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Formula for calculating the output voltage of a pickup coil

  1. Feb 25, 2017 #1
    It should be possible to calculate the output voltage in millivolts of a pickup coil
    at a given frequency, based on the number of turns of wire, the wire resistance / meter,
    the inductance of the coil in Henrys, coil capacitance, the BH max in Joule, or flux density in Tesla,
    and volume of the core magnet,

    I haven't had any luck finding an equation online, so maybe there are some wily elektro-maths-wizards here
    who can help me find a way to predict the outcome of a pickup rewiring project, and work out
    the final signal voltage output at 220 Hz of a coil with 7000 turns of 0.05 mm insulated copper wire
    with a cross section area of 0.00196 mm³, and a DCR of 8.75 Ω/m, wound around a 1914 mm³ alnico 5 magnet
    with a BH max of 44 J/m³, and flux density of 1.25 Tesla.

    Any helpful ideas would be well received.

    Cheers!

    ~E:K
     
    Last edited: Feb 25, 2017
  2. jcsd
  3. Feb 26, 2017 #2

    jim hardy

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    I'm afraid you're going to have to measure it.

    Voltage is number of turns X rate of change of flux

    and i don't see a way to predict by how much the flux of your magnet will get modulated by whatever it is you're measuring.

    cartridge_mag.jpg

    e = n X dΦ/dt

    What's the area of your 7000 turn coil?
    By how much does your 1.25 Teslas fluctuate in response to that 220 hz something you're measuring?
     
    Last edited: Feb 26, 2017
  4. Feb 26, 2017 #3


    Thank you Jim!

    The problem is.. there is nothing to measure until the pickup coil is actually removed
    and rewound with a different wire gauge (0.05 mm),
    and then there is no going back. It would be a lot of wire to waste.

    Ideally, I would like to have some approximation of the mean output voltage with moderate plectrum attack, before winding the coil;
    But I can see now, this may not be as straight forward a calculation as I imagined.

    The area of the coil is 1078 mm² including the area of the magnet, or 654 mm² excluding the magnet area.
    126 turns x 56 outer wraps = 7056 turns.

    220 Hz is the frequency of a vibrating (guitar) A-string at the 12th semitone (A3).
    This was just a random frequency, it could be anything within guitar range (82 Hz to 1.3 kHz).

    Since the duty cycle is 50%, the rate of change would be 440 polarity reversals per second at 220 Hz.
    The amount of flux change would depend on how hard the string is attacked, which affects the amplitude of the string,
    and in turn the amplitude of the signal in the coil, and the intensity of the magnetic field around the coil and into the magnet.
    The only certainty is that it would be two full flux reversals per cycle.

    1/440 = 0.002272 s x 7056 turns = 16.036

    This figure is much too large to be signal voltage.
    It should be something in the 100-300 mV range.

    Is there anything here to work with?

    Cheers,

    ~E:K
     
    Last edited: Feb 27, 2017
  5. Feb 27, 2017 #4
    So, at 7056 turns, the rate of flux change would need to be 1 per 0.000042517 s to arrive at 300 mV,
    equal to 23520 half cycles (11760 Hz), which is over 3 octaves above the highest fundamental frequency on guitar.
    I suppose I should also be counting harmonic frequencies in the calculation, since they also affect the flux change.

    ~E:K
     
  6. Feb 27, 2017 #5

    Nidum

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  7. Feb 27, 2017 #6

    Thanks Nidum

    Internet "How-To" guides are fine for DIY novices, who simply want to have a go at something.
    but, In terms of engineering design, they're quite useless.

    These tutorials are typically written (or video demonstrated) by "enthusiasts" with only the most basic
    understanding of how an inductive transducer works.

    Most enthusiasts still believe in the urban myth that DCR is the best indicator of pickup output,
    which is an absolutely false and misguided approach. A coil with 5000 turns of 0.04 mm wire
    can have the same DCR as a coil with 15000 turns of 0.07 mm wire, and these two coils will have radically different outputs.

    DCR is simply the easiest value to measure with a multi-meter;
    However, DCR is completely valueless for an inductor, where voltage is directly proportional to the derivative of the current.
    Since the derivative of a constant (DC current) is always equal to 0.

    The only real practical use for DCR is measuring the length of wire wound on a coil,
    assuming the diameter of the wire is known, without cutting off 10 meter of wire to take a measurement
    with an Ohm-meter.

    I've been winding pickups for many years, and I'm pretty good at it.
    I can easily calculate most design variables before winding a coil, such as inductance, DCR, nominal impedance, resonant frequency..
    However, I've never found a formula for calculating signal voltage, and this is why I decided to seek out such a formula.

