Does anyone know whether it would be possible to dissasemble this kind of 16 mm illuminated switch to change the internal resistor for the LED to work better with a lower voltage? Thanks.

 

It should be tricky to find the catches the first time, but not after you find the way it is held tight.
If many at hand, perhaps risking sacrificing the first will be less of a headache later.
Start prying out the blue section.

 

sure ... they come in packs of 10 , so dismantle one to see where the resistor is positioned ,

then , on the next switch you can just make a small hole/s in the casing where needed to get electrical contact with the resister , probably best to leave the original in place and put one of the right value in parallel with it

One end of the resister must be connected to one of the terminals , so you only have to make one hole and one connection inside the casing..

I'm assuming dismantling the switch will break plastic and prevent reassembly.

 

Use these https://www.idec.com/language/english/catalog/Switches/A6Series.pdf#page=4. Lamps are available for 5V.

I either used these or incandescent rectangular versions that were converted to LED, but at 24 VAC. The LED color should match the lens.
What's really nice is that you can label the buttons by printing on transparency film and placing it under the lens.

 

Would be helpful if we knew what voltage you want to operate the switch at.

Once you've found the way to open them so that they can be reassembled, you can simply piggy-back a parallel resistor to bring you into the correct current for the LED.

Do you understand how parallel resistors change the current? If so - good. If not, we can explain that part of it too.

 

why not just boost voltage for driving LEDs at 12V or whatever they are meant for? this way you don't have to hack switches.

 

Wouldn't this work? Two coin cell batteries between the ground terminal of the switch and chassis ground? The switch would be working at 6 volts but the current path through the coin cells would still see 12 volts, 12 mA. No?

[edit] forgot to draw in the polarity of the coins. The small end is the negative. In the circuit, the LED will still see 12 mA while the switch controls six volts to the load. If 5 volts is desired as a control, then 11 mA will be the calculated current through the LED. The coin cells should last quite a long time.

Keep in mind, I think in simple terms. This should work. Whether it's practical or not - I don't know. Depends on how circuit is being used. If it's infrequently and for short periods of time I'd imagine this should last quite a long time, without need to modify the switch at all. Otherwise, if you truly want to modify the switch, piggy-backing another 1KΩ resistor on top of the existing 1KΩ (1K is assumed) should drop the operating voltage down to 6 volts. The difference at 5 volts is negligible.