Measuring pickups Capacitance?

[Helmholtz]

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.

[Helmholtz]

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.

[Mike Sulzer]

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.

[Helmholtz]

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.

[Antigua]

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.

[Mike Sulzer]

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.

[Helmholtz]

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.

[Helmholtz]

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.

[Mike Sulzer]

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.

[Mike Sulzer]

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.

[Helmholtz]

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.

[Antigua]

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.

[Helmholtz]

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?

[Helmholtz]

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.

[Mike Sulzer]

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.