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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.

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    \$\begingroup\$ Can't. Not without a schematic. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 20 '13 at 20:15
  • \$\begingroup\$ The crackle is due to wear and dirt (old age) which results in poor contact between the wiper and the track (intermittent contact). I'd go with a faulty pot. Most volume controls work by forming a potential divider circuit with the wiper being somewhere between 0 and 100% of the incoming signal. \$\endgroup\$ – JIm Dearden Nov 20 '13 at 20:26
  • \$\begingroup\$ It's possible that the pot is being used in the feedback of the opamp such that a larger resistance yields a higher volume. In the simplest case they could have an inverting opamp and the pot could be your feedback resistor. Typically (as others have mentioned) you would attenuate the input signal with the pot, but based on the behavior you're describing it sounds like they're putting it in the feedback loop. Without a schematic there's no way to be 100% sure. As others have commented the "crackling" is 99.9% due to dust. \$\endgroup\$ – Doov Nov 20 '13 at 20:46
  • \$\begingroup\$ Oh perhaps I misunderstood your description -- I thought you were suggesting that when the the two pots didn't realize equal loudness (and that the 10k was louder than the 5k). If that's not the case then what I posted above isn't possible. \$\endgroup\$ – Doov Nov 21 '13 at 7:31
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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.

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  • \$\begingroup\$ Thank you so much for taking the time to give this detailed answer, it almost makes sense (need to reread a little bit again). It's clear you have a great understanding of this subject and I thank you for sharing it \$\endgroup\$ – user33026 Nov 20 '13 at 22:50
  • \$\begingroup\$ Just for clarification (Ive no academic background in electronics) the 'ratio' produced by R2/(R1+R2) is basically describing how much of the original signal (voltage?) is allowed to pass through and how much is bled off to ground/earth? So In a simple circuit (battery, lightbulb, pot) the value of the potentiometer has no bearing on the value of Vout, it could be 5k or 500k??? Coming back to my original note about now having finer grain control over the volume, it would that the only explanation for the is the quality of the new pot, and that its ohm value plays no part In that??? MANY THANKS \$\endgroup\$ – user33026 Nov 21 '13 at 0:24
  • \$\begingroup\$ @user33026 Bingo! You've clued in to the fact that in an idealized circuit, the size of the potentiometer doesn't matter one bit. Why? Because in an idealized circuit, we imagine that it's being driven by a perfect voltage source, and that the load has a very high input resistance. In reality, it matters because the signal source is less than perfect (does not like driving loads that are too small), and the input driven by the pot has some finite impedance (and so it shorts out the bottom of the voltage divider if it is too small). Plus large resistors cause more noise, which matters too. \$\endgroup\$ – Kaz Nov 21 '13 at 0:42
  • \$\begingroup\$ @user: You've got the right idea about how the pot scales the inputs voltage. However, it won't work that way with a lightbulb since the lightbulb draws substantial current. The nice linear ratio is only valid when you are drawing no current from it. Whatever is downstream from your volume control is designed with high input impedance, so the approximation of no current is good enough. The low impedance of a light bulb messes up the linear ratio so that you'd get the whole light range in the last little bit of pot travel. \$\endgroup\$ – Olin Lathrop Nov 21 '13 at 13:43
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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.

schematic

simulate this circuit – Schematic created using CircuitLab

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