Linear Drivers?

[noisillator]

OK, here's another possible option, but it's a technology I haven't seen mentioned elsewhere in these threads. Anyone have any experience with linear drivers like these? Can this approach perform well for our use?

http://www.piclist.com/techref/io/st...step/index.htm

[jalessi]

Noisillator,

Contact Phil at http://pminmo.com/

He has many different designs on his site, he is very familiar with many of these designs.

Jeff...

[Jeff-Birt]

From the website - Unlike old style linear drivers or chopper drivers, the Linistepper doesn't need a power supply that is 2 or 3 times the rated voltage of the motor. You can use a supply with 5 to 10 volts over the motors maximum, and our active current regulation will still give very good performance.

Evidently they managed to warp the laws of physics! The increased drive voltage is needed to overcome the motors inductance. It allows you to build up the magnetic field faster, i.e. better performance.

What they are claiming above is simply NOT possible.

[noisillator]

Originally Posted by Jeff-Birt Evidently they managed to warp the laws of physics! The increased drive voltage is needed to overcome the motors inductance. It allows you to build up the magnetic field faster, i.e. better performance. What they are claiming above is simply NOT possible.

Perhaps the difference is that the driving power is being delivered as sine waves. I would expect the inductance of the motors to have less effect than when square waves or stepped sines are applied.

[MechanoMan]

Originally Posted by noisillator Perhaps the difference is that the driving power is being delivered as sine waves. I would expect the inductance of the motors to have less effect than when square waves or stepped sines are applied.

Nope. Total rubbish. The problem with motor inductance is that when steps change, the motor current will not rise to a new current instantly but ramp up with dI/dT=V/L. It'll eventually reach the target current and regulation will stop the current rise. But the dI/dT ramp-up time is the same for fast or slow steps, when the steps are fast the time spent in dI/dT becomes a significant part of the cycle and the average current for the whole time it spend in that step goes down. Nothing about a linear drive changes that.

The linear drive in these plans is very inefficient, and needs a big power supply and heatsink. The inefficiency makes it impractical to use with a more powerful drive. Heck, at 48V with a 3VDC stepper coil, a 3A stepper would generte 135W on the driver! Per motor! And the power supply must also be rated at greater than the sum of the motors. So >12A for 4 axis of 3A motors (switchers don't have that limit). The extra cost of a big-amp power supply and heatsink won't make this as cheap as imagined.

The PIC cannot step smoothly at very high speeds and cannot make straight, angled lines perfectly. See the steps MUST be resolved to PIC instruction cycles, which seems to be set at 4MHz. With no microstepping, with like 10 instructions needed per step, you'd have a 2000rpm (100ipm on a Taig) limit. Sounds great at first. But at 18x microsteps, we're down to 111rpm and 5.5ipm. But that's not the real problem. The problem is that Mach 3 won't be providing integral numbers that agree with the clock rate on the PIC. Given the way microcontrollers check the inputs in loops that take several instructions, asking for 5 ipm will result in some pulses being longer than others because the Step pulse comes in at the start of the code loop sometimes and sometimes at the end.

That inconsistency in pulse width can be very detrimental to torque.

It does "work", just not competitive with modern designs and wouldn't perform anywhere near useful on a CNC mill.

It's also UNIPOLAR. Right there that means 29% of the rated torque cannot be achieved because it only uses one winding at a time.

[pminmo]

There are a whole bunch of truths and mistruths that people take from one application and tag it to another application. If I had one fault with the cnczone, that would be it.

Mechanoman did pretty much nail the highpoints. The linistepper is inefficient and because of that requires more power supply than pwm current control methods. That inefficiency also will manifest itself in heat dissipation. One of the reasons you can't have too high of coil voltage to supply ratio is simply the power dissipation of the transistors.

Regarding "sine waves" Sin-Cosine relationship is used in virtually all microstepping stepper drivers regardless of linear or pwm.

The linistepper does work, it's quiet because there is no pwm switching going on. (Although I'm constantly baffled why people worry about stepper squeel because when your machining you can't hear anything over spindles cutting.)

Unipolar motors do have less pull out torque than bipolar. BUT the issue then turns different depending on drive method. I did a video on rapids to illustrate driving a 6 wire motor unipolar, bipolar half coil and bipolar series to illustrate some inexpensive drive performance differences. http://pminmo.com/which-stepper-motor

Bottom line, choose motors and drivers together considering the mechanics of the machine your going to use them with.

[MechanoMan]

Originally Posted by pminmo Unipolar motors do have less pull out torque than bipolar. BUT the issue then turns different depending on drive method. I did a video on rapids to illustrate driving a 6 wire motor unipolar, bipolar half coil and bipolar series to illustrate some inexpensive drive performance differences. http://pminmo.com/which-stepper-motor

But the limitation comes from the type of coil usage and has nothing to do with drivers.

A motor with unipolar wires can be configured as uni or bi.
In unipolar, only one half of a phase coil is used at a time. That straight out gives half the torque per amp.
But we're not done. Say a phase has a series resistance of 0.5ohms per a half-phase for uni usage and 1ohm when used in series as a bipolar.

