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Hi,
I am still not sure whether I understand the principles of a guitar's wiring. I understand the electro-magnetic energy conversion which produces the sound on the amp - all that is clear. But I get lost understanding the basic concept of current passing thru the guitar. I still seem to be stuck with the notion that current moves around one circuit like water passing down a pipe.
We have a current which (for the lack of a better explanation) is pushing the electrons to the pickup - so
1) how does the converted signal return back to the amp - does it return thru the shielding or is the result oblivious to my concept of water in pipes?
2) If I am connected to the shielding (which I am thru the convential string earth) then don't I add to the resistance to the signal and impair the returning signal?
hmmm ... let's start over .. and dont get too attached the the water in the pipe thing ...
it starts here: the magnet and coil of wire of the pickup produce an electromagnetic field (i.e. a static voltage potential that is not moving yet) .. it is just sitting there, minding its own business ... no signal is present, no current is flowing (well, none worth mentioning for simple purposes) ... then someone plucks a string - a metal string, at that - and we're off to the races .... the vibrating string induces an AC current in the pickup's wire coil because it disturbs/excites the E-M field (i.e. the voltage starts changing) ... so now current starts to flow and it is a broadband AC signal ... which means that there are different frequency components all flowing together at the same time ... on the 'positive' part of the AC signal, the signal flows 'out' of the guitar on the 'hot' side of the wiring to the amp where it gets converted into sound, then on the 'negative' side of the AC signal, the signal flows back into the guitar on the 'ground' side of the circuit .... then it reverses and starts again ... anyhere from a hundred or so times per second to some thousands of times per second ... like i said, dont hokld onto the water pipe idea too firmly, it really doesnt work like that ...
as for adding resistance, i doubt it is meaningful because you are in parallel with the 'better conducting' part of the circuit
hope this helps
t4d
Gotcha!
I believe I am mixing up the pickup signal with the current. I think I have nailed this down now to my simple terms:-
1) the electrical current is AC and the hot cable is the highway for the signal coming from the pickup to travel to the amp (just it goes back and forth - 50 times a second - herz).
2) the shielding does not carry the pickup signal, it just closes the circuit necessary for the current.
Wow! I think I have that now - let me know if this is wrong - thanks
This may make things worse or better depending upon how much of this is making sense, but to me the difference in AC and DC is key-
In DC, the electrons are actually flowing much like your water pipe analogy and it's a key concept because so much of electronics is based on DC-
But as stated above, pups create an AC current, and in AC the electrons dont really go anywhere- First they 'bump' the upstream electron which causes the next electron to bump the next, and then it swithes direction and bumps the downstream electron and you get the whole domino effect in that direction as well-
Because the electrons dont have to waste energy on a long trip (the energy is simply bumped from one electron to the next) AC is far more effecient than DC and the reason that we use it to run electricity accross long distances.
Zion makes a good point, The plumbing/water thing should limited to DC theory for the most part. AC is much more complex.
The shielding is just as important as the center conductor, whether it be the cord between the guitar and amp, or a quot;vintagequot; pickup with only two conductors (the center wire, and the outer shielding). You can ditch the outer shielding on modern pickups with four wires (and a seperate shield wrap)...but this is not practical in the real world due to noise problems.
As far as Question #1, you need not worry about the what the amp is doing, as the amp has quot;high input impedencequot;, which preserves the pickup's AC signal strength. I suppose you could say: the dog (the pickup) wags the tail (the amp). Since the amp only sucks a tiny fraction of the pickup's generated output...the rest is travels back to through the shielding on 1/2 of a theoretical string-swing.
The reason why AC is necessary for long distance electrical travel is that DC is limited to simple build up of resistances, and is only good for a few hundred miles at the most in Edison's day. James Westinghouse realized that AC power could be sent through step-up transformers/step down transformers over far greater lengths.
