only need to select a driver that is half the amps of the motor ?

So if I had a motor that was rated at 4a, would that be 4a total regardless of the number of phase’s or divided by the number of phase’s?

So in this case it is a two phase motor rated at 4a, two phase’s at 4a = 8a so I would need an 8a driver or am I right in thinking only half of the motor (one phase) is in use at any one time?

What happen's when the current is moving from one phase to the next are they both in use at the cross over stage?

Thanks,
:feedback:

Lee,

Drivers are usually rated per phase...


Only one phase is on at a time in full step, wave drive. two phases are on in other modes, but the currents in each phase are reduced..

Lee Roberts
29-08-2009, 01:34 AM
Drivers or motors ?

irving2008
29-08-2009, 01:41 AM
Drivers or motors ?both...

If you look at the drivers from DIYCNC for example they are 2.5A drivers... which is per phase

Lee Roberts
29-08-2009, 01:42 AM
"both..."

ok thanks mate !

Gary
29-08-2009, 08:36 AM
A motor and driver is rated as per phase, so if a motor is 4A, then get a driver that is rated at 4A or over.
To get any power that is useful from the motor you also need volts, so dont just look at the current.
As an example if you see a 10A driver that is only rated at say 30V, this will be completely useless for larger motors, unless you are happy watching paint dry.
In other words you may get the torque, but you wont get the speed.

irving2008
29-08-2009, 03:24 PM
Gary's point is well made. The bigger the motor the higher the inductance of the windings, therefore the longer it takes for the magnetic field to build up. even small cheap steppers can have big inductances. A higher voltage 'forces' a higher current through the motor at the start of the step when the inductive impedance is highest. A simple rule of thumb is:

volts = sqrt(32 * H^2) where H is inductance in milliHenries.

so a 5mH motor would require ideally sqrt(32 * 5 * 5) = 30v, while a 10mH motor needs 56v (hence big motors on cheap drivers is a waste of time)

This is partly why the bipolar parallel arrangement give the best torque at high speeds. For typical motor windings that are bifilar wound (as most are) this halves the inductance as compared to bipolar series connection which quadruples it (all relative to unipolar).

Lee Roberts
01-09-2009, 02:57 AM
So let’s take the PM542's (PDF (http://www.axiscnc.com/pdfdocs/pm542.pdf)), they have a Typical Supply Voltage of 36v with a MAX of 50v.

A late forum member told me I should supply them at the max (50v) is this the correct thing to do? Or would it be better to supply the driver with the Typical Rated Voltage?

I want to get the best speed I can from the motors, will supplying the driver at the MAX voltage mean the motor will also receive a higher voltage and give me better speeds?

If we take the Nema23 3Nm motor (PDF (http://www.axiscnc.com/pdfdocs/sy60sth863008bf.pdf)), the document said:

Bipolar Parallel 2.73v, 4.2a, 3Nm

So from that, i take it for the Motor to run at 3Nm Holding Torque it will require from the driver 2.73v and 4.2a ?

The other question i have is about the Motor, when a phase is "in use" is the Volts and Amps it is pulling from the driver divided by the amount of wires coming from the motor?

Gary
01-09-2009, 08:32 AM
The max voltage is 50V, but over this voltage and the driver will trip into error, so i suggest to use a 40V PSU.

Forget the 3Nm holding torque, it is not the holdiong torque that you need, it is the torque while moving.
Also a stepper driver is a constant current device, so the motor does not pull current, the driver controls the current.


So letís take the PM542's (PDF (http://www.axiscnc.com/pdfdocs/pm542.pdf)), they have a Typical Supply Voltage of 36v with a MAX of 50v.

A late forum member told me I should supply them at the max (50v) is this the correct thing to do? Or would it be better to supply the driver with the Typical Rated Voltage?

I want to get the best speed I can from the motors, will supplying the driver at the MAX voltage mean the motor will also receive a higher voltage and give me better speeds?

If we take the Nema23 3Nm motor (PDF (http://www.axiscnc.com/pdfdocs/sy60sth863008bf.pdf)), the document said:

Bipolar Parallel 2.73v, 4.2a, 3Nm

So from that, i take it for the Motor to run at 3Nm Holding Torque it will require from the driver 2.73v and 4.2a ?

The other question i have is about the Motor, when a phase is "in use" is the Volts and Amps it is pulling from the driver divided by the amount of wires coming from the motor?

Lee Roberts
01-09-2009, 01:41 PM
Hi,

Well yea I know it would give error going over 50v, still need the questions answered please people.

Isn't the holdiong torque the same thing as "torque while moving" ?

Gary
01-09-2009, 01:59 PM
Holding torues is at zero speed.
as soon as the motor starts to move the torque goes down.


Hi,

Well yea I know it would give error going over 50v, still need the questions answered please people.

Isn't the holdiong torque the same thing as "torque while moving" ?

irving2008
05-09-2009, 08:04 PM
So letís take the PM542's (PDF (http://www.axiscnc.com/pdfdocs/pm542.pdf)), they have a Typical Supply Voltage of 36v with a MAX of 50v.

