turbo411 Posted September 24, 2006 Report Posted September 24, 2006 I'm trying to understand how the basic circuit works for guitar wiring and would appreciate if someone can fill in the blanks on my narrative This is from the stewmac site A pot is essentially a variable resistor. In that sense there are three poles. The middle pole is the sweep and dictates the level of resistence. In this picture the input comes in at zero resistence (max volume) the third pole is connected to the ground so would be zero volume. The output is sweeper and moving the knob will move the sweeper between full volume and zero volume. The current also is sent to the tone pot on the first pole. What I dont understand is how this current gets back to the volume and ultimately the output jack?? Also what is the explanation of the sweeper being connected to a capacitor (which I understand to affect different frequencies) if the capacitor seemingly goes to the ground. I'm assuming there is another wire that Im missing. Please walk me through in a narrative fashion. Thanks! Quote
mammoth guitars Posted September 24, 2006 Report Posted September 24, 2006 The jack also has a ground connection and that is how the current path completes. The capacitor used on the volume across the middle terminal and the pickup or input connection (not to ground) is a high frequency bleed cap. When the volume is turned down there can be some high frequency loss and the bleed cap allows some of this high frequency to come through to the output jack via the middle terminal. The tone control uses a cap to ground to dump high frequencies to ground. Quote
JoeAArthur Posted September 25, 2006 Report Posted September 25, 2006 The current also is sent to the tone pot on the first pole. What I dont understand is how this current gets back to the volume and ultimately the output jack?? Also what is the explanation of the sweeper being connected to a capacitor (which I understand to affect different frequencies) if the capacitor seemingly goes to the ground. I'm assuming there is another wire that Im missing. Please walk me through in a narrative fashion. Thanks! No missing wire. The tone control network, consisting of the tone pot (variable resistor) and capacitor wired in SERIES, is connected across the volume control in PARALLEL - there is no need for any current to get "back" to the volume. The same current is split - some flows through the volume control to the center lug or sweaper and out the output jack... and the other flows through the tone network. The tone network removes the high frequencies at a rate determined by the value of the tone control capacitor and the resistance of the tone control pot - they are bypassed to ground AROUND the volume control. Quote
turbo411 Posted September 25, 2006 Author Report Posted September 25, 2006 (edited) So lets assume a current has frequencies 1,2,3,4 from the pickup some of the current goes through the volume pot and the resistence can "shrink" the 1,2,3,4 affecting the volume additionally, some current goes through the tone pot and the resistence and capacitor could hypothetically remove frequency 3&4. so now we have some level of current with frequencies 1,2,3,4 that go out to the jack via the output connecntion and we have current with frequencies 1,2 (3&4 were removed by tone pot) that goes where? If it is wired in parallel to the volume, it doesnt go back through the volume as you said. so how does this all flow together? The current modified for the tone has to go somewhere for it to have an audio impact... Im sorry but Im still having conceptual difficulty putting it together. Edited September 25, 2006 by turbo411 Quote
fryovanni Posted September 25, 2006 Report Posted September 25, 2006 (edited) Think of it more like opening flood gates or valves (using water flow in place of current). You have a water pump (pickup) and this is a closed loop of water pipe. You can divert some of that water back to the pumps input (bypassing the pipe that soaks your amp ). The key is the water flows it does not shrink or go away, voltage or pressure can be reduced or dropped as current tries to pass through the circuit. If you have a circuit with two or more paths (parallel) it will flow proportionately based on the impedence (resistance+reactance+capacitance) of each path. In the electrical field I see many people that get confused when they think of a grounds and neutrals and negatives. I think batterys make more sense to most people. The negative in a car is connected to the frame effectively making your cars body the path back to one side of the source(the battery negative). The possitive remains insulated from the body and is forced through various loads to complete the circuit and get to the frame (negative side of the source). We make things confusing when we start bonding (or "grounding" things for safety and to cut down on introduced power introduced from outside sources). I hope that makes some sense(I know that is not a good complete explanation). Maybe the circuits would make more sense if you draw them up in ladder diagram form. Peace,Rich Edited September 25, 2006 by fryovanni Quote
