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Post by roadtonever on Jun 15, 2011 17:16:37 GMT -5
I recently learned that parallel combined coils drop the total inductance in half thus any external RC network(cable capacitance, tone control) carefully tuned for each pickup individually will be off. This makes me wonder about the explicit scope of decoupling capacitors counteracting this effect. I'm mostly concerned with keeping the resonant frequencies intact rather than the height of said peaks, in a way simulating individual buffers passively to this effect. 1000Hz was used in the examples above and I wonder what this might mean to other frequencies.
Do the decoupling caps in series with the parallel coils make up common circuit? If so what is it called? Maybe I'll be lucky enough to find a handy calculator so my atrocious math skills won't get in the way.
Also I recently acquired an LCR meter, would the measured capacitance reading of the pickups be useful for these calculations?
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Post by roadtonever on Jun 15, 2011 17:35:19 GMT -5
Regarding the sound clips I posted earlier, the circuit around the pickups was very bear-bones with no tone control/s. With the cable I used the bridge pickup was contributing a crazy bright tone, So it's no wonder series sounded more pleasant. With the resonances carefully set de-coupled parallel sounds a lot warmer than earlier: www.box.net/shared/aarpydjbd3v4yu0nxs4iThis makes me see the series setting in a different light as having a lower and more accented low-mid resonanace(from the higher inductance?) beside the having a lower resonant frequency.
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Post by sumgai on Jun 16, 2011 1:32:51 GMT -5
I recently learned that parallel combined coils drop the total inductance in half.... Whoa there, Nelly! What you said is true only when both inductors are exactly the same value.... which is rare indeed, when looking at pickups. More correctly put, inductors in parallel are calculated just like resistors in parallel (or FTM, capacitors in series): 1/x = 1/a + 1/b + 1/c..... Expressed mathematically, that's "the reciprocal of the sum of the reciprocals" yields the final value. In less formal terms, we can combine two pickups in parallel like so: (L1 x L2) / (L1 + L2) That reads "Inductor 1 times Inductor 2, divided by Inductor 1 plus Inductor 2." Whenever calculating for more than two devices, it's probably easier to use reciprocals.... at least on a hand calculator. The point of all this was that one shouldn't just buzz off and assume that they can calculate the value of a compensating capacitor just because they already know the inductance for each one of a pair of pickups - they must do some math first, as just shown. HTH sumgai
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Post by roadtonever on Jun 17, 2011 3:13:29 GMT -5
Newey, that's really something how you presented the equation in a way that made it look like I even could make use of it. Thanks!
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Post by newey on Jun 17, 2011 5:05:50 GMT -5
Wasn't me, so let's give credit where due.
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Post by roadtonever on Jun 17, 2011 7:26:39 GMT -5
Oops! +1 karma to both of you anyway, I've gone from being clueless to having some clue thanks to you guys.
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Post by roadtonever on Jun 22, 2011 7:23:55 GMT -5
lets take your 3.2H pup, and lets say its DC resistance is 8k. Lets consider 1kHz The inductive reactance is 2Pi x 3.2 x 1000 = 20100, and this is 90 degress out of phase with the resistance. Combined reactance is Sqrt(20100^2 + 8000^2) = 21600 = 21.6k Could someone possibly dumb down the nomenclature on this equation? I'd like to see what happens when I use the pickups resonant Hz as a target and I hate to bug JohnH every time... EDIT: Actually the it's the combined reactance part that throws me off, I seem to be covered by various reactance calculators online for the first part
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Post by ashcatlt on Jun 22, 2011 16:48:45 GMT -5
It's vector math. Take a x-y grid. Start at the origin and draw a line along the x-axis as long as the resistance component. At the end point of that line (actually a segment) measure the phase angle and then draw a line along that angle as long as the inductive reactance component. Now connect the end of that segment back to the origin and measure (or calculate) the length of your new line. Since the phase angle is 90 degrees, you've got a right triangle, with the result as the hypotenuse, so we only need to consult Pythagorus once.
