Pot tapering

[jeroddumas]

I was reading the article about pot tapering that was located on another post. Can't find the post now though. Let's say I want to make a 2.5k rev log pot for my ssl9k. Would I take a 1k linear pot and the solder a 1.5k resistor across the ccw lug and the wipe? My real question is how to determine the tapering resistor size. Any help is appreciated, I am having a hard time sourcing the 2.5 rev log pot. I don't want to buy 10 from omeg just to do two preamps.

 

[rafafredd]

Forget about this. This is not doable. Have been discussed many times. The conclusion in the end is that it´s not doable. You will need a rev log pot or a rotary wsitch.

 

[JLM Audio]

It is true you CANNOT make a reverse log pot by adding a resistor to a log or linear pot.

You can roughly log or audio taper a linear pot with a resistor.

Joe

www.jlmaudio.com

 

[jeroddumas]

Thanks, I guess I will do a gain switch then.

 

[gyraf]

The thing is that you can actually do both a log- and a rev-log-approximation voltage divider by adding a resistor to a linear pot. No problem.

BUT

This voltage divider - that will work nice as e.g. a signal attenuator - does NOT have constant resistance like a "normal" pot - meaning that adjusting the voltage divider's division ratio also alters it's overall resistance.

Which again means that you can't really use this as a "predictable" variable resistance, only as a variable voltage divider.

And a neg-log variable resistance is exactly what we usually need when we're looking for neg-log pots; either for gain controls in transistor-pair input amplifiers or for frequency-determining pot for state-variable filters...

Jakob E.

 

[BYacey]

Has anyone ever used a FET as the resistance element and controlling this with a DC voltage in an antilog manner? This way a pot with a fixed resistor from the wiper to one leg of the resistor could be used as a variable three terminal voltage divider to drive the FET gate. I have never tried this, but can't see why it wouldn't work. Perhaps another method might be to use a light dependent resistor and an LED in an optocoupler setup controlled by a similar divider.

 

[PRR]

> can't see why it wouldn't work.

At some point in its "linear resistance" range, the maximum voltage across an FET has to be held to about 30mV or you get gross distortion.

The limit can be 10mV or 100mV depending on the FET and what you call distortion. But mike amps often have over 100mV at the input, and much more at the output.

Also for low noise and high gain, the required resistance value may be as low as 10Ω or so. That's very low for common JFETs.

 

[BYacey]

Perhaps the LDR might be a better way to go then, provided it can be made to work in the 5 to 2500 ohm range.

 

[AMZ-FX]

Let's say I want to make a 2.5k rev log pot for my ssl9k.

Here's how you do it...

Forget about this. This is not doable. Have been discussed many times.

:green: :green: :green:

regards, Jack

 

[alk509]

[quote author="AMZ-FX"]Here's how you do it...

[/quote]

Not so fast, Jack.... Isn't that the exact opposite of what's needed in the circuit? I think you need a pot whose resistance decreases fast when you start turning it from left to right, and slows down as you get closer to the right end of its rotation... The one you posted does this backwards (goes down slow at the left end, fast at the right end).

You want this:
L..........................................................................R
x
x
.x
..x
.....x
..........x
.....................x
...........................................x...............................x

Not this:

L...................................R
x
..............x
.......................x
............................x
................................x
...................................x
.....................................x
......................................x
......................................x

Granted, you can still use the pot you posted, if you use it backwards (with the scale on the knob - max. gain on the left, min. gain on the right).

In any event, what is this, the thirty-five-hundredth time we've been through this? Mr. Keen ought to take that article down and update/fix it. Didn't he just join us here at The Lab?

Peace,
Al.

 

[rafafredd]

Granted, you can still use the pot you posted, if you use it backwards (with the scale on the knob - max. gain on the left, min. gain on the right).

Then, just use a regular log pot backwards...
Like I said...

Forget about this. This is not doable. Have been discussed many times.

I just want to add that there is a mechanical way to do it. But I would contact OMEG instead...

 

[AMZ-FX]

Let's take a log taper pot and put it down on the bench with the knob facing you and the lugs going down. Connect an ohmmeter to the left lug and the center lug. With the knob all the counter-clockwise, it will read 0 ohms. If we turn up the knob to the midpoint of the rotation, it will read about 15% - 20% of the full value of the pot... i.e. with a 100k audio taper pot, the mid point is about 15k... so I clipped my meter on to a pot and guess what? Midpoint was almost exactly 15k... I repeated this experiment with 2 other audio taper pots on my bench and got similar results. From the midpoint to full on, the resistance will increase the remaining 80%-85% of the pot value. When graphed, it looks like this:

Okay... the linear pot with the tapering resistor is easy to calculate since the resistance is just the parallel value of the fixed resistor and the amount of the pot which is across it, which increases in a linear manner. Each 10% turn of the pot knob increases the resistance by 10% of the total value--or 1k with our 10k pot-- so we plot it like this:

One of these is log and one is anti-log... one of them MUST be appropriate for the circuit in question.

So, where is this wrong?

 

[rafafredd]

I think it should be reverse. Just verticaly mirror the pic you´ve posted... Exactly like Al posted above.

 

[jeroddumas]

So will it work for the 9k. I don't want to order 10 pots from omeg. I may just try a gain switch.

 

[soundguy]

I bet you could sell 8 pots in the black market over time, probably easier than building a switch.

just a thought.

dave

 

[alk509]

[quote author="AMZ-FX"]One of these is log and one is anti-log... one of them MUST be appropriate for the circuit in question.

So, where is this wrong? [/quote]

Anti-log is different from reverse log.

Read my previous post.

Peace,
Al.

 

[PRR]

> One of these is log and one is anti-log... one of them MUST be appropriate for the circuit in question.

You want the curve in red:

This is, sadly, not possible with a loaded linear pot, though I do not have a formal proof.

 

[AMZ-FX]

You want the curve in red

ah-ha, you want the inverse anti-log response! :grin: :grin:

Here is the Omeg page of pot responses: http://www.omeg.co.uk/lawe16ct.htm

No, a resistor mod won't do the inverse but there is another simple method of modifying pot tapers... let me dig out my info and see if it can do the inverse.

regards, Jack

 

[SSLtech]

Again, this has been covered in detail before. perhaps we chould have a distillation of the facts established earlier as a META?

Essentially, depending which value is on which axis, you either want an inversion or a reversal, but the fact is that you want about 10% full value resistance at the mid-position. With an external resistor influencing the curve, you'll always have more like 90% of the full value at the mid position. Worse than using a linear!

Keith

 

[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.