Something about pots bugging me

[Herr Masel]

I know the basic equation for a pot is (R2/(R1 + R2))*Vin = Vout  but what I don't understand (and what I can't find, I have no idea how to search for this) is how it physically works. Schematic explanations show two resistors with a wiper between them, when the wiper is fully clockwise it is connected entirely to resistor (or lug) three, and when counter-clockwise entirely to resistor 1. But what actually happens? When the pot is full on, how does it change the equation as opposed to it being say half on, or off? And in a volume control, which relies on watts (I think) why does increasing the resistance (maxing the pot) increase the volume if it is actually cutting the voltage - and thus wattage -  back?

[twabelljr]

A pot is actually 1 resistor. The value of the resistor is whatever the pot value is. The pot "resistor" lies between lugs 1 and 3. The signal or voltage usually flows into lug 3, through the resistor, then out lug 1. (or 1 thru 3 depending on how you wire it). Lug 2 (the wiper) is a tap into the resistor to pick up the signal at any point along the value of the resistor. As you turn the pot, you are actually moving lug 2 across the resistor to any value you choose between 0 ohms and the pots value. It only looks like 2 resistors because the schematic symbol shows the wiper (lug 2) in the center. Make sense?

[Herr Masel]

Yes, but, if the equation for calculating a pot's output voltage is (R2/(R1 + R2))*Vin = Vout (where R2 is the lower half and R1 is the top half of the resistor) how does moving the wiper across the pot change the values of R1 and R2? This is what I don't understand. Plus, if having a pot maxxed out means connecting the wiper completely to lug 3 isn't that the same as bypassing the resistor? I know the opposite is true but again the physical construction is a little confusing.

In effect/amplifier functions (and this is opening a new can of worms) is signal voltage or current based?

I think I understand the answer to the volume question,  because lug 1 is wired to ground so maxing the pot blends the ground out of the signal. I suppose having a jumper wire instead of a pot would yield more volume but it would obviously be unadjustable.

Thanks for answering my questions, I really appreciate it!

[zachary vex]

pots are made out of a strip of carbon (or other resistive material) connected at both ends to lugs 1 and 3.  lug 2 is connected to the metal wiper, and as you turn the wiper fully clockwise it (usually) shorts to lug 3.

along the way between lugs 1 and 3, the wiper can be connected to the strip of carbon at any point.  the pieces of carbon between the wiper and each end lug change length as you turn it.  in a linear pot, you can basically predict the resistance of each segment by looking at the position... if you're halfway up, each resistor formed by the two stips is equal.  at 75%, one side is 3 times the other in value.  it's pretty simple.

[AdamB]

basically think of it as a long piece of wire that when you put a current through it, has a certain known resistance (e.g. a 10k pot will have a wire of 10k of resistance...). Then, the wiper (lug 2) is just another piece of wire which can be connected to the resistance wire at any point. As resistance changes as you change the length of the resistance wire, thus changing the position of the wiper wire will give you a different resistance between lug 1 and the wiper, because moving it changes the distance of resistance wire between lug 1 and the wiper wire. You can either use it as a variable resistor in this way by not using lug 3 (so all your doing is changing the resistance by changing the position of the wiper and making the resistance wire longer or shorter), or as a divider - you can change the ratio of current flowing through lugs 2 and 3 by moving the contact.

I'm not sure the exact way that the pots used in stompboxes work inside, but it's on this principle. I've used those really big linear pots before, and that was simply what I've described above, except the wire is coiled along an insulator to save space.

-Adam

[R.G.]

But what actually happens? When the pot is full on, how does it change the equation as opposed to it being say half on, or off?

At the risk of being redundant, have you read "The Secret Life of Pots" at GEO?

Yes, but, if the equation for calculating a pot's output voltage is (R2/(R1 + R2))*Vin = Vout (where R2 is the lower half and R1 is the top half of the resistor) how does moving the wiper across the pot change the values of R1 and R2? This is what I don't understand.

Imagine a long strip of resistance from the signal source to ground. The "wiper" touches the strip of resistance at any point along the resistance. If the resistance stuff is uniform, consistent along the length and linear, then there is an equal amount of voltage dropped per unit length along the resistance strip. The current in the whole resistor is I = V/R, and does not vary at any point in the resistor - it's the same current. So the voltage at any point along the resistor must be the current times the remaining resistance to ground.

If we call the remaining resistance to ground "R2", then the voltage at the point we're looking is Vtap=I*R2, using "Vtap" for the volatage where we happen to be looking. For ease in measuring, let's just touch a meter probe to the resistance strip at the point we're looking. That's the wiper.

Now we know that the total resistance is composed of the resistance above the wiper and the resistance below the wiper. Let's call the resistance above the wiper "R1". We know that the total resistance R is equal to the sum of the two parts, or R = R1 + R2. So I = V/R, or I = V/(R1+R2). So  we get

Vtap = (V/(R1+R2))*R2, or Vtap= V *(R2/(R1+R2)), which is the standard voltage divider equation. And we made no assumptions about R1 and R2 other than they add to the total resistance R.

Notice that this is only true if there is no loading at the tap point/wiper. Otherwise, that loading must be taken into account to compute the output voltage. That is the basis of "tapering resistors".

The earliest "potentiometers" were just what I have described - long rods of carbon, or wire, or columns of water, with a sampling point that could literally be slid along the length of the resistance. Modern potentiometers are also the same, but the resistance is arranged in a circle for rotary pots.

Plus, if having a pot maxxed out means connecting the wiper completely to lug 3 isn't that the same as bypassing the resistor? I know the opposite is true but again the physical construction is a little confusing.

The resistor loading remains on the signal into the pot, but yes, the pot has no other effect. That is, the full signal voltage passes to the wiper as the wiper is connected to the full input voltage.

In effect/amplifier functions (and this is opening a new can of worms) is signal voltage or current based?

It can be either one, but it's usually primarily the signal voltage that matters.