The motor is ultimately rated by its case heat dissipation. The ability to dissipate case heat doesn't change with the wiring, but the heat generation is mostly due to simple I^2*R.

For a bipolar, say its case is rated to dissipate 10W. I=sqrt (10W/1ohm)=3.16A.

For a unipolar, I=4.47A. Larger... except torque per amp is half. So that brings us right to the official unipolar torque is 0.7x the bipolar torque. 29% less at the same point where the motor heat is at its max rating.

So also worthy of note is that the unipolar needs to deal in 41% higher currents to reach the maximum rating of the motor, and that is a problem since that linear drive is really bad for heat AND power supply load. However, even at that max, it's only going to achieve 29% of the torque.

[pminmo]

Originally Posted by MechanoMan But the limitation comes from the type of coil usage and has nothing to do with drivers.

I politely disagree. Not all drivers are equal, for example the MK unidriver gets more power at higher rpm's than conventional pwm drivers by switching step modes at certain step speeds. It's based on the same technique that Gecko uses in the G203V and G540. In addition, different drivers have different max voltage breakdowns. The driver that max's out at 24V won't acheive the same top end speed as the one that max's out at 50V all other things being equal.

A motor with unipolar wires can be configured as uni or bi.

Again I disagree. How a motor can be wired depends on the coil wiring configuration. A 4 wire motor can only be run bipolar, a 5 wire motor can only be run unipolar. A six or 8 wire motor can be wired in different configurations for bipolar as well as one configuration of unipolar.

In unipolar, only one half of a phase coil is used at a time. That straight out gives half the torque per amp.

While I don't disagree with the half of the phase coil, your looking at pull out torque. As the motor rpm increases, motor inductance comes into play. The inductance of 6 wire motor running unipolar is a fourth of the inductance of the same motor wired bipolar series. Thus for the same power supply voltage as you go up in rpms, the unipolar configuration will actually have better torque than the bipolar series configuration.

But we're not done. Say a phase has a series resistance of 0.5ohms per a half-phase for uni usage and 1ohm when used in series as a bipolar. The motor is ultimately rated by its case heat dissipation. The ability to dissipate case heat doesn't change with the wiring, but the heat generation is mostly due to simple I^2*R. For a bipolar, say its case is rated to dissipate 10W. I=sqrt (10W/1ohm)=3.16A. For a unipolar, I=4.47A. Larger... except torque per amp is half. So that brings us right to the official unipolar torque is 0.7x the bipolar torque. 29% less at the same point where the motor heat is at its max rating.

I don't disagree, but your resistive points are more pertenent to pull out torque. For total overall motor performance you also have to consider the inductive characteristics of a motor.

So also worthy of note is that the unipolar needs to deal in 41% higher currents to reach the maximum rating of the motor, and that is a problem since that linear drive is really bad for heat AND power supply load. However, even at that max, it's only going to achieve 29% of the torque.

I don't disagree with the linear drive assesment, but they do work. And for some peoples configuration and tools it fits them. As you increase the need for motor power, the less applicable they are. For example, you don't need the same axis power for an engraver that you do for a machine center. The linistepper might work for the engraver, and definitely won't work for a machining center. As to the original question about applicability to a Taig, they would work, but in all likely hood the user would eventually wish they had something better.

[MechanoMan]

Well, I'm saying this: low-speed torque is for certain lower with this "linear" unipolar drive. There's no way around that. Thus there's no basis for any claims of "full torque" or running cooler. PWM eddy current heating should not be very significant.

Bottom line is that the question is asked for specifically the context of a Taig Mill or Lathe. And I think we both have the same answer: no, it will not likely perform satisfactorily even for a hobbyist, and the price does not justify it. The low speed AND high speed torque are compromised by the lower voltage limits and unipolar drive type, nor will the PIC's stepping solution work well. The cost does not include the large heatsink and casing which make the solution end up much more expensive that it initially appears.

[noisillator]

Just to update, I decided against the linear drivers when the author of the boards informed me that running a 22V supply would be way too much. At that point, I realized the power the driver could deliver to the motor would be limited, and the system as a whole would be less likely to be satisfactory for the Taig.

What I'm planning at this point is to use the HobbyCNC drivers with their 205 oz-in (unipolar) motors. That's more holding torque than the bipolar motors I was considering (166-185 oz-in), so I don't expect problems. On the off-chance I'm not happy with the system, those motors could be used with a Gecko to produce 285 oz-in in bipolar. So, I would lose the investment in the inexpensive drivers, but nothing else. Of course, my plans to purchase the HCNC board could still change if I run into problems fabricating the dampeners. I have the materials; it remains to be seen whether I have the skill.

[pminmo]

Look at the Keling 8 wire motors KL23H276-30-8A and save yourself $5 per motor.

[noisillator]

Originally Posted by pminmo Look at the Keling 8 wire motors KL23H276-30-8A and save yourself $5 per motor.

Hmmm, just read a couple of your other posts - I'm gonna need to talk with you later about some PC boards I need to make. Simple stuff mostly, single sided glass boards for power supply components in my tube audio gear. I'm sure glad to see so many people exploiting the capabilities of these machines!

Jack
WB3U