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Load
Jun 20, 2019 16:25:36 GMT -5
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Post by Deleted on Jun 20, 2019 16:25:36 GMT -5
Can someone define Load to me Example Neck pickup is selected And the bridge pickup is coming in via a blend (Resister) But they say there is no load from the blend and bridge pickup!
To me , the blend and bridge pickup is defecting the total resistance 1/((1/Rn)+(1/(R+Rb))) Rn=neck DC resistance ... R = Blend resistance and Rb=bridge resistance
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Post by reTrEaD on Jun 21, 2019 10:12:36 GMT -5
Can someone define Load to me When it comes to defining one pickup as a load, in relation to another pickup, that gets very VERY messy. You certainly need to take more into account than the dc resistance of the pickup (and consider the internal inductance and capacitance) but more than the complex impedance, the pickup is also a signal source, so that has a huge effect on the 'load' it presents to the other pickup. At the fundamental frequency of the string, the signal from both pickups will be in-phase, so all we would need to do is determine the relative amplitude and use that along with the impedance each presents to do some calculations. But tone is all about harmonic content, and the harmonics each pickup sees will vary wildly in both amplitude and phase relationship, due to the location on the string they are sensing and the specific harmonic we consider. I couldn't begin to define that, other than to say ... It's complicated.
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Load
Jun 22, 2019 8:19:01 GMT -5
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Post by thetragichero on Jun 22, 2019 8:19:01 GMT -5
i see what you did there!
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Load
Jun 22, 2019 10:22:46 GMT -5
Post by sumgai on Jun 22, 2019 10:22:46 GMT -5
'bunny,
In our case (that of guitarists and their gear), we can't stop at just resistance or impedance, we also need to consider the current being generated by the pickup, and the Back EMF (VL) being generated by these signal sources. And of course, since we are speaking of frequency, we are also throwing another variable into the mix, known as reactance. Then we need to add the interaction between various frequencies (harmonics), and not just at any given single point in time, but over a time span as current rises and falls. (Remember, voltage and current are not in lockstep with each other, there is a finite time period of lead and/or lag between them.) Woah!!
Like reTrEaD said, it gets complicated - in a hurry!
Without going into a bunch of math and lots of diagrams, it should be sufficient to say that the bottom line is, we don't care about one pickup loading another. It's a fact of life, we can't do anything about it, so we might as well ignore it, and just get on with those things we can control. In other words, pick and choose your battles... life is always better when you can keep it simple!
HTH
sumgai
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Load
Jun 23, 2019 5:31:16 GMT -5
Post by perfboardpatcher on Jun 23, 2019 5:31:16 GMT -5
Can someone define Load to me Example Neck pickup is selected And the bridge pickup is coming in via a blend (Resister) Electrical load: any electrical component (or:anything) that draws current? It's also important to realise that a guitar pickup is a sensor. Signal amplification is what matters and not how much current goes to whatever is connected up to the guitar. So yes, bridge pickup + blend resistor is loading the neck pickup but I personally find it more convenient to look towards it as voltage attenuation. What happened to the output level (voltage), what happened to the tone (signal level across the whole frequency spectrum)? But they say there is no load from the blend and bridge pickup! To me , the blend and bridge pickup is defecting the total resistance 1/((1/Rn)+(1/(R+Rb))) Rn=neck DC resistance ... R = Blend resistance and Rb=bridge resistance Testing by multimeter seems like a legit Karl Popper to me, good work, A'SB !
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Jun 30, 2019 2:10:04 GMT -5
Post by perfboardpatcher on Jun 30, 2019 2:10:04 GMT -5
two 9 volt batteries with same voltage and internal resistance in parallel: one can calculate the current per battery: I_1=9/(2*R_batt) and then I_2=-(9/(2*R_batt) because of the direction of the current. Overall current flow through the two batteries is zero.