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Post by sumgai on Jun 23, 2011 0:56:02 GMT -5
Could someone possibly dumb down the nomenclature on this equation? Yes, but then I'd be wrong too. (This is gonna work out in the end, trust me.) John's first equation is correct, where he calculates the pickup's reactance. But his second one.... my oh my, tsk, tsk. I think that John made a simple Australian mistake (he forgot to pick up the correct, Made in USA mirror....) when he said "Combined reactance is....". More than likely, he meant to say "Combining the two, we get a total impedance of...." whereupon the rest of the formula is correct. Why'd he do all this mumbo-jumbo? Because 'reactance' is defined as opposition to the AC signal, as it passes through a component. (Said component being either a cap or a coil.) However, this opposition is not equal in value across the frequency spectrum. For a coil (an inductor), the signal meets more and more opposition as it increases in frequency. (For a capacitor, the effect goes in the opposite direction - declining frequencies meet more and more opposition.) This is why John specified a frequency of 1KHz, just for illustrative purposes. Calculating a series of reactance points with various frequencies can show us how a given component will behave in a circuit that's carrying an AC signal.* For the nonce, suffice it to say that too much inductance will result in a lot of reactance, and for our guitars, that will translate to less and less treble in our output. Might be a good thing, might not.... it all depends on how you define your tone. The resistor does have to be considered in the final analysis, but fortunately that's easy - it opposes all frequencies with equal fervor. OK, enough of that nonsense, let's answer your question. The formula for reactance in a coil is 2πfL, which we would read as "2 times pi times the frequency times the inductance". Please note, the frequency is in Hertz and the inductance is in Henries. Make any changes accordingly as you work with multipliers like kiloHertz, or dividers like milliHenries. So John said "2 * 3.141592 * 1,000 * 3.2". That's in the same order as I just spelled out the formula, I hope that was easy to follow. Next John said (after my correction) "The square root of the sum of the reactance squared plus resistance squared", which tells us how much impedance we'll have. You can think of that number as the opposition to expect your signal will encounter at the designated frequency. The reason for all of this VooDoo will become apparent as John next explains impedance matching, when you feed one device into another. ;D HTH sumgai * As ash correctly points out, plotting a series of such points on a graph results in a straight line. Magically, this straight line will be the hypotenuse of a triangle. Even more magically, we don't care about phase relationships, the formula works no matter what phase angle the components happen to be related to each other. Pythagoreas is always correct.
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Post by roadtonever on Jun 23, 2011 3:54:20 GMT -5
"The square root of the sum of the reactance squared plus resistance squared" That looks like something I can input to windows calculator or Google! In the not too far future I'll put up some clips I'll make comparing different target frequencies and straight parallel as well. Beside the bass I'm working on now my Les Pauloid is in the pipline for recieving simmilar treatment so guitar folks will be able to digest something from my studies as well.
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Post by ashcatlt on Jun 23, 2011 8:38:28 GMT -5
It just occurred to me that the angle between the resistive component and the "hypotenuse" might just tell us the amount of phase shift to expect at that frequency. Is that correct?
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Post by sumgai on Jun 23, 2011 17:29:39 GMT -5
ash, It just occurred to me that the angle between the resistive component and the "hypotenuse" might just tell us the amount of phase shift to expect at that frequency. Is that correct? I was afraid this might come up.... My first answer is this: Pretty much, the answer is 'no', because John drug a red herring across our path. You did note that neither of his equations included any terms regarding phase shift, yes? My second answer is a pair of questions: Why are you expecting any "phase shifting" at all, and why do you expect such to differ over varying frequencies? My third answer is: When speaking of AC, and particularly audio, there are several definitions of the terms 'phase', 'phase relationship', 'phase angle' and 'phase shift'. So now my question becomes, which definition are you most interested in? IOW, please explain what's got your curiosity notched up, and we'll see what we can discover, eh? HTH sumgai
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Post by JohnH on Jun 24, 2011 16:52:25 GMT -5
as John next explains impedance matching, when you feed one device into another. ;D HTH sumgai sumgai is doing a fine job of interpreting what I might have meant, so I'll pass on that
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Post by ashcatlt on Jun 24, 2011 20:47:38 GMT -5
You did note that neither of his equations included any terms regarding phase shift, yes? ...Except that he mentioned that the reactive component is 90 degrees out of phase with the resistive, and went on to calculate it as the hypotenuse of a right triangle rather than a straight line. I can't say that I completely understand why (or how) the reactive component lags behind the resistive, but I have heard it a number of times before, at least once from yourself. Well, my first clue was all the engineer/producers I've heard warning about "phase smearing" introduced by EQ's and stuff, and the fact that "linear phase" EQs are supposed to be super cool. Then there's 5spice, where I can graph the phase-per-frequency and watch it go all wonky around the cutoff of a filter like this. I thought I may have figured out how to plot that curve. Maybe not?
