I've tried my best to research this on Google, but I'm still not sure, so I thought I'd post it here.
We know that using a 1 meg pot (by having less resistance) will allow more highs to pass than the usual 500k pot used with most humbuckers.
We know that varitone-type switches use a variety of caps to specifically cut highs. AFAIK a varitone is placed after the vol pot, before the output jack. Here's a good page: www.blueshawk.info/varitone.htm
(For the sake of simplifying this example, pretend this guitar has no usual "tone control" w cap in parallel w the vol control... just a humbucker, 1 meg vol pot on "10", and a varitone... although I'll probably ask later what treble bleed cap I should use w a 1meg pot )
So considering that position #1 of a varitone is "straight" with no cap, and gives the full high freqs that the 1meg pot allows....
....what cap value for the 2nd or 3rd position of the varitone would I need to match the high freq attenuation I would have gotten had I used a typical 500k pot with this humbucker? Can that be solved mathematically? Is there a diff in capacitace between the 500k and 1meg pots that can be duplicated with a cap value in a varitone? (Thanks in advance)
Post by michaelcbell on Jun 5, 2007 19:40:25 GMT -5
Before I forget - welcome to guitar nuts! Given the knowledge you've displayed, you'll have fun here.
Unfortunately, I can't answer the capacitance question - I'll have to leave that for someone with a bit more schematic computing power. I can, however, clarify one thing - a 1meg pot has MORE resistance than a 500k, so it lets less signal bleed to ground, and more get where it's 'supposed' to go. This translates into a brighter tone for either volume or tone controls, but it isn't clear to me which you are speaking of in this example (I assume volume, but I'm not sure). Let me(us) know and I(we) will be able to help more.
what cap value for the 2nd or 3rd position of the varitone would I need to match the high freq attenuation I would have gotten had I used a typical 500k pot with this humbucker? Can that be solved mathematically? Is there a diff in capacitance between the 500k and 1meg pots that can be duplicated with a cap value in a varitone? (Thanks in advance)
that's hard to answer. The Varitone is a variable (well, selectable) cutoff frequency passive notch filter. The effect of using a higher resistance across a pickup (an AC generator) is to reduce the loading and hence to allow more harmonics to be unattenuated. The more that one loads a generator, the more "rounded" the response curve becomes.
The best bet is to get a copy of pSpice and run your own circuit simulations.
Thanks ChrisK - looks like I've got some work to do
BTW - would the usual .001uf be the correct treble bleed cap to use with a 1 meg pot, or should I use a different value? (is there going to be a lot more loss of highs as it's turned down - compared to a 500k pot?)
Thanks ChrisK. after a few days of thinking about this, I wonder if this could be done with resistors instead of caps? Let's make this as simple as possible: a way to switch between a "500K pot" sound and a "1meg pot" sound - can that be done like this:
humbucker -> 500k pot -> resistor (bypassable) -> output jack
Suppose the resistor (in SERIES) is on a bypass switch to bring it in or out. In bypass mode the signal is just going thru the 500k pot. What value resistor would I need to make this equivalent to making the pickup sound like its going thru a 1 meg pot to output? A 500k resistor in series?
(Or do I have that backwards- should I use a 1meg pot and add a 500k resistor in series to bring it down to 500k?)
Post by michaelcbell on Jun 9, 2007 14:59:18 GMT -5
I would strongly suggest not putting a 500K resistor in series with your signal, since it basically would do the opposite of what you're looking for. If you're looking for brighter sound, use the 1meg pot. If you're obsessed with the brightest tone, there is the possibility (in my opinion, a bad one) of cutting the top end of the track on the volume pot, which would give you a small dead spot around 9, but 10 would have no path at all to ground, giving you the brightest tone - then you'd be able to use whatever value pot you wanted for the rest of the volume attenuation (I'd still suggest 500k)
I can tell you that the higher the value of resistance of a volume pot the less control you will have over the volume in that you will turn the knob less before the volume is very low or nill. Likewise, the lower the value (you can go too low really load the pot down) the more control you will have. i.e. the Pot will turn farther to get a more gradual reduction in volume and hopefully be near 0 when you loose all sound. However you do effect the tone more.
My view is that a 1M pot provides very little beyond what a 500k pot will do in terms of treble, but is more sensitive to actually losing treble at lower volumes, due to the cable capacitance. Id suggest one of Wolfs latest switched solo circuits, to get both the volume and tone out of circuit when you want to, but with normal performace otherwise:
But to convert the sound of a 1m vol pot to that of a 500k - at full volume only - switch in another 1M fixed resistor between hot and ground (pot outer lugs). This will be in parallel with the pot, and 2 x 1M in parallel = 500k
.......... But to convert the sound of a 1m vol pot to that of a 500k - at full volume only - switch in another 1M fixed resistor between hot and ground (pot outer lugs). This will be in parallel with the pot, and 2 x 1M in parallel = 500k
Which works fairly well, until you realize that your taper curve is now parabolic!
Me, I'd just figure that turning the pot down a notch or two will do about the same job. If you don't use a treble-bleed circuit, then turning down a 1MΩ pot will not only reduce the volume, but knock off some of the highs too. Same tone as having a 500KΩ pot at full volume, although the output might be 1 or 2 dB lower.
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