Basic guitar components

[karpetking]

Hello again, i posted a thread earlier on 6 position rotary switches. i think i understand them now, thanks!

But i'm still learning about basic wiring, and how the components work.

i have drawn this scrappy diagram of how i understand it to work, with one pickup, 1 tone, 1 volume.



the red lines show circuit flow, not including the tone bit.

is this right? can anyone tell me anything else about this stuff?

and do capacitors only allow high frequencies through?

thanks again

[karpetking]

Hey, i just realised, when the tone is at 0, it is the same as having a capacitor in parallel with the pickup, right? just what does this do? how does it affect high frequencies?

[JohnH]

You could think of a capacitor as a resistance (correct term: Impedance) that becomes less with higher frequency. Putting it in parallel shunts more of the signal the lower the impedance. So that cap in parallel shunts not very much at low frequencies, but more at high frequencies, giving you a treble cut As you turn your tone from 0 to 10, a high value resistance is added to the parallel circuit, so less and less signal is shunted and the treble cut has less effect.

Thats how I think of it, a better explanation would need alot more words!

John

[UnklMickey]

well your drawing showing "flow" in the circuit is a bit incomplete. you didn't include the current through the volume pot from 0 to 10. but thats small anyway and no need to nitpick.

but, that inner loop showing electron flow from the wiper to 0 on that pot doesn't belong there at all.

moving on to capacitors.

i'm not gonna bog you down with lots of science and engineering details 'cause i can tell from your questions that you aren't quite ready for that yet.

let's do use a few conventional terms though.

you can think of a capacitor as having a reactance that decreases with frequency. put another way the higher the frequency, the lower the "resistance" (impedance) for a given capacitance.

also, for a given frequency the "resistance" decreases as the capacitance increases.

this still might sound a bit more complicated than you wanted, but i just can't bring myself to simplify it to "capacitors only allow high frequencies to pass through". it's not that simple.

capacitors allow ALL frequencies to pass through (except D.C.).

they allows highER frequencies to pass "more easily"


when the tone pot is at 0, most of the high frequency content is shunted to ground 'cause the reactance of the capacitor at the higher frequencies is low and the resistance of the pot is really low -- like zero.

we could go on to discuss the pickup as being an A.C. source with series inductance, parallel capacitance, and series resistance. and we could talk about those series elements forming a voltage divider with the capacitor in our tone control circuit. but let's leave that for much later, if ever.

my guess is you just wanted a quick understanding without gobs of details and math.

does this fit the bill?




btw: you can draw your capacitor as two parallel lines ----||-----


U.M.

[UnklMickey]

JohnH,

thats so funny! your post came up while i was busy writing "a lot more words". wrote part of it, had some coffee and had some visitors, then finished and posted. then i see your nice concise answer.

is this a case that proves "less is more"?

[karpetking]

thanks again, both of you. this has helped me. i understand quite a lot about audio

devices, but practically nothing about how they work, and i would really like to get to understand this.

i have some questions about your answers.

firstly,
in order to change the capacitance, the only way would be to physically change the actual capacitor component, unsoldering it or switching with some circuit, etc, is this correct?

second,
unklmickey wrote:-
you didn't include the current through the volume pot from 0 to 10. but thats small anyway and no need to nitpick. that inner loop showing electron flow from the wiper to 0 on that pot doesn't belong there at all.

i thought i understood this, but now i'm not so sure. doesn't the volume pot send some of the signal straight to ground? - why is it connected to ground? where does this
"current through the volume pot from 0 to 10" come from?

sorry if this sounds stupid, i'm just trying to understand it all.



i read on guitarnuts that the tone control wouldn't even work in an ideal world. i suppose i need to understand the other half of the circuit, the amp, a bit better too.



unklmickey
we could go on to discuss the pickup as being an A.C. source with series inductance, parallel capacitance, and series resistance. and we could talk about those series
elements forming a voltage divider with the capacitor in our tone control circuit.

do you know of a good place i could read about this, to save you some more of my initial stupid questions?

thanks again.

