Jack connection question

[David]

I have pretty much hung it up when it comes to circuit building and PIC projects.  I don't even think I've plugged a soldering iron in for at least five years.  I settled on my pedal set and my signal path a long time ago.  About the only things I want to do now is make my rig as light as possible, and the area near my pedalboard as neat as possible.  I have come to despise cables strewn all over the place.

I'm about ready to build what is probably my 25th pedalboard.  I'm going to construct a rectangular frame from 2 inch by 2 inch (nominal) poplar boards, miter sawn at 45 degree angles and glued together like a picture frame.  Over this I will attach poplar strips and fasten my pedals with wood screws through bike chain links.  I will raise the front (the side away from me) of the pedalboard with another strip of poplar glued under the frame.  I plan to attach an aluminum strap on each side.  On the left side, there will be jacks for a wall wart and 4 quarter inch mono jacks for signal input, amp send, FX send and FX return.  On the other side will be a single quarter inch mono jack, also for signal input.

There is a reason for the redundancy, and a method to my madness.  I intend for the signal input on the left to be used with a wireless receiver that will sit on top of my amp.  I will then have a snake to connect to the jacks on the left side, thus there will be no cords between me and the pedalboard.  If I'm not using an FX loop, I'll connect a patch cord between the amp send and FX send jacks and connect the amp input to the FX return jack.  The jack on the right I'm intending to use for direct guitar connection.

What this means is that I have two input jacks that I have to somehow parallel together and somehow wire so that only the one I have plugged a cord into is active.  I think I may have this figured out;  however, it sure does not strike me as normal practice.  I have these jacks with six connectors on them.  With the opening for the jack facing you, the connectors for the tip and ring on the jack are front and rear on the left.  It so happens that the connectors on the right side bridge to their counterparts on the left side WHEN NO CABLE IS PLUGGED IN.  This strikes me as the key to the whole thing.  It seems to me that this would work in the following way:  If I connect the tip of the signal input on the left so it goes through the middle pair of connectors on the signal input on the right and then to the first pedal in the chain, I would get signal into the effects because no cable would be plugged in to open the signal path.  The right input would be connected through the left "bridge" connectors in the same fashion.

Does anyone have experience with "switching" with jacks in this way?  Is it workable?  Have I missed something?

Thanks in advance.

[GibsonGM]

If I'm reading you right, David, you probably want to get some Switchcraft 12A switching jacks.   When the plug is not inserted, the signal is grounded,  When you plug in, you are opening the switch and allowing signal to pass. 

So, obviously, whatever input you are not using will be grounded, and therefore, silent...this is done all the time on amps and stuff that have more than one input, just like you want to do...

[David]

If I'm reading you right, David, you probably want to get some Switchcraft 12A switching jacks.   When the plug is not inserted, the signal is grounded,  When you plug in, you are opening the switch and allowing signal to pass. 

So, obviously, whatever input you are not using will be grounded, and therefore, silent...this is done all the time on amps and stuff that have more than one input, just like you want to do...

Mike, I must admit that I had not thought of that.  I would think this would work at least as well as what my fevered, out-of-practice brain was considering...

[ashcat_lt]

Wait, what?  How are we going to "ground the unused input" without shorting the whole thing?
Edit - well I guess you could run the tip of the receiver's jack to the ring on the side of the other jack that actually goes to the pedals and then jumper R>T on the switched side so that with no plug inserted it goes where it should, and when you plug in it gets shorted to ground by the sleeve of the TS plug, but then you're shorting the presumably low-Z output of the receiver.  That probably won't cause damage in a well-designed circuit, but it could make it malfunction, and if it can supply enough current to wiggle the "ground side", it could contribute a lot more noise than whatever capacitive coupling might leak through the switching contacts./Edit

The OP is the way it should be done.  I'd say have the switching on the side intended for direct connection.  The wireless receiver can stay plugged in, its signal goes to the NC switched lugs on the "guitar" jack and then through to the first pedal unless you plug into that jack in which case the receiver is completely disconnected from the circuit and can't really contribute noise.

[GibsonGM]

Wait, what?  How are we going to "ground the unused input" without shorting the whole thing?

The OP is the way it should be done.  I'd say have the switching on the side intended for direct connection.  The wireless receiver can stay plugged in, its signal goes to the NC switched lugs on the "guitar" jack and then through to the first pedal unless you plug into that jack in which case the receiver is completely disconnected from the circuit and can't really contribute noise.

