Potentiometer Volume Control

Question

Electronics theory newbie here (please be gentle):

So I have an old guitar amp that had a crackly volume control. It also had an annoying habit of going from no volume to "too loud" with the slightest twist past 0 (let's assume this was by design and not a fault of the pot for now).

I opened it up and it was a 10kΩ type B pot (linear taper from what I've read).

I went ahead and replaced it with a 5kΩ type B pot, my logic was that I could get finer grain control when twisting past 0, as a given rotation on a 5kΩ pot would change the ohms by a half the amount of a linear 10kΩ pot. The crackle went and I did indeed get finer grain control.

Now I'm puzzled why this worked in the first place? My basic understanding was that originally, whilst the potentiometer was dialed into zero, it would have had a 10kΩ resistance, which kept the volume so low to be inaudible, as the pot swept towards 0Ω, the signal faced less resistance and the volume increased, however this 'theory' cant be correct as the 5kΩ pot when dialed to 0 was also inaudible. If my 'theory' was correct then the (linear) 5kΩ pot should of had the same volume as the (linear) 10kΩ when dialed half way, which was VERY LOUD, it didn't!

I can only really think of of the following as to why this is happening:

The circuit is wired to "sweep backwards" i.e. 0Ω (on the other terminal) would be inaudible, turning the dial to 1 would INCREASE the ohms and this somehow INCREASES the volume. I do see a 4558d IC op-amp in the circuit, perhaps it has the logic to "inverse" the ohm reading passed to it? (if this "theory" was correct, I would now never be able to run the amp at full volume?)

Hope someone can help me understand what should be a basic concept.

Answer 1

What you are missing is that the pot is wired such that it produces a ratio of the input signal with the ratio varying from 0 to 1 accross the full sweep of the pot. This is why the 5 kΩ and 10 kΩ pots resulted in the same full volume.

The pot achieves this by being a resistor divider. It does not work by adding resistance in series with a signal. A resistor divider looks like this:

The output will be R2/(R1+R2) of the output. In the case of a pot, R1 and R2 are one continuous resistor with a mechanical wiper picking off OUT at some point along this resistor. The three pins of the pot are the two ends of this resistor and the wiper tap. Therefore, R2 will vary from 0 at no volume to (R1+R2) at maximum volume. Also, R1+R2 is always fixed, and is the resistance value specified for the pot. In your "5 kΩ" pot, for example, R1+R2 is 5 kΩ, which is the value of the physical resistor that the wiper slides over.

At half volume, for example with the 5 kΩ pot, R1 and R2 are each 2.5 kΩ. OUT is half of whatever signal is applied at IN. Note that since everything is ratiometric, you get the same answer whether the total pot resistance is 5 kΩ or 10 kΩ. This is why the volume levels didn't change.

The total pot resistance does matter in other ways to the driving circuit and whatever is using the signal at OUT. The 5 kΩ pot will require whatever is driving IN to provide twice the current than is necessary with the 10 kΩ pot. You don't know what exactly is driving IN and what its design constraints might have been, so it is best to replace the pot with one of the same value. It seems you got lucky in that whatever is driving IN can cope with the 5 kΩ load, but it could just as well have started clipping, otherwise distorting, or have the frequency balance different.

The crackling and the fact that you got sudden jumps in volume were due to the old pot being worn out. As pots age, dirt and oxidation accumulates on the surface of the resistor where the wiper slides over it. The resistor itself can also get worn down due to mechanical abrasion by the wiper. The wiper sometimes making good contact and sometimes not can sound like crackling, especially when the pot is being turned. Dead and worn out spots on the resistor can cause sudden jumps. These are all common failure modes of pots.

This is one area where construction quality makes a big difference. El-cheapo pots wear out a lot faster and may not be as well sealed against dirt or the materials are more prone to oxidation. If you want long-lived mechanical volume controls, you have to spend the money on good quality pots.

This is also one reason these things are done digitally nowadays. You can get a microcontroller to handle the audio stream digitally for less than the price of a top quality volume control. The digital multiplies inside the micro don't wear out, crackle, or drift over time.

Answer 2

The appliance of potentiometer is dividing voltage. You can bypass two pins of it by a resistor in order to have lower voltage. for example if the amplifier by potentiometer like below has higher output when it goes to down (to pin 3), You can bypass the lower pins by a resistor in order to decrease it's resistance faster:

In this schematic used two \$20 k\$ resistors to have the previous \$10k\$ equivalent when it is near to pin1.

^{simulate this circuit – Schematic created using CircuitLab}