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Thread: Measuring pickups at the input Jack, how much off?

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    Measuring pickups at the input Jack, how much off?

    Hi,
    you all know the situation when you're curious about a pickup that you just can't pull out of a guitar. Such as a vintage guitar or whatever reason. I wonder how much the readings will be off compared to a measurement that you do directly at the pickup leads.
    What is your experience?

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    ToneOholic! big_teee's Avatar
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    Since the Volume pot is loaded to ground?
    It would depend on the size of your volume pot.
    If you look at this schematic?
    Click image for larger version. 

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    Pickup in parallel with volume pot = ohms at jack.
    T

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    It's only Rock and Roll, but I like it!

    Terry

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    HI
    It depends many things. What are you measuring?
    Pot value, how many pickups, etc.
    Pu's resistance ? One PU, one 220 kΩ Vol pot, the meter shows about 5% lower than the PU itself. If I remember me right 1 PU+1 vol inductance meter shows kind of near right value.

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    Well in this current situation I'm using a Multimeter to measure DC resistance of one HB pickup that is installed in a stck. Gibson 335. I assume it is loaded w. 500 KOhm. I'm planning to also use my Extech to figure out other numbers like Impedance and Q factor. I'm afraid I didn't quite understood big tees answer. Still confused

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    ToneOholic! big_teee's Avatar
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    a neck humbucker with the pot dimed, should read 7k or so.
    Maybe more, maybe less.
    If two pickup guitar, flip the switch back and forth and compare them both.
    T

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    What I was trying to find out is how far off it is from a direct reading. I guess I'll have to make a test with one of my own guitars to know exactly how much it differs.

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    ToneOholic! big_teee's Avatar
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    I just tested my left hand les paul.
    I do low wind neck pickups 7.2-7.3k.
    It read 6.95k in a cool room, 500k Pot.
    7300-5%=6935
    So around 5% sounds about right.
    T

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    Cool info. Did you also compare other variables like Impdance? Just wondering if the 5% rule applies to this figure too

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    I'm off now. Happy New Year everybody

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    ToneOholic! big_teee's Avatar
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    Pretty primitive here, all I have is a DVM, and a gauss meter.
    The bridge measured arount 12500.
    It is wound full with 43, so it should read 13.2-13.5k.
    But, like I said this is in a cool 65F degree room.
    GL,
    T

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    It's only Rock and Roll, but I like it!

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    Old Timer J M Fahey's Avatar
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    Like Big Tee said, itīs quite straightforward
    You have 2 resistors in parallel: Pot end to end resistance, which you know, usually 250k or 500k, also realize thereīs some tolerance, a pot can easily vary as much as 20% , in parallel with coil DC resistance.

    So: Pot resistance in parallel with pickup DC R = resistance measured at jack.

    Knowing 2 values, you can calculate the third.

    But ... but ... I donīt know the exact pot resistance ... it may vary by up to 20% !!!!

    No big deal: since pot has 20 to 50 times higher resistance than pickup DCR, its error, in this parallel connection, becomes 20 to 50 times less important

    Which makes this assumption quite valid:just measure, deduct 5% and you are done.

    Inductance will be close too, what youīll find is somewhat lower Q than expected, because the pot resistance is "lossy".

    But Q is also negatively affected by wire DCR and you can do nothing about that, so ... donīt worry be happy

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    Juan Manuel Fahey

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    But Q is also negatively affected by wire DCR and you can do nothing about that, so ... donīt worry be happy
    The negative effect on resonance Q by a parallel resistance 100k to 300k is much greater than a couple of k series wire resistance..

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    Quote Originally Posted by J M Fahey View Post
    So: Pot resistance in parallel with pickup DC R = resistance measured at jack.

    Knowing 2 values, you can calculate the third.
    Just to elaborate, the resistance measured at the jack is:
    Rjack = (Rpup x Rvol) / (Rpup + Rvol)

    Solving for Rpup (assuming I remember high school algebra):
    Rpup = Rjack / (1 - Rjack/Rvol)

    Note that as Rvol gets larger, Rjack gets closer to Rpup.

    Plugging in some typical values:
    Say you have an 8K humbucker loaded with a 500K volume pot which is "off" by -20%
    Rjack = (8K x 400K) / (8K + 400K) = 7.84K
    Rjack differs from Rpup by (7.84K - 8K) / 8K = -1.96%
    Or, in other words, pretty close.