turbo411 Posted September 25, 2006 Author Report Posted September 25, 2006 (edited) Think of it more like opening flood gates or valves (using water flow in place of current). You have a water pump (pickup) and this is a closed loop of water pipe. You can divert some of that water back to the pumps input (bypassing the pipe that soaks your amp ). The key is the water flows it does not shrink or go away, voltage or pressure can be reduced or dropped as current tries to pass through the circuit. If you have a circuit with two or more paths (parallel) it will flow proportionately based on the impedence (resistance+reactance+capacitance) of each path. In the electrical field I see many people that get confused when they think of a grounds and neutrals and negatives. I think batterys make more sense to most people. The negative in a car is connected to the frame effectively making your cars body the path back to one side of the source(the battery negative). The possitive remains insulated from the body and is forced through various loads to complete the circuit and get to the frame (negative side of the source). We make things confusing when we start bonding (or "grounding" things for safety and to cut down on introduced power introduced from outside sources). I hope that makes some sense(I know that is not a good complete explanation). Maybe the circuits would make more sense if you draw them up in ladder diagram form. Peace,Rich So we have some pipes. The "water" is coming from the pickup and a portion is going to the volume pipe. The volume pipe then diverts some to the amp and some to the ground The other portion of the water goes through the tone pipe where the cap removes certain frequencies. Does this then travel back up through the same tone pipe up to the volume pot? Fundamentally, this altered current has to go up the same pipe that initially delivered it and through the volume because if I turn my volume all the way down I don't hear anything Thanks to all who have taken the time to help me Edited September 25, 2006 by turbo411 Quote
JoeAArthur Posted September 25, 2006 Report Posted September 25, 2006 So we have some pipes. Does this then travel back up through the same tone pipe up to the volume pot? Fundamentally, this altered current has to go up the same pipe that initially delivered it and through the volume because if I turn my volume all the way down I don't hear anything Water analogies do have their problems. They don't handle AC too well for one thing. There is only so much current coming from the pickup and this current contains many different frequencies. Regardless of frequency, the total current produced by the pickup from one wire is trying to find the easiest way back to the pickup - to enter in the other wire. Remember that current flow will only occur when there is a complete path. When it hits that split between volume and tone control, some of the current will go one way and some of it goes another. The tone control capacitor helps decide what goes where because it is frequency selective and presents an easier path for the higher frequency current and a more difficult path for the lower frequency current. The tone control resistor acts with the capacitor to determine how easy or difficult this path actually is... In general, some of the higher frequencies will find an easier path by going through the tone network and once through that, they can zip right back to the pickup. They have been "short circuited". As the frequencies get lower, the tone network path becomes more difficult. These go through the volume control, make it to gorung and then zip right back to the pickup. The volume control provides another choice of paths - the wiper that moves and determines what portion of resistance exists between both sides of the volume control. Some of the current takes the path through the volume control and back to the pickup and some of the current takes the path through the volume control wiper, down the center conductor of the guitar cable and into the amp. Once inside the amp, depending upon the amp, there may be other splits of current - but in the end, all of the current that traveled to the amp via the cable eventually meet up with one another and travel back through the cable... and eventually back to the pickup. Since this is AC, reverse the flow and the exact same thing happens, just in reverse. Quote
turbo411 Posted September 25, 2006 Author Report Posted September 25, 2006 Ha. thats funny how some things can be pulled out of a paragraph and seem so odd. Thank you very much for the explanation, that helped a lot! One more question to wrap up the basics, why are there two grounds? One ground that goes to the claw by the springs and one that is soldered to one prong on the amp output jack. Taking this knowledge one step futher In the picture below we have two tone knobs, one for neck , one for middle. Why are the wiring schemes different for each tone knob. One hs the input on an outside prong and the cap on the sweeper, the other has the input on the sweeper. Why the inconsistency? Thank you very much Quote