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Post by roadtonever on Jun 27, 2011 19:21:54 GMT -5
Is the effect of turning down either volume control in this diagram identical to what's being discussed in this thread? Ditto on the "fattening" resistors mentioned in the article. Sorry about the dumb-butt question just want to be sure If I understood correctly.
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Post by sumgai on Jun 28, 2011 0:32:06 GMT -5
Is the effect of turning down either volume control in this diagram identical to what's being discussed in this thread? Yeah, it's a real kissing-cousin, to be sure. As you change either factor, resistance or reactance, you change the impedance, pure and simple. While the Volume control is external to the pickup, its resistance must still be taken into consideration - it is in parallel with the pickup, and thus has an effect on the impedance, just as if it were the coil itself. In the case where two pickups are connected to the output at the same time, and each of them has a Volume pot, then the calculations can get hairy. Still, they can be broken down into simple steps where you do the calcs for one part of the circuit, then the other, then combine those results. Takes longer, but in the end you have a pretty accurate answer. HTH sumgai
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Post by sumgai on Jun 28, 2011 1:24:04 GMT -5
You did note that neither of his equations included any terms regarding phase shift, yes? ...Except that he mentioned that the reactive component is 90 degrees out of phase with the resistive, and went on to calculate it as the hypotenuse of a right triangle rather than a straight line.That's following the axes of the graph, not the angle of the phase shift. Simple vector math, as you yourself noted. In point of fact, while we can pin down the exact phase angle of a reactive component at any given moment in time, it's just that we can't use that datum for anything. Why not? Because an AC signal never "rests" at that point (whether it be 90 ° or otherwise) - it's always moving upward or downward! So why are we using "2π" as a multiplier in our calculations? Well, that's getting really deep, much more so than we usually go. Let me say only that it has to do with angular frequency, and the fact that it ends up looking, to us laymen, as if it were the RMS value of the peak value for one cycle (1 Hz) of a waveform. Let me off the hook here, and go do some Googling - it'll hurt your brain for a few days, but I swear - you'll recover from it! ;D This is gonna get real hairy, real fast. If you've heard me say some such, then it was probably in connection with that old technician's mnemonic: "ELI the ICEman". In an inductor, voltage leads current by 90 °. In a capacitor, it's the opposite - current leads voltage by the same amount. What's that mean? Not so simple. In effect, a component that reacts to voltage will 'see' the incoming voltage a bit before it understands that it's receiving power (voltage * current). That's important in that at low signal levels, we need to assure that the voltage is not hindered (opposed) any more than absolutely necessary. Ditto for high-level signals, we want current to be unhindered, because those components are current driven, and thus more sensitive to such. Now, when we talk about phase shift in an audio signal, we generally agree that combining the reactance of both coils and capacitors, we're gonna get a pretty ugly signal - the voltage will sometimes lead and sometimes lag the current, depending on the frequency at the moment. Largely, what comes out the end (the speakers) is all we care about - if we can hear it with clarity, we're happy. Since we can't get away from using caps and coils in our circuits, all kinds of phase-shifting is gonna occur within our amps, what-have-you. BTW, this includes speakers themselves, I think you can figure out why. Long story short, we can see and demonstrate phase shift in our laboratory instruments, but we can't really tell if it's "messing" with our signal, when we're listening to it. What I can say are two factoids I've gleaned over the years: A) No one has successfully "beat" the statistical average on any properly set-up listening test. By that I mean a double-blind test that eliminated all other possible sources of signal degradation. (Very hard to do!) B) In spite of all manner of super designs that spend thousands of bux to do away with caps and coils, in point of fact there are still "noticible" or "objectionable" artifacts in the final output of these designs. Probably not as many, nor as severe, but I put it to you - was that reduction in degradation due strictly to the elimination of reactive components, or was it due to overall better engineering principles and practices? I'd posit the latter myself. Not many designs over the years have taken the middle ground - leave in the reactive stuff, yet eliminate all other sources of crapola. Makes it kinda hard to draw comparisons, don't it? And as for "linear" and offshoots (ultra, etc.), I submit that we can get that without spending next year's vacation money. As far as I've tracked it over the years (and that's only peripherally), the figures have stopped trending downwards (meaning 'better and better') about two decades ago - we're way below the threshold of what most listeners can detect, in terms of both frequency response and clarity of the signal, even with good quality headphones. IOW, it's linear enough already.... HTH sumgai
