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>>And here is why:

Rtotal = R1 x R2 / (R1 + R2)

If we have a 7k neck and a 9k bridge pickup on their own they are 7k and 9k respectively. Together they are 7 x 9 / 7 + 9 = 63/16 = 3.93K

Hi, noob here, thought I'd introduce myself by starting an argument...

I don't agree that connecting 2 pickups in parallel (in phase) gives you half the output... although we discuss output in terms of resistance as above, that is mostly because the resistance is easy to measure, and
more resistance = more turns = more output.
So suppose you have two pickups side by side, but not wired together. One end of each coil is at ground. The other end of each coil has some voltage on it, which is generated by the moving strings... and if they were identical coils etc, it would be THE SAME voltage on both. SO now connect the two in parallel by adding a wire between them. No current flows in that wire because the things it connects were already at the same voltage. So the output stays the same--- because they are BOTH generating voltage, neither pickup loads the other one.

If you imagine some funny arrangement where one pickup is NOT under the strings, then it would just act as a load on the other one and you WOULD get half the output...

PeterS wrote:>>And here is why:

Rtotal = R1 x R2 / (R1 + R2)

If we have a 7k neck and a 9k bridge pickup on their own they are 7k and 9k respectively. Together they are 7 x 9 / 7 + 9 = 63/16 = 3.93K

Hi, noob here, thought I'd introduce myself by starting an argument...

I don't agree that connecting 2 pickups in parallel (in phase) gives you half the output... although we discuss output in terms of resistance as above, that is mostly because the resistance is easy to measure, and
more resistance = more turns = more output.
So suppose you have two pickups side by side, but not wired together. One end of each coil is at ground. The other end of each coil has some voltage on it, which is generated by the moving strings... and if they were identical coils etc, it would be THE SAME voltage on both. SO now connect the two in parallel by adding a wire between them. No current flows in that wire because the things it connects were already at the same voltage. So the output stays the same--- because they are BOTH generating voltage, neither pickup loads the other one.

If you imagine some funny arrangement where one pickup is NOT under the strings, then it would just act as a load on the other one and you WOULD get half the output...

Erm, dude? Less pickup resistance leads to it getting loaded down quicker by the AMP. The input of the amp applies the load to the coils. Less resistance means easier to get loaded down. Hence the whole "Hi" and "Low" inputs on some amps.

EDIT:

I understand where you are coming from on this and why you might think what you think, but you are leaving out the other end of equation. Higher pickup resistance correlates to higher impedance, and higher voltage pushed into the input stage of the amp. Too much impedance and your frequency response suffers.

But honestly, you could just read some of the very informative documents over at seymour duncan.

Less pickup resistance leads to it getting loaded down quicker by the AMP. The input of the amp applies the load to the coils. Less resistance means easier to get loaded down

Um, that's backwards.

I think you're getting confused about how voltage really exists as a physical quantity. It needs a resistance to exist as a potential difference to ground, obviously the ultimate resistance is infinity where you get maximum coupling of the energy, but we don't have infinity resistance in our pickups, it is finite and defined (as impedance, a frequency dependent quantity or reactance, I know, but still real and finite), and when we combine them that resistance is affected by that combination. Hence the potential difference at the combination of the pickup is affected by the combination of the pickups.

You'll also note in a subsequent post I detail that it is not half the output (at least volumewise), and the model is a simplistic explanation at best, being as the impedance is variable across the frequency of the guitar signal.

But hey, if you don't like the science, just flip that toggle.