AndrewCE

In one of my cheaper guitars, there is a large screw that is screwed into the body of the guitar from the electronics cavity. the grounds from the pickups/pots are soldered to this screw, which likely is acting as a ground instead of the bridge.

does that clear up the last bit?

That's probably to connect the cavity's shield (paint or aluminum foil/tape) to ground.

By the way, I did the treble bleed mod on my American Deluxe Ash Stratocaster today. It works! I was surprised; it was my first electrical guitar mod. I thought for sure I would screw something up.

No question; just reporting.

Oh, btw, I used a 133K resistance in series with 1200 pF capacitance. If I did it again, I'd go with slightly higher capacitance. Maybe 1400 pF.

uh oh. i did mess up. I accidentally converted wrong and put 1,000,200 pF capacitance. Oops.

But for some reason I still want a bit more treble coming through. Maybe I should just lower the series resistance.

And I did notice that the volume taper changed. Oh well.

By the way, I did the treble bleed mod on my American Deluxe Ash Stratocaster today. It works! I was surprised; it was my first electrical guitar mod. I thought for sure I would screw something up.

No question; just reporting.

Oh, btw, I used a 133K resistance in series with 1200 pF capacitance. If I did it again, I'd go with slightly higher capacitance. Maybe 1400 pF.

Here is a really good article from Guitar Nuts on what is really going on with "string grounding"...

Your Body is a Bucket 'o Noise!

This was a very enlightening answer to this question that we do, including me, just do without really realizing why, just the effects...

In fact we are not using the string to "ground" to our body at all...at least not in a simple sense. Your bodies attract noise towards the guitar and by touching the strings we ground ourselves to the common ground in the amplifier and such...but the article does a much better explanation than I can give.

or is it just so you don't get electrocuted?

Actually, it is largely the grounded strings that puts you most at risk. Because you are holding metal strings in both hands and have them electrically wired...it there is a fault or a mismatch in electrical equipment...it can often be fatal.

In a typical electrical accident...you may touch something and the power will try and run to ground through your body. Rubber soled boots will often "save you" as the power wont make the circuit to the (literal) ground through your feet. However, holding a guitar and wired to the amplifier ground, should a mic become live say, the power will run through your lips to the nearest ground, the guitar strings...

Why it is fatal is that the current surge through both hands grounded to the strings puts the current directly through the heart, stopping it instantly and you are dead!

One good idea to combat this is to get a powerboard with a circuit breaker on it (not just an overload breaker) and use this especially in strange places where you don't know the wiring is safe...and make sure all your amps and effects are connected to it...that way it will instantly switch off if you should say, reach over to adjust a light with one hand while holding a guitar and it will break before you get a fatal shock (hopefully)...unless of course you play without shoes and play in a puddle of beer...you may still risk yourself then!

thanks,

i had actually read that exact article a long time ago, but i forgot what it said and I couldnt find it again

so basically my body is an electrical shield and by grounding it, the EMI that i'm picking up gets sent to ground

yes you will generally need to ground the strings. Some Active pickups and piezos are exceptions.

why exactly do the strings need to be grounded? does it affect the sound? or is it just so you don't get electrocuted?

and while we're on the subject, why does a guitar hum when you're not touching the bridge/strings, and then stop humming when you touch the bridge/strings?

I've heard of this "treble bleed" modification that keeps your guitar from going muddy when you turn down the volume. It basically uses a cap to bypass the pot and go straight to the output.

My question is: With this mod, when you turn the volume all the way down, do you still hear high frequencies coming through the cap? If not, why not? This seems like it would be a pretty big problem, but it looks like a fairly common mod.

No, you won't hear anything with the volume all the way down because all of the signal is shunted to ground via the pot.

oh. duh. k thanks

how did i overlook that

I've heard of this "treble bleed" modification that keeps your guitar from going muddy when you turn down the volume. It basically uses a cap to bypass the pot and go straight to the output.

My question is: With this mod, when you turn the volume all the way down, do you still hear high frequencies coming through the cap? If not, why not? This seems like it would be a pretty big problem, but it looks like a fairly common mod.

As a passive circuit, an inductor with a series cap is more like a notch filter. If you want to play around with something interesting and make a resonant filter using an inductor, you could take the standard inductor based wah circuit and have an on-board "Q" filter kind of like what I did on my "SpankenStrat" - http://www.diyguitarist.com/Guitars/SuperStrat.htm

you mean like put just a cap and inductor in series (or in parallel, shunting to ground) to make a bandpass filter? to cut out the highs and lows? is that what you mean?

