"Ramp down" reverse log tapering....

[potul]

Hi DIYers,

First of all, let me introduce myself a little, as this is my first post in the forum. I  started in the DIY world some months ago. So far I have only built 3 stompboxes. Although I'm working in a IT/electronics company here in Spain, I'm a Mechanical Engineer, so no big experience on electronics...
I have been reading this forum for some months and I always found somebody else previously asked the question I wanted to ask, so I didn't need to post. But now, after using the search engine, I couldn't find what I'm looking for, so here is my fist post!

In October I finished building my Roger Mayer Octavia, and as usual I couldn't find a Reverse log pot. After reading "the secret life of pots" I decided to "taper" a linear one, but after some trials I couldn't get it work correctly. At the end I kept the linear with some offset resistor and that's it.

The point is that since then, there is something in my mind I don't see clear and I would like you to confirm if I'm right or wrong. IIRC, the RM Octavia Gain pot is a Reverse log, connected as a Variable resistor (no voltage divider here), and it's a "ramp-down" pot, so it's going from 100% resistance to 0% resistance (counterclockwise to clockwise). (Max gain when R=0)

In "the secret life of pots" I read

In the two terminal connection, the resistance through the pot is what we're interested in, not the voltage divider ratio.

It turns out that the tapering trick works here, but only partially. If we want to make a reverse log tapering pot, we're in! However, there is no way to get a simulation of a log taper pot in the two terminal connection. For that we have to buy real audio taper pots.

that's why I tried the tapering.

But here is where I have my question. Is this sentence true also when talking about "ramp down" pots? From what I see, if I want to emulate a Reverse log pot when used as a "ramp-down" rheostat, I want to get a curve similar to the red one in the graph:



The blue one is the "normal" reverse log curve, the resistance between wiper and lug 1. The red should be the resistance between wiper and lug 3, because both must sum always the total R of the pot. Is this assumption right?

If I'm not wrong, if I attach a resistor in parallel with the pot as suggested by R.G., I will end up with a curve similar to the yellow one, not the red one. And the yellow one is basically the response of a Log pot when used as a "ramp down" rheostat.

So at the end, instead of improving the response of the gain pot (I want big increase at the beginning of the travel, and more accuracy at the end), what I will get is the opposite effect, smooth control at the beginning but abrupt at the end. (is "abrupt" correct? Sorry, I need to improve my English)

So I would rather say, if you want to substitute a Reverse log pot working as a rheostat, the trick is good only if you want to use it in "ramp up" configuration. In "ramp down" configuration you cannot taper it, you will need to buy one

The good news are that the reverse thing is happening with Log pots. So if you want to substitute a Log pot working as a rheostat, the trick is good only if you want to use it in "ramp down" configuration.

Does it have any sense or am I missing any important point that will put the shame on me?

Thanks, and sorry for the long reading

Mat

[nelson]

If I understand you correctly, all you need to do is switch the tapering resistors.


http://www.diystompboxes.com/analogalchemy/emh/emh.html

Like this.

[potul]

hi nelson, thanks for the answer

If I understand you correctly, all you need to do is switch the tapering resistors.

Do you mean putting the tapering resistor in the other lug? this would not work, as this lug is the free one, not connected to anything. I would end up with a linear rheostat.
Maybe I didn't explain it correctly.

The original circuit is using the pot like this:


What I tried after reading "the secret life of pots" is:



But this is not working, is giving the incorrect response, the graph in yellow in my previous post.

If I do what you propose:



I think this will have a linear response, as the tapering resistor is attached to the not connected lug.

Mat

[J. Luja]

So I would rather say, if you want to substitute a Reverse log pot working as a rheostat, the trick is good only if you want to use it in "ramp up" configuration. In "ramp down" configuration you cannot taper it, you will need to buy one

yeah, you've got it right. unfortunately that happens to be the most common use of the reverse-log pot in effects

[potul]

I see.... so no chance to taper in my Octavia...
At least now I know I wasn't doing anything wrong, it's just the way it works.

wouldn't it be a good idea to update the "secret life of pots" with this piece of information?
Thanks,

Potul

[rasco22862]

hi, i´m with doing the roger mayer octavia right now, so there is any way to emulate a 10k reverse log pot? If not wich pot can make a good replacement.

Thanks

[potul]

After some testing what I ended up doing is using a linear pot (I don't remember the value exactly) and put a resistor in series. This way you improve the response of the pot, but you lose some "clean" margin at the beginning.

I suggest that you use a linear pot and just play with some values in series/parallel until you find your taste.

If I recall it correctly, I used a 10kpot, with a 10k resistor in parallel and a 2k in series. This way the max resistance is 7k instead of 10k, but you get a smother control at the end.

Potul

[rasco22862]

can i just buy a 10k linear pot and put a tapering resitor of 2K in RX:

[potul]

Well, according to my experience it will not work, that's why I started this thread a long a go. You will end up with a pot of around 7k, but the response will not be the same as an antilog, it will be more like a log.

Let's clarify one thing, the Octavia will work in any way if you have a pot with a similar value, no matter if it's linear, log or antilog. The problem will be the response of the pot, that will be smooth in one end and steep in the other.

Potul

By the way, I see you've visited Piso-tones.
They tell you in their article about the Octavia to use a linear pot with a tapering resistor, but I think it doesn't work the way it's expected. I tried to convince Calambres about it a long time ago and have the article modified but didn't succeed.