# Pot value question

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After all these years I still don't understand (exactly) how Vol Pots work. Whenever I read an explanation there's still something missing and I don't get it

My main question is

When the Vol is at 10 how/why does it matter what the Pot value is? Doesn't it by-pass the resistance?

I understand a resistor in a circuit. There's a wire, a resistor, then a wire. But a Vol Pot has 3 lugs. Ground, Hot, Pickup, and I think my understanding is lost somewhere in all that. Why the need for Ground? The Pickup goes to ground, isn't that enough?

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A volume pot is just a voltage divider, circuits of which are all over the net. What you are changing when you turn the pot is the value of the two resistors in a voltage divider. The pot on 10 means it goes through a zero series resistance, but there is still a large parallel resistance (the pot value) parallel to your output (amp's input). Because the pot value is not infinity and the amp's input impedance is not zero, the value of the pot will affect the signal to some degree. The only way the resistance is completely bypassed is by using a no-load pot, which actually just has a little switch that takes the pot completely out of the circuit and jumps the signal straight to the output.

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I'm not sure what you mean by two resistors and I'm completely lost when you're talking about amp input and output

However looking at a Pot's inner parts and thinking about what you've said, I might be getting the idea

Tell me if I've got it right

Power goes to the middle lug, then some to pickup AND some to ground

The position of the Vol Control determines how much goes to pickup and how much to ground

So when the Vol Control is on 10, there is still some current going to ground

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The position of the Vol Control determines how much goes to pickup and how much to ground

Very nearly. The volume pot determines how much of the pickup's signal gets passed straight through to the output jack vs how much gets "lost" to ground.

With vol on 10 the proportion of signal leaving the pot is much greater than that going to ground, and the result is that you hear the full output of the pickup. As the pot gets wound down towards zero, the ratio of pickup signal (ie, the sound of rock 'n roll) to ground (ie, the sound of silence) gets smaller, until eventually the pot is wound all the way to zero and 100% of the pickup's output is lost.

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You need to understand basic electronics like ohm's law. Your volume pot is connected in parrallel to the input of whatever you're plugging into.

I'm simplifying here because the actual circuit has R L C (Resistance, Inductance, Capacitance) components but in essence, if you have a 500k pot connected to and 500k input impedance circuit (like an Ibanez TS9), your nominal impedance as seen by the pickup is 250k.

If you connect into a 1M input (like a Marshall tube amp), the resulting impedance is 333K. That will affect the frequency response of the system. The guitar's controls are part of the entire circuit.

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• 2 weeks later...

Thankyou Curtisa for reminding me that the pickup makes the rock and roll. Thankyou all for chiming in, no offence but you made my first comments to be true to form LOL. However reading what you've all said and doing some homework I believe I have it worked out

The Indians send signals from the rocks above the pass and Davey Crocket rides around and says its Cool For Cats.....(ha ha)

The Volume Control depending on its position sends the signal to amp or to ground, or somewhere in-between. But there's always some goes to ground, so a regular pot when on 10 is not a by-pass.......(unless you have a very high resistant pot) And I have read numerous times that the lower the pot value, the more highs go to ground

And if you think about it, If there was no resistance, all would go to ground

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