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Post by 0wnyourtone on Feb 28, 2014 19:29:53 GMT -5
When editing audio files during a recording project, I have grown fond of a particular automated volume fade called the "S-curve" and it got me to thinking, what would this sound like in a guitar? I'd like to tap into what you guys think about accomplishing this; check out the image next post down - the blue line represents the curve I like. I am aware of adding parallel resistance to alter the taper and that this can be done on both "sides" of the pot (lugs 1-2 and 2-3). I'm wondering what component values would get me close to 500K and 250K total resistances at 10 on the dial, but give me something close to the "S" taper. I'm hoping you modeling gurus might help me out?  |
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Post by 0wnyourtone on Feb 28, 2014 19:33:05 GMT -5
Here is the image - the blue line is the "S-curve." Attachments:
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Post by JohnH on Feb 28, 2014 19:51:33 GMT -5
would that be in the form of a volume pot,? using three lugs, and with 50% at the centre?
I've thought a few times about using resistors to change the taper of things, and they always seem to work opposite to what this wants. ie, instead of small changes near the ends and steep in the middle, it would tend to lead to a quick fall/rise at each end, then the whole middle section at about half the resistance.
Instead, id suggest it may be possible to open up a linear pot of say 250k, and scratch at the edges of the track within the middle half of the track to raise its resistance, to create a steep slope in the centre and low slope per the original pot at each end.
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Post by 0wnyourtone on Feb 28, 2014 22:39:37 GMT -5
Yes, I am particularly interested in a volume pot having the same feel as this S-curve seems to give me in fade-outs and fade-ins in recording.
I have messed with parallel resistance to a degree as well and have noticed the same extreme drop offs towards the end of the dial.
I guess what I'm interested in finding out here is: could parallel resistors take a linear pot and give it this sort of S shaped taper?
So far as shaving the middle and getting the 50% at five on the dial so to speak, I'm not sure that's actually what I'm after. It's more the sound of the gentle slope at the top and bottom if that makes sense.
It occurred to me also, that the bottom resister from lugs 1 to 2 would have to work together with the total resistance of the pot to equal the desired parallel resistance of 250K or 500K.
So, let's say it were possible to get the general shape of the taper I am looking for using two resistors on the pot: this would seem to indicate to me that I would need a higher resistance pot than normal to compensate for the the bottom-half-of-the-voltage-divider parallel resistor making the total resistance lower than 'desirable.'
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Post by newey on Feb 28, 2014 22:45:55 GMT -5
0YT-
Sorry, I felt I should move this thread from the "guitars" section, which is really for reviewing axes, not for electronics discussions.
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Post by JohnH on Feb 28, 2014 22:49:04 GMT -5
So far as shaving the middle and getting the 50% at five on the dial so to speak, I'm not sure that's actually what I'm after. It's more the sound of the gentle slope at the top and bottom if that makes sense. Yes but, it would tend to do that. If you start with a linear pot, you are at 25% at the quarter point. If you take the middle half of the pot and manage to double its resistance, then, at the quarter point you are now at 1/6 of the new total resistance, or 16.7%. Not much different, but it is a change in the right direction. |
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Post by ashcatlt on Feb 28, 2014 23:15:51 GMT -5
So, I think this is easier with dual ganged pots. If you can find yourself a true log/antilog blend (not pan) pot it would be a simple matter of wiring the two wipers together and then using one outside lug from one gang and the opposite from the other. With a log/antilog pan pot I think you could probably just do the same thing and it would be maybe a little more extreme S action (I'd have to kind of draw that up, but I think it works...). Or you could use a dual linear with tapering resistors across one half of one and the other half of the other. Or you could just use a log/log and use the same outside lug of each gang. Honestly haven't really drawn any of these out, but I'm pretty sure one of them is going to be your easiest solution.
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Post by sumgai on Mar 1, 2014 0:16:29 GMT -5
By design, a pot is meant to taper in only one direction. It may go quickly or slowly, and by that I mean either an Audio or a Linear taper, but no pot I've ever met actually incorporated both an Audio and a reverse-Audio taper on the same resistance element. However, ash has the right idea - using a dual-gang pot with an Audio (log) and a reverse-Audio (anti-log) sections, you can simply tie the two wipers together, and get the desired effect. Just keep in mind that the total value you desire (250K, whatever) must be the value for each section - don't mistakenly order a pot with each element of double-size, thinking that you're gonna put the sections in parallel - that won't be the case. A few moments of cogitation should solidfy the concept here.  HTH sumgai |
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Post by JohnH on Mar 1, 2014 2:44:50 GMT -5
While these dual pot ideas will get the 'S' taper happening in terms of resistance ratios, I can see danger for Will Robinson...