Same goes for guitar pickups. Instead of current calculate voltages: 1) N pup as voltage source loaded by B pup, then 2) B pup as voltage source loaded by N pup. (voltages-impedances-thevenin equivalents)
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Load
Jul 1, 2019 13:49:54 GMT -5
Post by sumgai on Jul 1, 2019 13:49:54 GMT -5
.... Same goes for guitar pickups. Are you sure you wanna go down that path?
A battery transforms energy from a chemical reaction into an electrical action... it's all internal. A pickup tranforms a magnetic field into an electrical action, and thus it requires an outside stimulus - the string's vibrations. Those two different transformations are comparable only a rudimentaray way, but the way you espouse breaks down when the above points are considered. Reason being, a battery is DC, and a vibrating string is generating a frequency, which by definition (and after the transformation), is AC. Thus, reactance is going to play a rather noticible part in the equations governing load.
Which is why I didn't/don't want to get into it. This is one of thoses things that really require a large dose of advanced knowledge, far, far more than that needed by the average guitarist in order to mod their rigs. (Or even well-advanced pickers, to be sure.) You all can go at it, if it tickles your fancy, but I'd caution you to not be so surprised when your bench models break down in real life. (And come to that, when was the last time that any of you invested in some laboratory-grade test equipment, to check your final results? Still depending on your ears to see if there was an improvement? My my, tch, tch, tch. That won't win any Brownie points in my book.)
sumgai
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Jul 2, 2019 13:38:28 GMT -5
Post by perfboardpatcher on Jul 2, 2019 13:38:28 GMT -5
Are you sure you wanna go down that path?
The OP only asked to define "load" but since you're asking, yes, I already went down that path some while ago and constructed a spreadsheet for a 2 humbucker model with one voltage source per coil. Parallel humbucker I cheat (thevenin impedance) but single coil, humbucker and their series and parallel combinations are calculated with one voltage source per coil.
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Load
Jul 2, 2019 16:15:29 GMT -5
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Post by JohnH on Jul 2, 2019 16:15:29 GMT -5
The OP only asked to define "load" but since you're asking, yes, I already went down that path some while ago and constructed a spreadsheet for a 2 humbucker model with one voltage source per coil. Parallel humbucker I cheat (thevenin impedance) but single coil, humbucker and their series and parallel combinations are calculated with one voltage source per coil. Ok Im interested in that! What you describe there sounds similar to GuitarFreak. In that, I just use one voltage source per pickup, but can combine up to 3 together in series or oarallel. Depending on how you model coil impedance and guitar controls, it seems like there should be enough overlap between the two spreadsheets to each represent the same thing across a selection of configurations. Hence, if you are interested, there should be a useful opportunity for us each to have a QA check on our maths and code by each analyzing the same arrangement (within parameters that both can model) and seeing how they are the same or different. Any interest in that?
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Load
Jul 9, 2019 0:30:10 GMT -5
Post by sumgai on Jul 9, 2019 0:30:10 GMT -5
The OP only asked to define "load" but since you're asking, yes, I already went down that path some while ago and constructed a spreadsheet for a 2 humbucker model with one voltage source per coil. Parallel humbucker I cheat (thevenin impedance) but single coil, humbucker and their series and parallel combinations are calculated with one voltage source per coil. Well, that gets my attention... for not what you might call the right reason.
Thevenin's Theorem deals with replacing a source with a voltage-producing 'black box', i.e. a constant voltage. But pickups, by their very nature (that being magnetic), are current devices, not voltage. True, you can't have one without the other, but when it comes to analyzing components, it helps to remember that capacitors pass AC voltage, and inductors tend to resist changes in current - not voltage. Thus, I suggest that you try using Norton's Theorem, that of replacing a current source with a black box designed for such analysis. Your 'cheating' should come closer to real-world results.
Or at least that's how we did it, back when I went to school (when dirt was still a novelty ).
HTH
sumgai
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Load
Jul 9, 2019 1:05:04 GMT -5
Post by sumgai on Jul 9, 2019 1:05:04 GMT -5
I'm gonna get up on my pedestal here, and define Load. Stay with me, it's gonna be a bit bumpy, and I can't guarantee that you'll be a happy camper at the end, but here we go.
