Footswitch with LEDs

robinbowes

So I've finally got my Vibrotone switching mechanism into a box with the switching broken out via a stereo jack socket so any dual footswitch can be used (or any ext. switching mechanism, eg. the remote switches on my G2).

The switching is very simply, and essentially looks like this:

-----------------o /  tip (slow/fast)

                  /

-----------------o    common

                  \

-----------------o \  ring (off/on)

When a circuit is closed, the connected relay closes (it's switching mains voltage).

The way I've designed the switching, there is +12v supplied to both the tip and ring. I'm trying to figure out how to use that voltage to light an LED.

All I can think of doing is putting the LED in-line with each loop, eg. something like this:

------+LED+------o /  tip (slow/fast)

                  /

-----------------o    common

                  \

------+LED+------o \  ring (off/on)

However, that means the switching won't work if the LED fails.

I guess I can live with that, but is there some other way to do this?

R.

ICBM

That's the right basic way of doing it, but you need to arrange it so the full current doesn't go through the LED - LED current is much smaller than the operating current for a relay.

If you measure the relay 'on' current, then use a resistor to produce a voltage from that just above the forward voltage of the LED (2V should be about right, for a red LED) and then put a small resistor in series with the LED itself to limit the current to whatever it needs, that should work. As long as the main resistor doesn't fail - it won't if you use a big enough power rating - then the switching will always work.

robinbowes

Aha, I see, so resistor in line with the switch to produce a drop of around 2v, and put the led/series resistor in parallel with that. Ok, I should be able to manage that. Thanks, R.

robinbowes

How does this sound:

Relay coil resistance: 360R

Total Voltage: 12V

Desired drop over resistor = 2V

Voltage across relay coil = 10V

Therefore, current in relay coil: I = V/R = 10/360 = 0.02777 = 28mA

To drop 2V over a resistor with a current of 28mA, R = V/I = 2/0.02777 = 72R

Sound reasonable?

ICBM

Pre-coffee, but yes I think so!

68R should be close enough.