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Footswitch wiring help needed

Randy

Silver Supporting Member
Messages
3,947
I've got a footswitch that is currently wired as shown below. One switch changes amp channels from clean to dirty, the other is a boost. My problem is that the LED's light when the amp is in the clean channel, and when the boost is off. I'd like to rewire it so that the LED's light when the amp is in the dirty channel and the boost is on. I should be able to figure this out but... can't :red Should the ground be in the middle position of each switch? Any help is appreciated.

 

8len8

Silver Supporting Member
Messages
14,541
I've had a similar problem in the past, and you can't fix it. It is determined by the design of the circuitry in the amp, not in the footswitch.

The only option is to put a more complex switch in the footswitch box and have a battery power the LED.
 

Randy

Silver Supporting Member
Messages
3,947
There must be a way to do it. I didn't mention it but I'm trying to make a replacement FS for an amp that is missing one. I know the stock footswitch works as I described it should. A fellow owner of the same amp was kind enough to send me pics of the guts of his FS. I just can't quite figure out how it's working. Their LED setup includes diodes and a resistor, mine doesn't need those (resistor is built into the LED);


 
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Randy

Silver Supporting Member
Messages
3,947
LED = diodes, right? <G>
No, two separate things.

When I emailed Two Rock for a schematic, they wouldn't give it to me, but did mention I'd need a resistor and said that they used a "diode chain to ground".

In the edited pic below, the LED is circled in yellow, and what I believe to be the "diode chain" is circled in red.



I'm not really concerned with what they're using for led's and diodes since what I have is working fine. It's just a matter of figuring out the wiring for the switches.
 

Kyle B

Silver Supporting Member
Messages
5,115
I agree you can't do this without more complex circuitry.

With a schematic of the AMP, you would be in a MUCH better position to seek help and figure this out right.

They told you that you need resistors because if you don't put them in series with your LEDs, the LED will melt.


Add a battery and a couple resistors, this would work for you. The 1N4148 diodes and their series resistors may or may not be needed (w/o knowledge of the amp internals it's impossible to tell). I'm guessing they ARE required, that the current which would flow through the LEDs in your original circuit turn on a transistor internally. It's hard to tell what value those original resistors are, but use the same thing. They LOOK like "brown green yellow gold" which would be 15k. That seems a little high though, the LEDs would not light with a 15k series resistor.


In this schematic, adjust "R" to change the LED intensity. Target 350 ohms for a RED LED, 275 ohms for BLUE (that will limit LED current to around 20mA).




You might add a power switch for the battery to disable the circuit when you're not jamming so you don't kill it by leaving it on accidentally.

If there's room inside the foot switch, consider using 4x "AA" cells instead. They'll last many many times longer than a 9V. (Resistor values then would be 200 ohms for RED, 125 ohms for BLUE)
 
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Randy

Silver Supporting Member
Messages
3,947
I agree you can't do this without more complex circuitry.

With a schematic of the AMP, you would be in a MUCH better position to seek help and figure this out right.

They told you that you need resistors because if you don't put them in series with your LEDs, the LED will melt.


Add a battery and a couple resistors, this would work for you. The 1N4148 diodes and their series resistors may or may not be needed (w/o knowledge of the amp internals it's impossible to tell). I'm guessing they ARE required, that the current which would flow through the LEDs in your original circuit turn on a transistor internally. It's hard to tell what value those original resistors are, but use the same thing. They LOOK like "brown green yellow gold" which would be 15k. That seems a little high though, the LEDs would not light with a 15k series resistor.


In this schematic, adjust "R" to change the LED intensity. Target 350 ohms for a RED LED, 275 ohms for BLUE (that will limit LED current to around 20mA).




You might add a power switch for the battery to disable the circuit when you're not jamming so you don't kill it by leaving it on accidentally.

If there's room inside the foot switch, consider using 4x "AA" cells instead. They'll last many many times longer than a 9V. (Resistor values then would be 200 ohms for RED, 125 ohms for BLUE)

Kyle, I really appreciate your putting so much thought into my problem and the very thorough response.

The 'hot' wires on the amp put out around 9 volts. I'm using Radio Shack 12V LED's which come with a resistor already in place. So while they are not as bright as they could be, they are bright enough for me and there is no danger of them burning out.

I thought TR must have used a resistor and diode combination to get the maximum brightness out of the LED while insuring they woudn't burn out. But that was just a guess. Perhaps it's much more complex as you suggested and there is some feedback from the diode chain that triggers something in the amp?

It just seems like that would be much more complicated then it needs to be. There must be something very basic I'm missing here;

I know the way the stock FS works is that one LED comes on when the amp is switched to the gain channel, and the other comes on when the boost is kicked in. Yet it seems like those are the 'open' positions on the switch - hot wire connected to nothing / circuit not completed. How can the LED's be coming on in that case?
 

Kyle B

Silver Supporting Member
Messages
5,115
Sometimes footswitches are A/C, sometimes D/C. Hard to say w/o knowing the schematic. Or at least being able to see what's inside those "magic" white globs!

>> Yet it seems like those are the 'open' positions on the switch - hot wire connected to nothing / circuit not completed. How can the LED's be coming on in that case?

If the circuit were not complete, the LED's would not be turning on :)

I think what happens (presuming your circuit is correct) is that when the switch goes to "open", then current can flow through the LED, lighting it up. The combination of the LED & it's resistor forces approx 9V to be seen from colored wire (red/white) to GND. When you flip the switch in the opposite throw, the LED/resistor gets shorted out. It now has 0 volts across it. Which ALSO means there is 0 volts now from colored wire (red/white) to GND. Somehow the amp's circuit can tell the difference between 9V and 0V being present across those wires.

If the footswitch is A/C it may explain why there are extra diodes (supposedly) in there. The LED would be active during the (+) part of the cycle, the other diode could be active during the (-) part.

Try setting your meter for A/C measure, see if you get an A/C voltage across those wires. I'm guessing the 9V you measured was DC, right?

Maybe you can ask your friend to send you a few more pictures of that circuit - preferably taken IN FOCUS (LOL) and from several different angles???
 

jamesdlow

Member
Messages
31
@Randy, did you ever figure out the exact wiring for the Two Rock footswitch?

No, two separate things.

When I emailed Two Rock for a schematic, they wouldn't give it to me, but did mention I'd need a resistor and said that they used a "diode chain to ground".

In the edited pic below, the LED is circled in yellow, and what I believe to be the "diode chain" is circled in red.



I'm not really concerned with what they're using for led's and diodes since what I have is working fine. It's just a matter of figuring out the wiring for the switches.
 






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