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Stepper resonance - not a problem
Joel Jacobs
I spent the weekend and early part of this week polishing up the software on
my chopper drive and took the plunge last night and hooked it up to my mill. The first move was real slow and it sounded like someone running a jack hammer. Most of the racket was coming from the handles that were still attached to the other end of the lead screws. I removed the handles and was much better. I ran it through it's complete speed range with a circular interpolation - only one axis moving though. There were a couple speeds that made a pretty loud 'buzz' - found out that was coming from the tooth belt. After a few cycles I went back to 0,0 and it had not lost any steps. I had been reluctant to use half-steps because every other step only one field energizes and I thought it wouldn't have as much torque - after thinking about it a bit, that's wrong. It still gets full power every other step and even if it couldn't move at all on the half step it would surely move on the next full step. So I tried half step and the machine moved very smoothly at all speeds. Ran several tests and even took some cuts on a chunk of aluminum. Never lost a step. I'm real happy with this driver. I'm running the motor at 4amps/phase and at motor lock it's drawing about 1 amp from my 40 volt supply. At 20-25 ipm feed it draws 2 amps from the supply. (8000 steps per inch). 20 ipm is the fastest speed that the field coils can reach 4 amps but the machine rapids at 25 ipm just fine. I think I'll put amp meters on my control, seems like a good indication of machine friction/cutting loads. That was all... Joel |
On 3 Mar, CAD_CAM_EDM_DRO@... wrote:
Message: 1I think you're a bit mistaken on a couple of things here. If the motor stopped on a particular half-position step as a result of insufficient torque, it may not move on the next "full" step either. That's the whole point programmers are trying to make with SMOOTH accelerations tied to hardware timers. Resonance is a problem that can cause missed steps, but "unsmooth" accelerations will also cause missed steps. The constant torque "secret" is to increase the current to the winding, when only one winding is on. Appropriately scaled, this will result in constant motor torque for each position. The theory is clear and well understood. For any given angle, the current relationship should be sin(a) cos(b). For simple half-step, this translates to increasing the current by 40% when only a single winding is on. Note that at no time does the current through the windings exceed the current in the both windings on position, so no motor overheating should result. ( Your 4 amp steppers, for example, will have 8 amps total in "both windings on" position and 5.6 amps though one winding only in the "half" step position ) If the motor is stopped in the one-winding on position, though, the motor current must be reduced, or the winding will overheat. I've been working with DanM on this for a bit, and Dan has found that going to quarter step solves almost all resonance problems, and that going finer than that, to so-called microstepping is really not necessary. Be aware that absolute positioning is not as accurate in any of the intermediate positions, and this position error is increased as the steps get finer and finer. ( Yet another reason to avoid uStepping ) Moral: Don't change your pulley ratios to keep the speed up. Alan -- Alan Rothenbush | The Spartans do not ask the number of the Academic Computing Services | enemy, only where they are. Simon Fraser University | Burnaby, B.C., Canada | Agix of Sparta |
Joel Jacobs
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----- Original Message -----
From: <beer@...> If the motor stopped on a particular half-position step as a result ofHi Alen, If thats the case, it probly wouldn't work in full step mode either. What I was trying to say was that if it had insufficant torque to move to the half step position, on the next full step - it would go from one full step to the next just as if it was full stepping to begin with. The constant torque "secret" is to increase the current to the winding,I had thought about doing this, my controller has the capability to adjust the current sense reference voltage from 50 - 100%, with a resistor change it could do 60/100. But are the motors ratings limited by the temprature rise - or core saturation? Note that at no time does the current through the windings exceed theMy controller does a power saver function where the current is reduced automaticaly if it has nor recieved a step pulse in 1 second - it resumes full power imediatly when a step pulse arrives. This is the second time If seen 1/4 stepping mentioned - could you explain how this works? Do you have to be able to set different current levels simultaniously when two coils are energized? Joel |
If thats the case, it probly wouldn't work in full step mode either.You might be right .. theory and the real world are rarely in sync. I had thought about doing this, my controller has the capability to adjustGood question, although neither may be important. Torque is roughly proportional to core characteristics, so increasing the current to maintain a constant torque implies that if the core hasn't saturated with both windings engaged ( at some particular torque level ), it shouldn't saturate at the same torque level with only one winding engaged. Just to be on the safe side, since my motors are quire big enough for my machine, I mentally derated them just a bit. Instead of running them at 5 amps per winding in full step mode, I run them in half-step at 4 amps with both windings on, 5.6 amps with one winding on. Again, though, this should not be necessary for either temperature reasons ( assuming a reduced current "stop" feature ) or for core saturation reasons. My controller does a power saver function where the current is reducedAs does mine, for the reason stated just above. This is the second time If seen 1/4 stepping mentioned - could you explainExactly, in the proportion mentioned - sin(a) cos(b). The relationship is actually the same for all modes, full, half, quarter, eighth, etc. Increasing the number of intermediate steps does mean increasing the number of discrete current levels. This usually ( but not necessarily ) means a microprocessor to assist things. It IS possible with discrete logic, but ... <G> Alan -- Alan Rothenbush | The Spartans do not ask the number of the Academic Computing Services | enemy, only where they are. Simon Fraser University | Burnaby, B.C., Canada | Agix of Sparta |
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