fryovanni Posted September 25, 2006 Report Posted September 25, 2006 (edited) Think of it more like opening flood gates or valves (using water flow in place of current). You have a water pump (pickup) and this is a closed loop of water pipe. You can divert some of that water back to the pumps input (bypassing the pipe that soaks your amp ). The key is the water flows it does not shrink or go away, voltage or pressure can be reduced or dropped as current tries to pass through the circuit. If you have a circuit with two or more paths (parallel) it will flow proportionately based on the impedence (resistance+reactance+capacitance) of each path. In the electrical field I see many people that get confused when they think of a grounds and neutrals and negatives. I think batterys make more sense to most people. The negative in a car is connected to the frame effectively making your cars body the path back to one side of the source(the battery negative). The possitive remains insulated from the body and is forced through various loads to complete the circuit and get to the frame (negative side of the source). We make things confusing when we start bonding (or "grounding" things for safety and to cut down on introduced power introduced from outside sources). I hope that makes some sense(I know that is not a good complete explanation). Maybe the circuits would make more sense if you draw them up in ladder diagram form. Peace,Rich So we have some pipes. The "water" is coming from the pickup and a portion is going to the volume pipe. The volume pipe then diverts some to the amp and some to the ground The other portion of the water goes through the tone pipe where the cap removes certain frequencies. Does this then travel back up through the same tone pipe up to the volume pot? Fundamentally, this altered current has to go up the same pipe that initially delivered it and through the volume because if I turn my volume all the way down I don't hear anything Thanks to all who have taken the time to help me Joe- You have to walk before you can run. DC is a more simplified way to look at current flow. When you try to through inductive reactance, capacitive reactance, and resistance as well as understanding the current is flowing back and forth with each cycle. It makes it harder to get a grasp on the basic concept. Especially when the concept of series and parallel circuits is clear yet. Water is not a perfect way to describe AC current flow, but it is something most people can grasp easyier. So we have a water pump (pickup). It draws water in one side and pushes it out the other. All the water that gets pushed out returns to the inlet on the pump. We are pushing water (this pressure or force would be voltage-difference of potential). The pump only sees all the stuff between its inlet and outlet as something restricing the flow (this is resistance or in the case of an AC circuit Impedance-which combines Inductive/Capacitive reactance and resistance). When the pump pushes water through the pipe it can only push as much water as the pressure will allow (this is work done or wattage- wattage equals voltage(pressure) times volume of water(current or amps). Again the pump and pipes are a closed loop all the water(current) flows through the loop but never leaves. Holding that thought. We can have a single pipe (one path for water to flow=Series) or we can T off and have multiple pipes (more than one path=Parallel). Now if we look at One pipe(series) and we know all the water that flows in a pipe never leaves. Then we know the same volume will pass any point in the pipe even where it is restricted (this tells us current or amps is constant in a series circuit). We know though that pressure is being reduced as we push water passed the restrictions in the pipe- eventually we have used all the pressure to get back to the pump. This tells us that pressure(voltage) is dropped across each restriction in the pipe (thus in a series circuit voltage dropped is addative v1+v2+v3=VTotal). Now if we place a T in the pipe and have two pipes for the water to flow things are a little different. The amount of water that will flow through a pipe that is small and restrictive will be less than the amount that will flow through a larger pipe (thus current flow in a parallel circuit is addative we may have 4units in 1 one goes through the small pipe and 3 go through the larger but the total is still equal to how much went in). Now no matter how many pipes we push water through the pressure on each pipe is the same (thus pressure(voltage) in a parallel circuit is the constant same for each pipe). We can have all sorts of combinations of series and parallel depending on where we place restrictions in the system or create new paths. The rules of flow do not change though. So now you look at your circuit. You