Load is best defined as one or more components that complete a circuit, but do not, in and of themselves, produce any current, voltage, or power (the combination of the two). That means, everything outside of the power source is part and parcel of the thing we call a "load" - they all have an effect on the source. And has been mentioned above, the source itself is never "perfect", so it also provides a load, to itself.*
That said, we often deal with a "raw" source as just that, and don't consider what's going on inside of it - we can't do anything about it, so why bother. But if one really needs to do such an analysis, one could look up both Nodal Analysis and Mesh Analysis.... and that's just for starters.
Like reTrEaD said, it's not just messy, it's VERY messy. Indeed.
HTH
sumgai
* Which explains, in simple terms, why batteries go dead, seemingly when they've done no work (never been hooked up to anything) - internal loading.
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Load
Jul 9, 2019 4:17:15 GMT -5
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Post by JohnH on Jul 9, 2019 4:17:15 GMT -5
Load means lots of different things. But for electoral purposes, I think of a 'load' as a generally passive circuit or device that draws power from the circuit being 'loaded'. The load could be resistive, reactive or highly non-linear. But, so far as it acts as a load, it doesn't generate any power itself.
But, a given device such as a pickup, could act as both a load (in its effect on other pickuos) and as a source.
And on pickups, we definately can and do model them as voltage sources connected to a network of impedances, and learn much in the process
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Load
Jul 9, 2019 10:31:16 GMT -5
Post by reTrEaD on Jul 9, 2019 10:31:16 GMT -5
But for electoral purposes, I flunked out of that college. (Don't you just love autocorrect?) John, I reckon you can do a very accurate job of modeling the interaction of pickups that way ... at the fundamental frequency present on the string. At the fundamental frequency the phase relationship of the voltage sources will be either 0 degrees or 180 degrees and the amplitude will be rather predictable. But can your software properly model the AC voltage source of each pickup's harmonic content in both amplitude and relative phase?
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Load
Jul 9, 2019 10:49:53 GMT -5
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Post by JohnH on Jul 9, 2019 10:49:53 GMT -5
electoral load! I love that idea!
Yes there's no insumountable problem about capturing the right phase and harmonic content. To get it right it has to model pickup positions and string vibration though. GuitarFreak gets this fairly right, for combos of similar coils connected to each other , and 'usefully good enough' for combinations of significantly different coils. Its been there for a couple of years, but the next stage to accurateltmy capture very different coils in combinations, and subtleties such as hoop and part bypassed coils, will need a rewrite do-over.
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Jul 10, 2019 14:50:48 GMT -5
Post by perfboardpatcher on Jul 10, 2019 14:50:48 GMT -5
The OP only asked to define "load" but since you're asking, yes, I already went down that path some while ago and constructed a spreadsheet for a 2 humbucker model with one voltage source per coil. Parallel humbucker I cheat (thevenin impedance) but single coil, humbucker and their series and parallel combinations are calculated with one voltage source per coil. Ok Im interested in that! What you describe there sounds similar to GuitarFreak. In that, I just use one voltage source per pickup, but can combine up to 3 together in series or oarallel. Depending on how you model coil impedance and guitar controls, it seems like there should be enough overlap between the two spreadsheets to each represent the same thing across a selection of configurations. Hence, if you are interested, there should be a useful opportunity for us each to have a QA check on our maths and code by each analyzing the same arrangement (within parameters that both can model) and seeing how they are the same or different. Any interest in that? Sorry, John. At this moment I'm more into tweaking gnu octave code which I use to measure. The 'SoundCardSetup' program that aquin43 mentioned in 'Digital Impedance Measurement' didn't work on my Linux distro but was a good starting point to start programming myself. That and the gnu octave manual and mathworks examples. It's just a matter of dividing it up into smaller projects and go with the flow. I managed to create a sequence of sine tone frequencies (48, 8 octaves of the whole tone scale) which I use as test signal. This approach enables me to save the results in a 48 x 3 matrix and plot the results whenever I want. You should seriously take a look at gnu octave. The 'Guitarfreak' seems to me already good as it is.
