Determining a pickup's DC resistance from the output jack

[antigua]

This is a copy paste from TDPRI, I wanted to also post it here to save the info for future reference.

Suppose you want to know DC resistance of a pickup without having to open the guitar up. You can select the pickup, with the volume at 10 and measure the DC resistance from the end of the guitar cable, but the DC resistance will read a little low, since some current is bypassing the pickup though the volume pot. That leakage around the pickup can be factored out with algebra in order to get an approximate DC resistance of the pickup. 

If you assume the volume pot is a true 250k resistance, you can approximate the pickup's own DC resistance with math:

pickup = ( measured_R * 250000 ) / ( -measured_R + 250000 )

so if the resistance measured 5.800k, it would become

pickup = ( 5800 * 250000 ) / ( -5800 + 250000 )
pickup = 1450000000 / 244200
pickup = 5937.755938
pickup = 5.94k ohms

If the vol pot was actually 200k, the pickup's resistance would estimate higher at 5.973k ohms, and if the volume pot was actually 300k, they pickup's resistance would be 5.914k ohms, but assuming the pot is within 10% of the nominal value, the calculation will be accurate to within a 100 ohms.

The tone pot's resistance isn't important since the series capacitor reads open. 

If the pots are 500k, of course you just use 500000 ohms in the math instead of 250000.

[newey]

Back in 2007, ChrisK's original post on this was entitled "Brain scanning through a nostril", later retitled as "Discerning Strat-type resistances" and a parallel thread "Discerning LP-type resistances". Both are from our Reference section, and linked in the thread here:

guitarnuts2.proboards.com/thread/3186/discerning-strat-type-resistances

[Yogi B]

Aug 17, 2021 22:30:08 GMT -5 newey said:

Back in 2007, ChrisK's original post on this was entitled "Brain scanning through a nostril", later retitled as "Discerning Strat-type resistances" and a parallel thread "Discerning LP-type resistances".

An important exception to the method described within Discerning LP type resistances (which JohnH notes in a follow up post, in the other thread) is that parallel treble bleeds (i.e. the tapering resistor, or more generally any 'clever' wiring mods) will throw the measurement of the volume pot + pickup series quartered resistance, and subsequently give incorrect results.

Additionally ChrisK's math in "Discerning LP-Type Resistances" is just a tiny bit non-straightforward for my liking: the measured quarter value, R_max, is unnecessarily quadrupled to become R_tot before being plugged into the formulae (any multiplicative factor, when introduced consistently, is ultimately cancelled out).

Thus, duplicating Chris's definitions, if we measure:

• R_max, the maximum resistance value during a full sweep of the volume pot (measured at around midway for a linear pot, or 80% maximum for a regular log pot) which is equal to the quarter the value of the series combination of the pickup & pot;
• and R_parallel, the value of the parallel combination of the pickup & pot (when at maximum volume)

Then the resistance values for the pot & pickup are the two solutions for R in the equation:R^2 - 4\cdot R_\text{max} \cdot R + 4 \cdot R_\text{max} \cdot R_\text{parallel} = 0
That is:R = 2 \cdot \left( R_\text{max} \pm \sqrt{R_\text{max} \cdot (R_\text{max} - R_\text{parallel})} \right)
Technically, as we arrive at a single equation with multiple solutions, without further real-world limitations being introduced we don't know which result is which. However, once such limitations are established we can confidently state that the larger of the two results is the value of the pot, thus leaving the remaining smaller result to be the DCR of the pickup.

[antigua]

Thanks for the link, I did a Google search on this to see if the math was already presented, but I found nothing, so I posted. I was hoping to keep it simple, too though. A lot of people just want the cliff notes. 

[newey]

Aug 18, 2021 12:28:21 GMT -5 antigua said:

I was hoping to keep it simple, too though.

Simple is always good. However, there is a bit of difference in the focus here, antigua, in that your post is simply to find the DC resistance of a given pickup without dismembering the guitar. ChrisK was addressing the method more as a first step in troubleshooting wiring issues. To that end, he included the part about rotating the tone controls to see if there was perhaps an issue there, whereas (as you note) a properly-functioning tone control should show open due to the cap.

[JohnH]

For GN2, ChrisK was the first to have the insightful realisation that this assessment is possible.

On this thread:

guitarnuts2.proboards.com/thread/3186/discerning-strat-type-resistances

I added a post at the end to step through what I think is the simplest way to do it.

[ms]

Simple is good, but simple is even better if it leads to a concept that make it easy to remember how to do it.

Resistance describes what a resistor does, but conductance describes it just as well, and each is just one divided by the other. Put resistors in parallel and their conductances add.  This is the explanation for  equation with the "one overs" that ChrisK gave in the thread that Newey referred to.

So here is a description in words of what Antigua did in the first post here:

We have the pickup resistance in parallel with the pot resistance.  So convert the measured resistance to a conductance.  Now convert the known pot resistance to a conductance.  The latter is added to the value we want so just subtract it from the measured conductance.  This give the conductance of the pickup. Convert that to a resistance and you are done.

In [66]: 1./(1./5800. - 1./250000.)
Out[66]: 5937.755937755938

[antigua]

Aug 20, 2021 7:34:05 GMT -5 ms said:

Simple is good, but simple is even better if it leads to a concept that make it easy to remember how to do it.

Resistance describes what a resistor does, but conductance describes it just as well, and each is just one divided by the other. Put resistors in parallel and their conductances add.  This is the explanation for  equation with the "one overs" that ChrisK gave in the thread that Newey referred to.

So here is a description in words of what Antigua did in the first post here:

We have the pickup resistance in parallel with the pot resistance.  So convert the measured resistance to a conductance.  Now convert the known pot resistance to a conductance.  The latter is added to the value we want so just subtract it from the measured conductance.  This give the conductance of the pickup. Convert that to a resistance and you are done.

In [66]: 1./(1./5800. - 1./250000.)
Out[66]: 5937.755937755938

That's pretty simple, probably easier to remember too. 

pickup = 1 / ( 1 / measured_R ) - ( 1 / 250000 ) )

pickup = 1 / ( 1 / 5800 ) - ( 1 / 250000 ) )
pickup = 1 / (  0.000172413  - 0.000004 )
pickup = 1 / 0.000168413
pickup = 5937.76
pickup = 5.94k ohms


I'd include the unnecessary parenthesis so that a person doesn't have to even think about the order of operations, but I also think they help make it easier to visually see the relationship between the values. 

[JohnH]

All good if you know the pot. But if your not sure, or want to measure exactly what it is as well as measure the unkown pickup, the method I set out will let you quickly home in on both unknowns with minimal maths.

[ms]

It's good that there is more than one way to look at it.  One way: Assume that you do not know the pot value,  (Rpot = 1/Kpot), K for conductance.  Then we have switch positions to determine Kn + Km + Kpot and Km + Kpot.  Subtract and you get Kn.  The others are found the same way from other switch positions.

[ashcatlt]

“Brain Surgery Through a Nostril”