Pots type, connection and sweep

Started by Plexi, October 17, 2017, 12:42:43 PM

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Plexi

Hi forum!
Time ago I've read something about it... (I don't remember where...)

Tweaking some mods over the Voodoo Lab OD, connecting backwards the drive pot, found that the sweep can change depending the "input" lug you use.
Placed an A500K pot there, from lug 2, to ground on lug 3: we could say that we have  a 100% log sweep.

BUT! if I use any other connection, like in over lug 3, and lug 2 to ground, it isn't a log sweep?
Shouldn't be the same in all connections?
Bad/cheap pot sweep?
To you, buffered bypass sucks tone.
To me, it sucks my balls.

Fancy Lime

Hi Plexi,

I'm not 100% sure I understand what you mean. But let's go through the configurations of a log pot (lugs numbered left to right in top view):
1=gnd, 2=out, 3=in: log pot as voltage divider
1=in, 2=out, 3=gnd: rev log pot as voltage divider working in reverse
1= gnd, 2=in, 3=out: rev log variable resistor to ground (forming a voltage divider with the preceding in-line resistance), log resistor in series after that
1= out, 2=in, 3=gnd: log variable resistor to ground (forming a voltage divider with the preceding in-line resistance), rev log resistor in series after that
1=in, 2=gnd, 3=out: not useful as it is without further resistors or capacitors because it shorts all audio to ground
1=out, 2=gnd, 3=in: not useful as it is without further resistors or capacitors because it shorts all audio to ground

Does that help at all?

Cheers,
Andy
My dry, sweaty foot had become the source of one of the most disturbing cases of chemical-based crime within my home country.

A cider a day keeps the lobster away, bucko!

Plexi

Thanks Andy!

To clarify a bit more: I'm using as variable resistor.
Gain pot here:
http://2.bp.blogspot.com/-oa_3QLG4vFA/T07uGkD9vTI/AAAAAAAAAAk/pYOSpuJU3e8/s1600/VoodooLabOverdrive.png

So:
1=in, 2=ground: log sweep
2=in, 3=ground: log sweep
If I use a A500K or B500k, the gain sweep increasing is subtle.

3=in, 2=ground: rev log sweep?
If I use a A500K this way, the gain sweep increase abruptly in the las 10% of the sweep.
To you, buffered bypass sucks tone.
To me, it sucks my balls.

Fancy Lime

#3
Hi Plexi,

just to avoid the confusion I used to have with this for novices reading this: I am referring to lug numbering as in this picture:
http://www.tdpri.com/attachments/potentiometers-front-back-png.193158/

Quote1=in, 2=ground: log sweep
This means you have minimum resistance (theoretically 0Ω, in reality a bit more) when the pot is all the way to the left. In the middle position you have about 10% and all the way to the right you have 100% resistance. This is log resistance operation with reverse travel (max gain is left).

Quote2=in, 3=ground: log sweep
This means you have 100% resistance when the pot is all the way to the left. In the middle position you have about 90% and all the way to the right you have minimum resistance. This is rev log resistance operation with normal travel (max gain is right).

So these two configurations are very different.

Quote3=in, 2=ground: rev log sweep?
No, this is the same as the second configuration. Resistance is the same back and forth.

For a variable-resistor-to-ground type of gain control I find rev log pot wired like your first option to be the best characteristic. A linear pot (type B) is slightly worse in this position than a rev log (type C). A normal log pot (or audio taper, type A) does almost nothing for most of the way and then makes a big jump in the last 10% of the way; not very practical.

If you do not have a rev log pot, you can take a linear one (type B) and add a resistor. To get 500k "almost rev log", take a B1M pot and connect it like this:
1 to C3, 2 to 1 and to a new 1M resistor, 3 to ground and to the other end of the new 1M resistor. To get something closer to real reverse audio taper you would need a B5M pot and a 550k resistor, but B1M and 1M is a good start. B2M pot with 680k resistor works as well. You can use this program to calculate these curves:
http://www.diystompboxes.com/smfforum/index.php?topic=118649.msg1105106#msg1105106


Hope this helps and good luck,
Andy
My dry, sweaty foot had become the source of one of the most disturbing cases of chemical-based crime within my home country.

A cider a day keeps the lobster away, bucko!

Plexi

Thanks Andy!
I'll try again over proto and see how it goes
To you, buffered bypass sucks tone.
To me, it sucks my balls.

Plexi

Some update here:
The B1M+1M resistor en parallel trick isn't working: almost the same sweep as with the B500K pot.

Avoiding the direction of the increasing/decreasing ohms and gain; the only way I've found that works great, is to enter to pin1, and ground to pin 2 from an A500K pot.
I could change the label from "gain" to "cleaning"  ;D

I'm waiting to arribe some C500K pot.
To you, buffered bypass sucks tone.
To me, it sucks my balls.

Fancy Lime

Hi Plexi,

if you can get a C500k pot, that is the simplest and best solution. The B1M + 1M variant is between a B500k and a C500K pot in terms of characteristic, but closer to linear than to rev log. B2M + 680k would be almost half way. B5M + 550k would be very close to C500k. The whole trick with parallel resistors is mostly interesting if you need rev log characteristic with a value that is not available in rev log (or more expensive), which is sometimes the case because rev log pots are much less common in most applications than rev log.

Cheers,
Andy
My dry, sweaty foot had become the source of one of the most disturbing cases of chemical-based crime within my home country.

A cider a day keeps the lobster away, bucko!

Plexi

Sure..! Thanks
That's why I'll try one.

My doubt was why the speed change when the pot is backwards.
I should try another one, or test carfully the values that make the sweep with a multimeter
To you, buffered bypass sucks tone.
To me, it sucks my balls.