[UnklMickey]

Looks like I’ve opened quite a can of worms here. I’ll do my best to be concise, yet establish some conventions, and get accurate information out here.

First, on conventions, this discussion is getting a bit loose because we’re mixing two of them. In your drawing your red lines infer electron flow, because you drew them leaving the negative terminal of the source (the pickup). Of course we know that the pickup is essentially an AC source, but for the purpose of evaluating the circuit, it’s OK to do that. Incidentally electron flow is also known as contemporary current. (contemporary as in the present time.)

Later you describe part of the signal as going to ground through the volume pot. This infers flow from positive to negative in your diagram. That is known as conventional current. (conventional as in we have met and we are in agreement.) this was the common convention in the early days, and is still used sometimes today because it’s easier to use when describing hole flow in semiconductors. (before you ask: no, it doesn’t have anything to do with your girlfriend.) Either electron flow, or conventional current is acceptable for use in analyzing a circuit. We get all messed up when we try to mix them.

Since you used both of them, I guess I have the option of choosing one. Today I’m gonna choose contemporary current. We’re also going to treat your pickup as a DC voltage source. Of course we know that it isn’t, but it will be much easier for us to describe it that way for now.

All of the current leaves the positive terminal of your source. For now let’s ignore the tone pot and focus on the volume control. With no load on the output all of the current goes through the pot from 10 to 0.
That current is equal to the source voltage divided by the resistance of the pot. The voltage at the wiper arm is equal to the total voltage times the resistance from the wiper to 0, divided by the total resistance. Hence if the pot was set so the resistance was equal from wiper to 10, and from wiper to 0, the output voltage would be ½ of the source voltage.

If we put any load on the output, we have to re-analyze the circuit, but we can use the same method. In this case the new current would be equal to source voltage divided by the total resistance. Since the total resistance is equal to ( the parallel combination of the load resistor and the portion of the pot from the wiper to 0) plus the resistance from wiper to 10.

This whole thing is getting a bit wordy so I’m gonna cut to the chase here: unless there is a source in the output there are only two current paths:

1 -- From the source + , through the pot (10) out the wiper arm, through the load, through the ground wire and to the source –

2 -- From the source + , through the pot (10) out (0) through ground wire and to the source –





As to why the pot is connected to ground: the ratio of voltage would be even more dependent on the load resistance. With a 500k pot and 500k load and the volume pot at 0 the output would ½ the source voltage. With a 1meg load it would be 2/3 of the source voltage. With a 50k load it would be 1/11 of the source voltage.

AND the output voltage would never be zero unless the load was 0 ohms. That would mean you could never turn the guitar completely off (not an issue for some folks).



Here’s a link JohnH brought to GN2 back in april.

www.buildyourguitar.com/resources/lemme/index.htm

It discusses source/resistance/reactance modeling

About changing the value of the cap: really small caps can be of a variable type. but those have values in picofarads. they have movable plates. remember those little transistor radios from the '60s? Variable capacitors were used to select the frequency. For values in the nanofarads or tenths of microfarads like you use in a guitar tone circuit they would be big, clumsy, expensive, and act like antennas for hum and noise. So we just select the corner frequency by chosing the right value. (yes, you can use a switch to select different values) And the setting of the pot determines the amount of attenuation above that frequency.


BTW:

I was wrong when I said the current through the pot was small. Obviously, if there is a light load on the output, the current from 10 to 0 is the LARGEST current in the circuit!

U.M

[karpetking]

Thanks again. I've been a bit busy, but looks like i'm going to have to read a lot more to understand things. i vaguely remember the different ways of looking at a circuit from physics classes at school. I'm going to have a look at some electronics stuff. thanks!!
Oh, when you were talking about a strat quacking, is this the sort of sound heard on "under the bridge" (i know that it is a jaguar or jazzmaster on the video, and probably in the studio) and "coming right along" (by the posies)?