A late forum member told me I should supply them at the max (50v) is this the correct thing to do? Or would it be better to supply the driver with the Typical Rated Voltage?

I want to get the best speed I can from the motors, will supplying the driver at the MAX voltage mean the motor will also receive a higher voltage and give me better speeds?

If we take the Nema23 3Nm motor (PDF (http://www.axiscnc.com/pdfdocs/sy60sth863008bf.pdf)), the document said:

Bipolar Parallel 2.73v, 4.2a, 3Nm

So from that, i take it for the Motor to run at 3Nm Holding Torque it will require from the driver 2.73v and 4.2a ?

The other question i have is about the Motor, when a phase is "in use" is the Volts and Amps it is pulling from the driver divided by the amount of wires coming from the motor?

Lee, Gary has answered some of your points but let me throw in my 2p worth.

Electronic devices should never be run at the absolute maximum voltage quoted. Depending on the circuit this might either be the protective trip voltage or the voltage at which damage can occur due to internal breakdown of the device (which then results in high peak currents frying the device unless there is an external curent limit). In this instance its the protective trip. If you were to feed the driver with 50v there is the possibility that back-EMF from the motor when pushing energy back into the power supply (on the overrun/braking cycle) would cause the power supply voltage to rise and trip which would be inconvenient at best and catastrophic to the workpiece at worst! If your power supply is really good, with a regulation of better than 2% then you might get away with running it at 48v or so. But I'd stay below 45v on these drivers.

From the motor torque curves (http://www.slidesandballscrews.com/pdf/steppermotors/SY60STH86-3008BF.pdf) on Gary's site you can see that at 500steps/sec (150rpm) the torque is already down at 2Nm. Below 150rpm the torque will tend towards the holding torque but the relationship isn't linear. You will notice from these torque curves that the curve for 60v/4.3A is flatter/more linear than that for 40v/4.3A but the difference isnt really apparent below 2500steps/sec (750rpm). In other words if you are never going above 750rpm on the motor then 40v will perform as well as 60v. If you have only got 40v drivers then this information can be used to factor any gearing to keep the motor in the sweet spot.

The stepper driver is a constant current device. whatever the motor winding resistance the driver will never allow more than the rated current through the winding. For a driver set at 4.2A this is the maximum it will allow. The voltage rating of a stepper motor is irrelevant, it is merely the steady state voltage across the windings at the rated current due to the winding resistance. With a chopper or PWM current controller it has no meaning as the driver will switch the drive voltage on and off at high frequency to maintain the average current at the set value. Since it is being switched the winding inductance has a part to play as this controls the rate at which the current can change in the winding. This is where the higher voltage comes into play as a higher voltage can 'force' a higher current in a shorter time therefore maintaining the torque at higher speeds.

I'm guessing the meaning of your last question... refering to the data sheet for the motor again, when wired bipolar parallel the first phase has A and /C connected together and /A and C connected together. When that phase is energised, 4.2A of current flows from the driver along the connecting wire to the A + /C junction where it splits - 2.1A into each winding - and recombines at the /A + C junction, where 4.2A of current flows back to the driver. Therefore every one of the 4 wires connecting the driver to the motor carries 4.2A and needs to be rated for that.

Hope this helps...

Lee Roberts
06-09-2009, 12:18 AM
Hi Irving, thanks for taking the time to go into details I appreciate it.

Now I don’t have one of the motors on the data sheet to hand, but I’m confident that this will be the same…

On the motors I have got here in front of me ( two different manufacturers ) the wire's coming from the motors is only 16/0.02, this motor is 4.2a in bipolar parallel setup.

I’ve looked at a few different websites selling the 16/0.02 equipment wire and they all seem to say that 16/0.02 wire is rated at 3a, if that is true then the wire on these motors is under rated for what the motor could potentially be doing ?

irving2008
06-09-2009, 01:55 AM
Hi Irving, thanks for taking the time to go into details I appreciate it.

Now I donít have one of the motors on the data sheet to hand, but Iím confident that this will be the sameÖ

On the motors I have got here in front of me ( two different manufacturers ) the wire's coming from the motors is only 16/0.02, this motor is 4.2a in bipolar parallel setup.

Iíve looked at a few different websites selling the 16/0.02 equipment wire and they all seem to say that 16/0.02 wire is rated at 3a, if that is true then the wire on these motors is under rated for what the motor could potentially be doing ?

Assuming an 8-wire motor, the motor wires only carry 2.1A, remember that the 4.2A is for bipolar parallel operation with 2 windings in use, 2.1A per winding. 16/0.2mm cable is rated for continuous 3A capacity. For the cables connecting to the driver, which are carryig 4.2A you should use 24/0.2mm which is rated at 6A continuous

Lee Roberts
06-09-2009, 05:33 PM
Ahhh yes 2 windings ! Sorry !

Ok but if the cables connecting to the driver are carrying 4.2a, is that devided by 4 wires if the motor is in a bipolar parallel operation ?