DC and tranformers do not mix out of practicality! With DC, as soon as the input voltage stops rising in the primary winding...the electrical field induced in the secondary windings collapses, and pushes wave of current through the rest of the circuit ONE TIME ONLY! As long as that same DC input is there, or declines to zero...there will be no more current in the transformer's secondary output. AC and transformers are a practical match, as the varying voltage inputs are needed to get past the quot;single pulsequot; limitations of DC in tranformers.
pickup current is DC...
Your pick up is the water tower. Your amp supplys the superior ground and pushes the output current through AC/DC and then AC/DC again.
Shielding blocks or reduces the signal radiated outside of the positive wires that carry the signal from getting back to the pickups.
All Electricity produces electromagnetic radiation, which comes through a pickup as static or noise. The best hot setup is like a water pipe pumping fresh water through an ocean of salt water, but without sheilding, the hotter the signal, the more noise that is possible.
Originally Posted by MesaDCLPpickup current is DC...
Pickup current is definitely AC. Also, I believe you are all making this more difficult by refering to quot;currentquot;. A guitar pickup is an AC voltage generator. Current, is a by-product of that voltage. Voltage is the amount of quot;pressurequot;. Current is the number of electrons pushed by that pressure.
A guitar, by itself, has no ground. The amp does, by virtue of its power plug. When we plug a guitar into an amp, we quot;referencequot; one side of its circuit, and thus, call it ground. The voltage actually, flops back and forth at the frequency of the string vibration. One side of the pup pushes, one side pulls. Since the input of an amp works on voltage, the current is not really an issue. (The input impedance of most amps is around 1 meg.)
This is really a deep subject, but basically, think in terms of voltage, rather than current.
Artie
Originally Posted by man_in_a_blackboxHi,
We have a current which (for the lack of a better explanation) is pushing the electrons to the pickup - so
1) how does the converted signal return back to the amp - does it return thru the shielding or is the result oblivious to my concept of water in pipes?
I find it easier to think of it the reverse way. The ground acts as vacuum that sucks the electrons to the amp from the positive lead. That is, it is the ground that is responsible for moving the electrons through the circuit.
I don't know if that is technically correct, but I like my analogy.
I am still not sure whether I understand the principles of a guitar's wiring. I understand the electro-magnetic energy conversion which produces the sound on the amp - all that is clear. But I get lost understanding the basic concept of current passing thru the guitar. I still seem to be stuck with the notion that current moves around one circuit like water passing down a pipe.
We have a current which (for the lack of a better explanation) is pushing the electrons to the pickup - so
1) how does the converted signal return back to the amp - does it return thru the shielding or is the result oblivious to my concept of water in pipes?
2) If I am connected to the shielding (which I am thru the convential string earth) then don't I add to the resistance to the signal and impair the returning signal?
hmmm ... let's start over .. and dont get too attached the the water in the pipe thing ...
it starts here: the magnet and coil of wire of the pickup produce an electromagnetic field (i.e. a static voltage potential that is not moving yet) .. it is just sitting there, minding its own business ... no signal is present, no current is flowing (well, none worth mentioning for simple purposes) ... then someone plucks a string - a metal string, at that - and we're off to the races .... the vibrating string induces an AC current in the pickup's wire coil because it disturbs/excites the E-M field (i.e. the voltage starts changing) ... so now current starts to flow and it is a broadband AC signal ... which means that there are different frequency components all flowing together at the same time ... on the 'positive' part of the AC signal, the signal flows 'out' of the guitar on the 'hot' side of the wiring to the amp where it gets converted into sound, then on the 'negative' side of the AC signal, the signal flows back into the guitar on the 'ground' side of the circuit .... then it reverses and starts again ... anyhere from a hundred or so times per second to some thousands of times per second ... like i said, dont hokld onto the water pipe idea too firmly, it really doesnt work like that ...
as for adding resistance, i doubt it is meaningful because you are in parallel with the 'better conducting' part of the circuit
hope this helps
t4d
Gotcha!
I believe I am mixing up the pickup signal with the current. I think I have nailed this down now to my simple terms:-
1) the electrical current is AC and the hot cable is the highway for the signal coming from the pickup to travel to the amp (just it goes back and forth - 50 times a second - herz).
