DC Resistance of Combined Humbuckers

BillDL

I have a Squier Affinity STARcaster that has the stock open coil humbuckers fitted.  The bridge measures 13.4 kOhms DC Resistance and the neck 8.3 kOhms.  I was quite surprised by the difference, but they sound great and balance out quite well.  What surprised me more was that with the 3-way open toggle switch in the middle position it reads only 5.2 kOhms.  I expected it to read around 10.

Despite this I don't get much, if any, volume drop when both pickups are selected.

I don't have any twin humbucker guitars with this much DC Resisance difference between them, but my understanding was that on a normally wired twin humbucker guitar it would measure rougly half of the combined resistance when both are selected.  In this case 21.7 kOms / 2 = 10.87.  I am measuring using a disassembled ordinary guitar jack into the output socket and just probing the hot and ground connections.  I haven't looked inside the body to see if the pots are wired in some funny way.

Is this extreme drop in DC Resistance expected and normal?

If not, what would be the most likely explanation?

I really like the range of sounds on this guitar and the volume and tone pots work really well, so I don't intend swapping the pickups or changing any of the electrics.  I'm just curious because I've never had as much of a mismatch as this before.

Funkfingers

Ah, the joys of electro-magnetism.

Resistances connected in parallel = combined total divided by four. 

BillDL

Thank you for that @Funkfingers

Now that you've explained this I vaguely recall somebody showing how to read the resistance of volume pots without opening the control cavity.  From memory he selected one pickup, probed the output socket, then turned the associated volume put until he saw the resistance peak.  I think he then multipled that figure by 4 and deducted the pickup's resistance value from the total.

prowla

Resistors in series: R = R1 + R2

Resistors in parallel: 1/R = 1/R1 + 1/R2

R1 = 13.4

R2 = 8.3

1/R1 = 0.074626866

1/R2 = 0.120481928

1/R1 + 1/R2 = 0.195108794

1/R = 0.195108794

R = 1/0.195108794

R = 5.125345606

BillDL

Thank you @prowla.

Normally I have a mental block when I see anything that vaguely resembles a formula, but that one's easy enough to follow.  I was terrible at maths, algebra, fractions, trigonometry, etc at school and it has left me with a bit of a void when it comes to some calculations.  For me counting down the score in a darts game of 301 or 501 while being impatiently watched by others, or trying to work out what double I need to finish while the other player is tapping his foot and rolling his eyes, is quite challenging.  I suppose you could say that I know what needs to connect with what in a control cavity and what changing the connections results in, but I don't fully understand the formulas used that allow you to arrive at certain figures.

ICBM

prowla said:

Resistors in series: R = R1 + R2

Resistors in parallel: 1/R = 1/R1 + 1/R2

This is the correct formula .

BillDL said:

Now that you've explained this I vaguely recall somebody showing how to read the resistance of volume pots without opening the control cavity.  From memory he selected one pickup, probed the output socket, then turned the associated volume put until he saw the resistance peak.  I think he then multipled that figure by 4 and deducted the pickup's resistance value from the total.

Yes, that’s correct - because when the pot is at the resistance peak, effectively you have half the total resistance value of the pot plus pickup on each side, in parallel - so to find the total, multiply by four.

For the pickup resistance it’s simpler - just turn the pot up full. You have to then deduct the pot resistance using the formula prowla gave for a really accurate value, but the error is quite small because the pot value is usually around 50 times the pickup resistance.

BillDL said:

The bridge measures 13.4 kOhms DC Resistance and the neck 8.3 kOhms.  I was quite surprised by the difference, but they sound great and balance out quite well.

I really like the range of sounds on this guitar and the volume and tone pots work really well, so I don't intend swapping the pickups or changing any of the electrics.  I'm just curious because I've never had as much of a mismatch as this before.

The ‘mismatch’ is more likely because the bridge pickup is wound with finer wire, which has higher resistance. DC resistance is at best a fairly poor guide to output, and then only because it gives a rough guide to the number of turns *assuming similar wire gauge*. The actual output is determined by the number of turns and the strength of the magnet. It is true that you need finer wire in order to get more turns on, so the higher DC resistance usually indicates both, rather than just a lot more turns - but each is roughly equally important, so a pickup with twice the resistance will most likely have only around 50% more output.

On top of that, the human ear is very non-linear, and 50% more output *voltage* does not mean 50% louder - it’s actually only about 1-2dB. This will usually produce a reasonably good volume balance between the two pickups, because the string excursion is much less near the bridge than near the middle of the string.

BillDL

I remember you writing in another thread about the fact that a pickup's output strength cannot be measured by DC Resistance alone, and is is only a very rough indication of likely output strength based on common wire gauges and bobbin sizes.  You spoke about inductance, gauge of wire and number of winds being a large influence on the actual output strength, after which I went and looked up inductance.  The following blog page was as helpful as it could get for an electronics dunderheid like me, but the embedded video on that page went somewhat over my bald head.

https://www.seymourduncan.com/blog/latest-updates/inductance-what-it-is-and-why-it-matters

The most I've really had to consider the cross sectional area of wire was in working out whether the new cable I had to run for a higher kW electric shower would exceed 25m and would have to be thicker cable.

Funkfingers

BillDL said:

I vaguely recall somebody showing how to read the resistance of volume pots without opening the control cavity.  From memory he selected one pickup, probed the output socket, then turned the associated volume put until he saw the resistance peak.  I think he then multipled that figure by 4 and deducted the pickup's resistance value from the total.

That was probably @ICBM

BillDL

It most likely was.  If somebody trawled through all the threads of this forum snatching all the technical answers and suggestions made by the highly knowledgeable stalwarts and grouped them together it would make an expectionally comprehensive and in-depth guide to hardware, electronics and build/repair of guitars and basses.

Funkfingers

If anybody performed a word frequency analysis on this forum, the top returns would probably be:
Why is everything so expensive?
R8
Harley-Benton
It needs a Bigsby
Chappers*

* Auto spell checker wanted to change this to Choppers. Interesting.

PrecisionGuided

For two resistances in parallel the resulting resistance will never be higher than either resistor.
You've got your 8.3K path and also the parallel path via the 13.4K reducing the total a bit more.
Product over sum (8.3 x 13.4)  / (8.3 + 13.4) gives 112.22 / 21.7 = 5.17K

Dave_Mc

Funkfingers said:

If anybody performed a word frequency analysis on this forum, the top returns would probably be:
Why is everything so expensive?
R8
Harley-Benton
It needs a Bigsby
Chappers*

* Auto spell checker wanted to change this to Choppers. Interesting.

You really need "with a pickup swap" to come directly under "Harley Benton" in terms of the frequency! 

EDIT: And then under that, probably "as good as a custom shop Gibson"...

BillDL

One of the very frequently used phrases I see on most guitar related discussions is the nonchalant act of "throwing in" some different pickups, for example "After I throw in a set of Seymour Duncans it will play like a Gibson".

I've just looked back at my original post and slapped my forehead.  I have no idea why I typed that I expected the combined DC resistance to be around 10K.  I didn't.  I did expect it to be quite a bit higher than what it is and closer to 8K, so I don't know how I got 10K into my head.  Obviously now that this has all been explained to me it makes absolute sense.