bobbo
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Post by bobbo on Sept 26, 2008 12:03:40 GMT -5
Hey there, this seems like the perfect place for me to further my knowledge.... I'll soon be at the wiring stage for a homebrew Telecaster copy and I'm planning to use a Bill Lawrence Q-filter for tone control. For those interested: "The Q-Filter is an LCR network - a 1 henry, low Q noisefree inductor in series with a .02 micro farad cap in parallel with an 8 kilo ohm resistor."The diagram below seems to take care of everything nicely. The one question I have is regarding the wire running from the tone pot casing to the volume pot casing. Is it necessary for grounding, since the guitar will be rear-routed with no metal control plate, or is it superfluous? While I'm here, any other problems / suggestions would be welcomed. Thanks![glow=red,2,300][/glow]
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Post by ashcatlt on Sept 26, 2008 12:21:04 GMT -5
Welcome to the NutzHouse! I'm sure one of these folks around here has played with something like this and will be much better able to advise than I. I'd like a little clarification. You said "The Q-Filter is an LCR network - a 1 henry, low Q noisefree inductor in series with a .02 micro farad cap in parallel with an 8 kilo ohm resistor." The diagram shows a box labeled "Q-Filter" and then shows a couple of parts across those two lugs of the pot. One of them looks like the resistor. The other is hard to make out. Looks to me more like a diode than anything, but that can't be right, can it? Is the box there just the inductor, while the difficult to discern part is the cap? In any case, the answer to the main question is yes (sort of). As drawn now, the "Q-Filter" is trying to find ground through the tone pot shell. If this is not connected to the volume pot shell where ground can actually be found, it won't find ground (it'll find wood) and won't work as intended. The "sort of" part comes in that you could just as easily connect the "Q-Filter" to the grounded lug of the volume pot (or anywhere else it can find ground) and leave the tone pot shell un-grounded. This defeats the (questionable) shielding effect of the tone shell itself, and might add a little bit of noise.
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bobbo
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Post by bobbo on Sept 27, 2008 7:00:02 GMT -5
Thanks for the reply. Sorry for the poor image quality. The unclear section of of the diagram is a resistor and a cap in parallel. It seems (unfortunately I've found conflicting evidence and haven't been able to contact Bill Lawrence himself for clarification) that in the diagram I posted the box marked 'Q-filter' is, like you said, just the inductor. The effect, apparently, is : "like a two- or three-band EQ controlled by a single knob -- the entire tonal balance or timbre of the instrument gradually changes as the knob is turned. When you get the knob toward the counter-clockwise limit, the sound gets close to an acoustic." If it does do as claimed (and I've got no reason to doubt Bill Lawrence's theory) then it sounds like an excellent and inexspensive mod, and certainly worth a try.
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Post by flateric on Sept 28, 2008 16:16:09 GMT -5
This little LCR looks similar to the old sweet switch PRS added to Santanas guitars when he used a really long lead on stage and got some unwanted extra capacitance/resistance which took out some of the mids. I think this circuit can be sort of looked at as a mid-boost/cut. Earthing the tone and vol pots together is good practice to eliminate hum, not essential to the circuit - but I wouldn't say it's superfluous. Be interested to hear how you get on with the Q-tone.
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Post by newey on Sept 28, 2008 17:13:10 GMT -5
Bobbo: Over a year ago here, there was this discussion of Gibby L6-S wiring, Bill Lawrence, the Q-Filter, etc.: L6-S wiring thread
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Post by ashcatlt on Sept 30, 2008 14:05:50 GMT -5
Any chance you could edit the title of this thread to something a bit more descriptive? Anyway, 5spice is fun! This is a band cut or notch filter. I've modelled the frequency response using JohnH's numbers for a PAF humbucker. Other pickups would change it a bit, but this shows us the basic idea. The Q Filter pot here is a 500K, as is the volume pot. I may have it wired backwards, but... I think I can see where it might sound like a piezo equipped acoustic. Don't think I'd be very happy with it. You might like it. I thought perhaps you could get a good "metal scoop" by changing some of the parameters. Turns out that increasing the value of either or both the inductor and capacitor will lower the center frequency of the scoop, but doesn't do a whole lot to move the peak at the high end. Note - the frequency response here has been limited to the area of the spectrum relevant to guitar. To the left it's just a flat line. To the right, it continues to slope down toward negative infinity.
