Mathematical Q (keep the brightness of a big output potentiometer)

[stonerbox]

So I was thinking about a Mr. Hammer post I read a long time ago. He would sometimes use a 500k output potentiometer and leave the volume at maximum, for good. Great if you are for example pushing a tube amp, a pedal or a Neve preamp. In the process it will not only result in moar volume but also a lot less duller sound due to not loading down the higher frequencies, like a 100k potentiometer would.

I used a 500k pot in one of my latest builds and the results are a very pleasant and present sound but the volume is just too damn powerful, of course. "Why don't you just swap the current potentiometer for a 100k and slap a good old treble bleed capacitor on it? It's a tried and fully functional solution to your problem"

Yes, I will but... I was wondering if anyone of you fine gentlemen could help me understand if and how to do the math here. I want the same amount of high frequencies to pass with a 100k+capacitor as a 500k potentiometer alone would. Could it be calculated? I bet it has something to do with one of my favorite subjects in the world...impedance   *shaking in fear*

[GibsonGM]

You'd probably want to figure out what frequency you're dealing with....what is cut when you turn down the 500k. Have fun!    LOL...yes, you'd need to know the impedance of what comes after the output pot, that is what's eating the highs.

OR...if you can GUESS at what frequency is getting cut, you can use: 

C=2.74 / 2pi F R    where C = the bright cap you want,  F = the 1/2 boost freq (800Hz - 5kHz typical), R = the pot resistance.  That will get you most of the way there.  No impedance calculation!   This is just a SHELVING FILTER response (you can look that up if need be).

Lastly:  What I do, if a bright cap becomes necessary, is just 'decade' the cap.  Try .1u,  .01u, 1u.   You'll find which is more in the right direction....then you can maybe try .12u.....then .15u, .22u.....no math, thanks!!       In not more than 5 or 6 tries you'll probably get the one you want! 

[Rob Strand]

I want the same amount of high frequencies to pass with a 100k+capacitor as a 500k potentiometer alone would. Could it be calculated?

Without any specific info or measurements a starting point would be to reduce the cap by a factor of 5 = (500k/100k).  If however the pot drives a cable (which has capacitance) then it could be that leaving the same cap value is better.
That's reduced the range of possibilities to within a factor of 5.  With careful listening your ear should be able to narrow that down.

[PRR]

> lot less duller sound due to not loading down the higher frequencies, like a 100k potentiometer would.

I don't understand this.

[stonerbox]

> lot less duller sound due to not loading down the higher frequencies, like a 100k potentiometer would.

I don't understand this.

Probably my layman terms I guess but to be frank, neither do I. I however do know that with increased resistance in between ground / signal equals more presence and treble.
If I find Hammers original post on 500k potentiometers I'll post it. He explains things better than me.

I want the same amount of high frequencies to pass with a 100k+capacitor as a 500k potentiometer alone would. Could it be calculated?

Without any specific info or measurements a starting point would be to reduce the cap by a factor of 5 = (500k/100k).  If however the pot drives a cable (which has capacitance) then it could be that leaving the same cap value is better.
That's reduced the range of possibilities to within a factor of 5.  With careful listening your ear should be able to narrow that down.

Yeah, I could do it by ear but then I wouldn't learn much. I'm confused about "that leaving the same cap value is better" as I don't use a cap for the 500k pot.

You'd probably want to figure out what frequency you're dealing with....what is cut when you turn down the 500k. Have fun!    LOL...yes, you'd need to know the impedance of what comes after the output pot, that is what's eating the highs.

OR...if you can GUESS at what frequency is getting cut, you can use: 

C=2.74 / 2pi F R    where C = the bright cap you want,  F = the 1/2 boost freq (800Hz - 5kHz typical), R = the pot resistance.  That will get you most of the way there.  No impedance calculation!   This is just a SHELVING FILTER response (you can look that up if need be).

Lastly:  What I do, if a bright cap becomes necessary, is just 'decade' the cap.  Try .1u,  .01u, 1u.   You'll find which is more in the right direction....then you can maybe try .12u.....then .15u, .22u.....no math, thanks!!       In not more than 5 or 6 tries you'll probably get the one you want! 

Damn you Impedance! One day I will actually have to learn what makes you tick and why.. oh well!  Thanks for the equation Gibson, I'll make good use of it!

[GibsonGM]

If you don't want to to it by ear, then use the formula.  Read up on shelving filters (net search).   The formula doesn't use impedance, so you're ok for now!    Just resistance.  But not to say that impedances don't affect the highs, when dealing with a volume pot - they do!     

In the volume pot case, the source impedance - say, the output Z from an opamp - is working with the pot and making a voltage divider with a frequency-dependent component - the highs go first.    You COULD just keep the 500k and put a bright cap on THAT, you know!  Then you'd still have maybe 300k there, plus the cap for highs....I'd try that before going to 100k, because as you turn down, you're now dealing with maybe 30k!      Anyway, the formula will get you going, and so will your ears    

[Rob Strand]

Yeah, I could do it by ear but then I would'nt learn much. I'm confused about "that leaving the same cap value is better" as I don't use a cap for the 500k pot.

You can't actually do better unless you pin down the world around it like what resistive and capacitive loads.
Even after you do that you will find you can only match it at one position on the pot.

Here's how to match a divider:
http://emcesd.com/rcdvdwg.jpg

The response is flat when Tc1 = Tc2.

