Deleting volume pot for dummies (me)

[Smash]

I dont know that i actually will embark on this, but it should teach me something anyway. In reality, i have no neck pickup. So i use no tone pot or switch. I just go pickup to volume to output. A few basic questions

What does a (volume) pot's resistance actually do when all the way up?

What does a pot's resistance do when turned down vs up?

What does a tone pot do when maxed?

What would a tone pot do when maxed if you removed the cap?

If i have a 500kohm pot on volume, how would i replace that with a resistor?

If i have a 500k ohm pot on volume, what would happen if i put a resistor on the first lug (where i hook up hot wire from pickup) connected to ground?

What if anything would happen if anything if i were to put a resistor in the hot line not going to ground? So hot wire from pickup to resistor or then hot wire from jack to resistor.

Why are two resistors in parallel less resistance than one?

What does a 250 vs 500 vs 1 meg actually do?

What is the difference between a resistor and a capacitor?

 

[7704A]

Why are two resistors in parallel less resistance than one?

In short, because now there are two paths for current to flow through. Per Ohm's Law V/I = R, and since V stayed the same while I got bigger, R is now smaller. V is the voltage across the resistor(s), I is the current through the resistor(s), and R is the resistance of the resistor(s).

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Derivation for two resistors R1 and R2 in parallel with voltage V across them:

Our relations between resistor values, currents, and the voltage:

I_total = I1 + I2;
I_total = V/R_equivalent
I1 = V/R1;
I2 = V/R2;

Substituting into the top equation so we don't have any "I" variables:

V/R_equivalent = V/R1 + V/R2

Dividing out the voltage:

1/R_equivalent = 1/R1 + 1/R2

Or with some re-arranging:

R_equivalent = (R1 * R2) / (R1 + R2)
= R1 * [R2 / (R1 + R2)]
= R2 * [R1 / (R1 + R2)]

I wrote the answer three different ways to make the next step more obvious.

To convince yourself that R_equivalent is always smaller, look it this way: [RX / (R1+R2)] where RX is either R1 or R2 will always be smaller than 1 since it's basically RX divided by (RX + plus something). So now you have two different ways of writing R_equivalent:

R_equivalent = R1*(something smaller than 1)
R_equivalent = R2*(something smaller than 1)

So R_equivalent will be smaller than R1, and smaller than R2.

 

[7704A]

What is the difference between a resistor and a capacitor?

Capacitors let more current through (for a given voltage) at high frequencies, and less at low frequencies. At DC no current flows, and as frequency gets higher and higher capacitors get closer to being a short circuit. Resistors always let the same amount of current through for a given voltage regardless of frequency. Note that I'm talking about ideal components. Real world ones have imperfections that make this all not quite true.

 

[Smash]

In short, because now there are two paths for current to flow through. Per Ohm's Law V/I = R, and since V stayed the same while I got bigger, R is now smaller. V is the voltage across the resistor(s), I is the current through the resistor(s), and R is the resistance of the resistor(s).

-------------------------------------------------------------------------------------------------------

Derivation for two resistors R1 and R2 in parallel with voltage V across them:

Our relations between resistor values, currents, and the voltage:

I_total = I1 + I2;
I_total = V/R_equivalent
I1 = V/R1;
I2 = V/R2;

Substituting into the top equation so we don't have any "I" variables:

V/R_equivalent = V/R1 + V/R2

Dividing out the voltage:

1/R_equivalent = 1/R1 + 1/R2

Or with some re-arranging:

R_equivalent = (R1 * R2) / (R1 + R2)
= R1 * [R2 / (R1 + R2)]
= R2 * [R1 / (R1 + R2)]

I wrote the answer three different ways to make the next step more obvious.

To convince yourself that R_equivalent is always smaller, look it this way: [RX / (R1+R2)] where RX is either R1 or R2 will always be smaller than 1 since it's basically RX divided by (RX + plus something). So now you have two different ways of writing R_equivalent:

R_equivalent = R1*(something smaller than 1)
R_equivalent = R2*(something smaller than 1)

So R_equivalent will be smaller than R1, and smaller than R2.

Ok so if we use 470kohm resistor, and we put two in parrallel, would it be around 250k or would it go up to the 940?

 

[Smash]

Capacitors let more current through (for a given voltage) at high frequencies, and less at low frequencies. At DC no current flows, and as frequency gets higher and higher capacitors get closer to being a short circuit. Resistors always let the same amount of current through for a given voltage regardless of frequency. Note that I'm talking about ideal components. Real world ones have imperfections that make this all not quite true.

So if you were deleting a pot, would you use a resistor or capacitor?

 

[Smash]

In short, because now there are two paths for current to flow through. Per Ohm's Law V/I = R, and since V stayed the same while I got bigger, R is now smaller. V is the voltage across the resistor(s), I is the current through the resistor(s), and R is the resistance of the resistor(s).

-------------------------------------------------------------------------------------------------------

Derivation for two resistors R1 and R2 in parallel with voltage V across them:

Our relations between resistor values, currents, and the voltage:

I_total = I1 + I2;
I_total = V/R_equivalent
I1 = V/R1;
I2 = V/R2;

Substituting into the top equation so we don't have any "I" variables:

V/R_equivalent = V/R1 + V/R2

Dividing out the voltage:

1/R_equivalent = 1/R1 + 1/R2

Or with some re-arranging:

R_equivalent = (R1 * R2) / (R1 + R2)
= R1 * [R2 / (R1 + R2)]
= R2 * [R1 / (R1 + R2)]

I wrote the answer three different ways to make the next step more obvious.

To convince yourself that R_equivalent is always smaller, look it this way: [RX / (R1+R2)] where RX is either R1 or R2 will always be smaller than 1 since it's basically RX divided by (RX + plus something). So now you have two different ways of writing R_equivalent:

R_equivalent = R1*(something smaller than 1)
R_equivalent = R2*(something smaller than 1)

So R_equivalent will be smaller than R1, and smaller than R2.

Ok. I see it now. Reminds me of completing the square

 

[Smash]

Capacitors let more current through (for a given voltage) at high frequencies, and less at low frequencies. At DC no current flows, and as frequency gets higher and higher capacitors get closer to being a short circuit. Resistors always let the same amount of current through for a given voltage regardless of frequency. Note that I'm talking about ideal components. Real world ones have imperfections that make this all not quite true.

So the pickup would see a load with either, but the capacitor would cut highs based on the ohms, where as resistor would just cut output?

 

[SpiderWars]

I use this calculator especially if it’s more than two resistors.

A pot is just a variable resistor. A tone pot without the cap (but with a wire instead) is just another volume pot.

Connect both ends of two resistors and the two resulting ends are then in parallel. Connect the resistors end-to-end (just one end) and the two resulting ends are in series. You can just add them up when in series but use the calculator for parallel.

 

[7704A]

Ok so if we use 470kohm resistor, and we put two in parrallel, would it be around 250k or would it go up to the 940?

Around 250k. If all the resistors in parallel have the same resistance, the equivalent resistance for the parallel group is just the resistance of one resistor divided by the number of resistors in parallel. It's possible to work this out from the formulas I posted above.

 

[Smash]

Around 250k. If all the resistors in parallel have the same resistance, the equivalent resistance for the parallel group is just the resistance of one resistor divided by the number of resistors in parallel. It's possible to work this out from the formulas I posted above.

Yeah. It took me a second but i have ohms law pretty down now. Thanks. That is a big step

 

[PatF]

If you're deleting the pot you can just wire the pickup directly to the output jack.