Pot tapering

[verbos]

[quote author="alk509"][quote author="AMZ-FX"]you want the inverse anti-log response![/quote]

No, you want a reverse log law.

[quote author="AMZ-FX"]Here is the Omeg page of pot responses: http://www.omeg.co.uk/lawe16ct.htm[/quote]

It's important to remember that in America the letter designations for pot laws are different: "A" law is logarithmic, "B" is linear and "C" is reverse logarithmic.

Peace,
Al.[/quote]

All this anti-log, reverse log, inverse anti-log, reverse inverse stuff is nonsense. There are only 3 types. Log, anti log (or reverse log) and linear. The thing is, with either of the nonlinear ones, you have both curves (one on each side of the wiper) the only problem is which sides they're on. When you introduce the parallel resistor, you can only get the curve that appears on one side of log and anti-log but it's the wrong side for what you want to do in a filter circuit.


This A, B, C stuff has driven me crazy for a while now. Because depending on who you ask the A and B change between linear and log. Everyone seems to agree that the reverse log is a C taper. I like the idea of A being log , B being linear and C reverse log, because B really is in the middle of the three.....

BTW. Small Bear Electronics in the US has reverse log pots for a decent price. (around $3) http://www.smallbearelec.com/Search.bok?category=Potentiometers+and+Trimpots
both in single and dual versions. Sadly they are not PCB mount, lik eyou want for the Gyraf projects. They will work though.

Mark

 

[PRR]

> Small Bear Electronics in the US has reverse log pots

Only in 100K, 500K, and 1Meg.

For mike-amp gain-set, we usually need much lower impedances. For transformerless, ideally below 200Ω, though as a practical thing we use 5K or 2.5K or 1K and accept the increased noise at low gains. If can only get high-R rev-taper pots, then we might as well make the other resistor adjustable. Then we use a standard audio pot. Dirty pot-wiper becomes a big problem (gain goes infinite if wiper loses contact) but I've used a lot of gear that way and it works fine.

 

[Learner]

You can use a dual gang linear to do it but it might cost the same as a rev log pot?

http://home.comcast.net/%7Esbernardi/elec/og2/partsub_pots.html

 

[Bryson]

http://www.geofex.com/Article_Folders/potsecrets/potscret.htm

 

[alk509]

[quote author="Bryson"]http://www.geofex.com/Article_Folders/potsecrets/potscret.htm[/quote]

OK, goddamnit, I'm gonna go shoot myself now!

Peace,
Al.

 

[SmG]

[quote author="PRR"]
This is, sadly, not possible with a loaded linear pot, though I do not have a formal proof.[/quote]
This isn't a formal proof, but more of a common-sense observation. You have to consider that throughout about 60% of the rotation of the pot, the effect of its own resistance has to be swamped by the relatively low-value resistor between the end and the arm to achieve the result in the graph. And as you continue to turn it, it continues, and swamps the entire resistance. It's simply the fact that you need so much of the track to have such a small part in the end result that makes this impossible, I think. This works fine 'the other way up' simply because you don't need those small resistance values.

 

[SSLtech]

[quote author="Learner"]You can use a dual gang linear to do it but it might cost the same as a rev log pot?

http://home.comcast.net/%7Esbernardi/elec/og2/partsub_pots.html[/quote]
No.

Bear in mind that we want to use the pot as a 2-terminal variable resistance in both feedback gain (for a preamp) and R/C timing (for equaliser) applications, and all of these approaches NEED three terminals, and are only usable for a reverse-log approximation of potential division.

In summary, if you don't have about 10% resistance at 50% rotation, what you have is useless. None of these tricks can do it.

Keith

 

[Learner]

[quote author="SSLtech"]
Bear in mind that we want to use the pot as a 2-terminal variable resistance in both feedback gain (for a preamp) and R/C timing (for equaliser) applications, and all of these approaches NEED three terminals, and are only usable for a reverse-log approximation of potential division.

In summary, if you don't have about 10% resistance at 50% rotation, what you have is useless. None of these tricks can do it.

Keith[/quote]

Right, thanks for pointing that out Keith! :thumb:

As for the op amp gain, I guess you can still use the attenuator to control the gain by placing it at either the input or the output instead of Rf and Rshunt from -ve terminal.

As for the EQ, you can probably use the rev log potentiometer to control a current controlled transconductance amplifiers such as the LM13700 or THAT VCAs to replace the Fc resistor in the EQ. It might be a longer way around buying the rev log pot....... :?

But the pot is not in the audio path, also easier to than getting a dual gang rev log often needed in a filter circuit eg. SVF EQ......