Announcement

Collapse
No announcement yet.

Differences in sound of the same pickup wound to the same DCR with different wire AWG

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

  • #16
    Originally posted by John Kolbeck View Post
    If you end up with a higher impedance, as a result of using thinner 43 AWG in place of 42 AWG for the same number of turns, won't this also lower the Q factor of the pickup?
    These are some notes I've made along the years:

    The key factor in pickups winding is always Inductance - DC Resistance and stray ( or parasitic ) Capacitance are inevitable "side effects" caused by the wire's Resistivity not being zero.

    The basic equations are F0 ( resonant peak ) = 1/(2pi*(Sqrroot(L/C)))

    ( where F0 is the resonant Frequency in Hertz, L is the pickup's Inductance in Henries and C is the pickup's stray Capacitance in Farads ).

    As you can see, DC Resistance does not appear in the above equation ( =it does not affect the resonant peak ).

    DC resistance has some relation with the Q factor instead ( or Quality factor )

    Q=(1/R)*(Sqrroot(L/C))

    The bigger the Q factor, the narrower the pickup's bandwidth ( more "focused" around Fo ).

    This comes from this ( approximated ) equation:

    BW=F0/Q

    Where F0 and Q have the same meaning as before and BW is the bandwidth in Hertz.

    An increased DCR makes the divider ( 1/R ) bigger in the Q equation, resulting in a lower Q, this, together with the higher Inductance ( which grows with the number of turns SQUARED ), will most likely result in a hot pickup, at the cost of a very low resonant Frequency and a large bandwidth, so ( all other factors being equal ) it will probably sound loud, but bassy and muddy ( low definition ) with a reduced "attack".

    Hope this will clarify things for you.
    Pepe aka Lt. Kojak
    Milano, Italy

    Comment


    • #17
      Originally posted by LtKojak View Post
      The basic equations are F0 ( resonant peak ) = 1/(2pi*(Sqrroot(L/C)))

      ( where F0 is the resonant Frequency in Hertz, L is the pickup's Inductance in Henries and C is the pickup's stray Capacitance in Farads ).
      Oops, typing error. I think you meant:
      F0 ( resonant peak ) = 1/(2pi*(Sqrroot(L*C)))


      I grabbed this from some random site, just because it's a little easier to read (for me, anyway):


      Edit: It might be worth mentioning (for the billionth time) that when wired in a guitar, the cable capacitance (and tone cap, if turned down) will be much greater than the pickup's stray capacitance- so will determine the "in situ" resonant frequency. Then again, that might confuse the issue.

      Whatever,
      -rb
      Last edited by rjb; 02-06-2016, 05:06 PM.
      DON'T FEED THE TROLLS!

      Comment


      • #18
        Originally posted by LtKojak View Post

        Q=(1/R)*(Sqrroot(L/C))

        The bigger the Q factor, the narrower the pickup's bandwidth ( more "focused" around Fo ).
        Since R, resistance, is a determining factor for the value of Q, where the higher the resistance, the lower the Q, is it fair to say that for a given pickup, wound to say, 7000 turns, that the pickup with 43 AWG and the higher resistance will sound "less focused" than one wound with 42 AWG, on account of the increased resistance?

        Comment


        • #19
          I agree with the answers to this thread's original question. Alone, DC resistance is totally meaningless. In fact, I don't like to use DC resistance as a measure of anything concerning the sound of the pickup. It can be very misleading as to the final sonic pallet.

          To me, wire gauge and turn count are first for a particular style of pickup (P-90, HB, Strat, etc). Then magnet type, mass, and strength. Then TPL and layering pattern. Then insulation type and thickness. Then alloys. Oh, don't forget potting, tension, slightly changing coil geometry, flux field patterns, etc, etc.

          All of the technical measurements like Q Factor, Resonant Peak, Inductance, Band Width, Capacitance, and so on, are really the result of the pickup design and the things I just mentioned above.

          Asking for a certain DC Resistance in order to get a specific sound, by itself, is a fool's game. I really wish guitarists would get hip to this. But, as long as manufactures tout DC resistance and articles are written about why a certain DC resistance gives a certain sound, we must be prepared to educate in a kind and non-disparaging manner.
          =============================================

          Keep Winding...Keep Playing!!!

          Jim

          Comment


          • #20
            Originally posted by Jim Darr View Post
            Alone, DC resistance is totally meaningless.
            I see people say this a lot on various forums, and I think Bill Lawrence said this, but there are a couple caveats. One is that this never a lone fact, we always have an awareness that pickups tend to be wound with 42 or 43 AWG, and while it can be wound with thinner or thicker wire, also taking into consideration the described tonal profile of a pickup, being either "hot" or "vintage", so it's possible to infer from the target tone which wire gauge was used to achieve a given resistance, and then further estimation can be made from there.

            The second point is that at resonance, the impedance is at it's minimum, and the wire resistance is all that stands in the way, which is why the Q would be higher for a lower resistance, all other things being equal. It can be hard to describe the sound of a higher versus lower Q though, since they way you would characterize the sound would differ depending on the resonant frequency. If the resonant frequency is in the mid range, a high Q might manifest as a honking or nasal sound, while if the resonant frequency is higher, it could make for a piercing, shrill sound. It ends up being a balancing act of not too much, not too little. You can roughly model the effect of a higher or lower Q by emphasizing a particular frequency, above all others , with a multi-band EQ.

