1. You are right, I am sorry. Did not read carefully. Impedance always starts horizontally with DCR from 0 Hz. But the bandpass transfer response is different and must have 0 signal at 0 Hz.

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2. Originally Posted by Helmholtz
And here is the more important part of my comments, dealing with measuring the PU's transfer response:

Measuring PU transfer response requires access to an input port. Inserting a signal voltage source in series with the inductor part as typically done in simulations is not possible in real life. Instead, the well accepted method is to use the PU coil as secondary in a current transformer arrangement. The idea is to inject a current into the PU coil (inductance) via a coupled external coil driven by constant current and measure the resulting voltage across the PU terminals. Mind that driving the external coil directly by a (low impedance) voltage source would load down the PU and change its frequency response.
The induced constant current in the PU coil produces a voltage across its inductance, rising proportionally with frequency and consequently the PU shows a typical bandpass behaviour.
The main requirement for the external primary circuit is that the drive current must stay constant for all frequencies to be measured. This means that not only the self-resonance of the field coil has to lie far above the highest frequency of interest but also that the impedance of the field coil stays negligible compared to the total series resistance (279 Ohms in the example). With the values given in the article the corner frequency for this requirement is around 1.2kHz. Above this frequency the drive current drops with 6dB/octave and distorts the measured frequency response as can be seen in the PU responses of figure 5. The cure is to increase the L/R ratio by a factor of 20 or more.
The current in the field or exciter coil creates an ac magnetic field that induces a voltage, not a current, in series with the pickup coil. This follows from Maxwell's equation or the law of magnetic induction. Current flows if there is a load on the coil, such as the coil capacitance or the loading caused by eddy currents in the cores, etc. (A so called current transformer is a tightly coupled transformer driven from a high impedance so that the current in the secondary is related to that in the primary by the turns ratio. The very loosely coupled situation here does not behave that way.)

You describe one way to make the current through the field coil independent of frequency: make the inductive reactance low compared to the dc resistance of the coil across the whole useful frequency range. A better way is to drive the coil with a current source, that is, a circuit with an output impedance much higher than the impedance of the coil at any useful frequency.

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3. Originally Posted by Mike Sulzer
The current in the field or exciter coil creates an ac magnetic field that induces a voltage, not a current, in series with the pickup coil. This follows from Maxwell's equation or the law of magnetic induction. Current flows if there is a load on the coil, such as the coil capacitance or the loading caused by eddy currents in the cores, etc. (A so called current transformer is a tightly coupled transformer driven from a high impedance so that the current in the secondary is related to that in the primary by the turns ratio. The very loosely coupled situation here does not behave that way.)

You describe one way to make the current through the field coil independent of frequency: make the inductive reactance low compared to the dc resistance of the coil across the whole useful frequency range. A better way is to drive the coil with a current source, that is, a circuit with an output impedance much higher than the impedance of the coil at any useful frequency.

Thanks for your last few posts, you cleared up several things that were not clicking for me before, especially about there being a voltage, but not necessarily a current unless a load exists on the pickup. I know that's a basic idea, but I was slow to connect the dots.

And the other point, about the resistance preceding the reactance at low frequencies, I modeled this with LTSpice. The different plot lines indicate three steps of resistance, 10k, 20k and 30k ohms, with the flat, low frequency portion extending further for each increase in step (as well as lowering the Q factor).

Something I'm confused about though, are the circumstances under which +6dB...-6dB/oct slopes emerge, as seen in "raw" bode plots, and the above LTSpice model, but only when the AC voltage source has been placed outside of the pickup.

When a pickup is modeled as with the AC source inside the pickup, as shown below, there is a 0dB/oct line, then the resonant peak, and then a -12dB/oct slope:

In practical testing, both the exciter / field coil method, as well we putting the pickup in series with the function generator, both yield +dB...-6dB/oct slopes, as seen in the first simulation. Driving the pickup with a series voltage obviously puts the voltage source outside of the pickup, but shouldn't the exciter coil method place the voltage inside of the pickup? Why does this testing method not result in a 0dB/oct..-12dB/oct plot, as is seen in the second screen shot?

