Modifying RF Light Switch

[AdamUp]

Hi Everybody,

I hope this is the right forum to place this question. 

I just bought a Remote RF Light Switch from Radioshack.  It has a little remote that you can use to turn a lamp that is plugged into the receiver block on or off.  The receiver plugs directly into an outlet, and then the lamp plugs into the receiver.  Basically, when the remote's "on" button is pressed, the  circuit is completed by the receiver box and the light turns on.

You can see the unit here: http://www.radioshack.com/product/index.jsp?productId=2103886&cp=&origkw=light+switch&kw=light+switch&parentPage=search

What I would like to do is just use the circuitry inside the box as a remote on/off switch.  I don't need it to turn anything on.  I just want the thing to complete a circuit when I use the remote.  Here's how the reciever works.  There is a little electromagnet that, when activated, pulls a metal tab against a terminal to complete the circuit.  Voila!  The circuit is completed.  Obviously, this unit needs power to operate the electromagnet, and I really don't want to plug the thing into a wall outlet (which is where it usually draws power from).  Is there a way I can power the unit with a 9-volt battery?  Here are some pictures of the circuitry:






Thanks for reading this!  If you can figure this out, I'll forever be in debt!

Thanks again!

~Adam

[ante]

Hi Adam,

Welcome to our forum!

The “metal tab / electromagnet” as you call it is a relay. How about the black thing on the side, is it a DC jack or am I wrong? Can this be a way to power the unit, or is it used for something completely different?  :)

[AdamUp]

Hi Ante!

Thanks for the welcome.  I realized what the electromagnet was after doing some research last night....haha...I guess I never understood that auto relays looked like that on the inside. 

You're right, the black thing on the side is a DC jack.  I was hoping to not have to plug the unit into a wall to get it to work.  Rather, I wanted to attach a 9v battery to power the unit.  The relay has 24v written on the side, so I'm assuming that 9v might not be enough to power it? 

Do you think buying a 5 volt relay from Radioshack and soldering it into the board would work? 

Thanks again for the welcome and for helping me out on this!

~Adam

[SM2GXN]

Hi AdamUp!

It depends on what the rest of this device want  ;D a relay can be forced to activate by for a short time supply a voltage close to 24 volts, this voltage can probably drop down to 12 volts and still keep the relay activated so it might work with a 9 volt battery, try it.
Well that just my 2 swedish kronor  ;D

[AdamUp]

Thanks for your reply!  I tried hooking up a 9v to the prongs that would usually go into a dc socket to no avail.  I was wondering, though, if I connected the battery at a different point on the board if I could get it to work.  I have a feeling there is some stepping down of the power before it gets to the RF circuit which is what I need to activate with the remote.  If I could find the point on the board after this stepping down has occurred, I could connect the power there.  Do you think this would work? 

Thanks!

~Adam

[ante]

Is this device run from a wallwart, not from power coming directly from the outlet? I doubt that the circuit will work properly in all stages with a 9v battery. I don’t believe they would use different voltages for the RF circuit and the relay latch circuit, it would be unnecessary complex and expensive. But you could try to use two 9v batterys in series (18v), which might work. Just don’t expect two 9V PP3s to keep the relay closed for a very long time!  ;)

[SM2GXN]

You are right Ante most probably the RF part need more than 9volts but somehow they will initially need more than 12 volts for the relay to activate, some battery powered circuits I've seen had some kind of momentarily higher voltage supplied to the relay to activate it and than dropping the voltage just to keep it activated thus decreasing the current through the coil and increase battery life as well.
Adam can you see if there is any tracks that goes to components other  than the relay, I'm suspicious about using a 24V coil in a device connected to a 12volt battery.
Ante don't you think that18volts could be to much for the circuit to handle unless there is an regulator somewhere?

[ante]

The 18Volts is for the relay side of the circuit and the battery supplies the rest of it. I think the battery might supply some of the circuit and some of it is supplied from mains this can be the reason for the 24V relay! Perhaps the circuit is made this way to save on battery current, but I am not sure. I tried to follow the tracks on the pictures but it was too small (or maybe it’s my eyes)! A few things to clarify the basics:
The battery; is it inside the unit or is it connected to the DC-jack? If not, what's the purpose of the DC-jack?
Does the relay engage even if the unit is lying on the table or only when it’s in the mains socket? :-\

[AdamUp]

Hi everybody,

Thanks for the thought you've put into this.

There is no battery in the unit.  I just want to be able to power it using a battery.  The RF *and* relay circuits come off of the tracks coming from the prongs going into a DC wall socket. 

Does this help?

Thanks!!!
~Adam

[ante]

Hi Adam,

Ok, I got confused by the info in the radioshack link you provided? Then what is the purpose of the DC jack, could it be the power the unit from a wallwart? The pictures are not clear enough (at least not for me) to see where the tracks goes.

[AdamUp]

Hi again,

Yes, the DC jack is to be to power the unit from the wall.  The 12v battery in the radioshack link is to power the RF remote that signals the unit to turn the relay on/off. 

Any ideas?
Thanks Ante!

~Adam

[ante]

Hi Adam,

OK, I’m kind of confused here sorry! :-[  Right or wrong; the unit works of the wall if you feed it through the DC-jack and in the wallsocket without the need of external power?!?!  :-\

[AdamUp]

Hi Ante, it's my fault for misunderstanding AC/DC:

There really isn't a DC jack on the device.  The box plugs into an AC outlet, and then a lamp or other device plugs into the port that I have been (incorrectly) calling a DC jack.  Really, it's just a pass-through jack that supplies AC power from the wall to the device that's plugged into the jack when the relay is activated.

~Adam

[ante]

That explains part of it! There must be a big capacitor (X1 oor X2 type maybe) and some resistor / zener diode rectifier/voltage divider close to where the mains voltage enters the unit.