True Bypass Looper with Amp Channel Switch Help

[Hexjibber]

Hi there,

I am looking to build a true bypass loop pedal that also switches the channel on my Jet City 100HDM amp at the same time as engaging the loop. The Jet City uses what I understand to be a fairly standard mono jack method of shorting the sleeve to the tip to change the channel. The amp also powers the LED somehow? I've searched through the forum and found some relevant posts and cobbled together a rough diagram, I was hoping someone may be able to help confirm that I'm on the right track. I've built many pedals over the years but the realm of amp channel switching is quite alien to me and I don't quite understand it to be honest.

I started out presuming I may need a 4PDT but after copying the wiring from a picture of the Jet City footswitch into my diagram, it seems a 3PDT is all I need?

Here's my diagram;



And here's the guts of a Jet City footswitch which I have effectively tried to copy into the true bypass loop using the middle lugs;



Would very much appreciate if anyone has the time to take a look.

Many thanks,
Graham

[slacker]

That looks good to me, it will turn the effect on when the Amp LED is off, whatever channel that is.  You might want to use a plastic jack to isolate the amp switching from the audio path, this will prevent any possible ground loops and will stop anything nasty happening just in case the sleeve of the amp footswitch jack isn't ground.

[Hexjibber]

Thanks for the reply 

Ah, it's supposed to do the opposite, when the LED for the amp is on (which switches the amp to the "Overdrive" i.e. louder channel), the bypass loop should also be engaged. Basically I need to go from the "Normal" amp channel and no loop, to the "Overdrive" channel and the loop.

The way I understood it was when the footswitch is set to the bottom and middle rows of lugs, the input signal goes through to the output via the jumper, and the tip and sleeve of the amp jack will be shorted (which will set the amp to "Normal" via a relay switch). Then when the footswitch is on the top and middle rows, the input goes through the loop before coming back to the output, and the LED circuit is completed via the tip and sleeve.

Have I got that wrong then? For the amp switching it seemed to just be a case of copying the LED circuit from the picture which only uses two lugs of a DPDT, my thinking was I would just use the central row of lugs on the 3PDT to achieve the same result.

Any insight would be really helpful, thanks!

Graham

[slacker]

The amp switch works the other way round to what you said. When the middle and bottom lugs are selected, the centre lug, with the black wires going to it is connected to the bottom middle lug, which has nothing connected to it. There's no connection between the centre and top middle lug, so current can flow down the red wire through the LED and back through the black wire, so the LED is on.
When the middle and top rows are connected the top middle lug, with the red wire connected to it is connected straight to the centre lug, so current flows down the red wire straight to the black wire, the LED is shorted out and will be off.
You just need to swap the red wires to the centre bottom lug then the loop will be active when the LED is on.

[Hexjibber]

Hey Ian,

Thanks for the detailed reply but also simple explanation, I really appreciate it! I understand why I had it backwards now. Will give this a go and see how I get on.

Also thanks for the tip on using a plastic jack connector for the amp switch, will be sure to use one.

Cheers!
Graham

[TEAPOT.CUP]

Not to hijack an old thread, but Slacker I have sent you a PM about an old post. Not sure if they get notified or not.