Fader Puzzle - any ideas?

[Mark Hennessy]

EDIT: Split from this thread

If you think this approach is somewhat crude, you might like to ponder on the fact that many log law pots are nothing of the sort! Essentially, the track is laid down in two overlapping linear  sections of different values so that the selected portion of the track increases slowly in value until about the halfway stage, then increases rapidly to maximum!

Yes, absolutely true, and a bit of a shock when you realise this

An interesting sub-category is the humble fader on a sound desk. The desks I've played with - which tend to use Penny and Giles faders - are surprisingly good. With an audio test set connected to the desk, and with the fader moved to each of the markings on the scale (+10, 0, -10, -20, -30, etc), the level is surprisingly accurate - even down at the low levels. Like, within a dB at worst. Amazing given that the operator doesn't really look at the numbers, and uses their ears (and meters) when mixing.

Here's a puzzle for you. A couple of years back, a trainee brought me a fader - made by Alps - that was stamped 10k. And it measured 10k from end to end of the track. No problem... So, you'd expect that wherever you moved the fader to, the two resistances from the wiper to each end of the track would add up to 10k, right? Certainly, that would be the case with a regular 10k potentiometer...

This fader was different. The resistances from wiper to track ends were higher than you'd expect.

No-one at her base could explain this. So she brought it to Wood Norton to see if we could. We got there, but it took a while, and we were a little bit surprised at the outcome.

So, any idea what Alps were doing? And why, from the point of view of the manufacturer, were they doing this?

It's not a trick question - I still have the fader, and I can provide real measurements on Monday when I back at work if needed.

[Radio Fixer]

Mark, how about a fixed resistor between the wiper and its output terminal?

Gary

[ppppenguin]

Gary's explanation is a simple and obvious one but I can't see any benefit from doing it.

Did the 10K fader measure 10K end to end at all settings? That certainly won't be true if law bending resistors are fitted.

So a chain of reasoning that's goes nowhere at the moment. A professional fader needs an accurate log law, not the horrible 2 part track on an ordinary pot. You can't readily grade the resistivity of the track by changing its composition, at least I don't think you can. You can do it by changing the width of the track. Alternatively you can use law bending resistors. None of this really explains Mark's conundrum.

A slightly wild idea is that the track is a bit like the teeth of a comb. The bulk of the track is vaariable width to give the correct law.Then "teeth" stick out from the track to make a sort of multiway switch, remeniscent of the old stud faders. The teeth would add resistance in series with the wiper giving the effect seen.

[Mark Hennessy]

Afraid it's not that simple, Gary, there are no law-bending resistors, and hence, the 10k track resistance doesn't alter with fader position...

Jeffrey's idea about additional series resistance is an interesting one. For that to have the desired effect, the input impedance of the following stage would have to be very well defined, which isn't the usual way to design a post-fader amplifier. Usually, you feed the fader into a high-impedance amplifier so as not to alter the carefully established law of the fader.

So, not that... Anything else?

[ppppenguin]

We don't know how big the discrepency was. A small amount is easily explained by incidental series resistance. A large amount isn't.

As an aside, law bending resistors require a low impedance drive for accurate results.

[pwdrive]

Did the pot have two tracks?

Lawrence.

[Terry]

As an aside, law bending resistors  require a low impedance drive for accurate results.

Are you sure about that, Jeffrey?

Surely if you add the series resistance to the pot value when calculating the value of the law bending resistor, you should be able to find a value that gives the required law over the part of the curve that represents the pot.

Obviously, if you don't take the series resistance into account, you will get the wrong answer!

[ppppenguin]

If you add the series resistance when doing the calculations you're fine. If the source has a highish and rather indeterminate resistance then not.

[Mark Hennessy]

Did the pot have two tracks?

Lawrence.

We wondered if that was the case, and spent quite a lot of time eliminating that.

At first glance, it looked like that might have been the answer, but no. There are two black tracks in parallel for the whole length of the fader travel, and the contact springs of the fader join the two together at a point determined by the position of the fader. Careful examination with very sharp DMM probes showed that the track was just that, and the other track had zero ohms across the whole of its length. So, not that...

There isn't much left to dream up now - and I'm not sure I can give any more information without giving away the answer. The strongest clue I can give is to repeat something I said initially: "And why, from the point of view of the manufacturer, were they doing this?"

About to go out for the day, so look forward to seeing further thoughts later

Mark

[Alan Douglas]

Is the geometric center of the track different from the shaft location?