Resistors and current question...

[PowerChordHack]

Wow that pic looks awful. That's a PUP coil and a resistor, by the way.

R in series versus R to ground:
Is the current reduced by the same amount in both examples?
Or more specifically, would the PUP volume be reduced by the same amount?

Say R=250K in both if it helps....

 

[walterw]

lose the lines on the right of each picture, they shouldn't be closed-off boxes.

the parallel resistor would also not be to ground in your example, that's not established; it would just go all the way to the bottom line, essentially connected to both ends of the coil.

i can't whip out the scientific answer here, but i know that they are different, in that they affect the sound in different ways at different frequencies.

thing is, they're essentially opposite in effect; a big parallel resistance has less attenuation, while a small series resistance has less attenuation.

so if they were both like 10Ω, the series one would sound the same while the parallel one would be pretty much shut off. if they were both 250kΩ, the parallel one would be at normal volume (that's what fenders live at) while the series one would be shut off.

 

[StratoCraig]

Is this really a simple electrical question, or are we really talking about an electric guitar? It makes a difference.

In simple electrical circuits (a guitar isn't as simple as it seems), series resistances just sum, while parallel resistances equal the reciprocal of the sum of reciprocals.

In an electric guitar, resistance is really just one term in the equation of an RLC circuit. It is simply not the case that the pickup generates a signal and then the pots "attenuate" it. Instead, increasing the resistive load causes more signal to be generated. This is why higher-valued pots cause a hotter signal. It's not a matter of a larger parallel resistance somehow attenuating less than a smaller parallel resistance.

 

[John Coloccia]

The simple answer is no, and you don't have to do any analysis to figure it out. Replace the resistors with a 0 ohm resistor (a wire). If I use my imagination and fill in where you're taking the output from, in case 1, you have full output from the pickup, and in case 2 you have 0 output from the pickup.

As a general rule, it's sometimes really helpful to probe the extremes as a sanity check.

 

[walterw]

It is simply not the case that the pickup generates a signal and then the pots "attenuate" it. Instead, increasing the [parallel] resistive load causes more signal to be generated.

yeah, this is a concept i came to understand here on TGP from the likes of jef bardsley and others; while it's a useful shorthand to think of these components as "attenuating" a signal, or "bleeding it to ground" in the case of tone caps, in reality the signal that's supposedly "removed" never forms in the first place. the whole circuit is what creates the signal, and said signal "includes" all the components in its formation.

 

[ngativ]

In an electric guitar, resistance is really just one term in the equation of an RLC circuit. It is simply not the case that the pickup generates a signal and then the pots "attenuate" it. Instead, increasing the resistive load causes more signal to be generated. This is why higher-valued pots cause a hotter signal. It's not a matter of a larger parallel resistance somehow attenuating less than a smaller parallel resistance.

Maybe is not simple as that. You can certainly use the potentiometer to reduce the overall potential the overall circuit , hence to attenuate the signal amplitude . But you can not use it to generate more output since the only signal generator is the coil itself . It does't matters if the pot is 300000M Ohms , you are not going to "generate" anymore signal . The electromotive force is inherent to the inductor, not the resistance .

 

[StratoCraig]

Maybe is not simple as that. You can certainly use the potentiometer to reduce the overall potential the overall circuit, hence to attenuate the signal amplitude. But you can not use it to generate more output since the only signal generator is the coil itself. It does't matters if the pot is 300000M Ohms, you are not going to "generate" anymore signal. The electromotive force is inherent to the inductor, not the resistance.

I didn't say it was inherent in the resistance. But the pickup only produces as much signal as the resistive load allows. This does not imply that the pickup can produce as much energy as a nuclear reactor if you just give it enough resistance. I was just explaining why more resistance produces a hotter signal.

As we all know already, though, turning the volume pot down does not just reduce the signal's amplitude. You lose highs, too. The resonant peak frequency changes and the amplitude of the peak diminishes. In fact, frequency response continually changes as you turn the volume pot, even if you don't have a tone pot or tone cap. The treble bleed cap is an attempt to compensate for this, but it can lead to the opposite problem, where the guitar sounds tinny when turned down. These are the kind of details that get lost when people try to oversimplify how the circuit works.

 

[PowerChordHack]

got a handle on it...
using the water & pipes model and completely ignoring power theory, if the flow is 1 gallon / minute and if the hairball in the pipe (resistor) allows 1/3 of the flow to pass then:

* the flow on the left is 1/3 gal/min
* the flow on the right is 2/3 gal/min

/analogy

So the resistor on the left will "turn down" the PUP significantly more than the right.

maybe?

 

[StratoCraig]

yeah, this is a concept i came to understand here on TGP from the likes of jef bardsley and others; while it's a useful shorthand to think of these components as "attenuating" a signal, or "bleeding it to ground" in the case of tone caps, in reality the signal that's supposedly "removed" never forms in the first place. the whole circuit is what creates the signal, and said signal "includes" all the components in its formation.

