Why Do 250k Pots roll off more treble than 500K?

[triskadecaepyon]

Can anyone explain why 250k pots roll off the treble more than 500k pots?  A scientific explanation would be nice!  I really don't understand why it does that.

[Paul Marossy]

It's pretty simple - a 250K pot has a maximum of 250K resistance to ground vs. a 500K pot which can have a maximum 500K of resistance to ground. This means that a 250K will dump more of the high frequencies to ground than a 500K will. As far as why exactly this is so, I honestly don't know. I just know that this is how it is.  

[zachary vex]

if you're talking about pickup loading, that's pretty easily explained.  pickups are inductors (with some capacitance) besides being transducers.  the first resistance that the pickup encounters (to ground) is its initial load... it forms an LR combination which must follow the rules for frequency response.  the heavier the load (lower the R), the lower the overall frequency response, by simple math.

you could think of it this way... the system is an electrical generator.  the smallest movements of the strings are the high frequencies.  they barely generate any current in the pickup, so they are the weakest part of the signal.  when you load down the pickup, they're the first things to go.

one thing to keep in mind about pickup loading... it's very important to use the proper load on a pickup to get the classic tone associated with the guitar type/pickup type you are using.  using the wrong loading will change the resonance of the pickup, sometimes in a very significant way, and the flavor of the guitar can be damaged in ways much worse than losing some of the top end... also, the tone control may not have the same "howl" when turned down if you use the wrong loading.

[object88]

In a high-pass filter, you have a cap followed by a resistor to ground (as compared to a low-pass filter, which is a resistor followed by a cap to ground).  Caps block "low" frequencies, and pass "high" frequencies.  Of course, low and high are arbitrary terms... so what determines them?  That's where some math comes in, and I get really fuzzy.  So, I'll lift a little text from Everything2...

Filters are compared and examined by means of a "transfer function". A transfer function is simply a ratio of the output voltage (resulting signal), to the input voltage (original signal). The transfer function, or ratio of output to input, for the circuit in Figure 2

• is given by the following equation:
  
     H(ω) = (jωRC)/(1 + (jωRC)) = Vo/Vs
  
  The derivation of this formula is not difficult, but involves some basic circuit analysis. For information on how to derive this, check voltage divider (replace R1 with C = jωC -- This is because capacitors are complex circuit elements). ω = 2 * π * f, where f is the frequency and π is just Pi, or 3.14. This is more for convenience than anything. H is convention for transfer function, and is a function of ω. H is roughly equivalent to gain, as mentioned in Figure 1. This basic transfer function describes the functionality of this highpass circuit. Plotting H versus ω will provide a graph similar to Figure 1, but with much less of a hard corner at the cutoff frequency. The graph will take a much more gradual slope at the cutoff frequency. Note, that in the transfer function equation, if we evaluate for ω = 0 (frequency = 0), the gain (H) is 0 (or 0/(1+0) ). If we set ω = infinity, then the gain (H) becomes 1 (or infinity/(1+infinity) ).
  
  As was mentioned before, the actual graph of H will look more like a gradual slope, and less like Figure 1. Go ahead and graph it to see.
  
  The cutoff frequency (ωc)can also easily be determined:
  
     ωc = 1/(RC)
  [/i]
  
  
• Figure 2 is a simple RC circuit, with nothing more than a voltage source (Vs), a cap, and a resistor to ground.  Vo is taken across the resistor.
  
  If you can make perfect sense of that, well, you're far more clever than I.  Like Paul Marossy, I just know that's how it is.  

[triskadecaepyon]

Well, to explain things....

When I took out my 500K pots and replaced them with 250K pots, my Texas Special Pickups sounded like they lost all life in them.  I lost considerable highs and mids  I really hope that not all 250K pots do this much damage to your tone!  I think because their %  rating of error (i.e. like how you see 250K 40%)  that I am getting this... Most likely I'll go back to 500K.  Do most people use 500K with texas specials?

[Ed G.]

Well, to explain things....

When I took out my 500K pots and replaced them with 250K pots, my Texas Special Pickups sounded like they lost all life in them.  I lost considerable highs and mids  I really hope that not all 250K pots do this much damage to your tone!  I think because their %  rating of error (i.e. like how you see 250K 40%)  that I am getting this... Most likely I'll go back to 500K.  Do most people use 500K with texas specials?

Hotter pickups generally have more winds of coil. The more winds of coil, the more 'middy' the pickup gets, with less highs. By changing to the 500k pots, it compensated somewhat for the higher output and reduced highs of the Texas Specials.
I personally don't care for the sound of the texas special pickups. They went for, and got, the sound of an overwound strat pickup. Big deal. That's not the sound of SRV, at least not in my opinion.
Cesar Diaz, SRV's guitar tech, said he didn't like strat pickups to be higher than 6K of resistance or it goes to mud. Here's the article:
http://www.tonequest.com/ray.htm
Now, some pickups are designed to be used with higher resistance pots. The Bill Lawrence 500XL (as used by Dimebag Darrell) is an unbelievably hot pickup but it has lots of highs due to his exceptional design, and Bill Lawrence recommends 500k pots to be used with it.

[Satch12879]

In a high-pass filter, you have a cap followed by a resistor to ground (as compared to a low-pass filter, which is a resistor followed by a cap to ground).  Caps block "low" frequencies, and pass "high" frequencies.  Of course, low and high are arbitrary terms... so what determines them?  That's where some math comes in, and I get really fuzzy.  So, I'll lift a little text from Everything2...

Filters are compared and examined by means of a "transfer function". A transfer function is simply a ratio of the output voltage (resulting signal), to the input voltage (original signal). The transfer function, or ratio of output to input, for the circuit in Figure 2

• is given by the following equation:
  
     H(ω) = (jωRC)/(1 + (jωRC)) = Vo/Vs
  
  The derivation of this formula is not difficult, but involves some basic circuit analysis. For information on how to derive this, check voltage divider (replace R1 with C = jωC -- This is because capacitors are complex circuit elements). ω = 2 * π * f, where f is the frequency and π is just Pi, or 3.14. This is more for convenience than anything. H is convention for transfer function, and is a function of ω. H is roughly equivalent to gain, as mentioned in Figure 1. This basic transfer function describes the functionality of this highpass circuit. Plotting H versus ω will provide a graph similar to Figure 1, but with much less of a hard corner at the cutoff frequency. The graph will take a much more gradual slope at the cutoff frequency. Note, that in the transfer function equation, if we evaluate for ω = 0 (frequency = 0), the gain (H) is 0 (or 0/(1+0) ). If we set ω = infinity, then the gain (H) becomes 1 (or infinity/(1+infinity) ).
  
  As was mentioned before, the actual graph of H will look more like a gradual slope, and less like Figure 1. Go ahead and graph it to see.
  
  The cutoff frequency (ωc)can also easily be determined:
  
     ωc = 1/(RC)
  [/i]
  
  
• Figure 2 is a simple RC circuit, with nothing more than a voltage source (Vs), a cap, and a resistor to ground.  Vo is taken across the resistor.
  
  If you can make perfect sense of that, well, you're far more clever than I.  Like Paul Marossy, I just know that's how it is.  

Come on, it's not that hard.  It's basic circuit analysis with a little periodic math thrown in.  Try non-linear modal analysis of structures under dynamic loading.  That's a challenge

[Paul Marossy]

Thanks for that easy to understand explanation Mr. Vex. Pickups have always been mysterious to me, but I am learning more about how they work as I go on (meaning this sort of thing).