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  • #16
    Originally posted by alexoest View Post
    bbsailor (and anyone else who can help),

    Sorry to dig up this old thread, but I would like to build pickups of this type myself and I've been thinking about a couple of questions:

    Would it be possible to just have the primary loop under the strings and connect it by wires to a transformer tucked away in the guitar's control cavity? That'd give a cleaner look.

    Is there a particular reason that you're using a microphone adapter? Couldn't you instead have used any 100:1 current sensing transformer in series with the CSE187L?

    Here, I assume that sending the signal through a 100:1 transformer raises the impedance a hundred times. Is this correct? And if not, can anyone tell me the correct way to calculate the change in impedance when sending the signal through a transformer?

    Last question (for now): What if you instead of just one had about ten primary loops and then send the signal through the 500:1 transformer... would that be equivalent to the around 5000 windings of a traditional pickup?

    Please enlighten me. I am about to receive a couple of AS104 transformers this week and would like to know some more about the possibilities they give me.

    /Alex
    bbsailor (and anyone else who can help),

    Sorry to dig up this old thread, but I would like to build pickups of this type myself and I've been thinking about a couple of questions:

    It is very good that others are thinking about this!

    Would it be possible to just have the primary loop under the strings and connect it by wires to a transformer tucked away in the guitar's control cavity? That'd give a cleaner look.

    The short answer is yes but practicaly for only the bridge pickup that is physically closer to the control cavity. So the key question is how do you want to balance the tone of the pickup from the size of the primary loop. Put into scientific language, that means: what is the resistance of the primary loop? You need a primary loop less than about 1 milliohms or 1 one thousandths of an ohm. What is the distance that a continious loop must take to go around the pickup magnetic field (under the strings) and the loop moving inside to the closest place that the current transformer connection can be made? The answer is in your ears! Lower resistance primary loops favors lower frequencies.

    Is there a particular reason that you're using a microphone adapter?

    Yes!!! The output of the CSE 187L just about matches the input impedance of a Shute A95U matching transformer. But two CSE187L current transformers (CT) allow the outputs to be connected in either series, parallel or single CT, resulting to tonal changes. Where there are changes to be heard is totally dependent on the frequencies we are boosting or cutting compared to the organic sound of a high impedance pickup with 6 to 12 thousand turns of very fine wire. The low impedance pickup has a totaly different and wider bandwidth sound that may not please everyone. However, it has potential upper sound harmonics that can be amplified, if desired.


    Couldn't you instead have used any 100:1 current sensing transformer in series with the CSE187L?



    The low resistance of the primaty loop allows a vibrating string to produce anywhere between 1 and 10 milliamps at the plucked frequency. String mass, velocity and vibrating frequency all affect the output level. Changing the load on the CT will affect the tonal balance and the amount of gain needed to be recovered in the matching transformer method or in an active amplifier method.

    Here, I assume that sending the signal through a 100:1 transformer raises the impedance a hundred times. Is this correct?

    No! It changes by the square of the turns ratio. So the CSE187L with a 1 to 500 turns ratio, changes the impedance by 1 to 250,000. Assume that the output load of the CT is 2.500 ohms (like on the Les Paul Low Impedance pickup schematic). Then, the reflected load is 2500 divided by 250,000 or .01 ohms. But you want the source to impedance to be 10 times lower than the load impedance so that means getting a loop resistance in the neighborhood of 1 milliohm or .001 ohms.

    And if not, can anyone tell me the correct way to calculate the change in impedance when sending the signal through a transformer?




    Last question (for now): What if you instead of just one had about ten primary loops and then send the signal through the 500:1 transformer... would that be equivalent to the around 5000 windings of a traditional pickup?

    It is all about how much current is sent through the primary loop. Lower resistance means thicker wire and shorter distances to obtain the right tone to satisfy the ear.

    Please enlighten me. I am about to receive a couple of AS104 transformers this week and would like to know some more about the possibilities they give me.