I was thinking of it more as a tone control, using a cap an inductor in series to ground. Essentially, it's the Gibson "Varitone" circuit: http://www.blueshawk.info/varitone.htm

I like the on-board adjustable "Q" filter (wah) idea. That way, I get a resonant filter in conjunction with having all of the tone controls I am used to.

oh so basically the opposite of a wah; instead of resonant peak there would be a resonant valley; it's a mid-cut control

i was wondering why you would put a passive wah-filter on your guitar. that would take away so much volume!

As a passive circuit, an inductor with a series cap is more like a notch filter. If you want to play around with something interesting and make a resonant filter using an inductor, you could take the standard inductor based wah circuit and have an on-board "Q" filter kind of like what I did on my "SpankenStrat" - http://www.diyguitarist.com/Guitars/SuperStrat.htm

you mean like put just a cap and inductor in series (or in parallel, shunting to ground) to make a bandpass filter? to cut out the highs and lows? is that what you mean?

Still...if you want to try it, no harm there I suppose, but as a basic idea I think bluesy has given more technical reasons...but if you step back a little, there is a flaw in the concept and looking for a novel approach to something that seems not to require a better solution.

But, hey...maybe I am missing something...

well i wasnt suggesting the use of an inductor over a capacitor just for the possibility of that mod, i also wanted to learn a little more about what inductors and caps do. i've read all i can find, and i know how they work and what they do, but thinking creatively is often the best way to familiarize oneself with a subject. i'm just thinking outside the box here. finding another way to do the exact same thing seems pointless, but it can actually be pretty interesting. heck, i saw a book once with 200 different proofs of the pythagorean theorem! pointless, but with it you can study how all the fields of math relate to one another.

i know what a high pass filter is, but there are two ways to make it, either with a series cap or a parallel inductor

oops i accidentally said "variable cap" but i meant "variable resistor"

A volume control actually does have an effect on the signal at 10, though. Just like a tone control bleeds off some highs at 10, volume control bleeds off volume. Because it's impossible for the volume pot to have infinite resistance (unless its a no load pot). A series pot would be able to have 0 resistance and therefore you wouldnt lose any signal.

OK, you asked what they usually do - and the answer is, series capacitor.

The Volume pot should be a large enough value, compared to the other impedances in the circuit - like pickup impedance and amplifier input impedance, that is has no appreciable effect when set to it's maximum. Most other impedances in the circuit are 10's of k (like the 50k input on a guitar amp), hence volume pots are 250k or 500k. Being an order of magnitude bigger, it's effect should be small. So small, that the small loss is swamped by variations in gain in all sorts of other places in the guitar pedals and amplifier.

A series pot would indeed let 100% through at full volume, but it would have another serious problem. At zero, there would still be some sound because it cannot have infinite impedance. In fact, for a 500k pot into a 50k amp input, you'd get approx. 10% of the signal voltage still coming through. The design decision is that it is better to be able to turn it all the way down.

good point in that last paragraph. it makes me wonder if theres a way to have the best of both worlds. i'll get to drawing for now

oh you mean like, filling up the bucket over and over, like 500 times a second...

Only if you don't trust the mathematics. A time constant is just that, constant, doesn't matter how frequently you try to fill or empty it.

Another way to look at it is that as frequency goes up, the impedance of the capacitor (ac resistance if you like) goes down. So as frequency get's really high, the capacitor starts to look like a short circuit, and since it's shunts the ac to ground, the high frequencies don't get through. Hence it's a low pass filter.

only what if you dont trust the math?

i know how all that works, i just dont think a bucket is the best analogy for a capacitor

It just seems like it would act differently. In the series inductor, youve got a "traffic jam" of high frequencies. But in the shunting capacitor, you have a "rerouting" of highs; they don't build up, they get released instead.

side question: To remove lows, do they usually put a series capacitor, or do they use an inductor to shunt to ground?

Also, why dont they make volume controls in this series way? Instead of putting a variable cap to ground, put a variable cap right in between pickup and output jack. This might actually be better, since when Vol is at "10", the pot has absolutely no effect on the signal.

Yes, to remove lows, essentially a series cap is used - google "high pass" filter.

Not sure what you mean in that last paragraph. A volume control is a potential divider, i.e. a device that takes an input voltage (signal in this case) and provides, as it's output, some fraction of that signal. When a pot is at 10, it is not supposed to have an effect on the signal, because you have asked for 100% of the signal to be let through.

i know what a high pass filter is, but there are two ways to make it, either with a series cap or a parallel inductor

oops i accidentally said "variable cap" but i meant "variable resistor"

A volume control actually does have an effect on the signal at 10, though. Just like a tone control bleeds off some highs at 10, volume control bleeds off volume. Because it's impossible for the volume pot to have infinite resistance (unless its a no load pot). A series pot would be able to have 0 resistance and therefore you wouldnt lose any signal.

But the response curve is only the same in single pole filters?