have given the current three directions it can flow when it hits the volume pot. One it can go through the tone pot that has a variable resitor in series with a capacitor then to the "inlet" on your pickup (completing the circuit). Second it can go through a portion of the variable resistance of the volume pot then back to the "inlet" of your pickup (completing the circuit). Third though a portion of the variable resistance of the volume pot, then the jack, through the cable, through the amp, then back through the cable and jack, finally to the "inlet" on your pickup(completing the circuit. When the current hits the parallel branches of the circuit the current will split three ways. How much goes to each depends on how much resistance that branch has. I hope that kinda makes sense, or at least gives you a general picture of how current is moving. If that makes sense we could add more information to the picture. This is a very basic DC way of picturing flow or current paths (circuits). Peace,Rich P.S. A book on the basics of electricity may be a better way to understand this stuff. It could provide pictures and diagrams as well as a lot of fundemental concepts I am skipping. Edit: Tie those grounds together. They are showing you they are the same point effectively by showing the ground symbol. This link will give you info on the potentiometers and what is happening inside as it relates to the terminals-click Did a quick search for tutorials and info that may be helpful. Info Another page Edited September 25, 2006 by fryovanni Quote
Sambo Posted September 25, 2006 Report Posted September 25, 2006 If you get Koch's book on making electric guitars, it has a very good section on explaining all this kinda stuff. goes into alot of detail about all sorts. Take a look. S Quote
turbo411 Posted September 25, 2006 Author Report Posted September 25, 2006 Thank you for the fundamentals explanation, its becoming so much more clear! So now here are the questions I have left 1. If the two end options are essentially the ground or the amp, how does the current get back to the pickup, especially in the case of the ground? 2. Back to my post above (2nd picture) as to why the sweepers on the tone pot are wired differently. Thanks! Quote
JoeAArthur Posted September 25, 2006 Report Posted September 25, 2006 Thank you for the fundamentals explanation, its becoming so much more clear! So now here are the questions I have left 1. If the two end options are essentially the ground or the amp, how does the current get back to the pickup, especially in the case of the ground? 2. Back to my post above (2nd picture) as to why the sweepers on the tone pot are wired differently. Thanks! #1 - Ground is not really a "destination". In schematics, the ground symbol is really a short-cut of showing a common connection without having to draw all of the physical connections (e.g. separate wires or chassis). So considering ground as a common connection - that's how the current gets back to the pickup. The braid or shield in your instrument cable to the amp is part of the ground connection. #2 - Whenever two components are connected in series the order of connection doesn't affect the performance of the circuit. In the specific case of the pots connections for the two tone controls, the current doesn't know the difference between the wiper and the outside lug. You'll note that the same outside lug is used in both cases - how you turn the pot to either increase or decrease the resistance is the same in both cases... right? Quote
fryovanni Posted September 25, 2006 Report Posted September 25, 2006 (edited) Thank you for the fundamentals explanation, its becoming so much more clear! So now here are the questions I have left 1. If the two end options are essentially the ground or the amp, how does the current get back to the pickup, especially in the case of the ground? 2. Back to my post above (2nd picture) as to why the sweepers on the tone pot are wired differently. Thanks! 1. "Third though a portion of the variable resistance of the volume pot, then the jack, through the cable, through the amp, then back through the cable and jack, finally to the "inlet" on your pickup(completing the circuit." The only option is to complete the circuit. Connect the ground or "inlet" (the other side of your pickup coil which is your source). The amp is just another load in the ciruit. The current may flow through it but only if it is a complete circuit. 2. The middle terminal is the wiper. As long as you are connected to an outside pole and the wiper you can vary the resistance between those two poles. Take a second look at this linkclick Peace, Rich Edit: Maybe another way of thinking about the pickup and circuit.... You have a rope that you wrap around a pulley(the load-- amp, and all components in your circuit) you hold one side of the rope in each hand. You pull on it with your right (hot) hand and the pulley turns. Then you pull on it with your left(ground) hands and it turns the other way. You are the pickup or source. This is basically what is happening in an AC circuit. You pull with one hand it turns until you can't pull anymore (your arm is only so long) then you pull with the other hand and it turns again in the other direction. If you let go or cut the rope (opening the circuit) the pulley stops turning because you can't pull anymore. Hope that is not too goofy sounding. I have gone from pipes to rope . Edited September 25, 2006 by fryovanni Quote
