Ohms at the jack

[JTE]

What is noticeable to my ear is the output from the guitar when the the 5 position selector switch is in position 4 and 2, in between two pickups. Also, when a Blender circuit is put in it is noticeable to my ear when the all three pickups are on at the same time that there is a drop in output.

So today I did a test using a device and an ohm meter to measure the resistance off the jack. What I found is:

With the vol. at 10 there is a considerable drop in resistance with the 5 way switch in the 4 and 2 position, almost half the resistance value found in either of the two pickups.

With the vol. at 8 through 1, that noticeable drop in resistance when in the 4 and 2 position, decresses as the vol. is decreased.

Regardless of what position the 5 way is in, the resisitance at the jack climbs heavly as I turn the vol. from 10 to 8. Then as I turn the vol. from 8 down to 1 the resistance decreases.

What I want or the outcome I'm looking for is to have the same output level regaurdless of what position the 5 way is in, no decernable drop in output in position 4 and 2. So if someone can help me understand what the cause is it would help me to fixing it.

 

[jimh]

Simple ohms law.

http://en.wikipedia.org/wiki/Ohm's_law

A bit of light reading for you!

 

[-CB-]

Ohms Law.  Like Murphy's Law and Coles Law.

Coles Law - finely ground cabbage and mayo, pepper to taste.
Murphy's Law - finely ground Murphy with mayo (or salad dressing since Murph was an Irishman)
Ohm's Law - finely ground resistors with mayo - no pepper needed.

Make sure you have a RW/RP neck pickup.  If its not, you get considerable drop in volume in 2 and 4.  The hum will also increase in those positions. 

The resistance changes you note, are normal, since when you have two pickups in parallel, the total DC resistance will be less.  If the pickups have the same DC resistance, the combined resistance will be half.  Ohm's law applies when working with DC resistance.  However, even as the resistance goes down you have double the "input" from the strings moving thru the magnetic field, and the output remains more or less the same, excepting for peaks and notches in the response when the two pickups are combined.  DC ohms is not a measure of output, or a measure of anything really, except the DC ohms of the wire in the pickup.  Its good as a relative gauge of output when A/B comparing two similar pickups, but thats about all.

 

[drewfx]

The short answers for combining resistors in
series: R = R1+R2
parallel: R = (R1*R2) / (R1+R2)  - this is all you're seeing.

drewfx

 

[blue313]

The most important note here that bears restating is that, the ohm reading on a pickup has little to do with anything.  It's relevant when combined with a specific gauge of wire, with a specific type of insulation, wound a specific way, and with the same exact magnetic forces involved.

A few Seymour Duncan's to illustrate the point.
SH-1 - '59 Model Humbucker - Bridge:8.13 k
SP90-1 - Vintage P-90 Soapbar  - Neck: 8.22 k
SSL-5 - Strat True Single Coil - Tapped: 6.6 k  Full: 12.9 k


But the only way I've been able to control the drop is to roll back my volume to 7 or 8 (or use a volume pedal).  Then I can cut or boost anything I need.

 

[ocguy106]

Not meant to be a shot at the OP but as I was reading this I was working ohm's law in my head and thinking....hhmmm sounds like perfectly fine to me.

 

[JTE]

Thank you, if nothing less, you've helped me frame my question better. The ohms law certainly appies here measuring the wraps around the magnets. Hense the higher the resistance the more wraps that are on the pickup. Then the total resistance of two pickups combined (a paralell circuit) equaling R1*R2 / R1+R2 or a drop in resistance. 

What I should be, and am now, measuring are the AC millivolts at the jack. And again a marked drop in when the two pickups are combined. Hense a marked drop in output of the guitar.

For further clarification this is the pickup array: neck - 5.5K single coil, middle - 9.8K humbucker, bridge - 9.3K P90. The millivolt output average with a normal strum is 65 mVolts neck, 122 mVolts middle, and 63 mVolts bridge. At the 4 and 2 position the millivolt output is in the 15 to 25 mVolts.

So there it is and what I don't understand is why the AC millivolt output doesn't remain realativly constant when combining two pickups. Is it that the drop in resistance when combining two pickups in paralell reduces the magnetic field generateting the AC millivolt output? And, how do I fix it?

 

[-CB-]

They're out of phase.  You're not RW/RP on the middle pickup - its humbucker already.  Reverse the wires on the neck pickup, and the P90.

Depending on where they're grounded, yuo might have to deal with that too, but that will preserve your volume

 

[JTE]

CB, That did it. All position on the five way working, The guitar is running on all cylinders now. Here's is link to the Guitar that you helped make right, http://jteguitarworks.com/NewsLetter.aspx   Thanks for your help.

For those who maybe reading this and don't know what RW/RP is (Reverse Wound, Reverse Polarity) here is a link that helped me understand it. http://www.deaf-eddie.net/rwrp.html. I guess I've either been lucky up to this point, and/or/also, simply working with sets of pickups rather than individual pickups; cause I never had this problem before. Live and learn, yea.

 

[-CB-]

YW!~

 

[Mor Paul]

Interesting headstock. I'd hate to drop it though.