    See the posts and posed photos by bajaman on this forum.

    /Alex
    Last edited by bbsailor; 05-06-2010, 07:19 PM.

    Comment


    • #17
      Thank you. That's a very thorough and quick reply.
      I have an additional question to your explanation of the number of turns and the reflected load:

      ---Quote---
      No! It changes by the square of the turns ratio. So the CSE187L with a 1 to 500 turns ratio, changes the impedance by 1 to 250,000. Assume that the output load of the CT is 2.500 ohms (like on the Les Paul Low Impedance pickup schematic). Then, the reflected load is 2500 divided by 250,000 or .01 ohms. But you want the source to impedance to be 10 times lower than the load impedance so that means getting a loop resistance in the neighborhood of 1 milliohm or .001 ohms.
      ---Unquote---

      Does that mean that you must adjust two parameters: To get proper output impedance, you choose a transformer with the right turns ratio _and_ to get a suitable reflected load, you vary the resistance of the loop under the strings? Or is impedance bridging the same thing as ensuring a suitable reflected load?

      But in practical terms, would that mean that to raise the CT's asumed 2.500 ohms output impedance to 250.000 ohms, you'd connect it in series to a 1:10 CT? (since 250.000 divided by 2.500 is 100, and the square root of 100 is 10).

      I saw bajamans pickups in the other thread. Since I'm getting the same CTs (AS104), I expect to make a similar pickup from a piece of flat brass bar. Plus try out the thing I suggested with around ten primary loops of thick copper wire under the strings.

      /Alex

      Comment


      • #18
        Originally posted by alexoest View Post
        Thank you. That's a very thorough and quick reply.
        I have an additional question to your explanation of the number of turns and the reflected load:

        ---Quote---
        No! It changes by the square of the turns ratio. So the CSE187L with a 1 to 500 turns ratio, changes the impedance by 1 to 250,000. Assume that the output load of the CT is 2.500 ohms (like on the Les Paul Low Impedance pickup schematic). Then, the reflected load is 2500 divided by 250,000 or .01 ohms. But you want the source to impedance to be 10 times lower than the load impedance so that means getting a loop resistance in the neighborhood of 1 milliohm or .001 ohms.
        ---Unquote---

        Does that mean that you must adjust two parameters: To get proper output impedance, you choose a transformer with the right turns ratio _and_ to get a suitable reflected load, you vary the resistance of the loop under the strings? Or is impedance bridging the same thing as ensuring a suitable reflected load?

        But in practical terms, would that mean that to raise the CT's asumed 2.500 ohms output impedance to 250.000 ohms, you'd connect it in series to a 1:10 CT? (since 250.000 divided by 2.500 is 100, and the square root of 100 is 10).

        I saw bajamans pickups in the other thread. Since I'm getting the same CTs (AS104), I expect to make a similar pickup from a piece of flat brass bar. Plus try out the thing I suggested with around ten primary loops of thick copper wire under the strings.

        /Alex
        Alex,

        The Les Paul schematic has a 2500 ohm pot value and that looks like a bridging impedance on the current transformer. The output of two CSE187L CTs can be wired in series, parallel of single to obtain some tonal variation like how two coils on a humbucker pickup can be wired.

        Putting 10 turns under the string loop will increase the resistance and affect the tone. I have tried that by using AWG 16 and AWG 18 but the output did not increase very much over a single heavy AWG 10 wire loop. Put the microphone matching transformer (Shure A95U or similar) at the amplifier end of the guitar cable. The output of the CSE187L current transformers seems to match the input impedance of this transformer which is near 2000 ohms to bridge the output of a typical 150 to 200 ohm microphone to obtain the maximum voltage.

        Try any variation you want as these pickups are pretty quickly put together.
        Use this web link to calculate the resistance of the loop using a variety of metals. Resistivity Calc

        Joseph Rogowski

        Comment


        • #19
          Thanks once again for the help - and for the link.