At the centre position. I think the overall resistance of such an arrangement will include the low resistance side of one pot half, connected to the same of the other. Eg, if this was made with two 500k log/antilog pots, each half will be about 50k on the low side, so we might have just 2x50k = 100k in the centre position total, which is too low in a passive guitar circuit.
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Post by 0wnyourtone on Mar 1, 2014 9:01:30 GMT -5
would that be in the form of a volume pot,? using three lugs, and with 50% at the centre? I've thought a few times about using resistors to change the taper of things, and they always seem to work opposite to what this wants. ie, instead of small changes near the ends and steep in the middle, it would tend to lead to a quick fall/rise at each end, then the whole middle section at about half the resistance. Instead, id suggest it may be possible to open up a linear pot of say 250k, and scratch at the edges of the track within the middle half of the track to raise its resistance, to create a steep slope in the centre and low slope per the original pot at each end. So John, Rereading what you said, I get what you're saying with the scratching comment to a degree. I do need explained to me how thinning the resistance element increases resistance, however. If this scratching idea worked okay, I would also wonder about scratching the 'bottom' end of a reverse audio or 'top' end of an audio, as those tend to be the quicker-moving sections of the track? |
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Post by 0wnyourtone on Mar 1, 2014 9:24:30 GMT -5
While these dual pot ideas will get the 'S' taper happening in terms of resistance ratios, I can see danger for Will Robinson... At the centre position. I think the overall resistance of such an arrangement will include the low resistance side of one pot half, connected to the same of the other. Eg, if this was made with two 500k log/antilog pots, each half will be about 50k on the low side, so we might have just 2x50k = 100k in the centre position total, which is too low in a passive guitar circuit. I'm not sure, but I think it may be an incorrect assumption that the 50K would apply to where both tracks were setting? Wouldn't the antilog side have much more resistance toward the bottom of the dial? |
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Post by sumgai on Mar 1, 2014 16:09:41 GMT -5
'tone-y, No, John's correct. In the event that only 50KΩ are available on the log pot at 50% of the wiper's rotation, the same will (should!) hold true for the other portion, the anti-log pot. If the two wipers are hooked together as ash and I have specified, then the total overall resistance element seen by the guitar's circutry will be only 100KΩ. This would hold true even if each pot section measured 250KΩ overall. After all, it's only the last 25 or 30% or so of the element's total resistance that we want to use, at 50% of the wiper's rotation - the other side of that element, the one carrying 70% of the resistance, we don't want to see that in our circuit at all. The best way out of this is to buy 500KΩ or higher pots, but sad to say, I can't speak to the availability of such things.  HTH sumgai |
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Post by 0wnyourtone on Mar 1, 2014 19:24:20 GMT -5
'tone-y, No, John's correct. In the event that only 50KΩ are available on the log pot at 50% of the wiper's rotation, the same will (should!) hold true for the other portion, the anti-log pot. If the two wipers are hooked together as ash and I have specified, then the total overall resistance element seen by the guitar's circuitry will be only 100KΩ. This would hold true even if each pot section measured 250KΩ overall. After all, it's only the last 25 or 30% or so of the element's total resistance that we want to use, at 50% of the wiper's rotation - the other side of that element, the one carrying 70% of the resistance, we don't want to see that in our circuit at all. The best way out of this is to buy 500KΩ or higher pots, but sad to say, I can't speak to the availability of such things. HTH sumgai I see what you're saying Sumgai. I've got pretty good connections for "weird" pots, so I'm not too worried about that. I'm actually pretty curious about this scratching-the-resistive-track idea of John's; it would seem to me if a person got fairly adept at it, it could be quite an asset in tailoring pot taper. I'm still unsure as to how that works, so far as how the physical attributes of the track effect given resistance - all I know is, "if I scrape enough crud off, I can make a "no-load" doohickey." |
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Post by JohnH on Mar 1, 2014 20:20:33 GMT -5
I've tried a bit of pot carving, beyond just a full cut. I had some 500k pots that came in a only about 400k, and I was able to scratch them up to 500k quite easily. To do that, I was scraping only at the edges of the track, with the blade held at 45 degrees, virtually just taking off the excess leaving the main area untouched. I was working on the steeper-tapered half of the track taking that area from about 350k up to 450k, which is about a 30% increase.