    PS: I leave it to others to explain why a pickup's DCR will vary with temperature, and isn't very important in the scheme of the universe.

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    Last edited by rjb; 01-01-2019 at 04:01 AM. Reason: Substitute $3 word "elaborate"
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    Old Timer Leo_Gnardo's Avatar
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    Quote Originally Posted by rjb View Post
    PS: I leave it to others to explain why a pickup's DCR will vary with temperature, and isn't very important in the scheme of the universe.
    Simple enough, copper's conductivity / resistivity varies with temperature. So when my ever so sensitive crustomers claim their favorite pickup measures say 7.35 KOhms but they're all worried because they just remeasured 7.52K and golly what's wrong with it??? I have to ask "when did you make these measurements? The first in winter the second in summer? Yeh, I thought so... it's not a problem, now shut up 'n play your geetar!

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    You only know the value at the Jack 7k.
    If the reading is 7k at the jack and you presume the value of the pot is 500k, but it could be 300k.
    What is the value of the pickup?

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    Quote Originally Posted by big_teee View Post
    You only know the value at the Jack 7k.
    If the reading is 7k at the jack and you presume the value of the pot is 500k, but it could be 300k.
    What is the value of the pickup?
    Short answer: A little over 7k. (Assuming the pot is turned to 10.)

    Long answer: Rpup = Rjack / (1 - Rjack/Rvol)

    I guess your scenario is that you have a Gibson guitar and aren't sure if the volume pot is 500k or 300k.
    Let's look at the extremes. A 500k pot with +/-20% tolerance could be 500k+20%=600k and a 300k pot could be 300k-20%=240k

    For a 600k pot:
    Rpup = 7k / (1 - 7k/600k) = 7.08k

    For a 240k pot:
    Rpup = 7k / (1 - 7k/240k) = 7.21k

    Rounding off, I would say the pickup is in the range of 7.1k to 7.2k.
    Or thereabouts.

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    Last edited by rjb; 01-01-2019 at 06:46 AM. Reason: added "with +/-20% tolerance"
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    I like
    7000+5%=7350
    7.35K close enough.
    T

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    OK, 5% worst case. But even with a 240K volume pot, the error is only 3%. As you say, close enough.


    Beating a dead horse:
    The error will be higher for an overwound pickup and low value volume pot than for an underwound pickup and a high value volume pot.
    Example: My MIM Tele originally had a 14K humbucker in the neck(!!!), and a 250K volume pot.
    Rjack = (14K x 250K) / (14K + 250K) = 13.26K
    Rjack + 5% = 1.05 x 13.26K = 13.9K
    So, in this extreme case, the estimate of Rpup = Rjack + 5% would be a tiny bit low.
    But it didn't take any measurements at all to tell that the pickup sounded like mud!

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    Last edited by rjb; 01-01-2019 at 07:14 PM.
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    ToneOholic! big_teee's Avatar
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    I checked my Les Paul neck pickup yesterday.
    It is 7.3k I wound it myself. 500k pot dimed.
    It measures 6.95k in the guitar.
    6.95+5%=7.2975
    Close enough.
    T

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    It's only Rock and Roll, but I like it!

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    Quote Originally Posted by big_teee View Post
    You only know the value at the Jack 7k.
    If the reading is 7k at the jack and you presume the value of the pot is 500k, but it could be 300k.
    What is the value of the pickup?
    An easy to remember formula for that is the 'product' over the 'difference'

    i.e. 500*7/(500-7) for 500k
    or 300*7/(300-7) for 300k

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    Quote Originally Posted by Dave H View Post
    An easy to remember formula for that is the 'product' over the 'difference'
    Yup. D'oh.

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    ToneOholic! big_teee's Avatar
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    i.e. 500*7/(500-7) for 500k
    That looks real neat and everything, but it don't come out to my readings, with my DMM.
    TBC
    T

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    Quote Originally Posted by big_teee View Post
    That looks real neat and everything, but it don't come out to my readings, with my DMM.
    It gives the same answer as the formula rjb used. I can only think there's a measurement error. A 5% change (1/20) implies the parallel resistor is 20 x the pickup resistance i.e. about 145k not 500k

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