It's really simple. Depending of the voltage on the pedal, you'll need to do a little calculating. The formula is Voltage / resistor = current. The resistance you'll need depends on the LED. If you use a basic red LED, you won't want to get over 10 mA. So suppose it works on 9 volt, you do 9volt / 10 mA = 900 ohm. In that case you'd wanna use a 1kohm resistor. Make sure you use the correct polarity on the LED(the cut side is the cathode, or the -).

When you buy the LED, check out on the package, it should tell you the maximum current. Then it's only a matter of doing V / R = A, or V / A = R.

EDIT: Actually, it will depend on the circuit of your pedal. The way I told you, the LED is always gonna be on. If you can get me the schematic I can modify it for you.

EDIT2: What you can probably do is hook up the anode side of the LED+resistor on the OUTPUT of the switch you want the LED on. So if you want it on reverb, hook up the anode on the output of the switch, and hook up the cathode to the ground(usually the case is grounded). Remember the resistor and LED need to in series.
Quote by MH400
a girl on the interwebz?

You have 2 options.

1. Tits.
2. GTFO.

Last edited by Spike6sic6 at Jul 16, 2010,
The pedal is just a simple 2-button deal to switch channels on the amp and turn the reverb on/off. It doesn't take any batteries. I suppose I cold just open it up and slap a 9v battery in there with an LED and some resistors, which would be simple enough to do (I've done some circuit-building in a robotics class so I'm familiar with it) but I was wondering if there was a way to draw power from the amp for the LEDs. That might get a little complicated though.
Quote by guitar98765432
The pedal is just a simple 2-button deal to switch channels on the amp and turn the reverb on/off. It doesn't take any batteries. I suppose I cold just open it up and slap a 9v battery in there with an LED and some resistors, which would be simple enough to do (I've done some circuit-building in a robotics class so I'm familiar with it) but I was wondering if there was a way to draw power from the amp for the LEDs. That might get a little complicated though.

I see I see. You can use power from your amplifier, but that would require you adding an extra cable. The way I told you, your audio signal will be used to light the LED on, so it won't be on at all times, etc.

Open up your footswitch and tell me how many pins the switches have. I might have an idea for you.

EDIT: Look up if the switch is in series with the circuitry, or if it's going from the circuit to the ground.
Quote by MH400
a girl on the interwebz?

You have 2 options.

1. Tits.
2. GTFO.

Last edited by Spike6sic6 at Jul 16, 2010,
Here are the guts: http://imgur.com/bZt4H.jpg
6 pins per switch?

And I forget exactly which off-brand ebay seller I bought this from; it's not the Marshall one (I didn't feel like paying stupid markup for the same exact thing). Can you tell if it's in series with the circuitry from that picture?

EDIT: I just realized they'd have to be the same regardless of brand. I'm looking for that info now. And thanks a lot for your help.
Last edited by guitar98765432 at Jul 16, 2010,
Here's what I came up.

http://fr.tinypic.com/view.php?pic=28nh5e&s=3

I'm not 100% sure about how the switch is wired, but I think that should work. If you do the wiring exactly like the picture, you won't risk anything on your amp/footswitch. It might not work, but if it doesn't, it won't damage your other circuitry.

Values are as followed;

1.5v AA battery
150 ohm resistor
2x 0.1uF capacitor(do NOT forget those, or you'll risk your amp/switch)
red LED

The resistor has been calculated for a regular red LED.

EDIT: Oh wow it's actually much more simple then I thought it was. You could use the extra pins from the switches for the LED, but you need to make sure the 1.5v DC doesn't go to your amp/other gear(the capacitors on my schematic blocks the DC).

It's a bit hard to tell just looking at a picture, but I think you should do what I showed you. There is a small chance that the LED will do the exact opposite of what you want(no big deal); if it does, update me and I'll "fix" it for you.

EDIT2: If you are familiar with testing switches, you could use a multimeter(check-diode/beeper function) to figure out with pins get connected when you turn it on. Then you can figure out if those 2 contacts come in contact with the already-used pins. If you can find 2 pins that gets connected WITHOUT being connected to the already-used pins, you can use those easily, without the need for capacitors.
Quote by MH400
a girl on the interwebz?

You have 2 options.

