Possibly Retreading Old Waters?

[Lucius Paisley]

I think I came across a thread here that talked about 250k vs 500k pots, but the discussion had expanded to include a reason/reasons and possibly a graph that explained why a pickup on full volume using a 250k pot is not the same/won't sound the same as a pickup with a 500k pot on half volume.

 

[freefrog]

why a pickup on full volume using a 250k pot is not the same/won't sound the same as a pickup with a 500k pot on half volume.

Because with a 250k full up, you have no resistance between pickup and output but 250k from output to ground.

Now, if you set a 500k volume @ half value (which is not necessarily 5/10, of course), you have 250k from output to ground but also 250k from pickup to output.

A 500k TONE pot set @ half its resistance behaves like a 250k tone control full up. A volume pot doesn't.

I'll share a graph if time permits.

HTH.

 

[Chistopher]

Hopefully this will help you see it. At half volume on a 500k pot you still have 250k inline with the signal, which is what causes your treble to rolloff when you roll the volume down.

 

[Lucius Paisley]

I know a split coil isn't going to be exactly the same as a single coil, but in a humbucker guitar, a split coil will still have a 500k pot acting on it. As the volume is increased, would the behaviour still follow a pattern, such as

250k 500k
1 2
3 6
5 10
7 14
10 20

500k 250k
1 0.5
3 1.5
5 2.5
7 3.5
10 5

Which I guess if it were represented in a graph would look like?



Am I thinking along the right lines here?

 

[beaubrummels]

A 500k TONE pot set @ half its resistance behaves like a 250k tone control full up.

That doesn't sound right. When any tone control is full up, hardly any of the signal is going through the resistor track. When any tone control is @ half, half the signal is going through the resistor track.

 

[Spaghetti Bolo]

That doesn't sound right. When any tone control is full up, hardly any of the signal is going through the resistor track. When any tone control is @ half, half the signal is going through the resistor track.

A tone control is wired as a rheostat, literally a variable value resistor.

 

[freefrog]

That doesn't sound right. When any tone control is full up, hardly any of the signal is going through the resistor track. When any tone control is @ half, half the signal is going through the resistor track.

What SpaghettiBolo said.
Consequently, a pickup "sees" 250k between it and the tone cap going to ground, with a 250k pot full up as well as with a 500k set at half of its resistive value.
Therefore these two solutions give strictly the same result tonally speaking...

 

[freefrog]

Lucius Paisley, I'd be glad to help but I don't understand your last question. What would you want to do / to solve exactly?

 

[Lucius Paisley]

Lucius Paisley, I'd be glad to help but I don't understand your last question. What would you want to do / to solve exactly?

I want to see if a 500K volume pot is twice as "loud" as a 250K or if I've got my (metaphorical) wires crossed.

I saw this on Reddit...

https://guitar.com/wp-content/uploads/2016/07/Pic-1-Resonance-and-Pot-Value-Graph.jpg



Maybe this answers my question insofar as a pictorial representation is concerned, but I'm not 100% sure what I'm looking at.

 

[freefrog]

I want to see if a 500K volume pot is twice as "loud" as a 250K or if I've got my (metaphorical) wires crossed.

I saw this on Reddit...

https://guitar.com/wp-content/uploads/2016/07/Pic-1-Resonance-and-Pot-Value-Graph.jpg



Maybe this answers my question insofar as a pictorial representation is concerned, but I'm not 100% sure what I'm looking at.

As far as I can see, this graph shows the amplitude and level of a resonant peak with different resistive loads.

IOW, the blue line @ the bottom would be what you'd obtain from a 5k volume pot full up. A 250k volume pot set to leave 5k between output and ground would also put 245k between pickup and output, so the resonance would be different in amplitude, level... and frequency (depending on the RLC specs of the pickup itself).

 

[freefrog]

Botched 5spice sim below.

Upper screen = variation of volume pot VALUE from 1M to 5k (but this volume pot is always full up).

Bottom pic= variation of volume pot SETTING from 1M full up to 1M (1000k) set @ 0,005 (=5k).

Different enough to require different vertical scales...

FWIW.

 

[Lucius Paisley]

Botched 5spice sim below.

Upper screen = variation of volume pot VALUE from 1M to 5k (but this volume pot is always full up).

Bottom pic= variation of volume pot SETTING from 1M full up to 1M (1000k) set @ 0,005 (=5k).

Different enough to require different vertical scales...

FWIW.

Wow. SPICE. I haven't heard that name in years. I used to have a copy of SPICE 3. It's probably still on a hard drive somewhere.

Looking at the graph I posted, at 4dbV, the line differences between 100k to 200k and 200k to 500k look very similar. However, the difference then between 500k to 1M is approximately half of the previous two. I really wish I knew the math behind all this. Because THAT is interesting.

 

[freefrog]

Looking at the graph I posted, at 4dbV, the line differences between 100k to 200k and 200k to 500k look very similar. However, the difference then between 500k to 1M is approximately half of the previous two. I really wish I knew the math behind all this. Because THAT is interesting.

From 1M to 100k, pots values mostly affect the resonant frequency, which is usually beyond any fundamental note. So, it affects brightness much more than output.

It's so true that the action of a tone pot is mostly resistive from 10/10 to 3.5/10, typically. The tone cap really enters in the game and reshapes the resonance only when the tone pot is set at a very low resistance (albeit the presence of a tone cap can be detected in the harmonics much sooner and even with the pot full up in some situations... but that's another question).