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Post by ChrisK on Sept 30, 2008 14:35:38 GMT -5
JohnH's Spice models use linear steps on the pots. Most guitar pots used are audio taper, as are our ears. I tend to use audio taper steps as measured from audio taper pots. If one looks at my post on the The Passive High-Cut Tone Control, they will see this in the form of the .STEP PARAM { var} LIST aaa, bbb, etc, declarations. Using this method one can create steps as desired. When I get the time, I'll run some simulations of the "Q" circuit.
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Post by ashcatlt on Sept 30, 2008 15:11:45 GMT -5
I was going to suggest that a log pot might give better control in that area where the scoop really gets going. Also, I think a bigger pot (1M or so) will give us a more "natural" response when turned to the "off" position.
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Post by newey on Sept 30, 2008 16:23:01 GMT -5
Done! With apologies to Bobbo, but Ash is right, too generic a title for informing anyone whassitallabout
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Post by ChrisK on Sept 30, 2008 17:38:52 GMT -5
Interesting concept, this "Q" filter. Hmmm, Gibson "Varitone" anyone? Now, these run aboot $20. One can misuse a secret electronic device that only costs about $2 - $4. They're called audio signal transformers. Mouser, DigiKey, and many other's carry them. If you only connect the high impedance primary leads, and don't use the secondary, they become (gasp!) inductors. ( Transformers - more than meets the eye.....) ;D ;D These are specified at 1Khz. The one with a primary impedance (Z) of 10K Ohms will have an inductance value of; Z/(2*Pi*1,000) = 1.6 Henry. A 1 Henry unit will have a primary impedance of 6.28K Ohms. Gee, they have a 7K one for under $4. This is the Mouser link. Look aboot, they sell other electronical stuff too.
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Post by JohnH on Sept 30, 2008 18:26:58 GMT -5
What I find interesting is that it is providing a range of high frequencies beyond what is normally the limit set by the interaction of the pickup inductance and (mainly) the cord capacitance. So the sounds are likely to have a piezo/acoustic/buffer style of added zing. I think it could be a way of getting closer to an acoustic sound for certain songs, as an approximation on an electric guitar.
This higher peak seems to be due to the filter inductor, combining with the pickup inductance to create a new lower effective value as they combine in parallel – which leads to a higher frequency for the resonant peak.
I think there is more to be discovered with this style of circuit. Also, most guitars already have 2, 3 or 4 inductors in the form of their pickup coils, and I think there is much mileage in combining them via capacitors to shape the response, in various part-bypassed arrangements.
John
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Post by newey on Sept 30, 2008 20:39:29 GMT -5
Well, we seem to have lost Bobbo in this discussion since Ash answered his grounding question. But it has taken an interesting turn . . .
JohnH said:
And ChrisK noted:
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Post by ashcatlt on Sept 30, 2008 22:45:14 GMT -5
So, am I surmising correctly, that one could buy differing such audio signal transformers, of varying inductances (or is it "inductices"? "inducti?"), and arrange them around a rotary switch in a Varitone-ish way? Not only have a "Q-filter", but also R.S,T and U-filters? What sort of tonal variations would we be talking? Would it be adding more and more "piezoness" with every turn of the switch? That transformer ChrisK talked about? It's actually a twofer. Assuming, of course, the secondary winding has a different number of turns from the primary. Actually, could these two coils be wired in series and/or parallel to get us a total of 4 for the price of 1? Or will that not work for the reason that a transformer does? Not to my mind, at least not with the specified values. I hate piezo pickups for exactly this reason. Now, if we rearrange the circuit a little, we can get a bandpass filter like what angelodp is talking about in his thread. This might be interesting.