Another issue with the model which has the voltage source inside the pickup, is as seen in the second screen shot, that increasing the series resistance from 10k to 20k to 30k has no apparent impact on the 0dB/oct slope, as it does with the-6dB slope in the first screen shot.

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4. Originally Posted by Helmholtz
I have no values for the coupling factor. Generally a coupling factor below 100% introduces additional series inductance in the equivalent circuit. But why don't you just measure and compare? Call it loading down or not, in result the (high) frequency response will change.

You have to add the field coil's resistance to the series resistance for total circuit resistance Rtot. What matters is the Rtot/L ratio. It should be well above 150kOhm/H. In other words the corner frequency should lie well above the frequency range analysed and is given by f=Rtot/(2pi*L). This can be achieved by increasing series and/or coil resistance as well as by decreasing field coil inductance.
In that past, I had measured a pickup using an exciter, both with and without a resistor in series with the exciter. It made no difference in the plot lines, but the added resistance made the exciter coil weaker, reducing the S/N ratio. So if the coupling is lower, the series inductance is higher, but apparently not high enough to interfere with measurements.

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5. Originally Posted by Antigua
In that past, I had measured a pickup using an exciter, both with and without a resistor in series with the exciter. It made no difference in the plot lines, but the added resistance made the exciter coil weaker, reducing the S/N ratio. So if the coupling is lower, the series inductance is higher, but apparently not high enough to interfere with measurements.
I cannot comment on your measurements. I gave you all the info necessary to make sure you have constant exciter current over frequency. When the exciter current falls with increasing frequency, so will your PU output voltage and thus frequency response will deviate. This effect can be seen in figure 5 of the document, where the responses show a pseudo plateau starting around 1kHz, just as predicted by the formula. In reality the PU bandpass response gets (increasingly) steeper towards resonance.

You may verify the frequency (in)dependance of the drive current by measuring the voltage over a series resistor, via the second channel of the Velleman. The effect of a decreasing drive may be partly masked by the resonance of the PU.

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6. The current in the field or exciter coil creates an ac magnetic field that induces a voltage, not a current, in series with the pickup coil. This follows from Maxwell's equation or the law of magnetic induction. Current flows if there is a load on the coil, such as the coil capacitance or the loading caused by eddy currents in the cores, etc. (A so called current transformer is a tightly coupled transformer driven from a high impedance so that the current in the secondary is related to that in the primary by the turns ratio. The very loosely coupled situation here does not behave that way.)

You describe one way to make the current through the field coil independent of frequency: make the inductive reactance low compared to the dc resistance of the coil across the whole useful frequency range. A better way is to drive the coil with a current source, that is, a circuit with an output impedance much higher than the impedance of the coil at any useful frequency.
True, the primary effect of induction is electric field and voltage. But this causality rarely matters in real life situations, where there are current paths and the currents produce counterfields.
Any pair of coupled coils can be descibed as a (non-ideal) transformer. And a transformer is able to transform voltage and current as well as Z,R,L,C. Loose coupling causes reduced voltages and deviations from the ideal turns ratio relations. But the principle works nevertheless.
The idea behind the method is to generate a frequency-independent current in the inductance part of the PU. Constant current through the inductance produces a voltage across the inductance rising linearly with frequency, just like in real PU operation, where the voltage is induced via dPhi/dt.
This constant current is the input test signal and must flow through the load consisting of DCR and capacitance. The output signal is the voltage developed across the capacitance/terminals.
The method is best descibed via the current transformer principle. And if done carefully it works just fine, as can be most easily seen by the straight horizontal line behaviour of the integrated output voltage. My explanations are in line with Lemme and Zollner.

Your are right, generating the constant current via the constant current driven field coil is nothing but a constant current source. And it could be replaced by an (active) wide band CCS if there were direct access to the inductance part of the PU, which is not.
Feeding a constant current to the (output) terminals of the PU inevitably yields the two-pole/two-terminal impedance response. This is different from the quadripole transfer response revealed by the descibed method.
I don't pretend, though, that the transfer response reveals information not available from the impedance response.