          The Shure A95 U is pretty expensive here in Europe, so if possible, I'd prefer instead to use another CT to provide the last 100x multiplication of the impedance that you use the microphone adapter for. Would that be possible or does the microphone adapter do something that a CT can't?

          /Alex

          Originally posted by bbsailor View Post
          Alex,

          The Les Paul schematic has a 2500 ohm pot value and that looks like a bridging impedance on the current transformer. The output of two CSE187L CTs can be wired in series, parallel of single to obtain some tonal variation like how two coils on a humbucker pickup can be wired.

          Putting 10 turns under the string loop will increase the resistance and affect the tone. I have tried that by using AWG 16 and AWG 18 but the output did not increase very much over a single heavy AWG 10 wire loop. Put the microphone matching transformer (Shure A95U or similar) at the amplifier end of the guitar cable. The output of the CSE187L current transformers seems to match the input impedance of this transformer which is near 2000 ohms to bridge the output of a typical 150 to 200 ohm microphone to obtain the maximum voltage.

          Try any variation you want as these pickups are pretty quickly put together.
          Use this web link to calculate the resistance of the loop using a variety of metals. Resistivity Calc

          Joseph Rogowski

          Comment


          • #20
            Originally posted by alexoest View Post
            Thanks once again for the help - and for the link.

            The Shure A95 U is pretty expensive here in Europe, so if possible, I'd prefer instead to use another CT to provide the last 100x multiplication of the impedance that you use the microphone adapter for. Would that be possible or does the microphone adapter do something that a CT can't?

            /Alex
            Alex,

            You cannot use another CT to boost the output after the first CT. You can use two CTs on the ends of a single low impedance (AWG 10) string loop with a magnet in the center. However the input impedance of a CT is so low that it will not work to boost the secondary of the first CT. It is best to use a microphone matching transformer because of its wide frequency range and the range of the low impedance 150 to 300 ohms input souce impedance and the better coupling. You can use any microphone matching brand that is similar to the Shure A95U series.

            For the lowest noise, ground the metal case of the CSE197 CTs to the main audio ground. Make sure your strings are grounded also.

            Joseph Rogowski

            Comment


            • #21
              Thanks for the explanation. I'll have to get a microphone adapter, then. Apart from that, I'll try to connect the pickup to my sound card mic and line in to see if they suit the impedance of the pickup signal. They should be around 600 ohms and 50 K ohms respectively.

              Haven't got the CTs yet, so I won't be able to try out anything at the moment.

              /Alex

              Originally posted by bbsailor View Post
              Alex,

              You cannot use another CT to boost the output after the first CT. You can use two CTs on the ends of a single low impedance (AWG 10) string loop with a magnet in the center. However the input impedance of a CT is so low that it will not work to boost the secondary of the first CT. It is best to use a microphone matching transformer because of its wide frequency range and the range of the low impedance 150 to 300 ohms input souce impedance and the better coupling. You can use any microphone matching brand that is similar to the Shure A95U series.

              For the lowest noise, ground the metal case of the CSE197 CTs to the main audio ground. Make sure your strings are grounded also.

              Joseph Rogowski

              Comment


              • #22
                Bump!

                I'm following this thread for a practical purpose.

                I'd like to build a pickup to be placed under my guitar strings that have less then a 3/8 clearance.

                The guitar has a body cavity very near the bridge which occupied by active electronic for piezoelectric pickups.

                Hey Alex how have you progressed in your project?

                Thanks for the science!

                Comment


                • #23
                  Randall,

                  I haven't progressed much. I have made a pickup consisting of a single loop of thick copper wire through the AS-104 CT. I put it in the microphone input of my computer and the sound was OK. I use guitar effects software on the computer, so a pickup with an impedance suiting the mic-in is very usable to me.

