After reading a lot of nonsense on the subject of guitar pot values all over the place, I thought I would say a few words about it based in electrical theory. This dovetails nicely with my guitar mod article, and it helps to explain why I wanted to keep within 500k of impedance.
Definitions aka electrical theory primer
frequency - this is the number of oscillations that a signal makes in a given amount of time. Since the electrical generated by the pickups will eventually be amplified and turned into sound, frequency represents the pitch of the note. A high note has a high frequency, and a low note a low frequency. It's simply a measure of how high or low a note is for the purposes of guitar electronics.
voltage and current- voltage is the electrical signal generated by the pickup. The metal string vibrating near the magnetic pickup "induces" a voltage in the pickup. A voltage causes a current to flow through a circuit, provided that there is a return path. Voltage is a difference in "potential" between two points. It has the potential to cause current to flow. Basically, charge builds up on the positive side, and a negative charge builds up on the negative side. These want to equalize to zero, and the only way they can do it is through a conductive circuit. If there is no conductive path between + and -, then the voltage will just sit there and the difference in charge will never equalize (in theory).
resistance/impedance - as its name implies, resistance resists the flow of current in a circuit. The voltage has a certain force that will cause charges to move. Resistance makes it more difficult for the charges to move through the material. Materials have different resistance. Metal wire has very low resistance, which is why it is used in electric circuits. Plastic, wood, air, enamel, and wire insulation have high resistances, so very little (if any) current flows through them. Carbon is in between, which is why it's used in pots. You don't want the charge to fly around the circuit, otherwise you won't get a signal. The pots slow it down and allow us to see the voltage level.
Impedance is just resistance for AC signals. I will use resistance for both terms.
AC voltage - here's where the frequency/oscillation comes in. The moving string doesn't generate a constant voltage. It swings up and down. Half the time, the + and - are switched. Think of it as a sinusoidal wave. As you move through time, the voltage swings high and low, and the frequency is the speed of that oscillation. Constant voltage (DC) wouldn't sound like anything. The oscillation makes the sound when put through the amplifier.
Voltage Drop - This is something of a weird concept. But suffice it to say the following. If we assume that wire has no resistance, a circuit boils down to two things: a source of voltage, and a load resistance. The resistance determines how much current flows. The resistor has some length, and the voltage distributes evenly across the resistance as you move down the length. So close to the end of the resistor where it connects back to -, the voltage would be very small. Near the top, where it connects to the +, the voltage would be nearly as high as coming right out of the pickup. This is the idea behind using a pot for volume control. The pot's tap grabs the voltage at some point along the resistor, and the tap can move up and down the resistor. So we can get different levels. The total resistance of the pot does not change as the knob is turned. Turning the knob simply repositions the tap somewhere along the pot's resistor.
Pot Resistance Effects
A pickup can be modeled as shown below...
The circle with +- is the pickup generating a voltage. The looped component is an inductor. The two angular components are resistors. The + V.out - represents where we will measure the voltage. Actually, I will talk about measuring at the top of R.load. This is just the standard way to draw a pot. The inductor and first resistor are part of the pickup. This is because the pickup is not a perfect voltage source. The R.load is the equivalent load of the rest of the circuit. For our purposes, we will say that it is the resistance of the pot itself.
Note: resistive components connected back to back (in series) will act like a single resistor with a resistance equal to the sum of parts.
The inductor is a complicated component. The effect comes from the wrapping of the wire around the magnet. Its resistance changes with frequency. To a high frequency, it is a high resistance. To a low frequency it has low or 0 resistance. Because of the inductor, we have to look at the circuit from two perspectives: low frequency and high frequency.
At low frequency, the resistance of the inductor is negligible. So we have only the 2 resistors in the circuit with the voltage source. The internal resistance of the pickup will remain the same. Now what if we had a low resistance for our pot? Since the two resistors are connected in line (in series), they act as a single resistor with the sum of the two values. The lengths are also combined, so the voltage drop idea applies over the combined length of both resistors. If we measure the voltage at the top of the pot (turned all the way up) then we will get whatever portion remains to be distributed across the pot. So if the pot is a low value, there is not much left to be distributed, and we will have a low volume.
Now suppose we have a high resistance for the pot. Most of the voltage will be distributed across the pot itself. So if we grab the voltage at the top, it will be close to what we would see if the pickup were a perfect source.
You can think of the low frequency as the general case and apply it to the whole frequency range. The resistors affect all frequencies equally. So obviously we will lose a lot of volume by using a small pot value. And this is undesirable. But there's more...
As I said before, the inductor's resistance increases with frequency. So for a high frequency, let's say it becomes the same as the internal resistance. What happens with a small pot value? More resistance before the pot makes the signal strength even lower for high frequencies! So the higher the note gets, the softer the volume will be. This is what leads to the decreased highs in a humbucker. The construction of the humbucker has a higher internal inductance.
Now consider a high pot value. We will still lose some volume as we get into high frequencies, but the effect will not be as noticeable because such a large portion of the voltage is already distributed over the pot.
Considering High and Low
So to really get a picture of what will happen when we have a low pot value, we have to consider both high and low, and everything in between.
A low pot value will give us lower volume, but it will also change the whole sound of the guitar. You will lose a lot of highs, and some mids. The effect will start gradually, and you will lose more and more volume as you get into the higher notes. Eventually, you won't hear anything above a certain frequency. Of course, this could be higher than your audible frequency. I'm not sure.
Limitations of this explanation and Conclusion
I tried to make this accessible to those with no electrical experience. So I apologize if it is overly simplistic for some people. The concepts are sound. There is a lot of other stuff going on in a guitar circuit. Multiple pickups. Tone control. The effect of connecting a long cable. The effect of connecting to an amplifier. All of these things play a part in the way the guitar sounds. But those effects were not the focus of this article. I hope that it makes sense now why a really high volume pot (say 2M) would make a guitar sound very bright, and somewhat louder. Of course, at some point there are diminishing returns. To get as close to a perfect signal as possible, we wouldn't even have any pots in the circuit! Then only the amplifier will load the pickup. But then we'd lose volume and tone control.
It's sometimes hard to stay on topic. I hope that someone will benefit from this knowledge.
Thanks for reading.