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7. oabx00z.png

In practical testing, both the exciter / field coil method, as well we putting the pickup in series with the function generator, both yield +dB...-6dB/oct slopes, as seen in the first simulation. Driving the pickup with a series voltage obviously puts the voltage source outside of the pickup, but shouldn't the exciter coil method place the voltage inside of the pickup? Why does this testing method not result in a 0dB/oct..-12dB/oct plot, as is seen in the second screen shot?
Placing a constant voltage source in series with L does not correspond to the exciter method, where a constant current through L is generated. You should use a swept current source in parallel with L instead. This will give you the bandpass transfer response with +/- 6dB slopes and 0 output for 0 Hz.
You did simulate the low-pass transfer response instead.

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8. Originally Posted by Helmholtz
The idea behind the method is to generate a frequency-independent current in the inductance part of the PU.

That does not happen. The constant current in the exciter coil generates a magnetic field with amplitude independent of frequency. This generates a frequency dependent voltage in the pickup coil. The current that flows is frequency dependent because of that and because the load on the pickup is frequency dependent. At very low frequencies where the reactance of the coil capacitance is very high and the effect of eddy currents is negligible, this current is very small, only that which flows in the input resistance of the amplifier that senses the pickup voltage.

As long as the current in the exciter coil remains constant, the excitation part of the total magnetic field does not change, and so there is no interaction between the pickup coil and the exciter coil, except for the capacitance of the exciter coil. But if you use a tiny exciter coil, this is negligible. In my set up the constant current is a result of the 8 ohm resister, which has an impedance much larger than the tiny coil. Any coil resistance just adds to the resister, and the reactance value of the coil inductance is too small to change the current significantly at audio frequencies.

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9. That does not happen. The constant current in the exciter coil generates a magnetic field with amplitude independent of frequency. This generates a frequency dependent voltage in the pickup coil. The current that flows is frequency dependent because of that and because the load on the pickup is frequency dependent. At very low frequencies where the reactance of the coil capacitance is very high and the effect of eddy currents is negligible, this current is very small, only that which flows in the input resistance of the amplifier that senses the pickup voltage.
You are wrong. The current circulating in the PU will be independent of frequency if done correctly. This is a precondition for measuring the true bandpass response of a filter circuit. It can be verfied by simulation.

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10. Originally Posted by Helmholtz
I cannot comment on your measurements. I gave you all the info necessary to make sure you have constant exciter current over frequency. When the exciter current falls with increasing frequency, so will your PU output voltage and thus frequency response will deviate. This effect can be seen in figure 5 of the document, where the responses show a pseudo plateau starting around 1kHz, just as predicted by the formula. In reality the PU bandpass response gets (increasingly) steeper towards resonance.

You may verify the frequency (in)dependance of the drive current by measuring the voltage over a series resistor, via the second channel of the Velleman. The effect of a decreasing drive may be partly masked by the resonance of the PU.
I don't know how the "figure 5" plot came to exist, with the huge "hump", I've never witnessed that myself, except in the case of extreme eddy current losses. For example, it looks suspiciously similar to a Fiolter'tron:

Both my exciter coil and "driven" plots show the same curves, so I think I'm good to go. The only difference, as mentioned earlier, is that the exciter coil will also reveal eddy current losses with respect to the guitar string, where as the "driven" method with a series resistor only shows eddy current losses with respect to the pickup's coil.

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11. Originally Posted by Helmholtz
You are wrong. The current circulating in the PU will be independent of frequency if done correctly. This is a precondition for measuring the true bandpass response of a filter circuit. It can be verfied by simulation.
Can you describe the physical process by which a constant magnitude ac magnetic field, varying over the required frequency range, can excite a current in a coil independent of the frequency varying load on the coil? Even if you could, it would not be what you want to do. Filter circuits generate the correct bandpass when the source and load have the impedance for which they were designed. For example, rf passive filters might be designed to work from a 50 ohm source into a fifty ohm load.

The field generated by the coil, if small enough, is very similar to the ac field generated by a vibrating string. If you have any doubts that that generates a voltage, read MacDonald (Princeton). Very similar application of Maxwell's equations. So if the string generates a voltage in series with the coil, then we want our test to generate a voltage.

