What kind of signal does a gee-tar pickup generate?

[Hiwatt25]

So, being a new guy, what kind of signal comes out of a guitar?  I've always assumed that it was an alternating current and that the guitar cord had three conductors.  Am I waaaay off?  Figured, I'd better try to understand the first link in the chain if I'm gonna try and build something that effects it.  Doesn't the positive voltage get carried on one conductor, the negative on another and a third is ground?  If it turns out the guitar signal is DC I'm gonna fall out of my chair. 

Thanks again, all.  This site is rich in information.

Joe

[junkyjunky]

http://www.buildyourguitar.com/resources/lemme/

an interesting read. here's a snippet.

A real coil can be described electrically as an ideal inductance L in series with an Ohmic resistance R, and parallel to both a winding capacity C. By far the most important quantity is the inductance, it depends on the number of windings, the magnetic material in the coil, and the geometry of the coil. The resistance and the capacitance don´t have much influence and can be neglected. When the strings are moving, an AC voltage is induced in the coil. So the pickup acts like an AC source with some attached electric components

[amz-fx]

http://www.buildyourguitar.com/resources/lemme/

an interesting read. here's a snippet.

Interesting article!

Here's some more about the output of a guitar:  http://www.muzique.com/lab/pick.htm

regards, Jack

[Hiwatt25]

So, I'm getting close to that AH-HA moment.  Would you mind answering a couple more questions.  The first would be, can the positive and negative voltage generated by a guitar's pickup travel over a single conductor?  Always thought one conductor was the positive and one was negative.

And the second question is, what purpose does ground serve.  I always thought it was to move current safely away from the user if something in the circuit should short.  I don't get why so many components are attached to the chassis of the different stomps/amp and why you don't get a shock when you touch the chassis.

Like I said, I feel that epiphany is coming...

[Paul Marossy]

can the positive and negative voltage generated by a guitar's pickup travel over a single conductor?  Always thought one conductor was the positive and one was negative.

Since it's an AC signal, ground is the zero voltage point. The hot swings above and below ground at whatever frequency. Have you seen sine waves on a scope? That horizontal flat line at the mid-point of the sine curves is your ground.

what purpose does ground serve.  I always thought it was to move current safely away from the user if something in the circuit should short.  I don't get why so many components are attached to the chassis of the different stomps/amp and why you don't get a shock when you touch the chassis.

It does serve as a safety measure. It's the voltage potential above ground that is generally potentially hazardous. You don't get shocked when you touch the chassis of a tube amp or a stompbox because modern amps have grounded chassis' and ground is where zero voltage exists. Ground also serves the dual purpose of being the crossing point for the AC signal as well. There are some old tube amps that have a "hot chassis" that are potentially dangerous. I don't think there is too many of them out there that haven't beed converted to a grounded chassis, but they do still exist.

Does any of that make sense?

[scaesic]

Voltage is essentially a potential difference.

you have the same potential as a ground connection, so there is no potential difference between you and a ground connection, and therefore current will not flow from you to it.

it's analogous to gravity in this way, if ground potential is the top of a cliff, you are standing at the top of the cliff, the ground connection is also on the top of the cliff, you can move over to it no problem and touch it, however if you go to the edge of the cliff, there is a potential drop. The bottom of the cliff is analougous to a negative voltage, if you drop from the top of the cliff (ground) to the bottom (- voltage) then its bad, you'll fall and hurt yourself (get electrocuted). (unfortunately this only works for negative voltages, as gravity only acts negatively, unlike electricstatics, which can be both positive and negative)

in short, voltage only has meaning as the difference in potential between 2 things, if there is no difference between them, there can be no voltage/current.

its probably important to add that a.c only oscillates around zero because at some point in the circuit you've introduced ground as a reference. It is possible, and often done in gain pedals, to introduce the reference as some d.c voltage, then the a.c signal will oscilate around that voltage instead. (if the ground of your guitar was 5V then you're a.c signal would oscilate around 5V, although you'd get shocked everytime you touched the jack)

[Hiwatt25]

I think I'm following you.  So in the schematic for The begginer project, many of the resistors are connected to ground.  Is that that so current WON'T move past that given resistor? 