Right!!  LOL, duh...my bad.    Ashcat is totally right.  I was in la la land and thinking only of grounding the input when not in use.  You'd need a real work-around to get away with what I suggested, not the way to do it.

My preferred method of doing things like this is with an actual switch, either foot pedal or the kind you bend over and throw.  But still nothing wrong with using jacks to achieve the same end and it may be simpler to do - as long as the dope you ask tells you the right way, ha ha! 

[David]

Wait, what?  How are we going to "ground the unused input" without shorting the whole thing?

The OP is the way it should be done.  I'd say have the switching on the side intended for direct connection.  The wireless receiver can stay plugged in, its signal goes to the NC switched lugs on the "guitar" jack and then through to the first pedal unless you plug into that jack in which case the receiver is completely disconnected from the circuit and can't really contribute noise.

Right!!  LOL, duh...my bad.    Ashcat is totally right.  I was in la la land and thinking only of grounding the input when not in use.  You'd need a real work-around to get away with what I suggested, not the way to do it.

My preferred method of doing things like this is with an actual switch, either foot pedal or the kind you bend over and throw.  But still nothing wrong with using jacks to achieve the same end and it may be simpler to do - as long as the dope you ask tells you the right way, ha ha!

No dope here, with the exception of the original post...

[GibsonGM]

Nah, it's a good question, and you see how fast I found an easy, and WRONG, solution, ha ha!  These things take thought. 

If it were simple, it would be noisy and undesirable. 

I love switches

[David]

Wait, what?  How are we going to "ground the unused input" without shorting the whole thing?
Edit - well I guess you could run the tip of the receiver's jack to the ring on the side of the other jack that actually goes to the pedals and then jumper R>T on the switched side so that with no plug inserted it goes where it should, and when you plug in it gets shorted to ground by the sleeve of the TS plug, but then you're shorting the presumably low-Z output of the receiver.  That probably won't cause damage in a well-designed circuit, but it could make it malfunction, and if it can supply enough current to wiggle the "ground side", it could contribute a lot more noise than whatever capacitive coupling might leak through the switching contacts./Edit

The OP is the way it should be done.  I'd say have the switching on the side intended for direct connection.  The wireless receiver can stay plugged in, its signal goes to the NC switched lugs on the "guitar" jack and then through to the first pedal unless you plug into that jack in which case the receiver is completely disconnected from the circuit and can't really contribute noise.

Actually, Ashcat, that's not how I originally intended to do it...  but it's a whole lot better!!!!!!!!!!!!!!  I was going to have shielded cable crossing between the jacks.  I knew that was stupid and there had to be a better way, which is why I put up this post - and got these answers.

Slick idea, man...   

[David]

Taking this a step further, Ashcat's last post gave me an even cooler idea...

You know how I said I was going to jumper the "to amp" and "fx send" jacks if I wasn't using an FX loop?  Why should I waste a patch cord doing that?  I can use the jacks themselves to jumper between "to amp" and "fx send" until I break the bridged connections by inserting my snake cables.

[GibsonGM]

Nice, David - I like that idea.    Some people are really good at using jack switching to do some VERY interesting things - maybe you're going to be one of them!! 

[David]

Nice, David - I like that idea.    Some people are really good at using jack switching to do some VERY interesting things - maybe you're going to be one of them!! 

Nah,  I just "jack" around...   

[ashcat_lt]

Taking this a step further, Ashcat's last post gave me an even cooler idea...

You know how I said I was going to jumper the "to amp" and "fx send" jacks if I wasn't using an FX loop?  Why should I waste a patch cord doing that?  I can use the jacks themselves to jumper between "to amp" and "fx send" until I break the bridged connections by inserting my snake cables.

Yeah, I missed that part.  You're talking about exactly the same thing as the FX loop in a guitar amp.  Usually, the switching is done on the FX return, so that you can actually split the signal out the send and still hear the dry through the amp, but when you patch into the return, it breaks that direct connection and you hear only whatever is coming back from the loop through the amp.

You can look at almost any amp schematic, but I guess it would have the fx send's unswitched lugs connected to the input and the switched lugs of the return, and the amp input on the unswitched side of the return jack.