                  I am in the process of making a better pickup from a flat bar of brass. I expect it to look something like this:

                  [I can't make the picture appear, so here is a link:]

                  Picasa Web Albums - agoest

                  /Alex
                  Last edited by David Schwab; 08-04-2010, 01:56 PM. Reason: fixed image link

                  Comment


                  • #24
                    Originally posted by alexoest View Post
                    I have made a pickup consisting of a single loop of thick copper wire through the AS-104 CT. I put it in the microphone input of my computer and the sound was OK. I use guitar effects software on the computer, so a pickup with an impedance suiting the mic-in is very usable to me.
                    How will you solder the posts to the plates without melting the current transformer? At these low voltages, full metallic bonding is required.

                    I am in the process of making a better pickup from a flat bar of brass. I expect it to look something like this:

                    [I can't make the picture appear, so here is a link:]
                    Picasa Web Albums - agoest
                    You can upload images (for which you must "go advanced"), or provide a URL in an "[img]" bracket.

                    Comment


                    • #25
                      Originally posted by alexoest View Post
                      [I can't make the picture appear, so here is a link:]
                      You have to open the image in a new window to get the link of the image.
                      It would be possible to describe everything scientifically, but it would make no sense; it would be without meaning, as if you described a Beethoven symphony as a variation of wave pressure. Albert Einstein


                      http://coneyislandguitars.com
                      www.soundcloud.com/davidravenmoon

                      Comment


                      • #26
                        Originally posted by Joe Gwinn View Post
                        How will you solder the posts to the plates without melting the current transformer? At these low voltages, full metallic bonding is required.

                        You can upload images (for which you must "go advanced"), or provide a URL in an "[img]" bracket.
                        I'll use screws rather than rods. Though I considered tapping the copper rods with a hammer to make them expand in the holes to get good contact.

                        Comment


                        • #27
                          I'd also like to build one of these pickups. Very cool stuff!!

                          However, is there not a way to get around having to use the Shure A95U (or similar)? Could you not use a transformer with a higher turn ratio? And if that's the case, does anyone know of a good, easy-to-buy-from (preferrably off-the-shelf...) source?
                          Last edited by Chris Turner; 08-04-2010, 11:34 PM.

                          Comment


                          • #28
                            Originally posted by alexoest View Post
                            I'll use screws rather than rods. Though I considered tapping the copper rods with a hammer to make them expand in the holes to get good contact.
                            None of these mechanical bonding methods are good enough. Although they may work for a while, in a year or two the contacting surfaces will corrode slightly from atmospheric gases, enough to hinder operation.

                            This is why such things as the shorted turn in an induction motor is always welded together or stamped as a ring from sheet copper or aluminum.

                            If you look at the alumitone patent drawings, you will see that they have avoided the necessity to make mechanical contacts.

                            I suppose that if one gold plates the contact areas of the mating pieces, one can achieve a sufficiently low and stable contact resistance, so long as sufficient clamping pressure can be achieved and maintained over time.

                            Comment


                            • #29
                              Originally posted by Joe Gwinn View Post
                              None of these mechanical bonding methods are good enough. Although they may work for a while, in a year or two the contacting surfaces will corrode slightly from atmospheric gases, enough to hinder operation.

                              This is why such things as the shorted turn in an induction motor is always welded together or stamped as a ring from sheet copper or aluminum.

                              If you look at the alumitone patent drawings, you will see that they have avoided the necessity to make mechanical contacts.

                              I suppose that if one gold plates the contact areas of the mating pieces, one can achieve a sufficiently low and stable contact resistance, so long as sufficient clamping pressure can be achieved and maintained over time.
                              Thanks for pointing it out. I wasn't aware that the bonds were that critical.

                              What if I seal the bonds with lacqer? Wouldn't that keep the air away?

                              Comment


                              • #30
                                Originally posted by alexoest View Post
                                What if I seal the bonds with lacqer? Wouldn't that keep the air away?
                                Not long enough to matter.

                                Comment

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