Potentiometer swap to get more brightness question

[micguy]

In addition to changing the pots you should also consider changing the tone capacitor.

The value of the tone capacitor is meaningless with the tone control at full. You can literally use a capacitor with 1000 times the normal value, and it'll sound the same. If you want to change what it sounds like when the tone pot is at minimum, then by all means change the tone cap, but as the OP wants more brightness, I'm assuming he's got the tone control dimed.

Here's a plot of a circuit simulation: It's a G&L MFD P pickup (similar in impedance to most P pickups), and it's loaded with a volume control , tone circuit, a cable, and a 1 Meg input impedance amplifier. The green curve is the stock circuit - .047 uF cap. The red circuit is with a 47 uF cap - 1000 times bigger. Most of what you see is red - that's because the red curve lies exactly on top of the green curve.

 

[Killed_by_Death]

The capacitor won't even make a difference with a 500k Ohm treble cut pot fully clockwise.

The capacitor doesn't impart any tones to the timbre. In fact it's the other away around, it subtracts them, but it can't with that 500k Ohms in the way.

If you don't like how the 500k Ohm volume pot affects the timbre when rolled back, you just don't roll it back as far as you rolled back your 250k Ohm volume pot.

 

[Inconnu]

Lots of comments! Thanks!

I’ll try stuff in the weekend:
- bypassing the tone
- plugging a pu straight in the jack.

Both are free and will determine the rest (plus it’ll get me some soldering practice!)

 

[B-Mac]

Hi!

I’m trying to get more brightness out of my Lace Alumitone Bassbars in humbucking mode (bright enough in single coil mode).

What would give a better result between:
1) replace the 250k volume pots for 500k or 1m ?
2) use a no load tone knob (or install a tone bypass switch) ? - I’m currently using a Fender TBX tone by the way.
3) other options?

Although you don't have an Ibanez Artcore, you might benefit from this thread
IBANEZ ARTCORE TONE PROBLEM SOLVED !!!

 

[AdamR]

Remember going to the 500K volume pot will also make it brighter in single coil mode.

 

[briandavismurph]

The unit is Ohms, & change them all for 500k Ohm.
Boosting treble on an active pre-amp just creates hiss.

 

[briandavismurph]

The unit is Ohms, & change them all for 500k Ohm.
Boosting treble on an active pre-amp just creates hiss.

Do you think this will work on a precision bass and where is the best place to buy the 500 a pass

The unit is Ohms, & change them all for 500k Ohm.
Boosting treble on an active pre-amp just creates hiss.

If I’m looking for more brightness out of my precision bass do you think this would work where is the best place to buy 500 K pots

 

[Killed_by_Death]

The unit of measure is Ohms, MojoTone has them:

[Invalid or Expired Link Removed]

I'd recommend Mouser.com, but they're tougher to navigate & seem to have more constraints on quantity lately.

 

[Matthijs]

It wouldn't. With a 250k Ohm volume pot you're still bleeding some highs to ground.

If you want to be sure there is some brightness in the pickups, wire it straight to the output jack.
IF it's not bright then, it will never be bright w/o a pickup change or a pre-amp.

The bleed would be exactly the same as with 2 500k pots for tone and vol, wich amounts to 250k in total. So if you want to know what that sounds like thats the simpelest way. Conecting to ground would be one step brighter.

 

[Wizzu]

2 500k pots for tone and vol, wich amounts to 250k in total.

Uh? What?

Conecting to ground would be one step brighter.

What do you mean exactly by "connecting to ground"?

 

[Matthijs]

The total resistance if two 500 k pots in parallel is 250k. Equal to having only one 250k pot in the circuit.

The conected to ground bit is a brainfart, i meant to output.

 

[Wizzu]

The total resistance if two 500 k

Oh. I was wrongly reading "2500k" (two thousand and five hundreds) instead of "2 500k" (two five hundreds). I'm used to spaces for numbers groupings... I was wondering where you found 2500k guitar pots

Nevertheless, despite having wired and/or modified dozens of guitars and basses, I still can't get my head around the logic (most probably correct but out of my intellectual reach) behind considering both tone and volume pots as a combined parallel resistive load in a guitar. I tend to consider both as totally separate loads when it comes to the way they impact frequency response. I need an engineer to educate me. Oh, well.

The connected to ground bit is a brainfart, i meant to output.

Aaah, certainly makes more sense this way

 

[Matthijs]

Typing this stuff on the fly on my mobile doesn’t really help

 

[Inconnu]

The total resistance if two 500 k pots in parallel is 250k. Equal to having only one 250k pot in the circuit.