turbo411 Posted September 26, 2006 Author Report Posted September 26, 2006 (edited) Thank you very much to everyone for all the help. I now understand the fundamentals well enough to continue researching and learn about more complex wirings. Its easy to follow a diagram and solder the pieces but it was important for me to understand the whys. P.s. The rope example is a real winner! Edited September 26, 2006 by turbo411 Quote
wolf Posted September 27, 2006 Report Posted September 27, 2006 turbo411 2. Back to my post above (2nd picture) as to why the sweepers on the tone pot are wired differently. fryovanni's explanation is quite good but I'm wondering if another reason for that wiring is saving money. That diagram illustrates the way a Fender Stratocaster is wired. Just guessing but from what I have heard about Leo Fender, he liked to find ways to cut costs. (I do not know whether this was due to his being cheap OR so that he could sell his guitars as cheaply as possible. - Maybe a little of both huh? ) So, by wiring a guitar in that way, you save the cost of a capacitor and it is 2 less solder connections that have to be made. It's also reducing the chances of something going wrong with the finished product because of 1 less part and 2 less solder joints. Oh, and that diagram is from my website. wolf Quote
fryovanni Posted September 27, 2006 Report Posted September 27, 2006 Wolf, I didn't really think about the circuit design(although your thinking cost and simplicity makes sense). I was just trying to get the general concept of current flow across. I am not as much of an electronics guy(even though I do electrical work for a living). I mainly build acoustics to stay away from electricity . Peace,Rich Quote
Primal Posted September 27, 2006 Report Posted September 27, 2006 Does anyone else find these hybrid schematic/component layout diagrams confusing to read? :-P Quote
fryovanni Posted September 27, 2006 Report Posted September 27, 2006 Does anyone else find these hybrid schematic/component layout diagrams confusing to read? :-P I guess that depends on how you look at it. It is good in an age of disposable components. Where the repair person or installer does not (or couldn't ) fully grasp what is happening(I am speaking more to circuit boards and what have ya- not so much components). Kinda like less info is more (just give them what they need to hook it up). However if you are designing/troubleshooting/ or want to maodify it is a painful way of looking at the circuit (kinda half blind). Primal- You must be one of those people that doesn't like connect the dots and I suspect like to color outside the lines . Peace,Rich Quote
Primal Posted September 27, 2006 Report Posted September 27, 2006 Primal- You must be one of those people that doesn't like connect the dots and I suspect like to color outside the lines . Peace,Rich Oh, I definately like connecting the dots, however, when it comes to coloring outside the lines... well, lets just call that artistic expression. What I meant, though, is that IMO its easy to "trace" the signal going through a circuit when you are looking at a schematic rather than an assembly diagram. Quote
JoeAArthur Posted September 27, 2006 Report Posted September 27, 2006 Primal- You must be one of those people that doesn't like connect the dots and I suspect like to color outside the lines . Peace,Rich Oh, I definately like connecting the dots, however, when it comes to coloring outside the lines... well, lets just call that artistic expression. What I meant, though, is that IMO its easy to "trace" the signal going through a circuit when you are looking at a schematic rather than an assembly diagram. +1 And once you understand the schematic, you can wire it up regardless of component choices. Pictorials on the other hand require that you have the same component and easily grasping things like what is happening with the switching doesn't come as easy as with a schematic. The exact component thing is essential. As an example... I think it is stewmac - they have a 3 position telecaster style switch that has the same 4 lugs per side, but the common lug for the switch poles is completely opposite of what most pictorials assume. Some blend pots have the pot sections opposite. So you might wind up turning down the wrong pickup when you really wanted it to turn down the other based on the rotational direction. And I believe it is Ibanez... has those all poles in one row like some of the import switches but the actual switching is really unique. Quote
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