Interpreting Opinions for Gecko vs Cheap Drivers

[joeavaerage]

Hi,
yes it is either AC or DC input.

The input circuit is a bridge rectifier followed by smoothing capacitors. The smoothing capacitors play an important role in first stabilising
the input voltage but also in the ability of the drive to accept reverse braking energy from a stepper when it decelerates.

The drive it self is quite small and would and so the smoothing capacitors will likewise be quite small. You might reasonably expect therefore
significant high voltage excursions when the stepper is decelerating. If the capacitors were larger the voltage excursion would be less.

You may recall that I commented in an earlier post that the ability of a drive to gracefully accept reverse braking energy is a large determinant
in the reliability and overall quality of the drive.

The manufacturer has made the input circuit in this way that his product might appeal to the widest possible market.....you can supply it with
AC or DC.

Lets imagine that you had a great big linear DC supply (transformer/rectifier/capacitor). With a normal stepper drive, that is without the input rectifier,
any reverse braking energy would impress itself on the very large capacitors of the power supply. Because the capacitors are large you could
expect that the reverse braking energy would be well contained and safely controlled. With the drive you have linked to the bridge recitifier
at the input of the drive would prevent the reverse energy from flowing back out of the drive onto the power supply capacitors. Thus the reverse energy
would have to be accommodated by the much smaller capacitors in the drive itself, and being smaller you might expect a greater voltage excursion with
consequent reliability concerns.

If you intend on using a large linear power supply, as I very much recomend, then I would look for a drive that accepts DC input only as that would allow
the large smoothing capacitors in the power supply to absorb and contain the reverse energy.

Other than that the drives you have linked to look good and certainly the price is right.

When I built my mini-mill I bought and used Vexta 5-phase steppers. Being 5-phase means that the selection of stepper drives is fairly slim and I faced
the choice, some smaller 5-phase dives costing about $40 each verses genuine Vexta 5-phase drives with 230VAC input at about $140 (excluding
shipping) second hand each. I didn't really want to pay all that extra....but I did. The Vexta drives are absolutely superb, they can spin my steppers
to 3000rpm without losing steps and they have not given a spot of bother in the six years I've used them. All in all they have proved the old
saw....'you get what you pay for' So while you seek to economize while building your machine, which is perfectly understandable, but if you buy
'right' you will have the products for years and years and be an un-ending source of satisfaction. So my Vexta drives and steppers have proven to be,
despite the initial cost.

Craig

[rodw]

Very good explanation thanks

[Jandyman]

The explanation makes sense. It does imply that the AC input voltage specified is RMS, not peak to peak, otherwise you wouldn't get the voltage boost in AC mode. Is that usually how AC power supplies are specified?

But essentially what is being said is that the drive is improperly designed for braking under certain circumstances. And getting back to the application at hand, how does one determine whether those adverse conditions apply? I get that if you are moving a very heavy gantry very fast and you want to stop quickly, then this is an issue. But a number of factors would mitigate this:

1. A lighter gantry.
2. Lower maximum speeds.
3. Setting the max deceleration in the machine control.

So how do I know whether these sorts of things are a problem for me? And what would happen under these circumstances of braking too hard? I'd blow up the drive? I'm still not clear on that. I'm beginning to think that the only way is to buy some stuff and try it. Actual specifications on such things seem impossible to come by, so that the relevant analysis could actually be done.

Yes it's true that you will generally get a better product by spending more money. But that doesn't mean one has to buy a 1 ton diesel truck to tow around a landscape trailer. I still don't have a way of understanding whether my situation is a landscape trailer or a 10,000 pound travel trailer.

I would be building a wood cutting router - high end hobbyist for building electric guitars and basses - with a working area of about 48" by 17". That's pretty small. The construction isn't completely sorted in my head, but suffice it to say that I would make an effort to obtain the necessary stiffness without unnecessary weight. I'm thinking of using 1610 ball screws. And BTW, it seems to me that the needs in terms of drivers and motors are very different in this case between the motors needed to propel the gantry as compared to the other two dimensions. That is, *if* the primary concern is acceleration and deceleration. Which might not even be the case here. The limiting performance factor may end up being the stiffness of the frame.

So hard to get a feel for where I really am in this space.

[joeavaerage]

Hi,
to follow any given tool path the higher the acceleration/deceleration of the machine the better.