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Jul 10, 2019 14:58:00 GMT -5
Post by perfboardpatcher on Jul 10, 2019 14:58:00 GMT -5
The OP only asked to define "load" but since you're asking, yes, I already went down that path some while ago and constructed a spreadsheet for a 2 humbucker model with one voltage source per coil. Parallel humbucker I cheat (thevenin impedance) but single coil, humbucker and their series and parallel combinations are calculated with one voltage source per coil. Well, that gets my attention... for not what you might call the right reason.
Thevenin's Theorem deals with replacing a source with a voltage-producing 'black box', i.e. a constant voltage. But pickups, by their very nature (that being magnetic), are current devices, not voltage. True, you can't have one without the other, but when it comes to analyzing components, it helps to remember that capacitors pass AC voltage, and inductors tend to resist changes in current - not voltage. Thus, I suggest that you try using Norton's Theorem, that of replacing a current source with a black box designed for such analysis. Your 'cheating' should come closer to real-world results.
Or at least that's how we did it, back when I went to school (when dirt was still a novelty ).
HTH
sumgai
And if I follow your suggestion what are my chances to wind up having a worksheet that works as beautifully as the one I created myself? I mean if I compare with the measurements I do then my model doesn't seem that far off. And if eddy losses play up I will cheat a little bit. I confess.
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Jul 10, 2019 16:49:20 GMT -5
Post by blademaster2 on Jul 10, 2019 16:49:20 GMT -5
The OP only asked to define "load" but since you're asking, yes, I already went down that path some while ago and constructed a spreadsheet for a 2 humbucker model with one voltage source per coil. Parallel humbucker I cheat (thevenin impedance) but single coil, humbucker and their series and parallel combinations are calculated with one voltage source per coil. Well, that gets my attention... for not what you might call the right reason.
Thevenin's Theorem deals with replacing a source with a voltage-producing 'black box', i.e. a constant voltage. But pickups, by their very nature (that being magnetic), are current devices, not voltage. True, you can't have one without the other, but when it comes to analyzing components, it helps to remember that capacitors pass AC voltage, and inductors tend to resist changes in current - not voltage. Thus, I suggest that you try using Norton's Theorem, that of replacing a current source with a black box designed for such analysis. Your 'cheating' should come closer to real-world results.
Or at least that's how we did it, back when I went to school (when dirt was still a novelty ).
HTH
sumgai
Whether you decide to use Thevenin's or Norton's equivalent usually depends on the magnitude of the load impedance versus the source impedance. A much higher source impedance, which is possibly true for a pickup, tends to behave closer to a current source. However, ignoring the cable capacitance for the moment, the amplifier input impedance is usually much higher still than the pickup impedance across much of the frequencies. As such it is not an obvious choice, and more likely either one could be used. Like JohnH, I have generally used Thevenin models for guitar pickups whenever I did simulations. This makes sense to me with the signal being fed to a tube amp, which is also a voltage-sensitive device that draws very little current. Using that approach, along with the principle of superposition, should (in theory) yield a fairly good prediction of the response - provided the complex impedances of the sources/loads in the end-to-end circuit of the guitar/cable/amplifier are modeled correctly. As with any simulation models, the modeling part of it - correctly modeling the complex impedances of the system - is where the real challenge lies.
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Jul 15, 2019 10:49:08 GMT -5
Post by sumgai on Jul 15, 2019 10:49:08 GMT -5
As with any simulation models, the modeling part of it - correctly modeling the complex impedances of the system - is where the real challenge lies. True, that.
I recognize that not everyone thinks along the same lines when it comes to deriving a solution to a problem. I guess that devolves as much to one's education as much as one's experiences in the field. I can't say one way is "more" correct than another, but I'll bet dollars to donuts that if enough people participate in finding a solution, and do so as a group, then the final "answer" is probably going to be as close to correct as we can ask for.
