how do I wire in a resistor to replace a wide open pot?

I've searched all over and cannot find an answer for this. I have a circuit where the volume and tone pots are always set wide open full volume and treble. I need to replace these two pots with resistors or (or whatever mini circuit is required).

How can I do this? I know that a guitar with tone/volume at 10 is not the same sound as a pickup gong straight to amp but I'm not sure why?

I've searched all over and cannot find an answer for this. I have a circuit where the volume and tone pots are always set wide open full volume and treble. I need to replace these two pots with resistors or (or whatever mini circuit is required).

How can I do this? I know that a guitar with tone/volume at 10 is not the same sound as a pickup gong straight to amp but I'm not sure why?

In my experience the effect of having no pots is noticeable but not that dramatic. If you want to bypass a volume pot just jumper from the wire coming in directly to the wire going out. If you want to bypass a tone cap just clip the cap out of the circuit.

The volume pot can be replaced by a 250k resistor to ground. Maybe that's not a common value. The tone pot can be replaced by a (250k + .022mfd) series configuration to ground. I don't know if its a noticeable difference to just a 125k resistor to ground, no cap. I don't know if it's a noticeable difference to no cnxn to ground at all. All example values

I would venture a guess, that it alters the load impedance of the pickups ... a minuscule amt.

You can simulate a volume at 99% eg by a 1k resistor in series with the signal before the above configuration. You would connect the tone circuit to either side of that small resistor. You can choose between modern wiring and the fezz parka mod :lol:

Hope this helps have fun

THank you...I'm specifically trying to wire a jaguar from scratch but I want a rhythm circuit without pots but with the same sound as when the pots are turned up.

I understand your instructions but I'm a little confused in how to relate it to a jag wiring diagaram, there's that third connection between lug 1 on the volume and lug 2 on the tone...if I'm wiring in series that cannot exist in the new circuit? Is that ok? I'm not sure what the purpose of that connection is in a guitar circuit.

http://www.jag-stang.com/wp-content/uploads/2009/10/jaguarSchematic.gif

...

THank you...I'm specifically trying to wire a jaguar from scratch but I want a rhythm circuit without pots but with the same sound as when the pots are turned up.

I understand your instructions but I'm a little confused in how to relate it to a jag wiring diagaram, there's that third connection between lug 1 on the volume and lug 2 on the tone...if I'm wiring in series that cannot exist in the new circuit? Is that ok? I'm not sure what the purpose of that connection is in a guitar circuit.



Now I follow you. About all you could do is set the rhythm circuit of a Jaguar to where they sound good to you and measure the resistance across the pots then replicate that circuit in fixed value resistors.

That's a good idea, I hadn't thought of just measuring. What hung me up was that I thought turning them all the way up was supposed to equal no resistance. Thanks

That's a good idea, I hadn't thought of just measuring. What hung me up was that I thought turning them all the way up was supposed to equal no resistance. Thanks

Actually, it's just the opposite. Turning them up all the way has the greatest amount of resistance between the signal and ground. So, if you want to simulate the effect of a wide open volume pot you could just place a resistor of the same value as the pot between the signal and ground.

It seems that most of the advice here reflects a misunderstanding of how a potentiometer wired as a volume control works. If you want the effect of having the volume pot turned all the way up to 10, you want no resistor at all.

A volume control is not the same as a variable resistor. Try it sometime.

It seems that most of the advice here reflects a misunderstanding of how a potentiometer wired as a volume control works. If you want the effect of having the volume pot turned all the way up to 10, you want no resistor at all.

A volume control is not the same as a variable resistor. Try it sometime.

This is my understanding of how a potentiometer works.

http://cnx.org/content/m13778/latest/Graphic1.png

This is my understanding of how a potentiometer works.

That's right. In the case of a guitar volume control, V2 would be grounded, V1 would connect to the pickups, (via whatever switching is used), and Vout would go to the output jack (and would have the tone control hanging off of it too). So, in this case, max volume equals max resistance, or 250K. So, a 240K or 270K fixed resistor would be a good choice.

The tone control works slightly differently. The signal from the volume control connects to V2, and the cap connects to ground at one end, and Vout at the other. So, as the pot is turned toward the min resistance position, a ton of highs and mids are removed, making the tone bassy. As the tone control is moved toward the high resistance position, much less high frequency stuff is lost, so it gets very bright. So, it the desired tone is on the bassy side, replace the pot with a low resistance, say 220Ω or less. If you want the very bright tone, use the 240K.

It seems that most of the advice here reflects a misunderstanding of how a potentiometer wired as a volume control works. If you want the effect of having the volume pot turned all the way up to 10, you want no resistor at all.
Not true. You're forgetting Thevenin's theorem. Your statement would be true if the signal source (in this case, the guitar pickups) had zero source impedance. But in fact pickups do have substantial impedance, particularly at higher frequencies, and therefore the potentiometer introduces some loading on the pickup even at full volume.

Therefore the way to simulate a pot at full volume is to replace it with a resistor of the same value as the end-to-end resistance of the pot, exactly as waparker4 said.

Tantamo, I'm not familiar with Jaguar wiring - which two of the four pots in that diagram do you want to replace with fixed resistors? (1M and 50K on the left, 1M and 1M on the right ???)

-Gnobuddy

The 1m and 50k on the left. So the wire connecting the lugs 1 and lugs 2 between these two pots is irrelevant for what I want to do?

I ended up buying tiny trimpots anyway and I will just recreate the circuit in full after all (the guitar I'm putting together didn't have room for the real rhythm circuit).

The 1m and 50k on the left.
If you had decided to replace these two with fixed resistors, what you needed to do is:

1) For each pot, connect together the two wires that ran to the middle terminal and the top ("hot") terminal.

2) Run a resistor from those two joined wires to ground. 1 Meg to replace the upper pot, 47 k to replace the lower pot.

I ended up buying tiny trimpots anyway and I will just recreate the circuit in full after all (the guitar I'm putting together didn't have room for the real rhythm circuit).
That is a nice solution! Now if you decide you don't want the pots at 100% but rather at 90% you can go in and tweak your trimpots to taste, and then leave them at that setting.

-Gnobuddy