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Silly me, doing half a job as usual.Beat me to it.
Yeah, but I disagree with you about how much difference it makes.Do you guys understand what I am talking about now that I posted a diagram?
Yeah, but I disagree with you about how much difference it makes.
Go down to RadioShack and pick up a couple of 250K (or 500K if that's what's in your P bass) resistors. It should only cost a buck or two.
Solder one of them across terminals one and three of the volume pot on your P-bass. (The P, not the PJ)
If you can't solder, then just wrap the lead around real tight and make sure it's making good contact for the test.
I was curious about this one, and actually performed this experiment. I wired in a 250k resistor between the output and ground, and hooked that up to mini toggle switch, so I could quickly switch the resistor in and out of the circuit.
The affect of the extra resistive loading WAS audible, but just hardly. The volume was slightly decreased, and the treble was VERY slightly attenuated. If I hadn't been able to flick that switch back and forth I never would have been able to tell the difference.
My verdict? I spent way too much time on this pointless experiment.![]()
Alright Jimmy, I'll give you the challenge then.
Go down to RadioShack and pick up a couple of 250K (or 500K if that's what's in your P bass) resistors. It should only cost a buck or two.
Solder one of them across terminals one and three of the volume pot on your P-bass. (The P, not the PJ)
If you can't solder, then just wrap the lead around real tight and make sure it's making good contact for the test.
Now play your bass and tell me if it sounds different to you.
Do you notice any less output?
Were the volume and tone controls on full?
Were you using an amp, or did you go direct?
I play my bass direct through studio monitors, and I hear a big difference.
I moved the volume and tone controls all over the place, and it behaved just as the laws of physics demand. A small decrease in volume and treble.
So you've actually listened to the same instrument both with and without extra resistive loading?
Okay, just been doing some sums.
Suppose the P pickup induces a current of 1 amp (I know that's silly, but it's just a figure for modelling the effect - I could have picked 100A or 1mA and the following ratios would still hold true).
Right, think of a traditional P pickup setup here. I just checked the resistance of a pickup and it's in the order 8-12K. If the volume for the P pickup was maxed, and the connections are pretty much zero ohms, then all of the 1A would get to the output. Correct?
Now, let's conect the J into the circuit. It's true that some of the signal from the P will find its way to ground through letters D to C (in my diagram) in that case and this will reduce output. But if the resistance of the pot is 250K (J pickup on zero), the current going through that branch will be split in a way that depends on the resistance of the two branches of the circuit. I think about 10/250 of the current will find it's way through that branch, if I understand correctly (or thereabouts). That means instead of 1A flowing to the output, 960mA will make it. Is that correct?
If it is, I'd be quite surprised if the difference was really audible.
Current is proportional to voltage, though. Yes?We are dealing with a drop in voltage more than a drop in current.
Also, with the P maxed, it doesn't go straight through per se, there is a 250K ohm resistance to ground.
Current is proportional to voltage, though. Yes?
Understood about the 250K. Pretty high compared to the internal resistance of the pickup. And what you say would, of course, apply even if you took the J pickup out o the circuit completely.