If one side of the resistor is like 5 volts and ground is at zero won't the current flow to ground cause of the potential difference?  Sorry to be such a pest but I think I'm getting really close to a major fundamental breakthrough.   

Thanks again.

[Paul Marossy]

in the schematic for The begginer project, many of the resistors are connected to ground.  Is that that so current WON'T move past that given resistor?

Not exactly. Current will still go to ground, but it will be less than without the resistor. I kind of think of ground like a hot water return pipe to a water heater. It returns the tepid water back to the water heater tank after it's been heated to start the cycle over again. In an electronic system, ground is the reference for all voltages in the system. And it is the common return path for electrons to the power source, or at least that is how I see it.

This page from GEO is a good one to read: http://www.geofex.com/Article_Folders/How_It_Works/hiw.htm

[scaesic]

I think I'm following you.  So in the schematic for The begginer project, many of the resistors are connected to ground.  Is that that so current WON'T move past that given resistor? 


If one side of the resistor is like 5 volts and ground is at zero won't the current flow to ground cause of the potential difference?  Sorry to be such a pest but I think I'm getting really close to a major fundamental breakthrough.   

Thanks again.

yes, if you touch the side thats at 5v, there will be a potential difference between you and the circuit, but if you touch the side thats connected to ground there won't be a potential difference, the 5V's is 'dropped' across the resistor.

[DiyFreaque]

yes, if you touch the side thats at 5v, there will be a potential difference between you and the circuit, but if you touch the side thats connected to ground there won't be a potential difference, the 5V's is 'dropped' across the resistor.

And thus you have Ohms Law:  Voltage = Inductance X Resistance, or

V=I*R

Understanding this basic law will help you immensely.

Voltage is measured in Volts.
Resistance is measured in Ohms
Inductance (Current) is measured in Amps.

If you know two of the above properties, you can always calculate the third.  For example, if the 5V mentioned in the previous post is on one end of a 10K resistor, and 0V (ground) is on the other end of the 10K resistor, you can calculate how much current is passing through that resistor to ground.

You manipulate Ohms law to find the unknown - it's the same formula, just reorganized:

I = V/R

5V is being dropped across the resistor, so.....

5 volts divided by 10000 Ohms (which is what 10K is).

5V/10000 Ohms = .0005 or 0.5 mA

If you had an unmarked resistor, no ohmmeter, but were able to measure current and voltage, you would find that your resistance was 10K by rearranging the formula again:

R = V/I

5V/.0005A = 10000 Ohms

If your DMM's Ammeter and Ohmmeter worked, but for some odd reason the voltmeter did not work, then you could calculate the voltage:

V=I*R

.0005A X 10000 Ohms = 5V

Of course, Ohms law is much more useful than for just making up for a bad DMM =0)

For example, a voltage divider follows the same law.  You'll run into this a lot with stompboxes.  Notice that single supply op amps usually need a reference that is halfway between 9V and ground (IE, 4.5V).  This voltage is usually derived by putting two equal value resistors in series, and taking the voltage from the junction between the two resistors.

Say you have 2 10K resistors in series.  The total resistance is 20K (10K + 10K).  9 Volts is at the top of the two resistors and 0V (ground) is at the bottom.  The two resistors together are dropping a total of 9V. 

Remember the total resistance is 20K.  So we find out how much current is passing through the two resistors to ground:

9V/20000 Ohms = .00045 or 0.45 mA.

Now we know how much current is passing through both resistors, how much voltage is each resistor dropping? 

10000 * .00045 = 4.5V.  Each resistor is dropping 4.5 V.  So, if the first resistor is dropping 4.5V, then 9-4.5 = 4.5.  That means the voltage at the junction between the resistors is 4.5V, and there you go. 