[UnklMickey]

haven't heard of the posies. could be they haven't had a lot of airplay in the U.S. also could be i've actually heard them on the radio, but didn't know what i was listening to.

[karpetking]

I meant under the bridge by the red hot chili peppers, everyone's heard that, right? also their cover of castles made of sand, amongst other things, and the original, and,well, almost anything by hendrix that didn't have oodles of distortion on. the posies have been around a long while now - they are a us band

www.allmusic.com/cg/amg.dll?p=amg&sql=11:3ifuxqr5ldhe~T1

that keep on splitting up and reforming. if it is this sound that i'm thinking of, then i like it a lot. but i've never owned a strat. but i've wanted one from time to time. is it just me, or is a maple fingerboard more pleasing to bend notes on? should do a pole on fingerboards or something.

[UnklMickey]

i've heard that rhcp tune before but didn't know what the name of it was. that sound is kinda similar, but to my ears it almost sounds like a tele w/ a phase shifter or carefully adjusted flanger.

found the posies tune on amazon and listened to a sample. so much effects its hard to tell what that kind of axe. maybe a fat strat?

the quack is the sound that almost sounds like a phase shifter or wah. it's real suBtle. SRV and Jimi get that sound, sometimes even without effects.

BTW that link was so long that the whole thing didn't "take" when i clicked on it. so i just copied the whole thing as text and pasted it to the address window in my browser's address window.


U.M.

[karpetking]

hello, me again, back to bug you with my lack of knowledge as to complex issues that can't be explained easily.

am i right in thinking that impedance only occurs in ac circuits? is it where electrons "back up" on themselves when the flow direction changes, in the opposite side of the line in a wave type diagram? and does everything (a piece of metal/wire, etc.) have some amount of impedance?


as for the quack.....
i'll have to listen to those songs you mentioned more clearly.
I don't know how to make the whole link appear as a link - i tried a few times.
thanks again.

[UnklMickey]

does everything have a certain amount of impedance, you bet!

you can think of the relationship of impedance and and resistance as being similar to the relationship of velocity and speed.

when all the velocities are in the same direction, we just add the speeds together. but when you have say two velocities at right angles, you can't just add them together. they have to be added as vectors.

for instance if a boat was traveling east at 40 kph (relative to the water) in a river that was flowing south at 30 kph, then the boats velocity would be 50 kph - ESE. i only gave an approximate heading, i'm not gonna do the trig., but i think you get the idea.

i only used the velocity analogy to get you used to the idea of vectors.

when we describe the relationship between capacitors, resistors, and inductors, we also use vectors.

the inductive reactance and the capacitive reactance are 180deg from each other so we can think of them as being say, north and south. and we can think of resistance as being say, east. so if we plot out the inductive reactance, the capacitive reactance and the resistance of a network, we can determine it's impedance at a certain frequency. that impedance will have both a magnitude (ohms), and a direction (phase angle).

whew, this is starting to get too much like work!

i hope that gets you started in the right direction.



i know that there is a limitation on how many characters this board will support in hyperlinks. the only reason i pointed out what i did is so folks would understand why sometimes those links fail, when they just click on them. and how to get around that.

U.M.

[karpetking]

thanks again. phew.... reactance, eh? i'll have to read some more. this is getting complicated.

when you wrote-
i know that there is a limitation on how many characters this board will support in hyperlinks. the only reason i pointed out what i did is so folks would understand why sometimes those links fail, when they just click on them. and how to get around that.

i thought that might have been why you wrote it. i'm sure there should be some way to do it, though.

[Mustang]

Great job unklmickey. When can we talk about Smith Charts, Nyquist and Bode plots?