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Post by roadtonever on Jun 28, 2011 9:35:50 GMT -5
In the case where two pickups are connected to the output at the same time, and each of them has a Volume pot, then the calculations can get hairy. Still, they can be broken down into simple steps where you do the calcs for one part of the circuit, then the other, then combine those results. Takes longer, but in the end you have a pretty accurate answer. So in a dual volume scheme simply* following JohnH's equation, which you kindly broke down for me, wouldn't be the best practice? I was about to try this on my Les Pauloid... (*Or in my case trying it with different target frequencies; external resonant frequencies of each pickup, 1000Hz etc) EDIT: Reading some older post on the subject of adding series resistance as I write this. I understand not everyone will be inclined to mess with their in-between sounds in a permanent way in a two volume scheme where it might be convenient enough to turn down one of the volumes. Personally I kind of resent the straight parallel sound so I hope you can expand on the caclulations for different parts of the circuit you speak of. EDIT2: Do you think doubling the volume pot values is an OK way to counter the treble loss?
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Post by sumgai on Jun 28, 2011 21:51:22 GMT -5
[I]n a dual volume scheme simply* following JohnH's equation, which you kindly broke down for me, wouldn't be the best practice? I was about to try this on my Les Pauloid...
(*Or in my case trying it with different target frequencies; external resonant frequencies of each pickup, 1000Hz etc) When I said 'break it down into smaller pieces', I was referring exactly to John's formulas (reactance and then impedance). Do that for one pickup, pretending that the other doesn't exist. (See the next paragraph for more details.) Next do the same calcs for the second pup, pretending the first doesn't exist. Now you have two impedance values, right? Even if you used different target frequencies for the two pups, you still need only combine those two numbers for your final impedance at some specific Volume pot setting(s). (Remember to take into account how the two pickups are connected to each other and the output - are they in parallel or series?) Well, what I meant was, if you add (in parallel) the Volume pot's overall resistance to the pickup's resistance, then you have a "true" number to work with, in order to get the impedance figure. If you decide to turn down the pot, that's fine, all you're really doing is shorting some of the pot's resistance, and certainly that's acceptable. But instead of guessing, you should attempt to get a meter reading of that 'desired' resistance, and then use that in your calculations. Now if you've been paying attention, this will point out why we notice a slight drop in the treble region whenever we turn the pot down from maximum volume - we've shorted some of the resistance, and that changed the impedance, and that knocked our frequency point off from where we started. (Meaning, if you worked the calcs backwards, you would arrive at a lower frequency in Hertz.) All of which leads us to..... Well, yes, if you run wide-open most or all of the time anyways. Otherwise, when you turn down, you're gonna notice even more of a loss in the treble range of your sound. This isn't a deal-breaker, not by any means. After all, Leo himself started by putting 1MΩ pots into all of his guitars (and basses) for both Volume and Tone. It wasn't until years later that he started dropping the pot values. So the deal is, or to be more precise here, my advice is that you consider where you like to run your pots - wide-open or cranked down a bit (or a lot), then take a few measurements, exercise your calculator, then experiment with your caps. I think you'll find what you're looking for without too much trouble. Just remember, for purposes of this experiment, you don't need to be neat and tidy. IOW, you can run long leads and use alligator clips to hook up various capacitors as you play around. Unless you live/work/play/goof around in an extremely noisy environment, you shouldn't worry about what little buzzing you'll get from the longer leads that aren't in the shielded cavity. Also, if you start with a small-value cap, you can add another cap in parallel, and you get a larger value. Conversely, if you start with a large cap and want to go lower in value, you'd need to open one of the connections and put another cap in series with the first cap. So why not just start out with a smaller one in the first place? OK, I think I've about covered it all.... any questions? sumgai
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Post by ashcatlt on Jun 29, 2011 8:44:23 GMT -5
Well, sg, the point of this here thread is to add series resistance with the pickups, and your reply above pretty much ignores that aspect of the Volume control's action. Turning down the V pot doesn't just "short some of its resistance.". It also takes some of what was in parallel with the pickup and puts it in series. We are using all three lugs after all. I mean I think the gist of your post remains the same either way, but...
