Are Treble Bleeds Transparent at 10?

Discussion in 'The Custom Shop' started by Chistopher, Dec 1, 2017.

Do you think it makes a difference?

  1. Yes

  2. No

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  1. Chistopher

    Chistopher Junior Member

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    On another forum there is debate on whether treble bleeds (specifically a cap wired in series with a resistor a la Kinman) affect the tone of the guitar when the volume is set to 10. Plenty of well respected members with proven track records are on both sides of the issue. What do y'all think around here?
     
  2. rabidhamster

    rabidhamster Senior Member

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    Sure it does, but not everyone has that good of hearing anyway so whether it makes any difference to you is the question

    Lots of people that can't hear the difference can't hear those "mosquito" ring tones either....
     
  3. Chistopher

    Chistopher Junior Member

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    So you're saying it does, but it's completely inconsequential?
     
  4. guitarjoem

    guitarjoem Double Platinum Supporter MLP Vendor

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    With a no-load pot it would be switched completely out, otherwise there is a potential for some effect.
     
  5. Chistopher

    Chistopher Junior Member

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    If I used a no-load volume pot, then I'd have the issue of it being too bright. If I were fine with a no load pot I'd simply use a no load tone, they remove the interaction with the tone pot when they are clicked out of the circuit. The issue I keep hearing about though is the treble bleed affecting the lower frequencies when the pot is at 10.

    And before anyone says "you can't have a no load volume, it would cut the volume when you turn it to ten", it can be done with a dual gang pot.
     
  6. DarrellV

    DarrellV Likes > Posts Silver Supporter Premium Member

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    Like this?
    [​IMG]
     
  7. DarrellV

    DarrellV Likes > Posts Silver Supporter Premium Member

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    The answer is easiest explained if you remember that electricity is always going to take the path of least resistance.

    Also comparing water to electron flow makes the illustration easier too.

    In this drawing our valve 'pot' is wide open. A small amount leaks to ground through the fixed valve under the 'Out' pipe.
    This represents our pot value resistance. Usually 500K or something similar. Even wide open it always bleeds a small amount of current to ground as this fixed valve is doing.

    The bleed resistor is shown as another fixed valve at the top of the drawing. It is connected across the valve between the source (pickup) and the destination (output).

    Something else to remember at this point is that it takes a difference in pressure (volts) for any flow of water or electricity to happen.

    In this case the valve is wide open posing no resistance on the flow going through the valve (pot). So pressure is about equal on both sides of the valve, or slightly less on the output side because of the loss to ground through the pot.

    The point being in this condition a small amount of water would still flow through the bleed valve, but it would be immediately mixed back into the larger stream coming through the wide open valve. The effects would be insignificant. There would be a minuscule amount of current flow through the circuit, but with near equal pressure on both sides of the resistor, plus the impedance of the capacitor, it would be negligible in amount and effect.
    water valve.jpg

    You can see here too, that as the valve is closed it builds up more pressure on the input side.

    This imbalance causes current to flow through the resistor and capacitor more vigorously as there is a greater push behind it.

    If you imagine a water cooler circuit instead of treble bleed you can get a better idea.

    The incoming water is already heated to warm.

    The chiller works at a fixed rate, but at wide open very little water flows through it due to a lack of pressure, and what little does go through is mixed back into the main stream of warm water. Will it be enough for you to feel a difference? Not likely.

    Turn that valve down and it forces a greater proportion of warm water through the chiller circuit to be mixed back into the reduced flow coming out of the valve. The mix ratio has changed. You will feel it now!

    Hope this helps and doesn't sound completely stupid. I gave it a shot FWIW! LOL!
     

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