That's all I have... for now.
sumgai
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Aug 18, 2019 5:34:35 GMT -5
Post by Yogi B on Aug 18, 2019 5:34:35 GMT -5
The pickup is also a signal source, so that has a huge effect on the 'load' it presents to the other pickup. At the fundamental frequency of the string, the signal from both pickups will be in-phase, so all we would need to do is determine the relative amplitude and use that along with the impedance each presents to do some calculations. But tone is all about harmonic content, and the harmonics each pickup sees will vary wildly in both amplitude and phase relationship, due to the location on the string they are sensing and the specific harmonic we consider. While this is true, this is most useful if you are attempting to model pickups and their interactions from scratch, e.g. guitar (re)synthesis. I don't think it's all that useful when considering the effect that circuitry placed within a regular guitar has on the signal -- remember in the OP angelsbunny is concerned with the effect on loading due to blend control. Anything we do to try and blend out a pickup won't -- or at least, the aim is that it shouldn't -- do a great deal to change the overall distribution harmonic content that pickup contributes, i.e. the frequency of the nulls will remain constant, and the relative phase shouldn't vary a great deal either. Putting that another way, at any frequency -- irrespective of whether there is a difference in phase or amplitude between the output of the pickups, or not -- the maximum amount of loading occurs when we have no external impedance between the pickups, and any series impedance we do add to either pickup can only reduce the amount of loading. Edit: Scratch the above, it only makes sense from the perspective of the pickup we're not blending. From an electrical point of view I tend to think of loading as current division, i.e. a component that adds load to a circuit is one which provides a path for current to flow such that it can bypass the input of the following stage. However, dividing the current isn't the only way to reduce the amount entering the following stage, we can also reduce the total current by adding impedance in series with the signal, but now not only do you have a current divider but also a voltage divider. The latter being what occurs to the signal from the blended pickup. Yes the series resistance from the blend pot is reducing the current that makes it to the output, but this is usually insignificant due to the following stage having a large input impedance. What is more significant is that the added series resistance narrows the bandpass filter formed by the inductance and capacitance of the other pickup and anything else that follows e.g. cable capacitance. But point I was trying to make still stands -- the loading due to the impedance of a pickup always applies, independent of how the signals sum. The apparent loading due to phase cancellation is exactly that, a result of phase cancellation, as such we only need to consider the relative amplitude of pickups' outputs.
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Aug 19, 2019 11:11:30 GMT -5
Post by blademaster2 on Aug 19, 2019 11:11:30 GMT -5
The effect of loading, either by one pickup on the other or by other resistances in series and/or parallel with them, will always obey Kirchoff's Law and will include the principle of superposition. The trick will always be to model the non-resistive circuit elements in that model (capacitances and inductances, effect of eddy currents, non-linear and frequency-dependent values of many parameters including magnetization curves, temperature dependencies, et cetera).
Keeping things simple/possible to model, assuming all of the non-linearities and variations do not exist, still is useful but the need for the model to consider Kirchoff's Law and superposition cannot be ignored. This applies whether Thevenin or Norton equivalents are used.
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Aug 19, 2019 17:16:43 GMT -5
Post by Yogi B on Aug 19, 2019 17:16:43 GMT -5
That's the word I was looking for, and I agree. I know the principle, but I have a bit of a blind spot for the word because I more strongly associate it with quantum mechanics. So, more succinctly than my previous attempts: reTrEaD, I know you said load in inverted commas, but to me it reads like you are trying to say that because a pickup acts as a signal source, we need to accurately determine that signal in order to calculate the loading effect that this pickup's impedance has on the another pickup, which (assuming linearity of all the components) is contrary to the superposition theorem.
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Aug 20, 2019 10:09:25 GMT -5
Post by reTrEaD on Aug 20, 2019 10:09:25 GMT -5
reTrEaD, I know you said load in inverted commas, but to me it reads like you are trying to say that because a pickup acts as a signal source, we need to accurately determine that signal in order to calculate the loading effect that this pickup's impedance has on the another pickup I reckon you read that as-intended. And I don't believe I'm wrong.
