Potentiometer Wiring

[seten]

I was testing out different pots to decide which one to use as a volume control for an effect loop pedal, and while I thought I understood how a pot works and the different ways to wire it I seem to have a gap in my knowledge. I thought that I was wiring it as a rheostat and thus would only be using the wiper pin and one of the others; that way at one extreme the wiper is connected to the other pin and there is no resistance, its essentially a jumper. then at the other extreme its at max resistance and the signal should be muted or close to muted if the pot I was using didnt have enough max resistance to mute the signal. I didnt think there was any chance I would want to wire it as a voltage divider since it was a passive circuit and I had no voltage to speak of.

As you knowledgeable people probably guessed, that didn't work. It did turn down the volume, but at max resistance it was nowhere close to being muted, even when I switched from 10k to 100k. I tried connecting the first pin to ground, and thats what I needed. Now I want to know why! It makes sense to me that with the first pin grounded, this would mute the signal when the wiper is in the "max resistance" position, since its shorting the signal to ground. But why does it affect the rest of the sweep? With the first pin grounded and the knob at just under max resistance, the signal was much lower than without the first pin grounded and the knob at just under max resistance. This is baffling to me because I thought in this position the grounded pin is not connected to the other two in any way so I don't understand how grounded/not grounded should make any difference in the signal running through the other two.

I hope I have explained my difficulty. If the answer seems obvious please give it to me anyway, there is a very good chance I am missing something right under my nose.

[GibsonGM]

Hi Seten, 

If you look at this image:  , looking just at the pot....this is a typical volume control.  It is a voltage divider, when wired this way, using all 3 terminals & ground.  The way you described would be wired as a variable resistor, and that's not really what we need here. 

If you will, picture that above and below the wiper, it is as if there are TWO resistors in series, with the wiper as a 'tap' in between them.   

If you move the wiper "up" (really,to one end, of course), you will get a decreasing resistance between the input and wiper, with a larger resistance below it to ground.   This allows  more signal to pass to the output thru the wiper, and the larger resistance to ground prevents the signal from taking that path to ground.  At max rotation "up", toward the "hot end" of the pot, it's then like a jumper with a resistor to ground at that node, which doesn't bleed very much signal to ground; the signal passes out the wiper to output.

If we turn the pot the other way, we get a higher and higher resistance between signal and wiper, reducing the current flow thru the wiper.  And that resistance "below" the wiper to ground gets smaller and smaller, so more and more signal level bleeds to ground; finally, the wiper IS connected to that end of the pot, and the signal is grounded, so our output is silent.

We could just as easily use two fixed resistors to lower a signal level; the pot just allows us to adjust this ratio continuously.    If you take a meter and measure the pot, you will find that from one end to wiper has 1 resistance, and from wiper to the other end, another resistance.  That ratio will change as you turn the pot...potentiometer action!  It's all about that ratio....

HTH! 

[PRR]

> I didnt think there was any chance I would want to wire it as a voltage divider since it was a passive circuit and I had no voltage to speak of.

Think again. Doesn't matter if it is "passive". And you sure as have have a "voltage to speak of": your precious guitar signal! (Which happens to be too strong.)

Draw the whole circuit!! You put the 0-100K variable in series with ... what? If it is a 100K input loading the variable resistor, it will fade to half. You may be wanting more cut than that. If it is a 1Meg input it will fade to 90%. Hardly worth doing.

Volume knobs are almost "always" wired as 3-terminal potentiometers. We normally want to fade to "very very small", so "to zero". (There are exceptions but more in connection with feedback amplifiers.)

[seten]

I think I got it, thanks guys! I'm starting to get it. So even when the knob is all the way "up" there's still a tiny amount of signal bleeding to ground, its just so small that it's negligible?

[Myampgoesto12]

I think I got it, thanks guys! I'm starting to get it. So even when the knob is all the way "up" there's still a tiny amount of signal bleeding to ground, its just so small that it's negligible?

Essentially. That small amount of resistance to ground is a large part of determining the impedance from your guitar, or from the output of any pedal that has a volume control. Units that don't have a volume control (typically like a two knob chorus or flanger etc) typically still have a fixed voltage divider after the mixing stage at the output, thus providing an impedance.

So that really matters when using effects. A guitar hat has one 500k volume and one 500k tone, not counting the pickups themselves, that makes a total of 250k impedance. That's not horrible but if its plugged into a pedal with low input impedance like ~500k or less, then the signal can be loaded, and you loose some higher frequencies. Its not usually catastrophic, but its something to consider when using effects.