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AC/DC Relays

AC/DC Relays

AC/DC Relays

(OP)
I always thought I had a solid understanding of AC/DC relays, but something happened today and I'm completely confused.

Today, I saw a DC coil relay operate perfectly fine (as far as we could tell) with reverse polarity on the coil (+24V on A2 and an open drain input on A1, sinking).

I always assumed that polarity was critical with DC relays, causing current to flow in a specific direction through the coil, magnetizing the core/coil and pulling the armature in a specific direction. I would figure that reversing polarity on the coil would repel the armature, not making the contacts.

I then started thinking about AC coil relays, and while I'm familiar with the phase shift properties of the shading coil. I realized that polarity through that coil would shift back and forth. I know the shading coil causes a phase shift that is present to keep flux in the coil at all times.

So, now I wonder:

Should a DC coil relay work in reverse polarity? Why?

Despite the shading coil, why wouldn't the flux shift direction eventually, and cause a momentary state of zero-magnetism?

Thanks to any and all that can help me here. This has been somewhat humbling.

RE: AC/DC Relays

It seems that in general, a relay's coil always attracts the armature because the coil is an electromagnet and the armature is an unmagnetised ferromagnetic metal. In the same way that bringing either end of a permanent bar magnet near some paper clips causes the clips to be attracted to the magnet.

So in the DC case, either polarity will magnetise the coil, attracting the armature. In the AC case, both half cycles cause attraction, and the shading coil just provides some ride-through of the zero crossing.

RE: AC/DC Relays

As you know (or knew smile), the force that an Electro-magnet excerts on a piece of iron is proportional to flux squared.
And then, you also know that flux is proportional to current (as long as the magnetic circuit isnt saturated).

So, if current is i=I*sin(wt), then flux also follows I*sin(wt) and if you square I*sin(wt), you get I^2*(1-cos(2wt))/2 or I^2/2 -I^2*(cos(2wt))/2.

The first term is a constant (DC, that is) and the second one is AC with twice the excitation frequency. Twice the excitation frequency is more than a properly designed electromechanical system with shading coil can react to, so the force acting on the armature is I^2/2 which is a constant (DC) force, which happens to be the same force that a DC current with the same amperage would produce.

I leave a few other details, that I perhaps should have mentioned, for others to comment on.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

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