UtilityBeltFX
Well-known member
Sorry for the somewhat elementary question, but if I wanted to forego using a pot on a pcb, and just wire the pads so it's as if the pot is always turned all the way down to 0, how would I go about that?
Omitting a pot so its hardwired at 0?
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UtilityBeltFX
Well-known member
Thank you! It was bizarrely difficult to find this answer on the internet.
Robert
Reverse Engineer
Alternatively, if you want to wire the pot at maximum you'd place the jumper across pads 2 and 3.
If you want the pot at noon you'd install a resistor across pads 1 and 2, and another resistor from pads 2 and 3. (Both resistors 1/2 value of the original pot)
You can hardwire any position by changing the ratio of the two resistors, the sum of the two should always be equal to the value of the pot.
If the pot is wired as a variable resistor (only two lugs used, or two lugs connected together in the schematic) you only need a single resistor.
(50% of original pot value would be at 50% of rotation/noon, assuming the pot is linear)
If you want the pot at noon you'd install a resistor across pads 1 and 2, and another resistor from pads 2 and 3. (Both resistors 1/2 value of the original pot)
You can hardwire any position by changing the ratio of the two resistors, the sum of the two should always be equal to the value of the pot.
If the pot is wired as a variable resistor (only two lugs used, or two lugs connected together in the schematic) you only need a single resistor.
(50% of original pot value would be at 50% of rotation/noon, assuming the pot is linear)
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swelchy
Well-known member
yall know i'm a big fan of the zero resistor
You just need to visualise how a pot works. Pins 1 & 3 are fixed and pin 2 is connected to a wiper which goes from one end to the other. If a pot is turned up full it means pins 2 & 3 are virtually connected. All the way down to zero means pins 1 & 2 are virtually connected. I say virtually because there is usually a tiny bit of resistance between them but it's not enough to worry about usually.
So in a volume pot where pin 1 goes to ground and pin 2 goes to output when you turn the knob down (CCW) you are sending the signal to Ground.
So in a volume pot where pin 1 goes to ground and pin 2 goes to output when you turn the knob down (CCW) you are sending the signal to Ground.
Feral Feline
Well-known member
@UtilityBeltFX — what's the circuit? What do you want to minimise?
Note that's only true for a linear taper ("B") pot. For logarithmic or audio taper ("A", "C", "W"), the noon/50% pot position is not 50% of the pot resistance.
Robert
Reverse Engineer
Thanks for pointing that out. I sort of vaguely mentioned it at the end there but wasn't clear that it applied to the whole post.
Ginsly
Well-known member
This is somewhat related, so thought I'd ask in this thread... There's a 500K Vol control at the end of a fuzz I've been playing with. I'm considering adding more to the end of the circuit, but I'm not sure exactly how I would omit the Vol control and simply connect the full-on Output signal to the next stage. It's not a pcb, but a stripboard layout. The Output comes from Lug 2, Lug 1 goes to ground and Lug 3 connects to Q2's collector via a 22n cap. Not sure how this would translate without the pot!
Ginsly
Well-known member
Ah! That simple, huh? I figured the 500k resistance of the pot might have some bearing on Q2 (like in a Fuzz Foundry), but I guess not- thx man!
Feral Feline
Well-known member
Try this, too:
Set the pot so you get unity gain (or a bump or attenuation — whatever you want),
then measure the resistance of the pot and ...
stick a resistor of that value (closest) in between C7 and the next circuit.
ie maybe you want to overdrive the next circuit or ... whatever. After setting unity-volume, see how it interacts with the rest of your circuit.
