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| | |-+  Wire encoders to SS or Gecko ?
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Author Topic: Wire encoders to SS or Gecko ?  (Read 1056 times)
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Pythagoras
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« on: November 09, 2010, 01:56:25 AM »

I'm not sure what all the pros and cons of wiring encoders to a Smooth stepper VS wiring back to the Gecko drive.

Does wiring back to the SS mean the system is closed loop?
I can see that if I wired back to the SS this would mean that I couldnt use the multiplier optionon the GeckoDrive. Is this correct?

Does any one know the pros and cons of wiring back to the SS?

Benny
« Last Edit: November 09, 2010, 02:00:19 AM by Pythagoras » Logged

The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c). a + b = c.
Hood
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« Reply #1 on: November 09, 2010, 02:23:59 AM »

You need to wire the encoders to the Geckos or you will not have closed loop to the drive, connecting the encoder to the SS will not give you closed loop.
The encoder inputs on the SS are I presume really meant for MPG input although I do have my spindle encoder going back there, for two reasons. One is the Index pulse is used for  the spindle feedback to Mach and the second is  in the hope that one day the spindle may have some sort of synchronising with the Z although I dont think it will as room is tight on the SS now.
Hood
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Pythagoras
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« Reply #2 on: November 09, 2010, 04:05:01 PM »

Thanks Hood.
WOW, havent you been a faithful contributor to this forum over the years. Good onya mate.
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The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c). a + b = c.
Hood
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« Reply #3 on: November 09, 2010, 04:21:11 PM »

Benny, you started this site and what a bloody difference it made, no more Yahoo crap, I am not  masochistic Grin So big thanks go to you Smiley

Hood
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Pythagoras
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a + b = c


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« Reply #4 on: November 09, 2010, 04:52:18 PM »

Ha, The spotlight was on you, no deflecting praise please.Smiley
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The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c). a + b = c.
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