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Aug 20, 2019 10:36:10 GMT -5
Post by blademaster2 on Aug 20, 2019 10:36:10 GMT -5
If I have understood both statements correctly, I believe the assumption of linearity with signal and magnetic flux *would* permit superposition to be used and I would go this way myself. However if this assumption is not suitable then the level of signal and flux in the core from one pickup would change the behaviour of the pickup for the superimposed signals from the other, and superposition principles would not work accurately (unless new values for model elements were used to account for the effect of the nonlinearities, but that would require a different circuit model for each case and signal level).
All of these nuances are fun to discuss in theory, but to model these in any accurate way that produces results that can be trusted would certainly make *my* brain hurt. Nonetheless I have been surprised recently when changes in configuration do not produce similar expected sonic results between different guitars/pickups, so I admit that simplifying assumptions are not always accurate but I am too lazy/time-starved to dig deeper. I would turn to laboratory testing, and even more so to listening to the pickups themselves with a guitar to get the true sense of what the effects is. Ultimately, even frequency response curves can only go so far in predicting how something will sound - my favourite test apparatus is a guitar, amplifier and my ears.
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Aug 20, 2019 14:24:07 GMT -5
Post by perfboardpatcher on Aug 20, 2019 14:24:07 GMT -5
I have a question for you guys. Consider the next test setups.
A. A guitar is plugged into some experimental stompbox that's nothing more than a paf humbucker (with cover) mounted on top of the stompbox. The paf humbucker is merely serving as a load. The stompbox is connected to an amp.
B. Same as in (A) but now an excitation coil (driven by a function generator into a poweramp) is placed above the paf humbucker. The excitation coil and paf humbucker are placed to form a cross in such way that voltage is induced in both of the coils of the paf hb but also so that at the same time the 2 signals of the paf pup coils cancel each other out. So from the outside the paf will seem to function as a load. But on the other hand I would suspect there will be considerable eddy current losses.
Will there be a difference in sound experience between A and B?
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Aug 20, 2019 16:03:49 GMT -5
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Post by JohnH on Aug 20, 2019 16:03:49 GMT -5
I'll have a go, and Im going to guess thar there will be no difference between A and B, within the parameters of your 'thought experiment'
Its a thought experiment because you'd never get such a perfectly balanced humbucker and induction setup in practice, and so any real attempt would give some residual signal.
Thinking on which that would depend:
Despite various losses within this setup, I believe that its still essentially a linear system. eg It would not be any way near to saturating the cores. I don't think 'skin' effects are significant. Lets also assume the pickup components are not subject to mechanical rattling etc? Anything else potentially non-linear?
Also, studies by several experimenters in the last year or two have demonstrated good success by representing pickups with their losses, by various arrangements of linear components ie coils, lossy transformers, R and C etc
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Aug 23, 2019 12:42:25 GMT -5
Post by perfboardpatcher on Aug 23, 2019 12:42:25 GMT -5
I'll have a go, and Im going to guess thar there will be no difference between A and B, within the parameters of your 'thought experiment' I think that would imply, when going back to Angelbunny's initial question, that pickup as load = inductor. Couldn't it be for instance that the eddy current losses caused by the excitation coil affect the paf humbucker's ability to store energy in its magnetic field? Its a thought experiment because you'd never get such a perfectly balanced humbucker and induction setup in practice, and so any real attempt would give some residual signal. That's very well possible, I have never tried it so I cannot tell you for sure. The way I would do the experiment would be: settle for an appropiate frequency band like around 3kHz and experiment with slanting the excitation coil and sound checking for the least leakage. Another way to perform a similar test is to measure the "stompbox" pickup as load (impedance measurement) while at at the same time applying the excitation coil as described in (B) of my previous post. I'm sure there must be a way to this this type of test. Also, studies by several experimenters in the last year or two have demonstrated good success by representing pickups with their losses, by various arrangements of linear components ie coils, lossy transformers, R and C etc But representing pickups as a load to another pickup? These multiple coil models always seem to simulate the source impedance of the pickup as part of a voltage source.
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