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Post by sumgai on Jun 29, 2011 12:08:45 GMT -5
ash, Dagnabit, you've caught me taking a shortcut again! Yes, the short isn't direct. It goes through the components comprising the input stage of whatever follows next, usually a resistor, sometimes that and the 'control' side of a tube or transistor. These are not overly complex calculations, but they do involve a lot of knowledge of where to look, and why. It was thoughtless of me to assume that most players who aren't also EE's would want to do this kind of thing. As for that 'part of the Vol pot's resistance is now in series' thing, well, that's still in parallel with the pup, as the path between that partial resistance is only 'complicated' between it and the other lead of the pup, as per my paragraph above. i.e., it's in series with the remaining Vol pot's resistance, which is in parallel with the input circuitry. Nonetheless, I stand by my assertion that if you treat the Vol pot as the total and only part you can deal with, for calculation purposes, then you arrive at a meaningful figure, albeit it's technically incorrect. But recall, there is a large impedance mismatch here - the input parts of the following circuitry should register at least 10 times the impedance of what we're dealing with in our guitars. That means we shouldn't be off by more than 10%, and probably less, after all is said and done. Most Tone Nazi's won't be able to tell the difference, and when we really look at the whole picture, we realize that we have to allow for using various gear following our guitar anyways - it's all gonna affect our tone, and each piece of gear is different, some in small ways, others in large and noticible ways. We just have to deal with it, that's all. Gotta run, HTH! sumgai
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Post by roadtonever on Jun 30, 2011 3:47:47 GMT -5
any questions? Sure I'm full of them. I want to get a subset of clips out the way very soon. One thing regarding upping the vol pots to 1M to counter treble loss. I assume going all 2M pots would be the Tim Allen way? 1M vols 500K tones closer to the sweet spot? I usually run my controls all the way up except in the middle setting. For my coming activities I'll put up the diagram I'll intend to use along with a subset of calculated values and wait for an A-OK from you or another regular before going ahead. Just to make sure my contributions are at least somewhat meaningful and free from fundamental errors. With that in mind I'm officially taking requests! BTW the through strikes me of later on adding the center frequency of the low-mid hump, the one that's prominent in high output pickups, to the list of target frequencies to compare and record.
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Post by newey on Jun 30, 2011 5:30:00 GMT -5
In the sense of overkill, yes.
Raising pot values beyond 1MΩ is problematic, because you'll lose the adjustability. The pot will start to act more and more like an on/off switch. At about 3M, it will essentially be an on/off switch; at 2M the "travel" for adjustment will be so minimal as to be effectively useless.
If 333.33KΩ is your sweet spot, then yes (For 2 resistors in parallel, R1* R2 / R1 + R2. If it helps you remember, think "product over sum").