You inserted "[parallel]" into your quote from my previous comment. Resistance is resistance. It doesn't matter whether it's parallel. All that matters is the total resistance value seen by the rest of the circuit.

 

[PowerChordHack]

I should say the reason behind this is I want to add a "rhythm switch". A switch that drops the vol of a pickup slightly. This is for a guitar with one vol pot and two pickups that are quite a bit different in output.

 

[PowerChordHack]

(relatively) better drawing

 

[StratoCraig]

What kind of guitar is this, and how is it wired at present? Also, I'm not quite clear on whether you're trying to change the volume of one pup while both are active, or just one pup by itself. Does your guitar have a master volume knob, or volume knobs for each pup?

 

[ngativ]

I didn't say it was inherent in the resistance. But the pickup only produces as much signal as the resistive load allows. This does not imply that the pickup can produce as much energy as a nuclear reactor if you just give it enough resistance. I was just explaining why more resistance produces a hotter signal.

Yeah, i just thought that could be wrongly interpreted that way .

As we all know already, though, turning the volume pot down does not just reduce the signal's amplitude. You lose highs, too. The resonant peak frequency changes and the amplitude of the peak diminishes. In fact, frequency response continually changes as you turn the volume pot, even if you don't have a tone pot or tone cap.

Not necessarily since that frequency response dependence on the resistance has more to do with guitar cable than the potentiometer itself. The "impedance" resistance is inherently transparent to the frequency of the signal. The big picture is more about the combined impedance. If you reduce the resistance to ground, the electrons will simply avoid the capacitance of the cable and will go through the less resistive path to ground which is the potentiometer. The "second" resonant peak that appears thanks to the cable goes away , but the resonant peak of the 'of the pickup itself ,its pure 'voice', will still be there .

The coil itself will keep producing an EMF when you pluck the strings, even if you turn down the volume pot to 0 . Remember that the volume potentiometer is just a resistive voltage divider

 

[ngativ]

I should say the reason behind this is I want to add a "rhythm switch". A switch that drops the vol of a pickup slightly. This is for a guitar with one vol pot and two pickups that are quite a bit different in output.

Use a voltage divider for that .

 

[John Coloccia]

Maybe you can talk more about the behavior that you want? The specific configuration of the guitar and pickup selector? There's all sorts of neat stuff you can do. For example, you could replace the volume pot with a concentric volume pot (giving you two completely different volume controls) and either add a switch, or use a push/pull pot on a tone to select the two volume controls. That would be a cool setup, and one that I've never done before, actually.

Talk to us about what you need the guitar to do...in your dreams, how do you want the thing to behave?

 

[StratoCraig]

ngativ's idea will certainly work, but I like John's suggestion of concentric pots and a switch. It seems more flexible.

 

[walterw]

You inserted "[parallel]" into your quote from my previous comment. Resistance is resistance. It doesn't matter whether it's parallel. All that matters is the total resistance value seen by the rest of the circuit.

now you lost me; increasing that series resistor is gonna decrease the result, while increasing a parallel resistor is gonna increase the result (up until the point where it's the same as no resistor at all, obviously).

 

[PowerChordHack]

I thought of concentrics but they won't fit the design.

I think I have it figure out. I'll do some experimentation & see.
Man, I REALLY wish I still had access to a signal generator & a scope.

 

[StratoCraig]

Well, one of us has lost the other, definitely. I don't see how increasing series resistance can have an opposite effect to increasing parallel resistance.

Let's try going back to basics just to figure out what we're actually disagreeing about.

If you have two 62k resistors in series, they total 125k. If you increase one of them to 100k, they now total 162k.

If you have two 250k resistors in parallel, they total 125k. If you increase one of them to 500k, they now total 166.67k.

In both cases we have increased the value of one resistor and in both cases the total resistance has increased. I trust we are in agreement so far, since this is straight out of any introductory electrical theory textbook.

Now, if we are using these resistors as the load on a typical Strat pickup (replacing the standard volume and tone controls), in both cases the guitar's sound will get noticeably brighter when we increase the values as above, because the resonant peak amplitude will be higher. The RLC equation does not care whether you have resistors in parallel or not; all it cares about is the total, calculated as above. If the total is 125k, it doesn't matter whether you have a single 125k resistor, two 62k resistors in series, two 250k resistors in parallel, or some other combination that sums to the same total.

What result are you referring to that goes down if you increase one of two resistors in series, but goes up if you increase one of two resistors in parallel?

 

[walterw]

Well, one of us has lost the other, definitely. I don't see how increasing series resistance can have an opposite effect to increasing parallel resistance.

Let's try going back to basics just to figure out what we're actually disagreeing about.

right!

i'm just referring to the original crude examples, a resistor in parallel with the pickup vs. a resistor in series with the pickup.

increase the resistor in series with the pickup and the pickup gets quieter.