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bobbo
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Post by bobbo on Oct 1, 2008 9:22:30 GMT -5
Done! With apologies to Bobbo No problem, I'm glad to have sparked some discussion (even it is mostly by accident ) I've been without the internet for the last few days, but this certainly makes for some interesting reading, and a nice tip off on the budget version, thanks Chris! Unfortunatly my own electronics knowledge is not up to much so I don't understand the theory as fully as you guys, though I'm now certainly convinced to give this a try. Maybe spend a few days messing about with varying the inductance, and the cap and resistor values as well. I much prefer to gain a little knowledge and then physically do things to further understand the subject, than make my brain hurt with theory that I may have misunderstood. I'll certainly let you know how things get on when I get to it.
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Post by ChrisK on Oct 1, 2008 13:15:34 GMT -5
InducticatorsNope, doesn't matter. Yeppers. Well, if you connect two coils on the same core in parallel, each with "n" turns, you just have a coil with "n" turns, so the inductance is roughly the same. Since parallel is fairly meaningless (other than improving the "Q"), you'd have to settle for three for the price of one. Now, the primary and secondary may be center-tapped, so you have up to four coils to play with. If you look at the 42TU012 unit with a center-tapped primary of 10K and a center-tapped secondary of 5K, there is much to play with. The whole primary indicates an inductance of 1.6 Henries. Each half of the primary indicates an inductance of 0.4 Henries (1/2 the number of turns, 1/4 the inductance). The whole secondary indicates an inductance of 0.8 Henries. Each half of the primary indicates an inductance of 0.2 Henries (1/2 the number of turns, 1/4 the inductance). Various combinations of each winding, both one side and full, will render different combinations of inductance. Since each can be none, 1 sided, or the full winding, this would indicate 9 possible combination with one of them being pointless. This is without any degradation to parallelistic thinking. Also keep in mind that the actual inductance is related to the turns ratios realized. Here, you're on your own. I have an LCR meter that can measure the realized inductance, but I'm easily bored. Looking at the Inducticators link (it's at the beginning of this response - you did look, I hope), one can see that the driving factor in inductance is the number of turns (N). It always shows up as N2.This means that if you double the number of turns, the inductance increases by a factor of four. If one had a transformer with a primary of 6.8K and a secondary of 6.8K, one would have either a 1 Henry (for a single winding) or a 4 Henry (for two windings in series) inductor. Oh, and do mind your coil phasing. Interesting is. It will work for the reason that an inductor does. After all, a transformer is just an inductor with a terminal disease (quick trivia question; what is the minimum number of leads/terminals for a transformer, and why?).
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Post by wolf on Oct 1, 2008 13:17:39 GMT -5
If you have an old computer modem kicking around, the transformer from that should be about 1.4 henries. I've added that (along with a tone circuit) inside a volume pedal. If I can find the schematic I'll post it here. I believe I posted it a couple of years ago but that might be a little difficult to find. It's better to search my home computer - I'm at work now.
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Post by ChrisK on Oct 1, 2008 13:48:49 GMT -5
Hmm, well, don't try to search with a dial-up connection. ;D ;D
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Post by wolf on Oct 6, 2008 13:52:15 GMT -5
Okay, here it is: The resistors are there to stop those nasty sounding switch "pops" when you change the rotary switch position.
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Matt
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Post by Matt on Oct 8, 2008 16:12:59 GMT -5
So I have a question... In an attempt to understand the Q-filter i researched the LRC circuit a little and want to see if i understand it correctly. In wolf's diagram, when selecting the .02 cap, would the center frequency be around 290 hertz? Then when moving to a lower value cap the frequency would move up and vice versa? and since he's using a high value of resistance that means it will have a large bandwidth?
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