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12. Can you describe the physical process by which a constant magnitude ac magnetic field, varying over the required frequency range, can excite a current in a coil independent of the frequency varying load on the coil?
Yes, I can but don't bother to elaborate. No sense discussing with people who don't even consider my arguments.

The field generated by the coil, if small enough, is very similar to the ac field generated by a vibrating string. If you have any doubts that that generates a voltage, read MacDonald (Princeton). Very similar application of Maxwell's equations. So if the string generates a voltage in series with the coil, then we want our test to generate a voltage.
I never denied that the ac field generates a voltage. In fact, the voltage induced across the inductance of the PU images the voltage across the inductance part of the exciter coil (which is not directly accessible, though), namely a voltage rising linearly with frequency. And this means a constant, frequency independent current.
I definitely don't need private lessons on physics by amateurs.

This is my last post on this subject. I have wasted enough time trying to convince people who don't want to learn.

You may try to direct your questions to GITEC e.V.

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13. Both my exciter coil and "driven" plots show the same curves, so I think I'm good to go. The only difference, as mentioned earlier, is that the exciter coil will also reveal eddy current losses with respect to the guitar string, where as the "driven" method with a series resistor only shows eddy current losses with respect to the pickup's coil.
Well, if you don't care, I don't either - but will rely on my own measurements. Your "eddy current losses" are only partly real.

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14. Originally Posted by Helmholtz
Yes, I can but don't bother to elaborate. No sense discussing with people who don't even consider my arguments.

I never denied that the ac field generates a voltage. In fact, the voltage induced across the inductance of the PU images the voltage across the inductance part of the exciter coil (which is not directly accessible, though), namely a voltage rising linearly with frequency. And this means a constant, frequency independent current.
I definitely don't need private lesson on physics by amateurs.

This is my last post on this subject. I have wasted enough time trying to convince people who don't want to learn.

You may try to direct your questions to GITEC e.V.
No, the voltage is generated in series with the pickup coil, and this voltage is not directly accessible. The inductance of the pickup coil is the series leg, and its capacitance is the shunt leg of a voltage divider. This makes a resonant low pass filter. Thus the voltage across the pickup coil is the output of the filter, and it is flat at low frequencies, rises below the resonance and falls above it.

In addition, the voltage rises with frequency because the induced voltage is proportional to the rate of change of the flux through the coil. This has to be accounted for. Finally, the pickup coil couples to metal parts like a transformer with high leakage flux. This places a load across the coil that mostly affects the response in the region of the resonance.

The current that flows through the pickup coil is a function of these loads on it. It is very small at low frequencies since the load is a very high resistance, the preamp input. The current peaks at resonance because of the circulating current in the parallel resonating circuit, and then it falls at higher frequencies.

I am not an amateur.

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15. Originally Posted by Helmholtz
Well, if you don't care, I don't either - but will rely on my own measurements. Your "eddy current losses" are only partly real.
So are you saying that a thick conductive cover cannot reduce the signal reaching the pickup coil? It can. Both types of eddy current losses can be significant, and that is why it is necessary to make both kinds of measurements if the pickup has a thick cover.

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16. So are you saying that a thick conductive cover cannot reduce the signal reaching the pickup coil? It can. Both types of eddy current losses can be significant, and that is why it is necessary to make both kinds of measurements if the pickup has a thick cover.
I can only advise to make sure/control that the exciter current stays constant over the whole frequency range. Otherwise you might find unreal sag below the resonance. But I am repeating myself. This said, some PUs with strong eddy effect do show some real sag. But it is suspicious if this effect depends on the drive circuit.

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17. Originally Posted by Helmholtz
But it is suspicious if this effect depends on the drive circuit.
One example would be the base plate of a pickup, it's very close to the coil itself, so it figures prominently with respect to the coil, but it's very far away from the moving guitar string, so it would have little interaction with the AC magnetic field of the guitar string. A cover would be the other way around; impeding the AC field of the string more so than the AC field of the coil.