            Comment


            • #21
              Originally posted by John Kolbeck View Post
              we always have an awareness that pickups tend to be wound with 42 or 43 AWG, and while it can be wound with thinner or thicker wire, also taking into consideration the described tonal profile of a pickup, being either "hot" or "vintage", so it's possible to infer from the target tone which wire gauge was used to achieve a given resistance, and then further estimation can be made from there.
              This is simply not true... for the record, have you actually made any pickups, John?

              Originally posted by John Kolbeck View Post
              The second point is that at resonance, the impedance and the resistance are equal, which is why the Q would be higher for a lower resistance, all other things being equal. It can be hard to describe the sound of a higher versus lower Q though, since they way you would characterize the sound would differ depending on the resonant frequency. If the resonant frequency is in the mid range, a high Q might manifest as a honking or nasal sound, while if the resonant frequency is higher, it could make for a piercing, shrill sound. You can roughly model the effect of a higher or lower Q by emphasizing a particular frequency, above all others , with a multi-band EQ.
              John, I think you should re-read this very thread. You seem to have missed several very important bits of relevant info in posts #3, #10, #11, #13 and #14.

              Plus, not everything the great late Bill Laurence published should be taken as God's own words, channeled through him.
              Last edited by LtKojak; 02-07-2016, 08:25 AM.
              Pepe aka Lt. Kojak
              Milano, Italy

              Comment


              • #22
                Originally posted by John Kolbeck View Post

                The second point is that at resonance, the impedance is at it's minimum, and the wire resistance is all that stands in the way, which is why the Q would be higher for a lower resistance, all other things being equal.
                At resonance, the impedance is at its maximum. Look at the measurement of the impedance of any pickup, for example:SHn59.pdf

                This measurement is made by measuring the voltage across and current through the pickup and taking the ratio. Thus it is the impedance of the two terminal device without the influence of external capacitance. The simple description of the pickup is a parallel resonant circuit with the coil inductance and capacitance. Thus impedance must be maximum at resonance.

                Comment


                • #23
                  A possible answer to the OP?
                  Wind two bobbins with different size wire and try them.
                  Or better yet wind 4 bobbins, two with different wire gauge wire with same DCR.
                  Two with different wire gauge same amount of turns, with different DCR.
                  Make up your own mind what they sound like!
                  Everything else IMO, is just a big debate!
                  I think you will find the extra leg work to be worth the effort!
                  T
                  Last edited by big_teee; 02-08-2016, 04:10 AM.
                  "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
                  Terry

                  Comment


                  • #24
                    I was wondering another stuff regarding pickup output and number of turns:

                    Will 2 pickups have approximately the same output if they are winded with the same number of turns with different wire gauge? Let's assume that the 2 pickups are exactly the same in all aspects except wire gauge (one 42 and the other 43)

                    Comment


                    • #25
                      Originally posted by alexirae View Post
                      I was wondering another stuff regarding pickup output and number of turns:

                      Will 2 pickups have approximately the same output if they are winded with the same number of turns with different wire gauge? Let's assume that the 2 pickups are exactly the same in all aspects except wire gauge (one 42 and the other 43)
                      From Mike Sulzer's #22 above: The simple description of the pickup is a parallel resonant circuit with the coil inductance and capacitance.

                      They'll be close, but not identical. # turns will be the major determinant in inductance. Since you can wind more turns per layer with thinner wire, the capacitance is bound to be different in the 2 coils, probably less in the thinner wire coil; that will drive the resonance peak to a higher frequency. Also thinner wire will have a higher resistance but I expect that would be a small factor sonically. Wind away, I'll bet you can hear a difference.
                      This isn't the future I signed up for.

                      Comment


                      • #26
                        Originally posted by alexirae View Post
                        Will 2 pickups have approximately the same output if they are winded with the same number of turns with different wire gauge?
                        In a nutshell, no.

                        HTH,
                        Pepe aka Lt. Kojak
                        Milano, Italy

                        Comment


                        • #27
                          Ok, then how you would approximate the output of the second pickup with respect to the first one in terms of number of turns?

                          Comment


                          • #28
                            Originally posted by alexirae View Post
                            Ok, then how you would approximate the output of the second pickup with respect to the first one in terms of number of turns?
                            It looks like it is time for you to bend some wire.
                            Are you a real pickup winder, or just going through exercises in futility?
                            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


                            • #29
                              Originally posted by alexirae View Post
                              Ok, then how you would approximate the output of the second pickup with respect to the first one in terms of number of turns?
                              In overall volume, all else being the same (magnets, distance from strings, etc.) they should be equivalent. # of turns being the main item. Some difference in high frequency response would be expected, with the resonance difference.

                              Wondering, didn't you post on here over the last year or so with an avatar, you (?) with a blue guitar? I could have you confused with another similar MEF poster.
                              This isn't the future I signed up for.

                              Comment


                              • #30
                                Well I'm just asking all this questions to have more theoretical knowledge from "The Experts". I've winded just one pickup so far, but I wouldn't like to throw away some wire just experimenting if some people already knows what could happen if X Y Z thing is done in the P Q R way. Sharing experiences/discuss stuff is the main point of the forum, isn't it?.

                                Also any information registered here will be helpful in the future for other people that wants to know more about this wire gauge subject.

                                Comment

                                Working...
                                X