Somewhere under 100k of resistive load (with a volume pot measuring less than 100k), the overall output level starts to drop, affecting fundamental notes and harmonics altogether.

Explaining all this with maths quickly becomes very complex, since output level not only depends on LRC specs of the pickup + pot value + input impedance + capacitive load of the wiring, but also on magnetism, obviously...

Maybe you'll find something concretely useful in the video below? Doesn't come from me but I've found it meaningful (it also illustrates how the Q factor diminishes as pot value decreases, making the resonance less pointy/narrow and the sound therefore progressively more open/ less nasal).

https://youtu.be/HJKYIWGl_KI?si=jXe_jyFV9COAmuoV

 

[beaubrummels]

What SpaghettiBolo said.
Consequently, a pickup "sees" 250k between it and the tone cap going to ground, with a 250k pot full up as well as with a 500k set at half of its resistive value.
Therefore these two solutions give strictly the same result tonally speaking...

How does it see 250k to ground on 10 when the wiper is almost direct connected its own the input and the pot is not grounded?

 

[Chistopher]

I'll draw another equally wonderful diagram of my first to explain it a lil better.

 

[freefrog]

In answer to beaubrummels, let's try again to be clear. If it's not the case, attribute that to my health issues.

A tone pot is grounded through the tone capacitor, which is schematically a frequency-dependent resistor.

Typically (with conventional 100nF/47nF/33nF/22nF values) this tone cap / frequency-dependent resistor avoids to the signal to go to ground until a few hundreds of Hz but it doesn't block at all what is beyond the cutoff/corner/knee frequency : high mids and high frequencies go to ground as the resistance of the tone pot decreases and that's precisely why it behaves like a tone pot...

Obviously, the resonant peak of a passive pickup is typically located in the high mids or high range, somewhere between 1500hz and a few thousands of Hz.

So, it's directly sensitive to the resistance between pickup and ground, this resistance being due... to the pots.

A second resonant peak, much lower and due to the tone capacitor, appears only when the resistance of the tone pot is reduced to nothing (or almost), connecting directly (or almost) the pickup to ground through the cap and changing its properties of resonant filter...

I like the following pic when it comes to illustrate what I mean: https://nextgenguitars.ca/product_images/uploaded_images/tonepotfreq.png

Let's also share below a pic showing the electrically induced resonant peak of a real pickup with vs without tone pot. It's an old test (from 20 years ago) but that's what I had at disposal immediately and it clearly shows what is going on with a Strat bridge pickup VS a neck or mid one, for instance...



​And thx to Christopher for his contributions below or before.

 

[Chistopher]

Hopefully this helps you visualize it a little better. 250k resistance to ground is near as makes no difference to infinite for a passive guitar signal. Cable has capacitance as well, but I omitted it to make it easier to understand. V_R1 and V_R2 represent the voltage drop across each resistor, ie how much signal goes through each.

Also worth noting in my diagrams, in the first one I used a rectangle to represent the resistances as impedances, in this second diagram I used the symbols for resistors and capacitors

 

[beaubrummels]

Hopefully this helps you visualize it a little better. 250k resistance to ground is near as makes no difference to infinite for a passive guitar signal. Cable has capacitance as well, but I omitted it to make it easier to understand. V_R1 and V_R2 represent the voltage drop across each resistor, ie how much signal goes through each.

Also worth noting in my diagrams, in the first one I used a rectangle to represent the resistances as impedances, in this second diagram I used the symbols for resistors and capacitors

I think that's the opposite. On 10, the signal is not going through the resistor (or very little of it goes through the resistor).

 

[beaubrummels]

In answer to beaubrummels, let's try again to be clear. If it's not the case, attribute that to my health issues.

A tone pot is grounded through the tone capacitor, which is schematically a frequency-dependent resistor.

Typically (with conventional 100nF/47nF/33nF/22nF values) this tone cap / frequency-dependent resistor avoids to the signal to go to ground until a few hundreds of Hz but it doesn't block at all what is beyond the cutoff/corner/knee frequency : high mids and high frequencies go to ground as the resistance of the tone pot decreases and that's precisely why it behaves like a tone pot...

Obviously, the resonant peak of a passive pickup is typically located in the high mids or high range, somewhere between 1500hz and a few thousands of Hz.

So, it's directly sensitive to the resistance between pickup and ground, this resistance being due... to the pots.

A second resonant peak, much lower and due to the tone capacitor, appears only when the resistance of the tone pot is reduced to nothing (or almost), connecting directly (or almost) the pickup to ground through the cap and changing its properties of resonant filter...

I like the following pic when it comes to illustrate what I mean: https://nextgenguitars.ca/product_images/uploaded_images/tonepotfreq.png

Let's also share below a pic showing the electrically induced resonant peak of a real pickup with vs without tone pot. It's an old test (from 20 years ago) but that's what I had at disposal immediately and it clearly shows what is going on with a Strat bridge pickup VS a neck or mid one, for instance...



​And thx to Christopher for his contributions below or before.

That's showing the resulting resonant peaks, but it's not showing the actual resistance the pickup 'sees'.

I'll have to open a guitar and measure the tone pot to check.

 

[Chistopher]

Assuming R2 is 1 ohm and V1 is 1v, at 10, V_R1 = 1/250,001 = 000004v . So at 10, almost no treble gets cut to ground. From this we can tell that V_R2 will be very close to 1v, because V_R1 and V_R2 must add up to Vin.

Note that the numerator is the resistor you are not trying to find the voltage drop of.
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