Ad Widget

Collapse

Announcement

Collapse
No announcement yet.

Measuring pickups at the input Jack, how much off?

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • 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?

  • #2
    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

Name:	standard_2pickups.gif
Views:	1
Size:	9.7 KB
ID:	852561
    Pickup in parallel with volume pot = ohms at jack.
    T
    "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
    Terry

    Comment


    • #3
      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.

      Comment


      • #4
        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

        Comment


        • #5
          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
          "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
          Terry

          Comment


          • #6
            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.

            Comment


            • #7
              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
              "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
              Terry

              Comment


              • #8
                Cool info. Did you also compare other variables like Impdance? Just wondering if the 5% rule applies to this figure too

                Comment


                • #9
                  I'm off now. Happy New Year everybody

                  Comment


                  • #10
                    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
                    "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
                    Terry

                    Comment


                    • #11
                      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
                      Juan Manuel Fahey

                      Comment


                      • #12
                        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..
                        - Own Opinions Only -

                        Comment


                        • #13
                          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.
                          Last edited by rjb; 01-01-2019, 03:01 AM. Reason: Substitute $3 word "elaborate"
                          DON'T FEED THE TROLLS!

                          Comment


                          • #14
                            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!
                            This isn't the future I signed up for.

                            Comment


                            • #15
                              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?
                              "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
                              Terry

                              Comment

                              Working...
                              X