Going back to the transformer analogy, wouldn't eddy current losses be greater if the conductive plane was in between the coils, rather than being placed at far side of the transformer?

Within the next week or two I will compare driven versus externally excited plots of pickups with covers and substantial steal cores in order to quantify any difference.

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18. Originally Posted by Antigua
One example would be the base plate of a pickup, it's very close to the coil itself, so it figures prominently with respect to the coil, but it's very far away from the moving guitar string, so it would have little interaction with the AC magnetic field of the guitar string. A cover would be the other way around; impeding the AC field of the string more so than the AC field of the coil.

Going back to the transformer analogy, wouldn't eddy current losses be greater if the conductive plane was in between the coils, rather than being placed at far side of the transformer?
Yes, of course. But where is the link to the exciter coil circuit?

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19. Within the next week or two I will compare driven versus externally excited plots of pickups with covers and substantial steal cores in order to quantify any difference.
I may have misinterpreted you, as it is not clear to me what you mean by "driven". Both measurements, impedance as well transfer response require a drive (signal).

-The impedance measurement (direct current feed) is sensitive to eddy current effects in cores but much less so to eddy current losses in conductive covers.

-The field drive method for transfer response much better (more realistically) reveals the effect of lossy covers at medium and high frequencies by producing a drecrease (loss) in output voltage. After integration, this typically shows as a depression below resonance.
But this same kind of response with depression can be produced as an artefact, if the exciter current starts decreasing in this frequency range, caused by the increasing impedance/reactance of the exciter coil over frequency. In this case you need to increase the series resistor value and/or reduce exciter coil inductance.

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20. Originally Posted by Helmholtz
I may have misinterpreted you, as it is not clear to me what you mean by "driven". Both measurements, impedance as well transfer response require a drive (signal).

-The impedance measurement (direct current feed) is sensitive to eddy current effects in cores but much less so to eddy current losses in conductive covers.

-The field drive method for transfer response much better (more realistically) reveals the effect of lossy covers at medium and high frequencies by producing a drecrease (loss) in output voltage. After integration, this typically shows as a depression below resonance.
But this same kind of response with depression can be produced as an artefact, if the exciter current starts decreasing in this frequency range, caused by the increasing impedance/reactance of the exciter coil over frequency. In this case you need to increase the series resistor value and/or reduce exciter coil inductance.
But this kind of artefact would be seen without a cover as well, and so it is easy to determine if it is a serious problem.

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21. But this kind of artefact would be seen without a cover as well, and so it is easy to determine if it is a serious problem.
Right, but I am not sure if everbody cares to check. The depression might also hide below the resonance and cause a too low resonance peak (Q) and a high frequency drop-off steeper than -12dB/octave in the integrated response. I would recommend to use a low loss control PU with very high resonance frequency (low/medium impedance type) for this kind of check.
But it is easier to verify that the exciter coil current does not change over the whole frequency range.

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22. Originally Posted by Helmholtz
I may have misinterpreted you, as it is not clear to me what you mean by "driven". Both measurements, impedance as well transfer response require a drive (signal).

-The impedance measurement (direct current feed) is sensitive to eddy current effects in cores but much less so to eddy current losses in conductive covers.

-The field drive method for transfer response much better (more realistically) reveals the effect of lossy covers at medium and high frequencies by producing a drecrease (loss) in output voltage. After integration, this typically shows as a depression below resonance.
But this same kind of response with depression can be produced as an artefact, if the exciter current starts decreasing in this frequency range, caused by the increasing impedance/reactance of the exciter coil over frequency. In this case you need to increase the series resistor value and/or reduce exciter coil inductance.
Right, this is what is at issue. The Filter'tron plot in post #80 shows that depression in dramatic form, but we know it's caused by eddy currents, because other plots of Fender single coils are very flat before they reach resonances, for example:

Also, the plot in this post http://music-electronics-forum.com/t46007-4/#post493813 has a plot that gets to 16kHz without any dips. Also, that plot shows the effects of dramatic eddy current losses when the secondary coil was closed and continuous.

