Dual Outputs

[dpm99]

I'm upgrading my wife's Squier Bronco Bass. It has one pickup, and two pots (no switches). I'd like it to stay that way. I'm installing an active onboard amp circuit with a speaker, and would like to maintain the ability to have a passive output with a volume control and no tone. I'm having some trouble figuring out how best to do that. Here's basically what I'm trying to do:

All that crazy wiring out of the circuit is no problem. My issue is trying to figure out how to split the output between the passive output jack and the active amp circuit. As you can see in the "diagram," the amp circuit goes to a speaker and to a 3.5mm headphone jack. I don't want it to go to the main guitar output jack. That little circuit also has a built-in pot you can't see in the diagram that's directly connected to the circuitboard. When you turn the pot to zero, it turns off the battery (nice). It looks like this:

How might one accomplish this without complicating the control scheme?

The goal: One potentiometer controls the active amp circuit. One potentiometer controls the passive output. No other controls.

Here are some ideas I've had already...

1.) I could use a pan pot (commonly mistaken for a blender pot), and wire it in reverse, which I think would work. The problem is that if the pot is ever turned all the way toward the passive output, the active output won't work. That's inelegant, and might confuse my wife.

2.) I could use a push-pull switch. This really wouldn't be too difficult to do. I could set it to send the signal to one output when it's pulled up, and one when it's pushed down. But again, it's an inelegant solution. I can just see my wife calling me away from my work because her bass won't come on, and I have to pull the little knob up for her to make it work. She's not dumb, really. Highly intelligent, in fact. Just...I dunno. Girly? Not tech-savvy?

3.) I might be able to use what's referred to as a "true blender pot," such as the one featured here:

Acme Guitar Works 250K Blender Pot

The only problem with that is I can't find any good information on them, and can't figure out if they'd work for the purpose I intend.

4.) Just do like in the "diagram" and split the wire out in two different directions. I don't really think this would work though. I think the signal would find the path of least resistance, wherever that might be, and not do what I want it to do. Even if it did work, I'd have to turn up both volumes at once to make the amp work. Inelegant.

So if any of this makes sense to anyone, I'd sure appreciate some input. Anybody have any ideas?

Thanks,

David

[dpm99]

I think I may have answered my own question. It occurs to me that most of the devices I listed just interrupt an electrical connection. Can I just do this?

Or will it give me problems?

[borge]

having the signal going to the active circuit at all times will almost definately load down the output, whether it'll be a problem depends on your taste and the circuit.

I'd suggest using a switching output jack that'll dissconect the signal from the active circuit when a lead is plugged in. Assuming, of course, you won't want to use both the internal speaker and and external amp at the same time...

[dpm99]

having the signal going to the active circuit at all times will almost definately load down the output, whether it'll be a problem depends on your taste and the circuit.

I'd suggest using a switching output jack that'll dissconect the signal from the active circuit when a lead is plugged in. Assuming, of course, you won't want to use both the internal speaker and and external amp at the same time...

That's brilliant. I didn't know about those, but it seems simple enough from what I've read. I do this, right?

My understanding is that the volume pot wired to the circuit board clicks off the power when you turn it to zero. Wouldn't that effectively be doing the same thing, or is this different?

Thank you for your reply!

David

[borge]

It looks like your two diagrams are electrically identical,the only difference is the physical location of the batteries earth,it's earthed at the jack in the 2nd pic and the earth pad on the PCB in the first.

To disconnect the passive signal from the circuits input when the circuit isn't in use (ie when the bass is plugged into an amp) you need to break the connection between the left lug of the passive vol. pot. and the circuits input (it's a red wire) when a lead is plugged in.

Jack 'IX' in this link is what you need.

http://www.switchcraft.com/Documents/Jack_Schematics.pdf

wired:

top lug-left lug of vol. pot.

2nd lug- PCB input

3rd-passive output from centre lug of vol'pot'

4th-earth

this'll stop any loading of the passive output when plugged into an amp, but you still have the problem of the circuit being on at all times and draining the battery.

If you could get a hold of the circuit's schematic or the input impedance from the manufacturer you could get an idea of whether the loading will be an issue.

[dpm99]

Thanks. The amp's pot is one of those on/off volume pots. I spoke to the manufacturer about it, and he doesn't think it'll load down the passive circuit when it's clicked off. (That takes care of the battery problem too.) Nonetheless, I've decided to go ahead and use a push-pull for the passive volume, just in case there is a big difference in tone, even with that amp circuit turned off. I'll set it up where the passive line is always on, but the active line can be switched out of the circuit by pulling up on the volume knob. That way my wife can ignore it unless it's a necessity. I appreciate your advice and help with the switching output jack nonetheless.

David