C1 is the value you are trying to find  so

C1 = C2 * R2  / R1

Suppose you let the pot setting be k and it varies from 0 to 1.   R = pot resistance.
R2 = k*R
R1 = (1-k)R

ie.
R2/R1 = k / (1-k)

If your cable has capacitance it will appear as C2, then you need

C1  = C2 * k / (1-k)

So now the problem is clear if C2 is fixed then C1 must vary with the pot setting k

k       C1/C2
0.1   0.1111
0.3   0.4286
0.5   1
0.7   2.3333
0.9   9

If you set things up to match when the pot is at half position (ie. k = 0.5 and so C1 = C2) you will find the amount that it is off when you move the pot isn't too bad.

These calculations depend on C2.   If you don't know C2 then you can't proceed.   If C2 is the cable capacitance then it depends on the specifics of the cable and the length.  This will vary!

It's a non-idea situation full of variations you can only plonk a point in the middle and rely on the fact even large mismatches are only going to produce smallish deviations.

You get a similar problem with x10 oscilloscope probes,
https://www.elexp.com/Images/Importance_of_X10_Probes.pdf

Here C2 can vary from CRO to CRO and to some extent with the cable used for the probe.  R1 and R2 are fixed so you don't have to deal worry about variations in those.  So in order to get a match for a flat response the probes have an adjuster which trims C1 to match C2.   For a x10 probe R1 = 9*R2, C1 = C2/9.

[stonerbox]

You can't actually do better unless you pin down the world around it like what resistive and capacitive loads.
Even after you do that you will find you can only match it at one position on the pot.

Here's how to match a divider:
http://emcesd.com/rcdvdwg.jpg

The response is flat when Tc1 = Tc2.

C1 is the value you are trying to find  so

C1 = C2 * R2  / R1

Suppose you let the pot setting be k and it varies from 0 to 1.   R = pot resistance.
R2 = k*R
R1 = (1-k)R

ie.
R2/R1 = k / (1-k)

If your cable has capacitance it will appear as C2, then you need

C1  = C2 * k / (1-k)

So now the problem is clear if C2 is fixed then C1 must vary with the pot setting k

k       C1/C2
0.1   0.1111
0.3   0.4286
0.5   1
0.7   2.3333
0.9   9

What an excellent and truly asskickin' answer Rob, thank you!

If you set things up to match when the pot is at half position (ie. k = 0.5 and so C1 = C2) you will find the amount that it is off when you move the pot isn't too bad.

These calculations depend on C2.   If you don't know C2 then you can't proceed.   If C2 is the cable capacitance then it depends on the specifics of the cable and the length.  This will vary!

It's a non-idea situation full of variations you can only plonk a point in the middle and rely on the fact even large mismatches are only going to produce smallish deviations.

You get a similar problem with x10 oscilloscope probes,
https://www.elexp.com/Images/Importance_of_X10_Probes.pdf

Here C2 can vary from CRO to CRO and to some extent with the cable used for the probe.  R1 and R2 are fixed so you don't have to deal worry about variations in those.  So in order to get a match for a flat response the probes have an adjuster which trims C1 to match C2.   For a x10 probe R1 = 9*R2, C1 = C2/9.

I get it. It's a tricky situation where numbers ultimately are not my friend. For some reason I did not think about of things outside of the "box" influencing the frequency response. If I were to be Asperger-picky (no offense to anyone out there) I could just add a EQ or a buffer pedal after to correct any loss of highs in any give situation or cable length.
I'll probably go by ear from here on. I'll record the pedal with the 500k at the sweet spot (full volume) and then swap to a 100 or 250k and high shelve boost until it matches the response/sound of the 500k recording.

[anotherjim]

You won't lose highs due to cable capacitance with any value of output pot that is left on maximum. The point is to avoid adding resistance in series before the cable jack. A lower value pot should perform better at less than maximum in this regard.

You will lose lows if the out pot is too low in value for the DC blocking cap preceding it.

You may lose highs or lows or shift mids if the preceding stage reacts to load on the output. Classic would be a copy of a guitar amp tone stack which was designed to be loaded by a tube control grid and/or 1M volume pot. Cure - keep the 1M volume pot AND add a hi-z buffer to drive the world. The output volume control does not have to be at the end of the signal path.

[stonerbox]

I'll keep your notes in mind when I dive back into the circuit next week or maybe even earlier. We are getting a kick ass Shiba Inu pupper very soon so I'm not really certain of how much time I will have for the wonderful DIY-madness the coming months. Damn I love it so much. Thanks for chiming in Jim!

[Gus]

A link to a thread of a circuit showing what happens with two different values of an output volume control.
http://www.diystompboxes.com/smfforum/index.php?topic=96381.msg838899#msg838899

[stonerbox]

A link to a thread of a circuit showing what happens with two different values of an output volume control.
http://www.diystompboxes.com/smfforum/index.php?topic=96381.msg838899#msg838899

Big difference, maybe not on first glance but a closer look reveals that the 500k performance a lot better. Good solid post.

[Rob Strand]

So in order to get a match for a flat response the probes have an adjuster which trims C1 to match C2. 

It occurred to me most CRO probes will actually add a fixed C1 then place an adjustable C2 in parallel.    Wide-band voltmeters and the inside of CRO's tend to have adjustable C1's.   The idea is still the same.