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Post by mattmayfield on Aug 4, 2006 18:45:20 GMT -5
This is something I've been curious about for several years, ever since I first put a HB (PAF Pro) in the bridge of my Strat. I also wired a series/parallel switch and moved the middle tone control to the bridge.
The bridge tone control behaves normally when the PAF is in series, but when it is in parallel, the pot acts as a treble cut from 10 down to about 4, and then as it approaches 0 it lets some treble back in and seems to do a strange midrange notch sort of thing à la Boston or a stationary wah pedal.
It's kind of a hip sound (for Boston anyway) so I want to keep it, but I'm curious... Has anyone experienced this effect, or know what would cause it?
Matt
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Post by JohnH on Aug 4, 2006 19:04:20 GMT -5
Hi Matt,
Yes, that makes electrical sense. Its to do with capacitance (the tone capacitor in your control), inductance (the coils of your pickups) and resonance.
Usually, as the tone control is turned down, it is simply shunting more and more treble to ground through the capacitor, as the resistance of the tone pot is reduced. But as it approaches zero, another effect becomes significant, which is the tendency for a capacitor and inductor to form a 'tuned circuit' and resonate around a particular frequency. Which frequency depends on the values. The parallel setting has a much lower (1/4) the inductance of the series wiring, which pushes the resonance up into the mid range. With a series wiring, the resonance is at a lower frequency and may not be noticed.
While the tone pot is at medium settings however, its resistance damps this effect, so you dont hear it
I like to play with this effect, using very small capacitors to tune the resonance, changing the voice of the guitar, rather than just killing the treble.
cheers
John
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Post by ChrisK on Aug 4, 2006 22:16:19 GMT -5
I've done this by adding a small cap (0.001uF) to the unused lug on a tone pot. By using a 500k pot, which is at 250k around "8", one can turn the pot all the way up and create a peak in the response for some mid to high boosting.
Ain't nuthin' free, this occurs at the expense of frequencies higher than the peaking frequency.
In essence, the 1:4 ratio in effective inductance will raise the peaking frequency by a factor of 2 (1/[2π[LC]^1/2]).
Fender have solved this pesky effect by developing the "Grease Bucket" tone control.
I've found that by using a tone control in an ANALOG manner (as opposed to a BINARY manner), one can eliminate this pesky effect just by not turning the control all the way to "0".
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Post by sumgai on Aug 5, 2006 0:10:00 GMT -5
Hey Chris!
Check out this formula:
2 (1 ÷ (2 π (LC)1/2))
The Pi almost looks like an "n", but it can be seen for what it is. I concocted Pi by using & pi ;, all mashed together, and the divide symbol is & divide ;. The tag takes care of the exponent.
I wonder what else we can get to work for us, in terms of representing symbols.
sumgai
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Post by UnklMickey on Aug 7, 2006 10:58:28 GMT -5
or put it another way,
a line that looks like this in the editing window:
2 (1 ÷ (2 π (LC)[sup]1/2[/sup]))
looks like this, when posted:
2 (1 ÷ (2 π (LC)1/2))
of course to get a line that looks like this when posted:
2 (1 ÷ (2 π (LC)[sup]1/2[/sup]))
i had to use a line that looked like this in the editing window:
[noubbc]2 (1 ÷ (2 π (LC)[sup]1/2[/sup]))[/noubbc]
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