For example your machine is traveling towards a right angle corner. It will need to decelerate as it approaches the corner until
it nearly stops at the corner and then accelerate away from the corner. The higher the acceleration the faster and more accurately
the tool path can be followed.

Lowering the mass of the gantry reduces the torque required to produce a given acceleration but commonly the inertia of an axis is
dominated by the rotational inertia of the ballscrew and the rotor of the stepper itself. So lowering the mass of the gantry will have
at best a secondary effect on the total inertia of the axis but will badly reduce the stiffness of the axis.

Lowering the maximum speed has zero effect on acceleration/deceleration, it will affect the tiime in the acceleration/deceleration phase
but not the peak required torque.

That is, *if* the primary concern is acceleration and deceleration. Which might not even be the case here.

Acceleration/deceleration is the primary measure of the suitability or otherwise of the steppers.

If you really want to find out where your design is then you need to do the calculations. I have outlined the procedure here:

https://www.cnczone.com/forums/servo...ml#post2372136

This is the effective rotational inertia including the 110kg mass of my axis (of my new build machine):

Jeff= (1.13 + 5.252 + 0.69) .10-4 kg.m²

Note that the terms 1.13 and 5.252 represent the rotational inertia of the servo rotor and ballscrew respectively. The term 0.69 represents
the contribution of the linear momentum of the axis, and note that the axis weight in my new build machine is 110kg.Another way of looking at it is that
the linear momentum of the 110kg axis is only 9.7% of the total inertia. So even if I reduced the axis weight to zero....a non-sensical idea....still
the inertia would be 90%, or almost unchanged!!!

Craig

[Jandyman]

That was a hugely useful reply in terms of understanding the rotational inertia of the ball screws. I hadn't thought of that. That's an argument for belt drive, though I've found on my existing machine that belt stretch is a significant factor in overall "flex" of the machine.

It's still the case that in my case, the rotational mass is much greater for the axis that moves the gantry (the X axis in my case) than it would be for the Y and Z axis, so the quality of the Z drivers and motors logically would need to be higher for the X motors. One crazy idea is to use belt drive for the X, and ball screws for the Y and Z. It just so happens that because of the nature of what I'm cutting, that flex in the Y and Z (especially the Y, across the gantry) is an issue. This is because of the grain direction and the way the tool interacts with the wood.

Thanks for the link to the procedure, I'll try to deepen my understanding. It's still not clear what acceleration/deceleration performance I would need to achieve.

It's also not clear to me what the failure mode of the stepper would be in the case of not being able to dissipate braking energy. What would actually happen in this case?

[joeavaerage]

Hi,
in interests of stiffness and accuracy I wouldn't even consider a belt drive, stick with ballscrews.

It is true that ballscrews add rotational inertia. In the case of my new build machine I was able to secure some superb 32mm diammeter C5
screws fairly cheaply. What I hadn't considered was how much much rotational ineria that they have. It has by no means scuppered my design
but there is such a thing as 'too much of a good thing' with ballscrews.

In your case you are using 16mm diameter screws and as a first approximation the same length as mine. Because the first moment of inertia
varires as the square of the radius and likewise the weight per m reduces as the square of the radius I would guess the first moment of your screws
will be approximately 1/8th of mine.

The component of inertia representing the linear movment varies as the square of the screw pitch. Your screws are 10mm pitch
whereas mine are 5mm, but as a first approximation your gantry is 1/4 the weight of my axis, thus I would expect the linear component of
the total effective first moment would remain about the same as mine.

All up I would expect the first moment of your gantry axis to be approximately 1/4 of mine with the components relating to the rotor, the ballscrew
and the gantry mass to being 33%, 33%, 33%. Those ratios suggests a well balanced design, with good low inductance steppers it should perform
very well indeed.

In interests of tool path accuracy you want as high an accelertion as your steppers can reliably deliver. That may give your drives a hard time when
decelerating.....but after all thats what they are designed for. While preference would be for a driver not to have a bridge rectifier at the input
thereby allowing the power supply capacitors to moderate the reverse energy, I would not let that alone put me off what otherwise look to
be good drivers. Remember low inductance steppers are by far and away more important than the driver architecture.

