My new build will require a 4-button footswitches (clean/dirty, crunch/lead, solo boost and FX loop)
For sake of being able to find spare parts easily, I wanna stick with good old TRS (Stereo) 1/4" cables, which mean I'll need two cables and I don't get to have an extra wire to provide power to the LEDs (I need to use the common for ground on both, this way if I'm stuck with a 2-button footswitch, I can still use 2 of the options)
Now, without a way to "phantom power" the LEDs, I figure there are two ways to light them up:
- Put them in serie with the switch
- Put them accross the switch (parallel) along with the proper dropping resistor
Problem with the first option is that the relays will pull more current than the LED might like (the clean channel will require switching as much as 3 relays, so around 100mA).
2nd option seems pretty simple really: with the switch off, most voltage is dropped accross the dropping resistor and the LED, there is not enough voltage accross the relays to turn them on. With the switch on, the LED is bypassed therefore it turns off and full power is restored to the relays.
However, this relies on the fact that putting the LED and resistor in serie with the relays will still provide enough power to light the LED while not switching the relays. Does that make sense? Will it work?
For sake of being able to find spare parts easily, I wanna stick with good old TRS (Stereo) 1/4" cables, which mean I'll need two cables and I don't get to have an extra wire to provide power to the LEDs (I need to use the common for ground on both, this way if I'm stuck with a 2-button footswitch, I can still use 2 of the options)
Now, without a way to "phantom power" the LEDs, I figure there are two ways to light them up:
- Put them in serie with the switch
- Put them accross the switch (parallel) along with the proper dropping resistor
Problem with the first option is that the relays will pull more current than the LED might like (the clean channel will require switching as much as 3 relays, so around 100mA).
2nd option seems pretty simple really: with the switch off, most voltage is dropped accross the dropping resistor and the LED, there is not enough voltage accross the relays to turn them on. With the switch on, the LED is bypassed therefore it turns off and full power is restored to the relays.
However, this relies on the fact that putting the LED and resistor in serie with the relays will still provide enough power to light the LED while not switching the relays. Does that make sense? Will it work?
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