Volume Pot with Resistor to Ground

[turdadactyl]

If your volume pot (assume lug 1 to ground) is wired with a resistor between it and ground, presumably the volume won't go to zero.  But what other effects, if any, does this have on the circuit?  For example, does it change the output impedance?

For purposes of discussion, let's assume a common A100K volume pot and a 10K resistor (in case anybody needs to throw any math into the discussion).

[BluffChill]

If it's just being used for volume, probably nothing else will be affected. It would affect the taper of the pot if it's logarithmic, though.

[willienillie]

10K is 10% of a 100K pot, well within a common 20% tolerance, so likely no effect on the rest of the circuit.  Another 10% to think about, most A taper pots are "10A", meaning at 50% rotation, the clockwise resistance has increased to 10% of the total.  Your A100K pot would have 10K from wiper to pin 1 when turned half up.  So 10K fixed in series to ground would make your A100K volume pot basically a 110K volume pot that starts about half up.

[R.G.]

It's about now that all those algebra and other math courses start to look good. 

Here's the deal: a volume pot can be modeled as two resistors in series. If you call the part above the wiper R1 and the part below the wiper R2, then the output voltage at the lug is Vin * R2 / (R1+R2) and R1+R2 equals the total resistance of the pot. If you call the % rotation "p" then you can write the output voltage in terms of rotation:
Vout = Vin * (p*(R1+R2))/(R1+R2), which handily reduces to Vout = p * Vin.

Adding a resistor below the R2 merely means that R2 gets replaced in the equations with R2+R3, and the effective value for p can never be less than R3/(R1+R2).

Input and output impedance is a bit tougher. A voltage divider (pot or discrete resistors R1 & R2) has an output voltage equal to R2/(R1+R2) which acts like it's coming through a series resistor of R1 paralleled with R2, whatever the R1 and R2 values are. For all pots, the "output impedance" is maximum at exactly half the resistance rotation (that is, R1 and R2 are equal) and it's equal to 1/4 of the pot value. Any other position is a lower series resistance. (Look up Thevenin equivalents). Adding an R3 at one end - either end! - increases the equivalent series resistance by no more than 1/2 of the added resistor at the half-total-resistance point.

Notice I've been talking about equivalent series resistance. That's because a voltage divider has little use unless the loading on the pot is more than about ten times the equivalent series resistance. Otherwise, you have to look at the load impedance/resistance as part of this little resistor network. It appears in parallel with the resistance to ground.

If you're wondering about "tonal effects", there is only a gross generalization possible without discussing the pot, the added resistor AND the source impedance of what feeds it, AND the load impedance on the wiper. There are always bits of stray capacitance to every conductor, a few picofarads in most cases, but upwards of 10-20pf on ground flowed PCBs and such. Whether this causes an audio band effect or not depends on what the values of the pot, added resistance, source and load are. Generally under about a megohm, this can be neglected.

This stuff, and more, was the subject of a three-credit sophomre EE course back in 1971. It's probably all different now. 

[Rob Strand]

You might find this useful, see pic in reply #3,
https://www.diystompboxes.com/smfforum/index.php?topic=121339