Here the last set of graphs shows that cascading filters provides a different response, even if it still has the same cutoff freq:

http://www.dspguide.com/ch19/2.htm

so maybe the stellartone knob has multiple pole filters...

btw, i'm dont think your analogy applies well to AC cicuits and signals

That page is about DSP filters and is looking at recursive filters. Also cascading 2 single pole filters gives you a 2 pole filter, so the slope is doubled to 12db per octave.

I was only saying that there was no need to have multiple caps for a tone switching circuit to allow you to achieve all the desired cutoff frequencies (with single pole low pass filters), just vary the resistance.

The analogy demonstrates the time constant of the RC circuit, and that is exactly what determines the cutoff frequency, so yes, it does apply to AC.

well it's like you just said, all single pole filters can react the same by just varying the resistance; so the only useful mod would be to make multiple pole filters and cascade them at different switching positions; and that requires multiple caps

oh you mean like, filling up the bucket over and over, like 500 times a second...

What effect would there be if you put an inductor in series with your pickup (assuming a one pickup guitar)? I'm guessing it would increase the impedance and take away highs, but how would it act differently than using a capacitor to shunt to ground (like most tone controls)?

Me again yes it would act pretty much the same as a cap shunt to ground. It's another way to make a single pole low pass filter. It's not usually done this way in production because caps are cheaper and lighter than inductors.

It just seems like it would act differently. In the series inductor, youve got a "traffic jam" of high frequencies. But in the shunting capacitor, you have a "rerouting" of highs; they don't build up, they get released instead.

side question: To remove lows, do they usually put a series capacitor, or do they use an inductor to shunt to ground?

Also, why dont they make volume controls in this series way? Instead of putting a variable cap to ground, put a variable cap right in between pickup and output jack. This might actually be better, since when Vol is at "10", the pot has absolutely no effect on the signal.

Oh so it's the cutoff frequency.

Thanks, bluesy!

But that only tells you that the cutoff frequency will be the same in the two cases (halving resistance vs doubling capacitance). Is there a way to know that the entire frequency curve will be the same? Or at least approximately the same? I don't think there is.

Also, how do you know that there will be a 6 dB drop per octave?

The 6db figure can be calculated, but check any text or online reference for any single pole filter. Sometimes it is expressed in decades (I think it's 20db per decade).

How about an analogy? The capacitor is a bucket, and the resistor is a faucet restricting the flow of water (electricity). For a given size bucket, and faucet setting it will take X seconds to fill the bucket. Now get a bucket (capacitor) twice as big, and halve the restriction (resistor) - meaning the flowrate will be doubled. The bucket still takes the same X seconds to fill.

But the response curve is only the same in single pole filters?

Here the last set of graphs shows that cascading filters provides a different response, even if it still has the same cutoff freq:

http://www.dspguide.com/ch19/2.htm

so maybe the stellartone knob has multiple pole filters...

btw, i'm dont think your analogy applies well to AC cicuits and signals

Oh so it's the cutoff frequency.

Thanks, bluesy!

But that only tells you that the cutoff frequency will be the same in the two cases (halving resistance vs doubling capacitance). Is there a way to know that the entire frequency curve will be the same? Or at least approximately the same? I don't think there is.

Also, how do you know that there will be a 6 dB drop per octave?

What effect would there be if you put an inductor in series with your pickup (assuming a one pickup guitar)? I'm guessing it would increase the impedance and take away highs, but how would it act differently than using a capacitor to shunt to ground (like most tone controls)?

Is it possible to make a variable capacitor by just putting a cap in parallel with a variable resistor and varying the resistance? I think this will vary the capacitance, but will it change the impedance of the current?

If you're talking about switching between .022u and .047u, then yes, you should be able to hear the difference.

Now if you generally find guitars muddy sounding and never move the tone off 10, then maybe you're best option would be switching (maybe with a push-pull switch on the pot itself) the tone control in and out of the circuit. Removing the tone control will have a noticeable effect in the top end, crispness, or whatever you wish to call it. Give it a try and see if you like it.

theres a difference b/t those 2 caps, but is there a difference b/t a [.047u in series with a resistor] and a [.022u]? would they both act the same if the resistor value was just right?

OK, speaking only about values here, there's not a lot of point in having different cap values. The formula for what a tone control (which is really a low pass filter) does is

F = 1/(2*pi*R*C)

If you double the cap size, that's EXACTLY the same as halving the resistance. So just use one cap, and multiple resistors (or trimpots at different settings) and you'll hear exactly the same thing (and save the cost of all the other caps). If you are using a trimpot as well as a normal tone control, the R in that equation is the total resistance.

what does "F" mean? frequency? frequency of what?

i'm pretty sure that halving the resistance and doubling the capacitance don't give you the same frequency response curve. maybe at one specific frequency it's the same, but not the whole curve.