Greetings good people!

I have a simple circuit that uses a 12v DC wall socket adapter to power a Mac valve operating at 6 watts. Power switching is done through a mechanical air pressure sensing switch from AirTrol with a Cherry switch inside. This works fine. The only odd behavior in the system is an occasional chattering, rapid-fire on-off-on-off-on cycle that happens when the pressure switch gets near the shut-off level. I can fix this with a pressure damper in the air sensing line.

I recently put a 12v LED rocker switch in the circuit and the LED burns out after a day or so of use. Seems to happen when the system does the ratta-tat-tat on/off cycle. I checked the voltage on the power adapter and it's fine. I've replaced the switch 6 times in a week thinking at first that this was just a bad switch issue. But now I've tried 6 different switches from 3 different manufacturers.

Any ideas how to save the LED?

 

The LED should have a resistor in series.
The switch could include a resistor - can you give a link to the switch you are using.
For 12V the resistor should be about 1kΩ

 

Greetings good people!

I have a simple circuit that uses a 12v DC wall socket adapter to power a Mac valve operating at 6 watts. Power switching is done through a mechanical air pressure sensing switch from AirTrol with a Cherry switch inside. This works fine. The only odd behavior in the system is an occasional chattering, rapid-fire on-off-on-off-on cycle that happens when the pressure switch gets near the shut-off level. I can fix this with a pressure damper in the air sensing line.

I recently put a 12v LED rocker switch in the circuit and the LED burns out after a day or so of use. Seems to happen when the system does the ratta-tat-tat on/off cycle. I checked the voltage on the power adapter and it's fine. I've replaced the switch 6 times in a week thinking at first that this was just a bad switch issue. But now I've tried 6 different switches from 3 different manufacturers.

Any ideas how to save the LED?

My guess.
The valve may not have a clamp diode so the voltage from the inductive kick is the culprit.
A clamp diode across the coil would fix it if that's the case.

 

Please upload your circuit as 800x600 jpg or gif file and more details about the 12Vdc, if you have the link to the LED specification would be better.

 

Here is a link to the switch (with the LED) that I am using...
http://switches-connectors-custom.cwind.com/Asset/GRB112B802BR1.pdf

Any help would be hugely appreciated at this point.

 

If the resistor included in the switch then you can measure the current as:
+12V → ⊕-(A meter)- Θ → Pin.1, Pin.2 → R → ⊕--|>|--Θ → Pin.3 → 0V

If the watts is 6W then the I_Led = 6W/12V = 0.5A, it's quite big current.
The links didn't shows the current or watts of led, so where did you get the watts = 6W?

 

The 6 watts is the rating of the Mac valve that is being turned on and off by the switch.

 

The resistor is included in the switch in your link and is suitable for 12V:


As you are switching an inductive load I would guess that it was the back-emf generated by that switching which is killing the LED.
As ronv said (post #3) a diode across the coil would fix that.

 

Thanks for all of the help. Can you suggest what type of diode I would use?
For what its worth, attached is a picture of the lay out.

 

Thanks for all of the help. Can you suggest what type of diode I would use?
For what its worth, attached is a picture of the lay out.

1N4001 - 7 are about as inexpensive as any.
Place it across the coil with the cathode (band) going to the positive lead and the Anode to the negative side.

 

Thanks RonV.
The coil itself is inaccessible...the Mac valve just has two wire leads. Can I put the diode on the switch instead?

 

The best place to put the diode is across the two wires from the valve.

 

Try this:

 

Just wanted to thank everyone for the help. The diode placed as described by Tonyr1084 did the trick.

 

Followup question. In the attached drawing, I've identified two places where I can install the diode. Location "A" would be the norm but I'd rather install it at location "B". Will the diode still do its job and protect the LED in the switch if it is mounted in the "B" location?

 

Neither location is right. Position A is the closest to the right location, but you have it wired across the power supply and not across the solenoid.

When the solenoid shuts off - the magnetic field collapses. Changing magnetic fields generate electrical energy, called BEMF or Back Electro Motive Force. It's that force that is blowing your LED's out.

So if you take position A and redraw it so that the diode is across the first and last legs of the switch you'll be fine. I know the drawing I gave you showed the diode tied in at the wire nuts. You can move it back to the switch but it must be across those two leads. Give me a few minutes and I'll bang out another picture for you.

OK, here's the drawing:

 

Thank you again.
The odd thing is that I tested the diode as shown in the "A" location in my caveman drawing and after a couple dozen tests, the LED didn't burn out. That configuration is how you had it shown overlaid on my picture. The red wire (with the white bar side of the diode) went to the middle switch tab.

 

OK, however you had it wired, in the overlaid picture the diode was across the coil wires. In my more recent drawing the diode is once again across the coil wires. Your drawing shows the coil leads to the outside terminals on the switch. Looking back at your picture and your drawing, the two don't match. It's not easy to see details on your picture, but it looks like the blue wire (power) goes to the outer most leg to the right (your picture). If that's how the wiring is connected, then yes, your "A" diode location would be correct. But if you look at your hand drawn diagram your positive power supply lead goes to the center lead.

As long as the diode is across the coil wires you'll be fine. However your switch is wired.

Good luck to you.