In another case I stretched a 250k pot up to about 500k.
To do what I was describing here would need more significant surgery, and so would be a bit of an R&D project to see if this is possible while keeping a good track surface. A nice feature of this process is that it is quite easy to maintain extremely tight control what you are doing just by keeping a multimeter connected to the two track ends. You can virtually watch the resistance increasing ohm by ohm as you work.
I was thinking about doing something like this to create the ideal tone pot, one where the frequency response graphs are all nicely evenly spaced as you roll the control, instead of being uneven. I think the outcome for that would be a log pot, with the second quarter increased and also the fourth quarter increased, to make a 'super-log' pot, with no-load at the end.
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Post by sumgai on Mar 1, 2014 21:16:52 GMT -5
That "scratch away conductivity" thing does intrigue me.
Since as John points out that you're still changing the resistance of the whole track as you carve away, we're fortunate that there are 200KΩ pots available, which should work fine for arriving at 250KΩ. For those who want 500KΩ at the finish, there are also 300KΩ units, too. Both can be found in Linear and Audio (log) tapers.
I'd start experimenting on the Linear one first, but I'd also have an Audio pot on hand too, just to see what happens when it has been subjected to the knife.
I can also see where this would do interesting things to the overall tone as the pot is adjusted downwards. Especially if there's a treble bleed circuit involved....
HTH
sumgai
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Post by 0wnyourtone on Mar 1, 2014 23:00:08 GMT -5
Okay, so I'm going to start pot-carving in the near future.
John, when you say "edges" of the track, are we talking either end of the dial or the "shoulders of the road" so to speak? I would presume you're talking about carving out either end of the dial in your example of adding resistance.
What is it about thinning out the track that causes more resistance, if that's not too much of a loaded, lazy question?
EDIT: I just reread what you said early on, the way I'm thinking about the track (like a road) you were talking about narrowing the road. So, what is it about "narrowing the road" that yields more resistance? Or am I still mixed up?
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Post by JohnH on Mar 1, 2014 23:20:12 GMT -5
Okay, so I'm going to start pot-carving in the near future. John, when you say "edges" of the track, are we talking either end of the dial or the "shoulders of the road" so to speak? I would presume you're talking about carving out either end of the dial. What is it about thinning out the track that causes more resistance, if that's not too much of a loaded, lazy question? Its the shoulders of the road - a good description, which is the first target to attck when increasing pot resistance. That material is helping to conduct, but is not in contact with the wipers, so we can scape it off. Its the easiest bit to do. Lets say we had a pot that went from 0 to 50k at halfway, and 50k to 500k from mid-way to the end, and lets say we wanted to increase the last 25% of the track to something higher. we would scrape along each side of the track over than last 1/4 length of it. If that is not enough, then we need to thin out the surface of the track over that length, maybe with very fine sand paper. And on what that does: The track is covered in some kind of partly conductive compound that has a material property of Resistivity. The actual resistance of a piece of it in Ohms, is resistivity x length / cross sectional area. The resistivity and length of the track is a given for a particular pot (but the wiper position determines what length is active, and hence the measured resistance), but we can mess with the cross section of the track by reducing its width and/or its thickness, to increase overall resistance. |
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Post by 0wnyourtone on Mar 2, 2014 0:26:43 GMT -5
Okay, so I'm going to start pot-carving in the near future. John, when you say "edges" of the track, are we talking either end of the dial or the "shoulders of the road" so to speak? I would presume you're talking about carving out either end of the dial. What is it about thinning out the track that causes more resistance, if that's not too much of a loaded, lazy question? Its the shoulders of the road - a good description, which is the first target to attck when increasing pot resistance. That material is helping to conduct, but is not in contact with the wipers, so we can scape it off. Its the easiest bit to do. Lets say we had a pot that went from 0 to 50k at halfway, and 50k to 500k from mid-way to the end, and lets say we wanted to increase the last 25% of the track to something higher. we would scrape along each side of the track over than last 1/4 length of it. If that is not enough, then we need to thin out the surface of the track over that length, maybe with very fine sand paper. And on what that does: The track is covered in some kind of partly conductive compound that has a material property of Resistivity. The actual resistance of a piece of it in Ohms, is resistivity x length / cross sectional area. The resistivity and length of the track is a given for a particular pot (but the wiper position determines what length is active, and hence the measured resistance), but we can mess with the cross section of the track by reducing its width and/or its thickness, to increase overall resistance. Ah-hah, okay, I'm with you now. I suppose the last thing I'd like to familiarize myself with regarding pot tapers is how they are in fact constructed so far as the strip is concerned and what the typical measurements are %-wise at various intervals. Someone mentioned earlier how some log pots are just three separate sections of linear resistance, designed to approximate a log curve. I've seen graphs of this. I'd like to locate a good resource for the aforementioned data and start brain-storming on how I might start "sculpting pottery." - Probably more like breaking pottery, but either way, it should prove fun. |