2) the shielding does not carry the pickup signal, it just closes the circuit necessary for the current.
Wow! I think I have that now - let me know if this is wrong - thanks
This may make things worse or better depending upon how much of this is making sense, but to me the difference in AC and DC is key-
In DC, the electrons are actually flowing much like your water pipe analogy and it's a key concept because so much of electronics is based on DC-
But as stated above, pups create an AC current, and in AC the electrons dont really go anywhere- First they 'bump' the upstream electron which causes the next electron to bump the next, and then it swithes direction and bumps the downstream electron and you get the whole domino effect in that direction as well-
Because the electrons dont have to waste energy on a long trip (the energy is simply bumped from one electron to the next) AC is far more effecient than DC and the reason that we use it to run electricity accross long distances.
Zion makes a good point, The plumbing/water thing should limited to DC theory for the most part. AC is much more complex.
The shielding is just as important as the center conductor, whether it be the cord between the guitar and amp, or a quot;vintagequot; pickup with only two conductors (the center wire, and the outer shielding). You can ditch the outer shielding on modern pickups with four wires (and a seperate shield wrap)...but this is not practical in the real world due to noise problems.
As far as Question #1, you need not worry about the what the amp is doing, as the amp has quot;high input impedencequot;, which preserves the pickup's AC signal strength. I suppose you could say: the dog (the pickup) wags the tail (the amp). Since the amp only sucks a tiny fraction of the pickup's generated output...the rest is travels back to through the shielding on 1/2 of a theoretical string-swing.
The reason why AC is necessary for long distance electrical travel is that DC is limited to simple build up of resistances, and is only good for a few hundred miles at the most in Edison's day. James Westinghouse realized that AC power could be sent through step-up transformers/step down transformers over far greater lengths.
DC and tranformers do not mix out of practicality! With DC, as soon as the input voltage stops rising in the primary winding...the electrical field induced in the secondary windings collapses, and pushes wave of current through the rest of the circuit ONE TIME ONLY! As long as that same DC input is there, or declines to zero...there will be no more current in the transformer's secondary output. AC and transformers are a practical match, as the varying voltage inputs are needed to get past the quot;single pulsequot; limitations of DC in tranformers.
pickup current is DC...
Your pick up is the water tower. Your amp supplys the superior ground and pushes the output current through AC/DC and then AC/DC again.
Shielding blocks or reduces the signal radiated outside of the positive wires that carry the signal from getting back to the pickups.
All Electricity produces electromagnetic radiation, which comes through a pickup as static or noise. The best hot setup is like a water pipe pumping fresh water through an ocean of salt water, but without sheilding, the hotter the signal, the more noise that is possible.
Originally Posted by MesaDCLPpickup current is DC...
Pickup current is definitely AC. Also, I believe you are all making this more difficult by refering to quot;currentquot;. A guitar pickup is an AC voltage generator. Current, is a by-product of that voltage. Voltage is the amount of quot;pressurequot;. Current is the number of electrons pushed by that pressure.
A guitar, by itself, has no ground. The amp does, by virtue of its power plug. When we plug a guitar into an amp, we quot;referencequot; one side of its circuit, and thus, call it ground. The voltage actually, flops back and forth at the frequency of the string vibration. One side of the pup pushes, one side pulls. Since the input of an amp works on voltage, the current is not really an issue. (The input impedance of most amps is around 1 meg.)
This is really a deep subject, but basically, think in terms of voltage, rather than current.
Artie
Originally Posted by man_in_a_blackboxHi,
We have a current which (for the lack of a better explanation) is pushing the electrons to the pickup - so
1) how does the converted signal return back to the amp - does it return thru the shielding or is the result oblivious to my concept of water in pipes?
I find it easier to think of it the reverse way. The ground acts as vacuum that sucks the electrons to the amp from the positive lead. That is, it is the ground that is responsible for moving the electrons through the circuit.
I don't know if that is technically correct, but I like my analogy.
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