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Magnet operating temperature 1

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Clyde38

Clyde38

Electrical
Oct 31, 2003
533
OK, here’s my question. How do you calculate the temperature that a magnet can be exposed to without permanent loss? And, how do you calculate how much loss you’ve suffered based on the temperature that your magnet has been exposed to? In particular, I would like to use a magnet to activate a sensor over a given temperature range. This would constitute an open circuit condition.

What I think I understand: A BH curve that is presented by the manufacturer usually consists of the second quadrant information and in some cases the third quadrant. There are usually several curves consisting of various temperatures that make up both the normal and intrinsic portions of the curve. The permeance coefficient will help describe the operating line and operating point of the magnetic circuit and is based on the magnetic circuit described. The temperature coefficient of Br is usually given in percent per degree C. In rare cases, a graph that describes the typical irreversible losses for various working points depending on the temperature is provided. Example: percentage loss for B/µ o = 0, -0.05, -1, -2 etc. with various temperatures. If the operating point falls below the knee on the normal curve, irreversible loss occurs. The amount of recovery will follow a line parallel to the original line.
 
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Hello Clyde38,

You have a very good understanding of the issues involved. While it can be quite calculated, I've oftentimes found it easier to just bake a bunch of magnets at the maximum expected temperature to find out.

As you mentioned, if the operating point stays above the knee of the normal curve at the maximum temperature, then only reversible loss will occur and you can use the reversible temperature coefficient to calculate the loss while at temperature. In this instance, the magnet will return to full strength after returning to room temperature.

If the operating point is below the knee of the curve at the maximum temperature, then it gets a bit more complicated and it's tough to describe in a text-based forum such as this, but your understanding of it is good. The key thing (as you mentioned) is the recovery will follow a line parallel to the original. If you can measure the percentage difference in the Br (Room Temperature) and B(after drawing the parallel line), you'll have a good estimation of the percentage decrease.

If you can get your hands on one, Parker's book on permanent magnets does a great job illustrating this.
 
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Thanks MagMike. If you are referring to Rollin J. Parker's Advances In Permenant Magnetism, I have this and it's been a long time since I've read it. Quite honestly, until you mentioned it I forgot that I had it. I'll review, but I don't recall a method of calculation. Only empirical methods. I am blessed with those that beleive that if the material is rated to 150C, then operation at this temperature is just fine. Again I'll mention that this application is using the magnet to trigger an electronic sensor. This may be a Hall device or array etc. Again, thanks for you help.
 
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Yes, that is the book I was referring to. If you have the 1990 edition, check the section on Permanent Magnet Stability.

It's too bad your colleagues don't realize that while a given magnetic material may be "rated to 150degC", it will produce a different magnetic output at that temperature than it would at room temperature.

If it's not too late, you might want to consider SmCo permanent magnet material. It has a much smaller temperature coefficient than NdFeB. It's likely to have less of an irreversible loss too.
 
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The one point which I constantly battle is the permeance of the magnets in the circuit. The simple temperature rating of the magnet does not tell you the capability of the magnets when they are in their magnetic circuit. It could be better or and usually it is worse.

Mike
 
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