Quick impedance question

[eukatheude]

So, i was wondering. Since having 500K pots instead of 250K on, for instance, a J, can give you a brighter sound, wouldn't it be the same if i put a 750K resistor between the output wire and the output jack? Thanks

[Crater]

No, it wouldn't be the same. It would reduce your output (loudness) a lot, and would reduce some treble too.

You could replace the standard 250K or 500K with a 1 megohm (1,000 K ohms) pot, or use a Fender "no-load" pot that removes the resistance to ground entirely when the the control's on 10.

[line6man]

I'm not sure where you got 750k from, but for the average pickup impedance, that's enough to kill most of the output.

What you would do to go from 500k to 250k is place a 500k resistance parallel to the signal. Realistically, 470k or 510k resistors will be easiest to find.

[eukatheude]

Originally Posted by line6man

I'm not sure where you got 750k from, but for the average pickup impedance, that's enough to kill most of the output.

(500*3) - (250*3)

What you would do to go from 500k to 250k is place a 500k resistance parallel to the signal. Realistically, 470k or 510k resistors will be easiest to find.

So if you were looking at a J wiring, where would that be? (don't waste time drawing, it's just a curiosity)

[line6man]

Originally Posted by eukatheude

(500*3) - (250*3)



So if you were looking at a J wiring, where would that be? (don't waste time drawing, it's just a curiosity)

Where did the number 3 come from?

When dealing with resistances and impedances, just remember the series and parallel formulas.
Series: R_Total=R₁+R₂+...R_n
Parallel: R_Total=1/([1/R₁]+[1/R₂]+...[1/R_n])

The resistor can just run parallel to the jack.

[eukatheude]

Originally Posted by line6man

Where did the number 3 come from?

When dealing with resistances and impedances, just remember the series and parallel formulas.
Series: R_Total=R₁+R₂+...R_n
Parallel: R_Total=1/([1/R₁]+[1/R₂]+...[1/R_n])

The resistor can just run parallel to the jack.

3 pots.

How would i wire it parallel to the jack?

Not sure about parallel, if i had say a 100k and a 200k resistors, total resistance would be 1/[(1/100)+(1/200)] = 1/(0.01+0.005) = 1/0.015 = 66.666?

[line6man]

Originally Posted by eukatheude

3 pots.

How would i wire it parallel to the jack?

Not sure about parallel, if i had say a 100k and a 200k resistors, total resistance would be 1/[(1/100)+(1/200)] = 1/(0.01+0.005) = 1/0.015 = 66.666?

The tone pot in a VVT setup presents a resistance that varies with frequency. At high frequencies, it approaches the value you have dialed in. At low frequencies, it approaches infinity.

One side of resistor goes to the tip, and the other goes to the sleeve.

100k parallel to 200k is 66.667k. Be mindful of the "k" suffix, which means "*1000."

[SGD Lutherie]

Originally Posted by eukatheude

So, i was wondering. Since having 500K pots instead of 250K on, for instance, a J, can give you a brighter sound, wouldn't it be the same if i put a 750K resistor between the output wire and the output jack? Thanks

The series resistance would increase the impedance and dull the tone and reduce the output. The control pots are parallel resistance.

You can use a larger value pot, like a 1M if you want a brighter tone, or go active, which accomplishes the same thing since it isolates the pickup (buffers) from the cable and controls.

As much as some people say that active sounds artificial, it's the true sound of the pickup with minimal loading.

[eukatheude]

Thank you. I'll try putting a spare resistor and see what it does to my tone.

[wcriley]

Originally Posted by SGD Lutherie

As much as some people say that active sounds artificial, it's the true sound of the pickup with minimal loading.

Repeating for emphasis.