DC Relay from AC

James S

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Hi Folks,

I've got a 24 volt AC signal, and I need to control a 24 volt DC relay. (This has little to do with flashlights /ubbthreads/images/graemlins/wink.gif )

This is very low current stuff. I actually started putting some diodes together into a bridge, but I don't think that is really necessary. Can't I just put a capacitor across the lines to flatten it out a bit?
 

PhotonWrangler

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[ QUOTE ]
James S said:
Hi Folks,

I've got a 24 volt AC signal, and I need to control a 24 volt DC relay. (This has little to do with flashlights /ubbthreads/images/graemlins/wink.gif )

This is very low current stuff. I actually started putting some diodes together into a bridge, but I don't think that is really necessary. Can't I just put a capacitor across the lines to flatten it out a bit?

[/ QUOTE ]

Nope. Placing a capacitor in parallel with an AC signal can turn the capacitor into a confetti generator! /ubbthreads/images/graemlins/eek.gif

You were on the right track with the bridge rectifier. You can then place an electrolytic capacitor on the DC output of the bridge assembly and you'll be ok.
 

Floating Spots

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There is no need for a smoothing cap on a relay from a full bridge rectifier.
Actually, you could probably get by with a half wave rectifier (one diode).
Its not going to be clean, but it may work.
Adding a cap here would be of a lot of help in riding through the 8 ms of off time (and the rest of the decay time below the hold threshold).
 

JohnJ80

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Hi James (I find you here too - not just on the Xtension list!)

I think all you need to do is to put the diode in series with the relay. You can put a cap (appropriate voltage - the larger the cap value the better) across the relay to smooth out the ripple. This ought to work.

The diode needs to be able to handle a reverse voltage equal to 24V *sqrt(2). It also has to be able to handle the current the relay coil will draw. 24V AC is the RMS value so it should net out close to that with the DC of 24V after the diode and the cap.

If that doesn't work, then go buy a monolithic full wave bridge from Radio Shack and use that. However, there you will probably need to think about a current limiting resisitor (just to be safe).

J


J
 

James S

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Hi John,

Internet's a small world huh? /ubbthreads/images/graemlins/grin.gif

I saw your post in the thread on home automation and I've been meaning to drop you a PM and say hello, but things are a little hectic around here at the moment and I got distracted...

I see you've been spending a lot of time around here too /ubbthreads/images/graemlins/grin.gif

The device in question is actually getting connected to the AC/Heat in my office. I like to monitor those things and watch the money drip away at the end of each day /ubbthreads/images/graemlins/smile.gif The relay is controlling an X10 device that is currently not supported by XTension, but will be soon /ubbthreads/images/graemlins/grin.gif Gee, I hope Michael doesn't read this board too or I'll be in trouble.

But it's getting late, and it's getting warm in my office since I have the AC turned off so I'll need to get this sorted out, or just disconnected soon. I'm going to try just a single diode first and see what happens, there might be enough inertia in the relay to keep it on, or I can add some capacitance after that without fear.

The only reason I thought I could put the cap just across the power like that is because I've got an old 12v power supply from radio shack that is nothing but a 13v transformer with a capacitor across the output.
 

markdi

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I think just a diode is all you need
the magnetic inductance of the relay coil and the mass of the moveable contact arm will cause it to stay closed for your aplication.
 

James S

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A single diode didn't cut it, so I tore a full wave bridge out of an old powersupply and wired that in, and success.

The relay is working fine, but it does get a little warmer than I'd like, so I'll experiment with adding some resistance shortly.

Thanks everybody!
 

Brock

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A cap might also get the voltage down a bit and keep it cooler.

I have to wonder though, why wouldn't you just replace it with an AC version /ubbthreads/images/graemlins/wink.gif
 

James S

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[ QUOTE ]
why wouldn't you just replace it with an AC version

[/ QUOTE ]

because that would be the easy way out? /ubbthreads/images/graemlins/grin.gif

No, I just have a box full of these little relays and rather than order something new and pay for it and pay for shipping and wait and wait and wait I decided to do it now. thats all /ubbthreads/images/graemlins/smile.gif


So far so good, behaving rather nicely now.
 

Floating Spots

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[ QUOTE ]
Brock said:
A cap might also get the voltage down a bit and keep it cooler.


[/ QUOTE ]

Actually, on light loading, it would raise the RMS voltage.
At the same time it would lower the ripple.
 

JohnJ80

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The diode basically clips out one half of the AC sine wave. So, when you do the root mean square calculation you get a lower DC equivalent voltage. For example, in 110VAC, that is the RMS voltage (110V) if a full wave bridge was added. The RMS voltage for a half wave (single diode) would be lower. The peak amplitude of a sine wave is sqrt(2) (~1.414) larger so in 110VAC the peak voltage is 110V*1.414.

So, the reason your relay won't work is that the minumum DC voltage to actuate the relay is lower than RMS voltage supplied by the single diode (half waved rectified waveform). It really has nothing to do with the inductance of the coil etc...

If you add a capacitor across the output of this primitive power supply, you start to bring the DC voltage (RMS voltage up) - also known as reducing the ripple. The cap charges up as the voltages rises to the peak and then discharges in the lost half cycle. If the cap is large enough and the current pulled by the load is low enough, then you get less and less droop in the dead half cycle and the corresponding increase in the RMS or DC equivalent voltage.

Hopefully that all made sense.

J
 

Brock

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/ubbthreads/images/graemlins/smile.gif I know I do that WAY to often, my wife always say's, why don't you just buy a new one, but why when I can make it work with all these parts /ubbthreads/images/graemlins/wink.gif
 

JohnJ80

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No kidding. Its way more fun to fool around a figure it out than it is to do it the easy way. But, hey, its a guy thing.

J
 
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