The amount of resistance used by each resistor, if they are equal, can be anything - the voltage will always be half the supply voltage between them.  The resistance value used for both resistors determines how much current is flowing through the resistors.

If the resistance values are different, than the voltage will be less or more than half the supply voltage.  You This is how you derive different voltages in a circuit.  You calculate the voltages in the same manner.  This is Ohms Law.  It's the first step anyone takes in understanding electronics.

Cheers,
Scott

[Hiwatt25]

Thanks all for the input, one thing I'm still confused about though.  Does current flow from the positive to negative terminals on the battery or vice versa.  In the diagrams given in th Geofex article the arrows seem to indicate current flows from + to -.  I always thought it was the other way round.

[R.G.]

Does current flow from the positive to negative terminals on the battery or vice versa.  In the diagrams given in th Geofex article the arrows seem to indicate current flows from + to -.  I always thought it was the other way round.

Thank you, Benjamin Franklin.
Back when people were just discovering electricity, it was known that there were two kinds, positive and negative. But there was no way to call which was which. Benjamin Franklin's electrical experiments led him to decide that charges were carried by particles that had a positive charge, so he named the polarities and we've used that nomenclature ever since.

He was wrong. It's not carried by particles with a positive (as we'd say today) charge. It's carried by electrons with a negative charge. So to be perfectly accurate, DC electrical current is carried by electrons, and they flow from the more negative terminal to the more positive terminal. More on AC currents in a moment.

So the electrons move from negative to positive. It turns out that the electron tube is the only simple place where we can tell the difference. All the equations, all the devices, everything except vacuum tubes work exactly the same if we consider that electricity is carried by tiny positive particles. So the world decided back when they found out about electrons, they decided "Too bad. We already understand it the wrong way, we'll just keep it that way. It all works out the same." And that's what they did. It's like the QWERTY keyboard - it was too entrenched to change.

It mattered a lot in days when the only electronics were vacuum tubes. It matters much less today. The convention is that current flows from positive to negative. You kind of have to have both things in your head at the same time, like a little mental sticky note that says "oh, yeah, if electron ballistics are involved, you have to worry about electron flow..."

On AC currents: In AC current flows, it matters even less, because AC only flows back and forth. The electrons don't run from one place to another, they run back and forth. It's the electrical field that moves down the wires, not the electrons.

On a side note, I'm guessing that you may have had military electronics training. The military was big on electron flow teaching at one time, may still be. It was the source of endless confusion for the guys from the military that were attending EE classes with me.

[Paul Marossy]

Thanks all for the input, one thing I'm still confused about though.  Does current flow from the positive to negative terminals on the battery or vice versa.  In the diagrams given in th Geofex article the arrows seem to indicate current flows from + to -.  I always thought it was the other way round.

I've always thought of it the same way as that GEO diagram indicates. But this page is stating the opposite: http://www.phy6.org/Education/woppos.html

[scaesic]

yes, if you touch the side thats at 5v, there will be a potential difference between you and the circuit, but if you touch the side thats connected to ground there won't be a potential difference, the 5V's is 'dropped' across the resistor.

And thus you have Ohms Law:  Voltage = Inductance X Resistance, or

that's current, not inductance.

Inductance (Current) is measured in Amps

oh dear... this is pretty tragic...

current is measured in amperes (A), inductance is measured in henry's (H)
although inductance is related to current, it is defenitely not the same thing.

inductance is a physical property of an inductor, or any coil of wire.

current is a calculated property of a closed circuit.

[scaesic]

He was wrong. It's not carried by particles with a positive (as we'd say today) charge. It's carried by electrons with a negative charge. So to be perfectly accurate, DC electrical current is carried by electrons, and they flow from the more negative terminal to the more positive terminal. More on AC currents in a moment.

although in a material full of electrons, there may be one "shell" without an electron, in this scenario electrons jump into the vacant shell, and cause a moving vacant shell.

these are known as electron holes, and are conventionally reffered to as positive, compared to a moving electron, which is negative.