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Post by newey on Mar 2, 2014 1:14:57 GMT -5
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Post by ashcatlt on Mar 2, 2014 12:20:40 GMT -5
John explained it better, but I was going to say that thinner track leads to more resistance for much the same reason that smaller wire leads to more resistance. Anyway... Does this help at all? |
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Post by 0wnyourtone on Mar 2, 2014 17:55:13 GMT -5
I didn't even know that about wire, Ash! But, yes it makes sense. So, in my simplistic way of thinking about it, larger wire is a bit like a larger highway for the electron cars, freeing up traffic a bit? I learn in mental documentary cartoons from the 50's evidently. I can see Goofy driving his electron car down a wide autobahn wire.
At any rate, those graphs are helpful. I used to visit that "Secret Life of Pots" page all the time - been a while. I'll start piddling around with wafer tracks and see what I can come up with. Then, once I master the art of pot-carving, I'll make millions off custom Own-Your-Tone pots on fleabay!
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col
format tables
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Post by col on Mar 2, 2014 23:26:39 GMT -5
Someone mentioned earlier how some log pots are just three separate sections of linear resistance, designed to approximate a log curve. I've seen graphs of this. I'd like to locate a good resource for the aforementioned data and start brain-storming on how I might start "sculpting pottery." - Probably more like breaking pottery, but either way, it should prove fun. I think that 'someone' was me. It seems I may have overstated it, and at the same time understated the situation! Looking at the 'Secret Life of Pots' webpage, it seems that some (probably cheap) so-called "straight line 'audio' taper" pots are manufactured this way, but actually use just two (not three) linear sections to approximate log. Hmmm, but reading through just now, the text does actually seem to indicate 'three sections'. I assume this is where I read about this before - I know I've read that page in the past. I'm sure someone else (with more knowledge) will be able to clarify this for you. I don't see why it would be any more complicated than manufacturing a standard log pot, which is usually just three linear tacks in series, only approximating logarithmic scale. : |
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Post by 0wnyourtone on Mar 3, 2014 2:34:06 GMT -5
Instead, id suggest it may be possible to open up a linear pot of say 250k, and scratch at the edges of the track within the middle half of the track to raise its resistance, to create a steep slope in the centre and low slope per the original pot at each end. John (or anyone else who knows), This statement you made isn't adding up according to my attempted comprehension of how we're altering the track resistance: The "S" taper would require a steeper fall between 75-25%, meaning less resistance in that section, allowing the signal to flow "faster" there. So, wouldn't I want to trim the strip toward either end in order to increase the resistance at the top and bottom of the dial, yielding a gentler slope there and a proportionately steeper drop in that middle chunk? I hear spicy food will give you a steeper drop in your middle chunk too, if you're not careful. |
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Post by JohnH on Mar 3, 2014 3:40:22 GMT -5
Instead, id suggest it may be possible to open up a linear pot of say 250k, and scratch at the edges of the track within the middle half of the track to raise its resistance, to create a steep slope in the centre and low slope per the original pot at each end. The "S" taper would require a steeper fall between 75-25%, meaning less resistance in that section, allowing the signal to flow "faster" there. nup, its more resistance that you want in the middle section to do that. If it was less, imagine it to the extreme, where there is a fully conductive middle section with no resistance. Then there would be no change of output through that whole middle part. |
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Post by 0wnyourtone on Mar 4, 2014 1:08:12 GMT -5
The "S" taper would require a steeper fall between 75-25%, meaning less resistance in that section, allowing the signal to flow "faster" there. nup, its more resistance that you want in the middle section to do that. If it was less, imagine it to the extreme, where there is a fully conductive middle section with no resistance. Then there would be no change of output through that whole middle part. Ah, okay I see I was backwards, thanks. |
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