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Post by roadtonever on Jul 4, 2011 5:42:48 GMT -5
Here's my proposed circuit for the upcoming comparative sound clips: The cable and tone capacitance was chosen during a session involving a variable air capacitor. I settled on a 6.7kHz cutoff on the neck pickup and a 2.1kHz resonance on the bridge pickup via the tone control. I hope this doesn't deem the study as too unrepresentative. To me it's sort of a modern take on a Clapton woman tone setup(Clap used a treble booster during the Cream-era)easily achieveable via a wireless setup or a 5ft low-capacitance cable. The pot values are a reasonable compromise I reckon. The 1M volume is for countering the treble reduction of the series caps which will be added in the middle setting. Will likely stay at "5" with linear pot but sometimes turned up for a solo boost effect in individual pickup settings. Not shown in the diagram above is a DPST switch for engaging and disengaging R1 & R2 during testing. The tone control remains typical 500k which moderates the potential ice-pick on the bridge pickup due to it's higher inductane while keeping most of the highs on the neck pickup due to it's lower inductance. Off to make some calculations. I'll be back... EDIT:Some correction to typos and grammar first so I make some sense...
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Post by roadtonever on Jul 4, 2011 6:34:18 GMT -5
So for a start I'll consider 6.7kHz and 2.1kHz that I liked and tuned the capacitances to. Neck PU= 4H, 7.95kOhm DCR | target=6.7kHz | Inductive reactance=168.304kOhm | Total impedance=168.5kOhm Bridge PU=5.6H, 10.29kOhm DCR | target=2.1kHz | Inductive reactance=73.8528kOhm | Total impedance=126.7kOhm
2Pi x 4 x 6700 = 168304 2Pi x 5.6 x 2100 = 73853
Sqrt(168304^2 + 7950^2) = 168491.658 = 168.5k Sqrt(73853^2 + 102900^2) = 126659.684 = 126.7k
Now do I add 168.5kOhm as R1 and 126.7KOhm as R2 or do I need to add the pots to the equation? EDIT: As for songs I've considered to include for the clips, so far: 'Blackbird'. Some Hendrix rhythm, maybe 'Midnight'. 'Can't find my way home' for open position arpeggios and strumming. Something arpeggioey up the neck. The intro solo to 'Reeling in the Years'.....Maybe some Bach melody. Still taking requests!
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Post by sumgai on Jul 4, 2011 10:34:25 GMT -5
A quick glance says that adding a 1MΩ pot will reduce those figures about 15% for the Neck and about 10% for the Bridge. But you'd have to do the math to be sure, that's just an WAG..... More to the point, I think that going any further with your "beforehand" manipulations is not gonna pay off any big dividends.... I think you're at the point where experimenting with real-world components is now gonna give quicker results, just as it did when you were compensating for cable capacitance.... IOW, why is that soldering iron still cold?! HTH sumgai
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Post by roadtonever on Jul 4, 2011 13:35:12 GMT -5
Hahaha... Give me a break I'm waiting for the cavity shield paint to dry. I don't even have the 1M part so my plan is to record the 500k straight as reference until I recieve the part 1M. You don't wanna give me the equations? Please.
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Post by roadtonever on Aug 30, 2011 7:14:15 GMT -5
Well, I'm still not set up for making the sound clips I promised. To still my curiousity on the matter I made a model with the values sumgai suggested: The responce graph seems to show an interesting if unfavorable result: Tone at "10" Tone at "0"(Modelled as 0.1k) I'd love to keep tweaking the target frequencies and possibly the added series resistance but I can't figure how to factor in the pots resistance properly. Sumgai, wanna give me that formula? ;D
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Post by sumgai on Aug 30, 2011 16:10:47 GMT -5
rtn, Really busy just now, can't stop for longer than a quick lunch break, doing this only to let you know I'm not ignoring you..... I'll need to review this whole thread before I can set forth what you're asking for. Give me a day or so. Of course, John, ash, ace, or one of our other members, might beat me to it. Hang tight! sumgai
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Post by asmith on Aug 31, 2011 5:36:35 GMT -5
Go on then.
I've read through this thread but am only picking up the gist here and there. Correct me if I'm wrong. You want to add resistors in series with your pickups so that the inductance of each pickup has a harder time affecting the other?
And then, you want some kind of equation to figure out exactly what's going on with each inductance and resistance?
On a side note, I'm not sure what you mean in your schematic. R4 and R5 seem to be static resistors in parallel. Care to elaborate on what's going on in there?
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