1. Tits.
2. GTFO.

Last edited by Spike6sic6 at Jul 16, 2010,
Sweet. Thanks a lot man, I really wasn't expecting such in-depth responses. I'll post some pics when I finish this if it works. And thanks again, I really appreciate the help.
it's easier than whats been said above...

see how in the picture, theres a red wire connecting the top right lug of the left switch, with the top left lug of the right switch?

both of those points are ground.

the middle top lugs of either switch need to be connected to ground to signal the amp to switch channels.

both also have a voltage in them. it's just a matter of finding out what the voltage is, what the current is, and what size resistor you'll need based on that. then you make it so that when the channel is turned on, instead of directly connecting the wires with the switch, connect them with a resistor/LED in series.

then the only parts you need are 2 resistors and 2 LEDs

this way you don't need a battery inside the pedal..and it's the way most footswitches with LEDs work
Last edited by james4 at Jul 16, 2010,
Quote by guitar98765432
Sweet. Thanks a lot man, I really wasn't expecting such in-depth responses. I'll post some pics when I finish this if it works. And thanks again, I really appreciate the help.

My pleasure. I wish I could actually have the same footswitch, then it would be easier for me. But hey, if you do what I showed you on the .jpeg, it will most likely work.

The capacitors blocks the DC(make sure you put the exact polarity I put on the image), so there's no risk of risking the rest of your gear.

Oh and very important, the capacitors need to be electrolytic type. This is what they look like; https://www.egr.msu.edu/eceshop/Parts_Inventory/images/330%20uf%20electrolytic%20capacitor.jpg . The grey bar on it represents the cathode(the negative electrode).
Quote by MH400
a girl on the interwebz?

You have 2 options.

1. Tits.
2. GTFO.

Quote by james4
it's easier than whats been said above...

see how in the picture, theres a red wire connecting the top right lug of the left switch, with the top left lug of the right switch?

both of those points are ground.

the middle top lugs of either switch need to be connected to ground to signal the amp to switch channels.

both also have a voltage in them. it's just a matter of finding out what the voltage is, what the current is, and what size resistor you'll need based on that. then you make it so that when the channel is turned on, instead of directly connecting the wires with the switch, connect them with a resistor/LED in series.

then the only parts you need are 2 resistors and 2 LEDs

this way you don't need a battery inside the pedal..and it's the way most footswitches with LEDs work

Yeah, my lack of experiment with footswitches doesn't help. If there is indeed DC voltage inside the pedal, then yeah it's like you said.

I would need the switch myself to experiment and tell you how it should be done. Switches are hard to figure out unless you have them in your hand, with a multimeter.
Quote by MH400
a girl on the interwebz?

You have 2 options.

1. Tits.
2. GTFO.

Quote by james4
it's easier than whats been said above...

see how in the picture, theres a red wire connecting the top right lug of the left switch, with the top left lug of the right switch?

both of those points are ground.

the middle top lugs of either switch need to be connected to ground to signal the amp to switch channels.

both also have a voltage in them. it's just a matter of finding out what the voltage is, what the current is, and what size resistor you'll need based on that. then you make it so that when the channel is turned on, instead of directly connecting the wires with the switch, connect them with a resistor/LED in series.

then the only parts you need are 2 resistors and 2 LEDs

this way you don't need a battery inside the pedal..and it's the way most footswitches with LEDs work

So I would just test the voltage between the two top middle pins and then use Ohm's law to determine the correct resistors to use? And where would I put the LEDs and resistors? Between the top middle pins and the wires currently attached to them?
but what I'm saying is that there's almost surely DC in the pedal

TS just needs to know how to measure the voltage and current, and it wou;dn't be too tough to install at all

edit: that was to 2 posts up.

but yeah, the voltage/current between the red and black wires in the little diagram thing
Last edited by james4 at Jul 16, 2010,
Sweet. I can try everything out when my friend gets home to let me borrow his voltmeter and we'll see what works from there. Thanks a lot guys, I'll keep you updated.
Hey I tried just that but when I plug it into my amp my sound stays distorted, no matter if I press the switch or not. On my amp the light indicating the distortion channel is switched on is constantly lit. Also I don't get any reverb. The LED's light up though.

I'm using smaller than usual LED's, don't have the packaging so not sure how much A it can handle, I'm assuming 10mA. I've measured 11,3V on the channel knob and 7,3V on the reverb one so I've used a 860 and a 1,2k resistor.

Guess the resistors lower the current enough to make the pedal useless. Anyone have any ideas on how to get the LED's working without dropping the current via a resistor?
Last edited by mitchell.philip at Feb 10, 2014,
In most circuits the LED indicator does not form part of or tap off the primary path circuit (not without a buffer anyway), be it control signal (like the footswitch) or guitar audio.

You need to seperate the LED indicator circuit. Don't just tack it onto the existing switch/bypass wires on the footswitches.

The switches in the photo looks to be standard 2 pole 2 position footswitches. But only one pole in each switch appears to be used. The other pole looks to be free on each switch. You can wire up a battery and LED circuit to work on the spare poles. This keeps the footswitch operation and LED operation together, but keeps the actual circuits and voltages apart. You don't need capacitors and other junk then.

Investigate this.
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Last edited by Phoenix V at Feb 11, 2014,