Here's something I don't understand: why can't I get the field drive coil to generate a -6dB/oct slope by putting capacitance in parallel with the field drive coil? Shouldn't the voltage drop with frequency? I tried it and it seems to have no such effect.

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23. Here's something I don't understand: why can't I get the field drive coil to generate a -6dB/oct slope by putting capacitance in parallel with the field drive coil? Shouldn't the voltage drop with frequency? I tried it and it seems to have no such effect.
The natural high frequency slope of the field coil drive response (aka bandpass transfer reponse) is -6dB/octave, i.e. capacitive behaviour The integrator transforms this to the low-pass response and adds another -6dB/octave. The result is -12dB/octave.
What is the point of putting a capacitor in parallel with the exciter coil? What is the bandwidth of the integrator?

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24. Originally Posted by Helmholtz
What are all these needle artifacts? I don't see them in my impedance measurements. Also the high and low frequency slopes appear not to be correct. Did you measure as I proposed without the field coil?
We call them "spurs" in English. They look like external interference to me, probably from clocked digital logic with very short rise and fall times on the edges. Plotting amplitude versus linear (not log) frequency will reveal any harmonic ladders, which may be a clue as to source. Be suspicious of nearby test equipment.

The zigzag anomality is the result of an additional series and parallel resonance with higher resonant frequencies. I could show in simulations that such behaviour can be the result of partially shorted windings. Another explanation could be a very sloppy wind, where the winding is not carefully layered but outer turns are used to fill lower spaces thereby causing an uneven distribution of the distributed capacitance. I have not found a way yet to prove this idea wrong or right, as I am not winding.
One way to dig into this mathematically may be found in the System Identification literature, which is large, and mostly directed at things like modelling chemical plants the better to control them. But we can use their methods in miniature, to deduce likely equivalent circuits. Note that eddy currents will likely need to be modeled directly, as no lumped-component circuit does it justice.

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25. The current in the field or exciter coil creates an ac magnetic field that induces a voltage, not a current, in series with the pickup coil.
This is nonsense. Induced voltage always develops across the inductor and not in series. This follows from Maxwell's equations as well as Faraday's law of induction.

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26. Here's something I don't understand: why can't I get the field drive coil to generate a -6dB/oct slope by putting capacitance in parallel with the field drive coil? Shouldn't the voltage drop with frequency? I tried it and it seems to have no such effect.
Well, this is easy, if you understand resonant circuits. The exciter coil and the parallel C form a parallel resonant circuit. In the vicinity of its resonant frequency, coil current and coil voltage rise, while the outer current through the series resistor decreases caused by the increased impedance. The normal voltage decreasing effect of the added C only shows above this resonance.

The most probable reason for a high frequency drop-off flatter than -6dB/octace before and -12dB/octave after integration is noise floor of the Velleman. S/N ratio is improved by higher drive current. I use a power amplifier connected to the generator output to drive the coil.

You may also want to check the integrator for errors at low signal levels caused by offset.

BTW, the 6dB/octave slopes below and above resonance prove that the current through the inductance at low frequencies and the current through the capacitance at high frequencies must be constant (independent of frequency). Those who understand filters will know what I mean.

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27. Originally Posted by Helmholtz
This is nonsense. Induced voltage always develops across the inductor and not in series. This follows from Maxwell's equations as well as Faraday's law of induction.
No. Take the turns one at a time. Integrate the E field around each turn. (The E field has a non-zero curl.) This gives the voltage around each turn. The turns are all in series; so add up the individual voltages to give a single voltage in series with the coil.

If a voltage source appeared in parallel with the coil, it could drive the coil inductance as well as any external load. This is obviously impossible; it would be able to provide an infinite amount of energy.

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28. No. Take the turns one at a time. Integrate the E field around each turn. (The E field has a non-zero curl.) This gives the voltage around each turn. The turns are all in series
Correct. In consequence the sum of the voltages appears at/between the terminals of the real coil.

so add up the individual voltages to give a single voltage in series with the coil.
The conception of an ideal inductor in series with a voltage source is one valid model. But this voltage source needs to be controlled to give f-proportional voltage, as does the real inductor. (Simulations using a constant voltage source will not give the real PUs frequency response, which shows an output voltage proportional to string velocity (frequency*amplitude) below resonance.)

