bypass 2 resistors.. with one switch?

[effectsbay]

Hello

I'm interested in bypassing 2 resistors (different parts of the circuit and not directly connected) with one toggle switch. I thought I was going to be able to do this with a 3PDT toggle I have, but running into a wall. I can see how to do it with a 4PDT, but wanted to see what other ideas/options there are to accomplish this.

Thanks!
hank

[nocentelli]

Bypass = short?  Can you not use two poles of a dpdt toggle, each pole shorts a different resistor.

[effectsbay]

Bypass = short?  Can you not use two poles of a dpdt toggle, each pole shorts a different resistor.

Nope.. needs to bypass the resistor (like running a wire instead of a resistor). I want to do the emitter mod on a Big Muff NYC, which involves removing R12 and R16, but I want to put them back with a single toggle throw.

http://rkerkhof.ruhosting.nl/Taas/Mods/Big%20Muff.htm

Thanks!
hank

[nocentelli]

Same thing

[Mike Burgundy]

without a schematic in hand, lets see if I interpret this right:
situation normal, both resistors in place.
situation bypass, both resistors replaced by a wire (short)
Right?
okay, look at the first resistor. draw that wire parallel to it, joining the circuit at both ends of the resistor. Wire is basically a resistor with value 0, so you have R and 0 in parallel, equals 0.
Now cut the wire, insert one SPST switch (middle and one outer connector of one half DPDT or one third 3PDT) in that wire.
Switch closed: 0Ohm.
Switch open, parallel wire is broken and "not there" total resistance is R.

Do the same with the other half of the DPDT (or 1/3 3PDT) for the other resistor. Use the last 3rd of a 3PDT to switch a LED.

[effectsbay]

Okay.. I think I'm understanding. Check out this drawing. Am I close?



Thanks!
hank

[nocentelli]

You just need to ground the emitters: connect both middle lugs to ground, connect the upper left lug to Q2 emitter, upper right lug to Q3 emitter and leave both lower lugs unconnected. Leave the resistors in place; When the switch connects the ground direct to both emitters, the resistors are bypassed/shorted. When the switch flips the other way, the emitter resistors are the only path to ground.

[effectsbay]

You just need to ground the emitters: connect both middle lugs to ground, connect the upper left lug to Q2 emitter, upper right lug to Q3 emitter and leave both lower lugs unconnected. Leave the resistors in place; When the switch connects the ground direct to both emitters, the resistors are bypassed/shorted. When the switch flips the other way, the emitter resistors are the only path to ground.

Thanks. I almost have it. I don't understand when you say - "When the switch flips the other way, the emitter resistors are the only path to ground"

Should I just solder a wire from the emitter leg to the switch? Is that how the resistor comes into play when the switch is off, since the trace is still connected to the resistor? The switch will short it to ground?

Thanks!
hank

[mth5044]

It looks like your drawing would work, as long as each of the bottom lugs is a connection to ground. It's not very explicit.

[nocentelli]

Should I just solder a wire from the emitter leg to the switch? Is that how the resistor comes into play when the switch is off, since the trace is still connected to the resistor? The switch will short it to ground?

Yes. Since you are connecting both emitters to the same point (ground) you could just use a SPST switch with one lug connected to ground and the other to both emitters.

[Mike Burgundy]

Please, please make sure you understand what the components *do* and how that translates into a schematic.
I have a very strong suspicion building stuff from layout diagrams severely hinders understanding of *what* exactly is going on and what should go where.
Take a look at this:


See the square block on the left? That's a resistor. Also sometimes represented by a wiggly line.  Over that is a switch. Ignore the bit on the right, I just needed the left bit of the picture.
This is *exactly* what we are talking about, for ONE resistor and HALF your switch.
Think about what terminals on the switch you need to use.
A DPDT (Double Pole Double Trow - that means TWO switches in one housing, both selecting between TWO options) is two SPDT's (Single pole double throw, ONE switch selecting....right) and has its middle pole alternating between the outer poles.
The switch in the drawing above is a Single Pole Single Throw - it chooses to connect to one pouter terminal, or nothing at all. In other words, the switch is ON or OFF. Note: this is exactly the same as a SPDT, with nothing connected to one outer terminal. Thats it, ignore one terminal and you have made an SPST out of an SPDT.
Look at your switch layout, and think about where the connections go - in both switch states. Wait, hold on - I get it now - I was reading it wrong. I tend to see a line as wire, even when it's called "R12". Woops, apologies.
Okay - that should work, but may not be the best universal way to do it, and youre leaving the resistors floating. I'd parallel them with the short. You can even leave the original R's in-circuit that way.