Why cant a PJ sound like a real P

[line6man]

What sort of wiring are we talking about here? Got a diagram?

If you really want a diagram, I'll draw one, but it's just VVT versus VT.

 

[steamthief]

Mine sounds like a P, a/b'ed with my P-bass. OP, everyone has an opinion. Use your own ears and decide.

 

[JTE]

My experience with VVT P/J and stock P is that when you solo the P PUP, it's indistinguishable from a regular P bass. Unless you're comparing other variables. Like comparing a US VS '62 with an alder body to a MIM P Special with the active EQ. Or maybe listening with your eyes?

A California P Bass Special compared to a Fullerton era VS '62 Precison, recorded with the same rig (Eden WT-400 and a Hartke 210XLT) at the same session couldn't be told apart.

The loading from the second pot might be noticable, depending on the rig, but I'd be hesitant to pin it all on the extra wiring because there's so many variable between two seemingly "identical" basses.

John

 

[JimmyM]

I think this business about the reduced volume is more about spec reading than real world usage. Most of my basses are cheap, with 250k pots, and I can't tell the difference volume-wise.

 

[BassyBill]

If you really want a diagram, I'll draw one, but it's just VVT versus VT.

Yes, please. Or links, maybe?

I can't see how the extra pot will decrease resistance and this will decrease output. From an electrical point of view, I mean.

In a standard VVT arrangement, the extra pot will not even be "seen" by the first pickup at all, as far as I can work out. If the P pickup is on 10 (or whatever), then the position of the J will make no difference at all to the electrical path of the first pickup's signal. If the second pickup is then turned to zero, the path of the signal for the first will be exactly the same as if the second pickup isn't there (as they're wired in parallel). I really can't see how the extra pot affects the signal from the P pickup in any way.

Please correct me if I'm wrong.

 

[line6man]

Yes, please. Or links, maybe?

I can't see how the extra pot will decrease resistance and this will decrease output. From an electrical point of view, I mean.

In a standard VVT arrangement, the extra pot will not even be "seen" by the first pickup at all, as far as I can work out. If the P pickup is on 10 (or whatever), then the position of the J will make no difference at all to the electrical path of the first pickup's signal. If the second pickup is then turned to zero, the path of the signal for the first will be exactly the same as if the second pickup isn't there (as they're wired in parallel). I really can't see how the extra pot affects the signal from the P pickup in any way.

Please correct me if I'm wrong.

Ok say you have a volume pot wired with the first terminal as an output, the second terminal as an input and the third terminal as a ground.
Regardless of the volume setting, you are always placing a resistance equal to the value of the pot from the output to ground.

Now you add in a second pot, which gives you a second resistor across from signal to ground...
When you place resistors in parallel, their resistance decreases.

So say you had a 250K pot in your P bass. That's 250K ohms from signal to ground. Now you add in a second 250K pot, and that is now two 250K ohm resistors from signal to ground, which drops the resistance to 125K ohms.

 

[JimmyM]

I see what you're talking about. You're talking about how when you have a P and J wired together that the output drops a little when both pickups are on full. That does happen, no doubt. But when the J pickup is off, the P's output comes back to normal.

 

[steamthief]

Or even when you roll the J off slightly, say to 90%

 

[BassyBill]

Ok say you have a volume pot wired with the first terminal as an output, the second terminal as an input and the third terminal as a ground.
Regardless of the volume setting, you are always placing a resistance equal to the value of the pot from the output to ground.

Now you add in a second pot, which gives you a second resistor across from signal to ground...
When you place resistors in parallel, their resistance decreases.

So say you had a 250K pot in your P bass. That's 250K ohms from signal to ground. Now you add in a second 250K pot, and that is now two 250K ohm resistors from signal to ground, which drops the resistance to 125K ohms.

I'm sort of following this and you obviously understand something I'm missing here. Surely when the pot is turned up full, the signal goes straight through?

Like I said, I'd get this better with a wiring diagram - or better still, a proper circuit diagram (or schematic, I think you call them over there).

 

[Darkstrike]

I see what you're talking about. You're talking about how when you have a P and J wired together that the output drops a little when both pickups are on full. That does happen, no doubt. But when the J pickup is off, the P's output comes back to normal.

The resistance still changes, even when the volume of the J is off, same as taking a P bass, and removing the tone pot completly sounds a little louder than the same bass with a tone pot installed and at 10.

Its not a big deal, IMHO, but it does exist, and a three way switch rather than seperate volumes negates it, as there is no extra resistance.

 

[Cyber Soda]

I think I can pull off a pretty good P tone while soloing the P of my P/J, which it is a Spector... but I don't know. Why not try to hear the difference for yourself?

 

[line6man]

I'm sort of following this and you obviously understand something I'm missing here. Surely when the pot is turned up full, the signal goes straight through?

Like I said, I'd get this better with a wiring diagram - or better still, a proper circuit diagram (or schematic, I think you call them over there).

Yes, the signal passes straight through, but you are also placing the full resistance of the pot from that signal to ground, which causes you to lose some output.

When you put in a second pot, that is a second resistor going from the signal to ground, which decreases the path to ground, causing you to lose more output.

I'll draw you a diagram in a few minutes...

 

[Darkstrike]

I'm sort of following this and you obviously understand something I'm missing here. Surely when the pot is turned up full, the signal goes straight through?

Like I said, I'd get this better with a wiring diagram - or better still, a proper circuit diagram (or schematic, I think you call them over there).

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[dsteinschneider]

If you like the J bass leave it as it and save up for a pbass. I recommend mid 90's mexican pbasses. I have a 72 ash body, maple neck pbass and a mid 90's mexican standard pbass with a rosewood neck. The newer bass is just as nice to play.

 

[JimmyM]

Well regardless, I hear little or no difference at all in output when I switch between a P and a PJ with the J off. Maybe others do, but I don't.

 

[BassyBill]

Right, we seem to be getting somewhere. Using the above diagram, I've sketched a crude schematic. Does this look correct? Please forgive the rough nature of it and ignore the letters in circles for now.

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[line6man]

 

[Passinwind]

Right, we seem to be getting somewhere. Using the above diagram, I've sketched a crude schematic. Does this look correct? Please forgive the rough nature of it and ignore the letters in circles for now.

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Nope. One lug of each pot goes to ground.

Edit: but the one on the tone pot does so through the cap, as you drew it.

 

[line6man]

Right, we seem to be getting somewhere. Using the above diagram, I've sketched a crude schematic. Does this look correct? Please forgive the rough nature of it and ignore the letters in circles for now.

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You are missing the connection to ground on the volumes...

 

[line6man]

Nope. One lug of each pot goes to ground.

Beat me to it.