I prefer Zollner's method of producing the induced voltage with a constant current source wired in parallel with the ideal inductor.

If done correctly both methods/models give the same simulation results and thus can be considered equivalent.

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29. Originally Posted by Helmholtz
The natural high frequency slope of the field coil drive response (aka bandpass transfer reponse) is -6dB/octave, i.e. capacitive behaviour The integrator transforms this to the low-pass response and adds another -6dB/octave. The result is -12dB/octave.
What is the point of putting a capacitor in parallel with the exciter coil? What is the bandwidth of the integrator?
The point of this is that I'm trying to see if the integrator can be removed from the testing process, by duplicating its effects by passive means, because it there's a way to get a similar plot without specialized equipment, it would make pickup testing more accessible to hobbyists like myself.

Just so you know, the integrator is between the pickup and the oscilloscope, as this is how Ken Willmott designed the device. I understand that Helmuth Lemme put the integrator in between the function generator and field coil.

I'm having partial success using this circuit on the field coil side. You can see from the simulation that I across L1 "field coil" slopes off past 1kHz, so that's good, but it's still far from being great.

Here are actual plots using an SSL-1, using the schematic above on the field coil side. It looks more like an integrated plot as the value of C is increased, but at lower frequencies it's not flat, and the noise increases as it's made flatter with higher values of C.

I notice that if R2 is a higher value, the slope drops off at a lower frequency, but doing that causes the field coil's strength to become too weak.

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30. Originally Posted by Helmholtz
Correct. In consequence the sum of the voltages appears at/between the terminals of the real coil.

The conception of an ideal inductor in series with a voltage source is one valid model. But this voltage source needs to be controlled to give f-proportional voltage, as does the real inductor. (Simulations using a constant voltage source will not give the real PUs frequency response, which shows an output voltage proportional to string velocity (frequency*amplitude) below resonance.)

I prefer Zollner's method of producing the induced voltage with a constant current source wired in parallel with the ideal inductor.

If done correctly both methods/models give the same simulation results and thus can be considered equivalent.
The sum of the voltages does not appear at the terminals of the real coil, although it can approach it at very low frequencies. The sum of the voltages is modified by the various impedances I mentioned before. In particular at the resonance, the output voltage across the coil is greater than the series voltage at that frequency.

Of course you can always use a current source, properly located. But this is just a distraction; neither of us was talking about this equivalent current source.

Of course the series voltage increases with frequency; that is what the physics tells us to do.

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31. Originally Posted by Antigua
The point of this is that I'm trying to see if the integrator can be removed from the testing process, by duplicating its effects by passive means, because it there's a way to do get a similar plot without without specialized equipment, it would make pickup testing more accessible to hobbyists like myself.

Just so you know, the integrator is between the pickup and the oscilloscope, as this is how Ken Willmott designed the device. I understand that Helmuth Lemme put the integrator in between the function generator and field coil.

I'm having partial success using this circuit on the field coil side. You can see from the simulation that I across L1 "field coil" slopes off past 1kHz, so that's good, but it's still far from being great.

Here are actual plots using an SSL-1, using the schematic above on the field coil side. It looks more like an integrated plot as the value of C is increased, but at lower frequencies it's not flat, and the noise increases as it's made flatter with higher values of C.

I notice that if R2 is a higher value, the slope drops off at a lower frequency, but doing that causes the field coil's strength to become too weak.
Did you read my post #96? It explains, why the C across the exciter coil is not a good idea. This method can never replace a real integrator.

The integrator can be placed between PU and scope as well as between generator and exciter coil. The latter arrangement provides a stronger signal for the integrator, which is generally good. But I am not sure, if the integrator is able to directly drive the low impedance exciter coil load. In any case I strongly recommend to drive the coil via a linear power amplifier. I use one channel of a 60W stereo amplifier.

For integration I use a passive integrator after the power amplifier, consisting of a huge air core inductor of around 20mH and a DCR of 2Ohms. This is wired in series with the exciter coil and takes care of the 1/f drive current.