    My goal is to design a pickup coil with a pre-determined (design) output voltage,
    without wasting over 1.5 km of expensive enamelled finewire on one-off experimental coils,
    just to take a measurement.

    Making a (measured) output voltage chart of various coils is a practical method, as long as every coil is using the same wire.

    0.05 mm is not a wire gauge commonly used in pickup winding, so there are no similar coils available to make a comparative analysis.
    The cross sectional area of the wire decides the current capacity and resistance.
    The wire diameter decides how many turns will fit in a given space.
    The number of outer wraps affects the capacitance, which, together with the inductance, decides the resonant frequency.

    f = 1 [COLOR=#black]x[/COLOR]√ LC
    [COLOR=#black].....[/COLOR]2π


    So there are a lot of variables in the wire alone, that affect the final electrical values of an inductor.
    Then there are the variables of the magnet, which also have a huge effect.
    There are too many coefficients involved to draw conclusions from simple DCR or voltage measurements.

    Cheers

    ~E:K
     
    Last edited: Feb 27, 2017
  8. Feb 27, 2017 #7

    Nidum

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    You could conceivably model a pickup system in ANSYS . It would be a major project though .
     
  9. Feb 27, 2017 #8

    Nidum

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    I guessed that someone had already tried FEM modelling -

    Google ' ANSYS Guitar Pickup ' and ' FEM Guitar Pickup '
     
  10. Feb 27, 2017 #9

    Nidum

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  11. Feb 27, 2017 #10
  12. Feb 27, 2017 #11

    jim hardy

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    Fabulous link, Nidum

    Elektrokanien is trying to calculate K .

    Magnetic fields are squishy things, even small current flowing in the pickup itself will reach back around into your formulas and affect flux.


    wow that's almost 3/4 inch diameter ! Way bigger than i envisioned.


    that's 1078/10002 = 0.001078m2 X1.25Tesla =.0013475 Weber of flux, 1.35milliWeber to slide rule accuracy

    So for every 1% that a 1.35 milliWeber magnetic flux gets modulated by the string at 220 hz sinewave
    variation in flux Φ(t) would be 0.01 X 1.35X10-3 sin(220pi X t)
    e would be n X dΦ/dt
    e = 7000 X 0.01 X 1.035 X 10-3 X 220pi X cos(220pi X t) = 65.3 X cos(220pi X t) volts, a 65.3 volt cosine wave


    Clearly the modulation is nowhere near 1% . Malcolm says
    Magnetic field out where the strings are is probably a lot less than a Tesla.
    https://www.kjmagnetics.com/blog.asp?p=surface-fields-101


    interesting thread, guys !
     
    Last edited: Feb 27, 2017
  13. Feb 27, 2017 #12

    jim hardy

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  14. Feb 27, 2017 #13
    Precisely!

    From the first equation : V(t) = P dL/dt + L dP/dt

    We find that :

    Output Signal Voltage = magnetic field strength H (A/m) x (length differential / time differential) + self inductance (H) x (magnetic field strength differential / time differential)

    The self inductance of 1585 m of 0.05 mm copper Ø wire = 5680 µH

    But It's a long road ahead to finding the value of K



    The level of precision involving magnetic fields can only be an approximation at best.
    I would be satisfied with even a roughly accurate range of voltage values, just enough to get an idea of what to expect.


    This is actually a very typical single coil pickup area.


    a 65.3 volt cosine wave, that's impressive.

    Now to find Vpk

    Vrms = Vpk • √ duty cycle



    Absolutely, the magnetic field drops quite sharply with every mm distance.



    It's getting more interesting all the time!

    Again, Thanks for your extremely helpful comments and calculations Jim.
    and thanks for the links Nidum.


    ~E:K
     
    Last edited: Feb 27, 2017
  15. Feb 27, 2017 #14
    The LM4562 is an amazing OpAmp!

    LM4250 is the most common OpAmp used in active pickups, such as the ubiquitous EMG-81, for example.

    Personally, I prefer passive pickups.
    When that 9 volt battery gets down to 70% charge, an active pickup starts to squeel like a pig.
    Although phantom power can remedy that.

    ~E:K
     
  16. Jul 13, 2017 #15
  17. Jul 13, 2017 #16

    Tom.G

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    What am I missing? As a first approximation, is induced voltage proportional to rate of flux change times number of turns? (As in a transformer.)
     
  18. Aug 8, 2017 #17
    induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil.
     
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