Trigger a relay via resistive load in AC circuit


by rp55
Tags: circuit, load, resistive, trigger
rp55
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#1
May24-13, 03:26 PM
P: 82
How does one go about triggering a switch (i.e. relay) via a resistive load in an AC circuit? For instance, if I have an AC relay (high impedance) and put the coil in series with the resistive load, it won't work: the resistive load will stop working. How does one go about triggering a relay (i.e. a switch in a different branch of the same AC circuit) via a resistive load without affecting the resistive load in any significant way? Am I missing something obvious perhaps? Dumb question?

I hope this didn't belong in homework section!
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rp55
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#2
May24-13, 03:28 PM
P: 82
btw this is a 120vac circuit with a resistive load of 9ohms (about 1600watts). I would like to have a switch come on in another circuit branch whenever there is current flowing in the resistive load branch.
skeptic2
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#3
May24-13, 07:56 PM
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Have you tried putting the relay coil in parallel with the 1600 W load?

rp55
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#4
May24-13, 08:54 PM
P: 82

Trigger a relay via resistive load in AC circuit


Thanks for the suggestion. I've thought about that but I can't get the magnetic field to be great enough to throw the relay switch. According to the math and simulations I've tried, while I get roughly 13amps on the 1600w resistive load, I'm only able to get 50mA on the relay coil. I believe it requires about 2 amps to have enough magnetic field to throw the switch. I could be incorrect there on how I'm understanding what I can find relevant in the relay specs:
-----------------------------------------------------------
nominal coil VAC: 120
must operate VAC: 102
max continuious VAC: 138

coil resistance +-10%: 2500 (i'm assuming ohms here)

coil power (at pickup voltage) 2.7VA

max continuous coil dissipation: 2.7VA

notes: relay may pull in with less than "must operate" value
------------------------------------------------------------

So that's all I can get from the relay spec sheet... my current understanding is that the relay coil must get enough current to throw the switch. Even in the simulator the current across the coil must reach the specified resistance for the coil before the simulation throws the switch, so I'm assuming I'm correct in that area.

No matter what I do, I can't get enough juice into that parallel branch to get the switch to throw (without simulating a huge source AC voltage).
rp55
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#5
May24-13, 08:58 PM
P: 82
And I tried putting caps in there (in various places, particularly the parallel branch you suggest along with the relay) since my understanding is the cap would cancel the effects of the relay reactance, but in simulations it doesn't seem to do much significant (still only milliamps across the relay).

And then so here is where I begin to get lost at this point.
milesyoung
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#6
May24-13, 09:13 PM
P: 427
So the relay is rated for a 120 V supply? That literally means you should apply 120 V across it for it to perform its function within specifications.

I assume you're just supplying it directly from the grid so the source impedance of the supply is nothing to worry about.
rp55
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#7
May24-13, 09:19 PM
P: 82
Quote Quote by milesyoung View Post
So the relay is rated for a 120 V supply? That literally means you should apply 120 V across it for it to perform its function within specifications.

I assume you're just supplying it directly from the grid so the source impedance of the supply is nothing to worry about.
Yes, the relay is definitely 120VAC rated. It specifically has an AC coil (has the D cup transformer type coil to support AC).

I guess I'm a bit confused in this area that you mention. I guess I'm unsure if it's merely voltage necessary at this point, or current. I.e. depends if it's in series or parallel and whether all components see the same voltage, etc. (hence my knowledge becomes a bit lacking).

Source will be grid, yes. Keep in mind I haven't tested this yet so I'm making some assumptions here (some of which may be incorrect).
milesyoung
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#8
May24-13, 09:27 PM
P: 427
All you need to do is provide the relay with the supply it's rated for. If you follow skeptic2's advice and connect your load and relay in parallel, both will have the same voltage across them (that's what defines a parallel connection).
rp55
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#9
May24-13, 09:42 PM
P: 82
Quote Quote by milesyoung View Post
All you need to do is provide the relay with the supply it's rated for. If you follow skeptic2's advice and connect your load and relay in parallel, both will have the same voltage across them (that's what defines a parallel connection).
Ok thanks that helps clarify things about the relays expectations. Thank you skeptic2 and milesyoung. And I understand that parallel circuits have the same voltage across all components whereas in series they do not, so that then makes sense. I must have got off track looking at a particular odd use of this relay in a known circuit where they do something tricky with it. It's in parallel, but not around the actual load. It's after the load around some closed switches which may have thrown me off.

I'll just have to give it a try and do the math and take measurements and see what it will do.
skeptic2
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#10
May25-13, 10:27 AM
P: 1,784
I commend you for thinking through the application before implementing it. Often people go straight to the implementation without thinking about the details first.
rp55
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#11
May25-13, 04:13 PM
P: 82
Quote Quote by skeptic2 View Post
I commend you for thinking through the application before implementing it. Often people go straight to the implementation without thinking about the details first.
Thanks for the encouragement, skeptic2! I got the simulation math to come out correctly now. It's so easy to be lost without a touch of knowledge guiding at times. I really appreciate the help!

I'm still struggling a bit with the formulas and trying to understand in an intuitive way (if that's possible) how impedance behaves in series and parallel circuits. But I'm trying to do all the math I can thru this circuit without merely jumping into a solution. It's so easy to skip the schoolwork and go for the solution. And the help I got from yourself and milesyoung has put me on the right track to do that. I appreciate that very, very much!
sophiecentaur
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#12
May26-13, 05:30 AM
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Relays work on DC, usually. That implies that you could improve matters by putting a diode in series with the relay coil (and possibly a reservoir capacitor in parallel with the winding). The Inductance of the relay winding must be limiting the amount of current it can draw to the 50mA you measured.
Windadct
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#13
May26-13, 10:04 AM
P: 532
Hello RP - I believe you are looking for a way to detect a condition : 1) The power IS applied and 2) there is (or is not ) any current in the load. The circuit is on AND working?

IN this thinking - you should see that you need to detect / sense 2 things. Ideally - this would be a CURRENT controlled relay - the only way to have current is if both conditions are true? -- is this accurate. Otherwise - am at a loss as to what you are really trying to do.
sophiecentaur
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#14
May27-13, 05:33 AM
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Quote Quote by Windadct View Post
Hello RP - I believe you are looking for a way to detect a condition : 1) The power IS applied and 2) there is (or is not ) any current in the load. The circuit is on AND working?

IN this thinking - you should see that you need to detect / sense 2 things. Ideally - this would be a CURRENT controlled relay - the only way to have current is if both conditions are true? -- is this accurate. Otherwise - am at a loss as to what you are really trying to do.
Yes, a detailed answer to this (implied) question could produce a useful answer to the OP. It isn't exactly clear what is needed. But there's bound to be an answer to this problem; there always is.


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