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32. Originally Posted by Helmholtz
Did you read my post #96? It explains, why the C across the exciter coil is not a good idea. This method can never replace a real integrator.

The integrator can be placed between PU and scope as well as between generator and exciter coil. The latter arrangement provides a stronger signal for the integrator, which is generally good. But I am not sure, if the integrator is able to directly drive the low impedance exciter coil load. In any case I strongly recommend to drive the coil via a linear power amplifier. I use one channel of a 60W stereo amplifier.

For integration I use a passive integrator after the power amplifier, consisting of a huge air core inductor of around 20mH and a DCR of 2Ohms. This is wired in series with the exciter coil and takes care of the 1/f drive current.
I did read your post, the -6dB slope comes after the resonance between the cap and the field coil, but I used very high C values, 1uF, so the resonant peak was very low, so that in and of itself didn't appear to be a problem. It appears to me, based on LTSpice modelling, that if R2 is a higher value of resistance, the roll off occurs at a lower frequency, but a more powerful amplification would be required to drive the field coil. The goal is to come up a more simple test procedure, and a power amp would add complexity, but at least it's a much (much much) more common device than an integrator circuit.

It sounds like your series inductor integration past a power amp is an easy setup to create, I'd be interested in seeing plots made with the integration, and pictures of the coil itself. I do keep a power amp on hand for testing purposes, I could give it a try.

Ken Willmott was of the opinion that putting the integrator between the pickup and oscilloscope was less prone to noise, but I don't have a strong opinion one way or the other.

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33. Here is a test showing the result of two different methods of bode plotting the transfer function of a pickup. The first method has the pickup hooked up directly to the function generator and oscilloscope, and treats the pickup as though it were just a low pass filter, a series inductance. The second method treats the pickup like a very poorly coupled transformer, where the inducing coil is a primary connected to the function generator, and the pickup itself is treated as secondary, hookup up to the oscilloscope.

This is what the test setup looks like when being directly driven by a function generator, and having the voltage difference measure across a 1meg resistor in series:

And this is what the external field coil test setup looks like. There is a tiny excitation coil wrapped around a wood stick:

For this test, I used an Epiphone 57CH PAF style humbucker with a brass cover, so that the "worst case scenario" of a brass cover that causes a high amount of eddy currents.

Here are four plots lines comparing difference measurements:

All four plots are made with the integrator in front of the first channel of the Velleman PSCU2000 to get -6dB/oct.

The two taller peaks (green and red) are with the brass cover removed, while the two lower peaks (blue and black) have the brass cover in place.

Of the two taller peaks with the brass cover removed, the taller green peak was made using the external field coil, while the shorter red peak was produced by driving the pickup directly with a function generator. For some reason, these two peaks show fairly different amplitudes, a difference of 3.5dB, with the external coil test method yielding a higher Q.

Then, of the two shorter peaks, for which the cover was in place, the higher blue line is the pickup being driven directly with a function generator, where as the lower black line was made using the external coil.

There are a few things going on. Interestingly the two testing methods show the greatest difference with the brass cover off, instead of on, which is the opposite of what I would have expected. The Q factors of the "cover off" plots are about 3.5dB apart at the peak. Even though there are differences in need of an explanation, they two test methods are more similar than I thought they would be, possibly suggesting that trying to mimic the geometry of a guitar string with an external field coil is not strictly necessary to create useful data plots, eddy currents and all.

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34. The sum of the voltages does not appear at the terminals of the real coil, although it can approach it at very low frequencies. The sum of the voltages is modified by the various impedances I mentioned before. In particular at the resonance, the output voltage across the coil is greater than the series voltage at that frequency.
This statement is correct. Mike Sulzer is/was right and I apologize. I was on the wrong track.

The total induced voltage/EMF would show at the inductor's terminals only in an open loop situation without any current. But any real world inductor like a PU is terminated at least by its own distributed capacitance. Just as you said.

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35. Is anybody going to wind a bifilar (two wires in parallel) pickup coil and measure its impedance curve in series configuration?

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