Basicly Irving i just want to know what the correct rating of the wire should be that i use to connect the drivers to each motor/XLR. Can i not just use the same rated wire as whats on the motors or is it better to step down as you go?

irving2008
06-09-2009, 10:52 PM
Ahhh yes 2 windings ! Sorry !

Ok but if the cables connecting to the driver are carrying 4.2a, is that devided by 4 wires if the motor is in a bipolar parallel operation ?

Basicly Irving i just want to know what the correct rating of the wire should be that i use to connect the drivers to each motor/XLR. Can i not just use the same rated wire as whats on the motors or is it better to step down as you go?Lee, the attached diagram might help.

Lee Roberts
09-09-2009, 12:35 AM
Thanks mate !

paul.hunaban
27-03-2010, 09:55 PM
I've just purchased the stepper noted in #8 [ Nema23 3Nm motor (PDF (http://www.axiscnc.com/pdfdocs/sy60sth863008bf.pdf)), the document said: Bipolar Parallel 2.73v, 4.2a, 3Nm], connected to my existing driver and hey ho smoke all around.

Iím very interested to know, did you decide on a driver as I clearly need to upgrade what I had ?

irving2008
27-03-2010, 11:26 PM
For those motors you could do worse than these drivers (http://www.slidesandballscrews.com/pm542-microstepping-driver-p-111.html?cPath=44)from Zapp. what drivers did you have before?

jonm
28-03-2010, 12:16 AM
lee
i have nema 23 3nm motors, on pm542 drivers from zapp driving them with 40v power supply , they work really well together , 4.2 amps parallel , tried to run those motors at 2.5 amps 24v, before i upgraded .
waste of time and money, to slow and motors stalling. now have plenty speed and torque .......... maybe have to upgrade machine now. at £35 you cant go wrong

paul.hunaban
28-03-2010, 10:46 AM
Irving, the drivers I have / had were common off the shelf units purchased from Australia, they’ve worked well but are limited to 40v 2amps output and I ran one against this motor at 24v with the windings in series, I thought I may get something out of them.
The reason for buying the bigger motor is to beef up what we already have and we like to experiment / learn.

John, many thanks for the info, it’s great to have positive confirmation of our needs.

irving2008
28-03-2010, 05:46 PM
Yes, but the term 'back-emf' is used to mean several things. Firstly, any motor creates a back-emf whcih resists the current that drives the motor and is a result of the changing, rotating, magnetic field - this is the voltage generated in the windings by the motion and would be apparent if the motor was driven by the inertial load (bad in a CNC machine because the load should be under control at all times!). Secondly, with a pulsed stepping current there is an induced voltage that occurs whenever the current is switched off and is a result of the magnetic field collapsing - it has nothing to do with the motion of the motor as the same back-emf is observed in relays and solenoids and ignition coils. It is the latter rather than the former that is the issue discussed above.

Lee Roberts
29-03-2010, 08:34 PM
On a different note.....Is everyone allowed gay themed avatar thingies or just the admin?

We dont have any admins here with an avatar, your more then welcome to have your own :).

1113562
30-03-2010, 01:22 AM
You are correct, to meet the holding torque (the force required to force the motor from its current step to the next or cause slip - not good) will require in your example 2.73V across the parallel phase winding which will draw a steady 4.2A. The parallel coil resistance = 0.65 Ohms in this case. However, due to its inductance when switching, the current will not get to 4.2 Amps immediately, there is a time constant due to L and R and your driver source resistance combined. The torque produced is a function of the current not the applied voltage. At time 0 at switching there is no current flow at all so no torque but current quickly builds up over time to meet the 4.2 Amps at time = infinity, it follows an exponential curve. To accelerate the time to get to full current i.e. to achieve maximum torque in a much shorter time (which will allow your motor to go faster) the trick is to apply a much higher voltage to the windings. This will cause a faster ramp rate and the coil will get to full current in a much shorter time but current will carry on going up so the driver needs to measure the current and turn off the voltage when it gets to full current. Then the current will begin to decay but the driver then zapps it with another voltage pulse to get it back up to full current again and so on thus retaining the nominal 4.2 Amps. The PWM type driver does the same except that it follows a current profile as defined by the microstep look-up table and not the motor maximum current for every step. I can send you or link you to a brilliant application note which describes this much better than me here if you would like. It is on the Microchip web site under the stepper motor applications section if you can't wait for my reply.

Cheers, John

Robin Hewitt
30-03-2010, 09:06 PM
Forget the 3Nm holding torque, it is not the holdiong torque that you need, it is the torque while moving.

I've been saying that for years, it's the torque against speed graph that counts, the one with the voltage next to it :whistling:

OTOH why mess around? I used all sorts of stepper drivers until Gary came up with his 220VAC jobbies and I haven't looked back since. Just make sure your motors can handle the lowest amp setting and fit a griddle to one axis so you can cook breakfast whilst cutting. Oeufs a la swarf. Crunchy :beer: