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Post by JohnH on Nov 17, 2018 14:19:35 GMT -5
I think that considering eddy current losses in the form of a poorly coupled transformer is a good conceptual model, with some physical credibility. It can be represented by an ideal transformer with additional components. From there, it is possible to derive equivalent models where the equivalent load is referred just to the primary, or just to the secondary, and the ideal transformer taken out. Here is a link that addresses this: electricalacademia.com/transformer/equivalent-circuit-transformer-referred-primary-secondary-side/The final simplified model shown on that page is this: ....which looks not a lot different to the 6-part model, once some winding capacitance is added.
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Post by ms on Nov 18, 2018 6:10:41 GMT -5
If you look into the secondary of a tightly coupled 1:1 transformer with winding resistances R and shorted primary, what you see is 2 x R, i.e. the primary and secondary resistances in series. Loosen the coupling and you add a series leakage inductance to that.
Arthur
In analogy to a pickup: If I measure the impedance of a transformer primary with the secondary open, I see an Rp in series with an Lp. If I put RL across the secondary, the magnitude of the impedance goes down because the RL appears in parallel with the inductance, but, yes, in series with the Rp. That is for tight coupling; loosen it and an L goes in series with RL
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Post by aquin43 on Nov 18, 2018 7:07:18 GMT -5
If you look into the secondary of a tightly coupled 1:1 transformer with winding resistances R and shorted primary, what you see is 2 x R, i.e. the primary and secondary resistances in series. Loosen the coupling and you add a series leakage inductance to that.
Arthur
In analogy to a pickup: If I measure the impedance of a transformer primary with the secondary open, I see an Rp in series with an Lp. If I put RL across the secondary, the magnitude of the impedance goes down because the RL appears in parallel with the inductance, but, yes, in series with the Rp. That is for tight coupling; loosen it and an L goes in series with RL The impedance falls because the current in the secondary due to the load induces a voltage in the primary, the imaginary part of which partially cancels the reactive voltage due to the primary inductance that was keeping the impedance high in the first place. The real part of the induced voltage causes a real current to flow which absorbs power from the source. This is an effective reduction in the primary inductance and the insertion of a resistance. All of this is in series with the secondary, which is why, in the extreme case of shorting the primary, the resistance seen never goes below the sum of the winding resistances (1:1 transformer).
Arthur
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Post by aquin43 on Nov 18, 2018 11:05:44 GMT -5
Calculating the loss components of the coil impedance when the losses are inductively coupled.
I thought it might be useful to show the whole calculation, which uses the circuit below.
The coil L11 is magnetically coupled to the eddy current loop L22 and R22
by mutual inductance M. Calculate the impedance seen between the terminals of L11.
For simplicity, reduce the circuit to a single loop by driving a current of 1A through
the coil and calculate as follows: (using W for Omega)
Subscript convention: V21 is voltage in 2 caused by current in 1, etc. V21 = s*M (voltage induced in L22) M s I21 = (-V21)/(s*L22+R22) (current in L22,R22 loop) M s - ----------- L22 s + R22 V12 = s*M*I21 (voltage induced in L11 by current I21) 2 2 M s - ----------- L22 s + R22 set VW12 = V12 with s = j*W) (put s=jW) 2 2 M W -------------- j L22 W + R22 R12 = realpart(VW12) (real part behaves as voltage across added resistance) 2 2 M R22 W -------------- 2 2 2 L22 W + R22 R12lim = limit of R12 as W goes to infinity 2 M R22 ------ 2 L22 Rcurve = R12/R12lim 2 2 L22 W -------------- 2 2 2 L22 W + R22 L12 = imagpart(VW12)/W (imaginary part behaves as a negative inductance) 2 2 L22 M W - -------------- 2 2 2 L22 W + R22 L12lim = limit of L12 as W goes to infinity 2 M - ---- L22 Lcurve = L12/L12lim 2 2 L22 W -------------- same as Rcurve 2 2 2 L22 W + R22 Rser = R12lim with M = k*sqrt(L11*L22) (k is coupling coefficient) 2 L11 R22 k ---------- L22 Lser = L12lim with M = k*sqrt(L11*L22) 2 - L11 k
So we have a positive resistor and a negative inductor in series with the coil, both with values that are modulated by a frequency dependent function with no phase shift, of the form 2 W -------------- 2 2 2 W + W / We
where We = R22/L22
The final result is quite simple in form. One would expect to see the resistance rise with frequency and the inductance fall.
I wrote a program that translates the measured amplitude and phase of the pickup impedance into resistance and inductance. With a pickup that has eddy current losses, the expected behaviour is observed. It can only be measured below about 1kHz because the pickup capacitance rapidly starts to interfere with the impedance calculation as the frequency rises.
Arthur
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Post by ms on Nov 18, 2018 17:26:56 GMT -5
I do not think that it make sense to require a negative inductance so that you can put it in series. Of course it makes sense to express the impedance in the form R(omega) +JX(omega), but if you want to make an equivalent model, you can have a positive inductance (in series with the resistance) in parallel with the primary.
It is possible to measure the eddy current effect at higher frequencies by making the measurements at enough frequencies so you can do fit for the parameters.
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Post by antigua on Nov 18, 2018 21:48:34 GMT -5
Here is a direct comparison between plotting the impedance by driving the pickup and measuring the difference across a 1 meg resistor in series, versus plotting the voltage across the pickup when driving the pickup with an exciter coil externally (poorly coupled transformer). There are four plot lines, each of the two methods, with and without the integrator. Note that the curves are nearly the same, no matter the method used, but the direct method, measuring the difference across a series resistor, shows a drop in impedance from the start, at 100Hz. The integrated plot line (blue) shows it very clearly. Meanwhile, the method that makes used of the external coil looks fairly flat from 100Hz, although it does show an initial increase and then a mild drop off past 200Hz that continues as the frequency increases. What this shows, IMO, is that measuring the impedance of the pickup across a series resistor, or in a transformer-like configuration, yield close enough results, as far as the resonance and Q factor are concerned, but that increased impedance between 100Hz and 1kHz creates a distracting slope where a flat line would be expected, which might mislead someone into thinking that the response curve of a pickup has a +10dB emphasis in the bass frequencies. It would be nice to find some way to flatten out the measurement across the resistor, as this testing rig is somewhat easier to assemble (not requiring a driver coil, or having to figure out where to place it) and would encourage more people to give it a try. It also gives way to better standardization, since there are less variables involved. The problem is, I'm not sure if there is a solution to this problem, or if it's inherent to the test method.
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Post by aquin43 on Nov 19, 2018 4:11:05 GMT -5
I do not think that it make sense to require a negative inductance so that you can put it in series. Of course it makes sense to express the impedance in the form R(omega) +JX(omega), but if you want to make an equivalent model, you can have a positive inductance (in series with the resistance) in parallel with the primary. It is possible to measure the eddy current effect at higher frequencies by making the measurements at enough frequencies so you can do fit for the parameters. I am not saying that the model should incorporate this particular negative inductance, which varies with frequency. The purpose of the calculation was to make clear how the measured inductance and resistance vary with frequency. The extra resistance at higher frequencies is the reason why the Q of the circuit is always lower than would be expected from measuring the dc resistance and low frequency inductance.
A model would include the mutually coupled loss, which would lead naturally to the negative inductance. There is a simple T network equivalent to a mutually coupled coil, so the simulator would not even need to be able to handle mutual coupling. There is also a frequency dependent loss between the string excitation and the coil, which would need to be included. This, again, would be dependent on a mutual coupling with the eddy current loop.
The pickup is a filter with a measurable amplitude response. The amplitude response can be emulated by many RLC networks, including the currently accepted on with parallel resistors and inductor, but none of these will have the correct output impedance and so will not interact with loads in exactly the same way as the real pickup.
Arthur
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Post by aquin43 on Nov 19, 2018 4:23:57 GMT -5
Antigua,
I would like to advise on the method for making accurate impedance measurements but I am uncertain of the capabilities of your test equipment.
What is the highest generator signal level?
What is the input impedance?
What is the maximum sensitivity?
Can you measure phase relative to the generator?
Arthur
PS Note that the impedance measurement does not include the frequency dependent forward loss between string and coil that is important in pickups with high eddy current losses.
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Post by antigua on Nov 19, 2018 11:43:02 GMT -5
Antigua, I would like to advise on the method for making accurate impedance measurements but I am uncertain of the capabilities of your test equipment. What is the highest generator signal level? What is the input impedance? What is the maximum sensitivity?
Can you measure phase relative to the generator? Arthur This is the device I use www.velleman.eu/products/view/?id=407512 , I think the specs for the unit might provide the answers. PS Note that the impedance measurement does not include the frequency dependent forward loss between string and coil that is important in pickups with high eddy current losses. As noted above, the fact that the Q factor between the direct impedance plot, and the external exciter method, are so similar, suggests to me that the eddy current losses caused in relation to the guitar strings are probably very tiny. It makes sense that the poor degree of magnetic coupling between the strings and the conductive metal pieces would mean a poor degree of eddy currents, since they can be thought of as relating like a transformer. The coupling between the strings and the coil is also poor, but the coil has a significant number of turns, and so a usable voltage is produced despite the poor coupling.
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Post by ms on Nov 19, 2018 12:46:50 GMT -5
I do not think that it make sense to require a negative inductance so that you can put it in series. Of course it makes sense to express the impedance in the form R(omega) +JX(omega), but if you want to make an equivalent model, you can have a positive inductance (in series with the resistance) in parallel with the primary. It is possible to measure the eddy current effect at higher frequencies by making the measurements at enough frequencies so you can do fit for the parameters. I am not saying that the model should incorporate this particular negative inductance, which varies with frequency. The purpose of the calculation was to make clear how the measured inductance and resistance vary with frequency. The extra resistance at higher frequencies is the reason why the Q of the circuit is always lower than would be expected from measuring the dc resistance and low frequency inductance.
A model would include the mutually coupled loss, which would lead naturally to the negative inductance. There is a simple T network equivalent to a mutually coupled coil, so the simulator would not even need to be able to handle mutual coupling. There is also a frequency dependent loss between the string excitation and the coil, which would need to be included. This, again, would be dependent on a mutual coupling with the eddy current loop.
The pickup is a filter with a measurable amplitude response. The amplitude response can be emulated by many RLC networks, including the currently accepted on with parallel resistors and inductor, but none of these will have the correct output impedance and so will not interact with loads in exactly the same way as the real pickup.
Arthur
I do not see why the concept of a negative inductance is useful in showing that the measured or apparent inductance drops with frequency. That is apparent from the the decreasing slope on a plot of the imaginary part of the measured impedance versus frequency after you "compute out" the effect of the C. I think it is possible to construct an emulator that gets the output impedance right from a measurement of impedance versus frequency. From the impedance at low frequencies you derive the coil resistance Rc and (low frequency, or true) inductance Lc. These in series become the series leg of a voltage divider. From the measurement impedance, compute a new impedance Zshunt, such that Z shunt in parallel with the impedance of Rc in series with LC gives the measured impedance. Lc in series with Rc is the series leg of a voltage divider, and Zshunt is the shunt leg. Now constructs an RLC network that is as close to Zshunt as you like. You need the full impedance, not just its magnitude. The attached plot shows the kind of impedance measurements that you need in order o do this. do asgardians sleep
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Post by aquin43 on Nov 19, 2018 13:04:59 GMT -5
Hello,
I had a look at the spec for the Velleman scope and it seems as if it should be capable of doing the measurement with some a assistance. Is the 10mV Bode sensitivity that for full scale - if so, it seems higher than the quoted scope sensitivity? The 8VP-P output is a little low, since a resistor of much more than 1M is required. At least 4M7 and preferably 10M. It appears in parallel with the pickup and so has to be corrected for. It is necessary to be able to output the bode data, gain and phase, preferably raw - not converted to dB. A little battery powered preamp is also necessary. If the scope sensitivity is high enough, this can be unity gain. A fet input OP Amp has to be used. At unity gain a TL072 will do, but something like an OPA1642 would be better from the point of view of phase. You always get phase shift problems long before bandwidth limits. The buffer would have no input resistor other than the test resistor so its input resistance would be very high; the input impedance would be just a small amount of capacitance.
In an earlier post in this thread, I described how the loading effect of the test resistor can be removed by calculation. This needs the phase information at any frequency other than the zero phase point near the peak. I use a language called gawk or this, principally because it makes reading lists of data trivially simple, Gawk is available free for Windows.
Concerning the effect of the eddy current losses on the coupling between the strings and the coil. The effect can be considerable. It takes the form of a downward step in frequency response and is the reason for the dip in response before the peak in many pickups. The field from the strings is about as tightly coupled to the eddy currents as the coil is. The effect acts in some ways as a pre-filter on the excitation of the coil. In the current standard model this dip is made part of the main coil response, but, although it is produced by the same eddy current loop, it is a separate effect that does not appear in the impedance plot.
Arthur
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Post by aquin43 on Nov 19, 2018 13:38:50 GMT -5
I am not saying that the model should incorporate this particular negative inductance, which varies with frequency. The purpose of the calculation was to make clear how the measured inductance and resistance vary with frequency. The extra resistance at higher frequencies is the reason why the Q of the circuit is always lower than would be expected from measuring the dc resistance and low frequency inductance.
A model would include the mutually coupled loss, which would lead naturally to the negative inductance. There is a simple T network equivalent to a mutually coupled coil, so the simulator would not even need to be able to handle mutual coupling. There is also a frequency dependent loss between the string excitation and the coil, which would need to be included. This, again, would be dependent on a mutual coupling with the eddy current loop.
The pickup is a filter with a measurable amplitude response. The amplitude response can be emulated by many RLC networks, including the currently accepted on with parallel resistors and inductor, but none of these will have the correct output impedance and so will not interact with loads in exactly the same way as the real pickup.
Arthur
I do not see why the concept of a negative inductance is useful in showing that the measured or apparent inductance drops with frequency. That is apparent from the the decreasing slope on a plot of the imaginary part of the measured impedance versus frequency after you "compute out" the effect of the C. I think it is possible to construct an emulator that gets the output impedance right from a measurement of impedance versus frequency. From the impedance at low frequencies you derive the coil resistance Rc and (low frequency, or true) inductance Lc. These in series become the series leg of a voltage divider. From the measurement impedance, compute a new impedance Zshunt, such that Z shunt in parallel with the impedance of Rc in series with LC gives the measured impedance. Lc in series with Rc is the series leg of a voltage divider, and Zshunt is the shunt leg. Now constructs an RLC network that is as close to Zshunt as you like. You need the full impedance, not just its magnitude. The attached plot shows the kind of impedance measurements that you need in order o do this. do asgardians sleepHello,
I am not advocating using a negative inductance in the model. The negative inductance occurs when you calculate the impedance of a coil coupled to a lossy inductive loop. My point is that the coupling is what is actually happening. It is the actual reason why the inductance falls and the resistance rises, so why not model it directly. None of the model components has to be physically constructed, so it doesn't matter what they are. I am trying to arrive at a model that reflects the underlying physics and if it needs mutual inductance, so be it. Behavioural matching of the frequency response seems to have been solved long ago.
By the way, your computed subtractions of the capacitive load are neat. I have found the calculation to become very unstable as I approach the resonant frequency. Even using data from a simulation, where the capacitance was known exactly, I found had an extreme sensitivity to the capacitor value.
Arthur
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Post by antigua on Nov 19, 2018 19:36:56 GMT -5
Hello, In an earlier post in this thread, I described how the loading effect of the test resistor can be removed by calculation. This needs the phase information at any frequency other than the zero phase point near the peak. I use a language called gawk or this, principally because it makes reading lists of data trivially simple, Gawk is available free for Windows. I might be able to add the phase angle, the steps involved a little more complex and I forgone it since I'm mostly interested in just acquiring a response curve for audio analysis specifically. But to confirm, are you saying that the high impedance curve that approaches 0Hz is related to the resistor, and that it can be factored out if the phase angle is known? Concerning the effect of the eddy current losses on the coupling between the strings and the coil. The effect can be considerable. It takes the form of a downward step in frequency response and is the reason for the dip in response before the peak in many pickups. The field from the strings is about as tightly coupled to the eddy currents as the coil is. The effect acts in some ways as a pre-filter on the excitation of the coil. In the current standard model this dip is made part of the main coil response, but, although it is produced by the same eddy current loop, it is a separate effect that does not appear in the impedance plot.
Arthur As a possible counter point, someone on another forum believe the droop is related to the exciter coil and has created a model with LTSpice that is able to recreate the effect www.tdpri.com/threads/physically-based-eddy-current-equivalent-circuit.846745/page-3#post-8531564I agree that the string is as tightly (or loosely) coupled to the steel parts as the guitar string (for the sake of argument), but the coil is far more substantial than the guitar string, both in physical size as well as the size of the magnetic field that they produce, and so even for the same degree of coupling, the amount of interaction between the coil and the steel parts will be a lot greater than the interaction between the string and the steel parts. I'm not sure if you understood what I was getting at before, but this plot shows measurements with and without an exciter coil, and you can see that the eddy current situation is nearly identical with and without. Unless this can be explained some other way, I don't think there's a debate to be had about it.
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Post by aquin43 on Nov 20, 2018 6:04:55 GMT -5
Hello,
I am familiar with the model that you refer to and, indeed, my own speculations are closely related to it. The calculations I have made are based on a pickup coil that is coupled to an eddy current loop.
The model proposed in that reference does not include any loading of the pickup coil by the the exciter coil. The exciter is driven by a current source and so is open circuit. This means that nothing the main coil does can influence the current in the exciter. Therefore, since magnetic coupling relies on changing electric currents (which, in turn, produce changing magnetic fields), the exciter cannot take energy from the main coil. This same condition will hold with a properly designed real exciter coil because, in order to have the exciter current depend directly on the generator voltage, the coil will be in series with a resistor that is much greater than its impedance - it will thus appear to be virtually open circuit.
Concerning the impedance plot with the integrator. Nothing is achieved by integrating the impedance plot, nor by expressing it in dB. Any pickup becomes purely resistive at low frequencies where L becomes unimportant, so the impedance curve becomes flat. If you integrate that flat curve you end up with a rise at the low frequency end that is purely an artefact of the processing and has no physical reality. If you convert the raw impedance plot back from dB to volts and scale it by multiplying by (1M0/generator-volts) you end up with a plot of the impedance of the pickup in parallel with one megohm. At the peak, the one megohm can be factored out but phase information is also required to do this accurately at other frequencies.
The purpose of the integrator is purely to process the naturally rising frequency response into a form that is more amenable to human intuition. The rising frequency response occurs in the first place because all magneto-electric coupling depends on rates of change of magnetic fields.
The coupling of the strings to the main coil is via the perturbations they make in the magnetic field. This can be considered as an alternating magnetic signal superimposed on a static field. The static field is always ignored, because only changing magnetic fields can induce voltages.
So, we have a magnetic signal, part of which couples directly to the main coil and part to the eddy current loop. The eddy current loop is also coupled to the main coil, so this constitutes a second path for the magnetic signal into the main coil. The eddy current path has a frequency response that levels off at high frequencies so, when the two are added together the effect is a certain overall coupling at low frequencies which falls to a lower level at high frequencies. This is the cause of the droop before the peak in a lossy pickup when viewed through the integrator. Your green Fidelitron curve is a good example.
Note that, while currents in the main coil can influence the string (think of the e-bow), the strings are so massive that the effect is easily small enough to be neglected, so the magnetic signal from the string can be considered as being independent of the pickup response or loading.
Arthur
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Post by ms on Nov 20, 2018 6:36:34 GMT -5
Hello,
I am not advocating using a negative inductance in the model. The negative inductance occurs when you calculate the impedance of a coil coupled to a lossy inductive loop. My point is that the coupling is what is actually happening. It is the actual reason why the inductance falls and the resistance rises, so why not model it directly. None of the model components has to be physically constructed, so it doesn't matter what they are. I am trying to arrive at a model that reflects the underlying physics and if it needs mutual inductance, so be it. Behavioural matching of the frequency response seems to have been solved long ago.
By the way, your computed subtractions of the capacitive load are neat. I have found the calculation to become very unstable as I approach the resonant frequency. Even using data from a simulation, where the capacitance was known exactly, I found had an extreme sensitivity to the capacitor value.
Arthur
Thanks. The capacitance is derived from data above the peak using non-linear least squares fitting. The model used includes eddy current parameters. This works well as indicated the very close values of the capacitances derived in the two cases with very different impedances resulting from eddy currents same coil, of course.
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