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Cabe Atwell
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Re: Encode
Cabe Atwell   2/8/2013 3:53:32 PM
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Tech novelty aside, what are the prices?

C

Nancy Golden
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Platinum
Re: Encode
Nancy Golden   2/8/2013 10:03:49 AM
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Great article - I particularly liked the mention of using hall effect sensors. Makes sense since historically they have been used as rotary position gear tooth sensors such as crank and camshaft sensing in automotive applications. They are usually a lot cheaper than optical encoders too.

Jon Titus
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Re: Strong magnetic fields
Jon Titus   2/7/2013 11:32:32 AM
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Hello, TJ.  As far as I can tell, the magnetic field for a Wiegand detector must be very localized, so magnetic interference shouldn't become a problem.  Resolvers get used with equipment that uses electric motors, so I bet the manufacturers have added shielding as needed.  Worth asking about before you buy a resolver, and worth the time to run some tests in proximity to specific equipment.

btwolfe
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Platinum
Does it work off-axis
btwolfe   2/7/2013 11:19:52 AM
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In my applications, I often need an axis with a hollow bore, so the position encoder either needs to be hollow. From what I can tell, the Wiegard effect only works when the rotating magnet axis of rotation is centered about the Wiegard wire. I'm sure I've misunderstood the tech in some way and hope that my understanding is wrong.

If my understanding is wrong and the axial rotations do not have to align, then I'd be inclined to merge an optical absolute encoder with a Wiegard sensor for counting rotations. You'd get the resolution of optical with the persistent turn counting of the Wiegard sensor.

TJ McDermott
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Strong magnetic fields
TJ McDermott   2/7/2013 10:08:37 AM
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Jon, terrific article.

How sensitive is a wiegland effect encoder to strong magnetic fields?

Cabe Atwell
User Rank
Blogger
Encode
Cabe Atwell   2/6/2013 4:24:53 PM
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I hope to see more of the Wiegand effect encoders in the near future. It might be a good alternative to optical encoders in a closed loop stepper motor system. Which, by the way, are getting cheap enough for anyone to buy.

 

C



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