The conected to ground bit is a brainfart, i meant to output.

On that logic, having 2 x 250k pots would equal 125k, right?

The bass I’m working on has 2 volumes and 1 tone, by the way... but the tone is a Fender TBX which is supposed to be « no load » in the center detent...

 

[Inconnu]

Remember going to the 500K volume pot will also make it brighter in single coil mode.

Which could be dealt with by adjusting the tone knob?

 

[Killed_by_Death]

The bleed would be exactly the same as with 2 500k pots for tone and vol,

Except this is a VVT configuration, so disconnecting one treble cut pot does not get you to where it would be with three 500k Ohm pots.

 

[James Collins]

Oh. I was wrongly reading "2500k" (two thousand and five hundreds) instead of "2 500k" (two five hundreds). I'm used to spaces for numbers groupings... I was wondering where you found 2500k guitar pots

Nevertheless, despite having wired and/or modified dozens of guitars and basses, I still can't get my head around the logic (most probably correct but out of my intellectual reach) behind considering both tone and volume pots as a combined parallel resistive load in a guitar. I tend to consider both as totally separate loads when it comes to the way they impact frequency response. I need an engineer to educate me. Oh, well.
Aaah, certainly makes more sense this way

Both pots are wired to ground. So if you turn the volume and tone all the way down, it is like creating an extra circuit to ground across the entire value of the resistor in the pot. Ignore the fact that you can change the resistor values and just imagine the situation where volume is maximum and drop the capacitor out for simplicity. The circuit would look like this:

Voltage = Current x Resistance
Current = Voltage / Resistance

In series the current is the same so you end up with something like this:
V(total) = IR1 + IR2 + IR3 + ... = I * (R1 + R2 + R3 +...)
R(total) = (R1 + R2 + R3 + ...)

Since the current is the same along the wire, the individual resistances are equal to the their sum. So two 500k pots in series to ground would be 1,000k to ground.

In parallel, voltage is the same across all of the parallel circuits. The currents change across each of the parallel circuits so you end up with something like this:
I(total) = V/R1 + V/R2 + V/R3 + ... = V * (1/R1 + 1/R2 + 1/R3 + ...)
1/R(total) = 1/R1 + 1/R2 + 1/R3 + ...

So for two 500k pots wired in parallel:
1/R(total) = 1/500k + 1/500k = 2/500k = 1/250k

 

[Wizzu]

Ignore the fact that you can change the resistor values and just imagine the situation where volume is maximum and drop the capacitor out for simplicity. So for two 500k pots wired in parallel:
1/R(total) = 1/500k + 1/500k = 2/500k = 1/250k

Thanks. This part, I get.

My mental thickness/density actually kicks in when the capacitor is present. And more precisely, when I try to consider - as I'm told again and again to do- that both pot values, combined, impact the frequency response from the RC low-pass filter (tone pot + capacitor), as if the volume pot resistance changed the way this filter behaves. THIS I don't get. Also, I don't hear it... With most "hi-fi-like" (high pitched resonance/cut) pickups -for instance Seymour Duncan Duckbuckers-, changing the tone pot (250→500→infinity) has of course a very audible impact. But changing the volume pot, well sorry, I never heard any difference on the way the RC (tone pot + capacitor) behaves. High values for the volume pot does seem to slightly increase highs (very subtly) in certain configs, but it does zero to what I hear when using the tone pot... seems to be exactly the same. Must be my ears.

Maybe there is a HUGE misunderstanding/confusion from my part when people more knowledgable than me, insist that I have to consider the combined resistivity of the pots when designing the circuit. Since not considering it doesn't seem to hamper the results for myself or my customers, I'm still (shamefully) "ignoring" it entirely... until someday, I see the light, maybe...

 

[James Collins]

The difference between the tone pot and the volume pot is that when you turn volume down, the resistance stays in the circuit. When you turn down the tone, the resistance leaves the circuit. Depending on how you wire the volume, it may affect the highs more.

 

[Inconnu]

Tried something today (free and easy) : unplugged the tone knob. To my biggest surprise, it made the sound clearer and the output is a little bit higher. I don’t know why, but it does it. The only thing is that in « single coil » mode, the sounds gets a little bit harsh. The tone knob is beneficial in single mode. So, I’ll try turning my (rarely used) bass cut switch into an on/off switch for the tone. All free, so worth trying out!