[UnklMickey]

mustang said:

Great job unklmickey. When can we talk about Smith Charts, Nyquist and Bode plots?

how about, let's not.

transmission line theory and feedback loop stability are a long way from the original post of this thread, even more so than phase vs frequency.

besides that stuff makes my head hurt.

the idea here was to give karpetking an introduction to some concepts that would help him understand how the components of his guitar fit together. and we're already a bit further off those basics than we need to be.


U.M.

[Mustang]

You're right. Anyway, as I said, great job. I've really gotten addicted to scanning the board for interesting stuff. Seems like a really good group of people providing input.

'stang'

[karpetking]

Hello again, ready for another stupid and possibly not too relevant question? (thanks for all the help, by the way)

I have been reading about basic electronics (including diodes and transistors, can't quite understand how they work - though i did pass a-level chemistry some years ago). My question is - according to ohm's law, the only thing separating voltage from being directly relative to current is resistance, right? so in a transformer, when one goes up and the other goes down, is this due to resistance? (some power is surely also lost, isn't it?).

i know it isn't too relevant, but it's bugging me. i'm not on to impedance yet - but have covered capacitors - it just says that they store electrons, so far.

[Mini-Strat_Maine]

karpetking said:

My question is - according to ohm's law, the only thing separating voltage from being directly relative to current is resistance, right? so in a transformer, when one goes up and the other goes down, is this due to resistance? (some power is surely also lost, isn't it?).

I'll take a stab at this, and then the guys who know more about it than I do (well, that's darn near everybody) can say if I got it right. ;D
If I remember correctly, transformers work by inductance: the winding of the one side induces current on the other side. The windings of the two sides dictate whether the transformer steps up the voltage or reduces it. (Like the "wall wart" that changes 110 volt house current to something that will power a stomp box without the need for a battery. There are also inverters that will change 12 volt DC automotive power to 110 VAC, but let's not even get into rectifiers and such. )

Re Ohm's Law, try this: "Twinkle, twinkle, little star. E equals I times R." So voltage (E), current (I), and resistance (R) are "directly relative" in that we can figure out one value if we know the other two. (I think that's algebra. I hate algebra.)

I hope this was helpful and not more confusing.

--
Doug C.

[UnklMickey]

that's pretty good Doug.

i wonder if Karpetking was asking a bit more. (his question was somewhat vague.)

karpetking said:

...My question is - according to ohm's law, the only thing separating voltage from being directly relative to current is resistance, right? so in a transformer, when one goes up and the other goes down, is this due to resistance? (some power is surely also lost, isn't it?)....

so, when you ask about voltage and current, one going up, and the other going down, are you referring to the voltage on the primary vs the voltage on the secondary, compared to the current through the primary vs the current through the secondary?

[karpetking]

I think so...
i mean, when the voltage level coming out of a transformer is higher than the level going in, then the current will be lower, but going on ohm's law, the only way for a given voltage to have different current values, is by varying resistance value (or something). so is this what happens? (with the resistance?). i'm not sure what i'm getting at, but something seems strange to me about current going up and voltage going down, and vice versa.

[Mustang]

karpetking,

Maybe it would help to think of a transformer as a 'power' device. For a specific chunk of metal, it has the capability to 'transfer' a certain amount of power. That's why they are rated in terms of 'VA' (volt-amperes). A transformer does 'step up' or 'step down' the AC voltage applied based on the number of 'turns or winding' on the primary side versus the secondary side. Then it stands to reason if the power into a transformer equals the power out, then the voltage times the current on one side of the transformer must equal the voltage times the current on the other side of the transformer. If the voltage on the secondary side is lower than the primary side, you would expect the transformer would be capable of providing a proportionately higher current on the secondary side. The limit to max power transfer is the size of the conductors used to wind the transformers and the ability of the transformer to radiate the heat. And you are correct there are 'losses' that must be accounted for. Transformer efficiencies are in the high 90 percent range and losses are due to heat generated and the currents required to magnetize the core.

Hope this helps.