Craig

[rodw]

Acceleration and Decceleration are essentially the same thing.
Remember that CNC use is a small subset of stepper motor useage. Many more are used in Industry.
Advanced motion modelling for steppers generally talk about a motion profile. Typically it might be 30% for a fixed length move over a specific time (lets say 10 seconds)
So for this design profile the motor would spend 30% of its time acellerating and 30% decellerating so its 3 seconds to get to speed, 4 seconds at speed and 3 seconds to stop.
In our case, what is the time period to use and what is the profile percentage?
But yesterday, we applied this model to a Z axis traversing a sheet of corrugated roofing iron at speed and following the profile. This meant the profile was determined by external factors.
The rotational inertia is critical. We tried a few NEMA23's and found they did not work due to the smaller rotor.
So then we tried the smallest NEMA34 we could find. It smashed our benchmark and had the advantage of being lighter than the large NEMA23. So from our model, we know the amps and miimum voltage the motor requires. So we can then select the driver and size the power supply. We need 1.6 amps, less than what the motor will draw at peak. We found a 3 amp driver.
So one solution might be to use one of the sophisticated european drivers that have a boost mode. We could then set the driver to run at 2 amps but then if the motion controller sensed a follwing error exceeding our requirements, we could tell the driver to apply the extra 1 amp available to boost performance. This is actually a use case where closed loop steppers actually have an advantage because the driver can add torque reserve at speed before steps are missed.

We never said this was easy, but that is one example where Gecko is now outclassed. Unfortunately, the brand I'm talking about is not available in the US.
The other frustrating thing we found is that in many cases, the drivers that can handle the amps and volts required come with all these expensive features for industry that are not needed for simple step and direction.

[rodw]

Hi,
in interests of stiffness and accuracy I wouldn't even consider a belt drive, stick with ballscrews.

Remember low inductance steppers are by far and away more important than the driver architecture.

Craig

This is not always the case on both counts.

In come CNC applications, rack and pinon drive becomes the best case. But you need to size the pinoon and gear reduction accordingly. Here belts offer less backlash than planetary gearboxes. But a single belt reduction might require a 450mm (18") pulley. Then you have to account for inertia and weight of the pulley. A 2 stage arrangement then gets up around the cost of a gearbox.

Planetaries add backlash so you need to loook at the pinion size to determine how much linear backlash exists and decide if thats acceptible for youe use case. In our case, the backlash was acceptible so gearboxes win the day.

After reviewing a lot of motors both cheap and expensive while validating our model, it seems to me that the inductance is less critical than people think. What I observed was that manufacurers designed a stepper to suit a specific voltage range. Maybe it was 48 volts, 70 volts or 90 volts for example. So the cheaper high inductance basically refused to accept more than 48 volts for example. I don't have any data to back this up it was just an observation.. rotational inertia trumps inductance in my books in a right sizing approach.

ONe motor refused to go over 50 volts until it was given a 1% profile and it basically imploded.

[peteeng]

"The component of inertia representing the linear movement varies as the square of the screw pitch" Hi Craig where does this come from? The linear inertia is Fi = Mass x acceleration. The screw force is (torquex2xpixefficiency/lead) The torque is set by the motor torque available at that instant and this torque is shared by the rotational and linear conditions. The rotational torque required is Jx alpha where alpha is the rotational acceleration and this is 2xPIx(delta - rotational velocity)/delta time of change. So not sure where the lead squared comes in?? Peter

[joeavaerage]

Hi Peter,
the derivation is in the link and furthermore it is the same as the Hiwin formulas you gave me to compare.

Yes......linear momentun is (mass x velocity) but the forumla I derived converts that linear momentum into its equivalent rotational inertia
in the form of J_eff, the first moment of inertia.

Note also the units kg.m²

Craig

[peteeng]

Hi Craig - Link? Peter

OK so the linear velocity (and linear acceleration, therefore inertia force) is related to the screw rotational velocity and that is related to the pitch vs rotational speed. All good. Peter

[joeavaerage]

Hi Peter,

https://www.cnczone.com/forums/servo...ml#post2372136

When I started thinking about the combination effect of linear momentum and rotational inertia in the above thread you were kind enough to PM me
some Hiwin design formulas. They too had the dependency of the first moment of inertia equivalent as to the square of the pitch.

If you follow the energy argument I took in the derivation the square dependency occurrs quite naturally.

I am long out of practice at deriving equations from first principles but I am quite glad I did, it turned out to be pretty simple in the end.

Craig