Correct way to create audio taper with linear pot by adding taper resistor?

[elipsey]

This article (which someone posted here the other day) explains it: The Secret Life of Pots

So when I try this with a real pot I can only get reverse log taper; in others words at 1/2 rotation, resistance from wiper to cold lug is more than 1/2 of the pot value, but I want it to be less. The taper resistor is smaller than the value of the pot, and it is connected from the "cold" lug (per article) to the the wiper. What gives? Did I fail at reading comprehension?

[J M Fahey]

Probably.
Recheck what you made.
Realize that if you have 1/2 carbon track from wiper (set to 5) and cold, and you wire a resistor in parallel with it, it will go down, not up.

As a side note, what will you use it for?
That kludge barely works as a volume control , and that IF itīs driven by a low impedance; anywhere else itīs useless.

[Tom Phillips]

Also be aware that, although you can mimic an audio taper with the method described in the article, you can't turn a linear pot into a part that is identical in all respects to an audio taper pot. This is because the overall end-to-end resistance of the altered pot changes as the pot shaft is rotated. Therefore, when using the altered part, two parameters (the overall pot resistance and the center lug tap resistance) are changing simultaneously. In a purpose built audio taper pot only the center lug tap resistance with respect to the end lugs changes. The effect on operation of your circuit depends on your exact overall circuit.

[elipsey]

Probably.
Recheck what you made.
Realize that if you have 1/2 carbon track from wiper (set to 5) and cold, and you wire a resistor in parallel with it, it will go down, not up.

Thanks. Yep, I'm OK with that.

As a side note, what will you use it for?
That kludge barely works as a volume control , and that IF itīs driven by a low impedance; anywhere else itīs useless.

I just tried it because I was thinking about it. I understand it would make the load smaller if I actually used it.

Here's an example -- 25KL w/ 6K8 taper resistor from cold lug to wiper. My intention is that, stated in the terms used in the article, b=R3:[R1+R2]=1/4, where R3 is the taper resistor, and R1+R2 is the total resistance of the potentiometer. In pic one, it's turned all the counter-clockwise ("down"), in pic two, it's turned to %50 of rotation, in pic three, it's turned all the way clockwise. I get 3.3K ohm at %50 rotation, and 4K at full clockwise (max). The test leads are on cold lug and wiper (I hope). Looks like reverse audio taper to me. What am I doing wrong here?

[frus]

Nothing, you are using (or rather, measuring) the pot as a variable resistor, not as a potential divider.
It's the last picture in Secret Life of the Pots, it clearly says that you can't get a log taper variable resistor, only rev-log

[elipsey]

Nothing, you are using (or rather, measuring) the pot as a variable resistor, not as a potential divider.
It's the last picture in Secret Life of the Pots, it clearly says that you can't get a log taper variable resistor, only rev-log

OK thank you, I get it now. I failed at reading

It is slightly confusing the the begining of the 3600 word article says:

the voltage division ratio of this rig is arbitrarily close to that of a log taper pot [...] Wow! No more waiting for volume control pots!

but the last paragraph says:

You can't simply put the tapering resistor from the CW terminal to the wiper and get a log taper pot emulation

[eschertron]

It's one of those stories with a surprise twist at the ending.

... no pun intended on the word 'twist'

[frus]

It is slightly confusing the the begining of the 3600 word article says:

Yes, but if you WERE using it as a potentiometer (volume control), then for instance in the middle picture, if it's really set mid way, you have 3.3k on one side and 12.5k on the other side. In other words, you have a 15.8k pot set to 20% resistance

If you set it at, say, 80%, you would have 20k||6k8 = 5k on one side, and 5k on the other side, so it's like a 10k pot set at 50% resistance

It IS a bit like a log pot, but its total resistance (= load to the preceding stage or pickup or whatever) varies a lot when you turn it. That's why Juan said it would work if it's driven by low impedance, because to a low impedance source, 25k could be similar load to 5k (wich are the min and max values of your "pot")