Convert 10K log pot to 5K rev log pot?

Started by spargo, August 05, 2010, 02:25:29 AM

Previous topic - Next topic

spargo

How would I convert a 10K log pot to a 5K rev log pot?  Yes, I've read the secret life of pots but want to be sure - I don't completely follow it.

Basically, I need a 5k rev log pot but don't have one.  A 5k normal log pot would also be acceptable as a second choice.  The current pots I have to work with are:

A10k
A100k
A250k
A500k

B1k
B5k
B25k
B100k
B1M

C500k

EDIT: Also, this is for the MOSFET boost if you're wondering how it is wired.

jasperoosthoek

If you add a resistor between the wiper and the "grounded" lug this will change the overall resistance as you change volume. That's a problem in some circuits. Where are you going to use it?

On the other hand, you can simply buy reverse log pots at small bear. If you order a very small quantity then you can ask him to put them in an envelope to reduce shipping costs.
[DIYStompbox user name]@hotmail.com

spargo

I'm using it as the gain knob in the AMZ MOSFET Boost.  Lugs 1 & 2 are grounded and lug 3 is connected to the circuit.

And smallbear is slow at shipping, so I'd prefer to not order a single pot from there and have to wait a whole week.

12Bass

Try the (linear) B25k with a resistor to bring it down to the 5 k reverse log range.  Starting out with a logarithmic taper would work against what you're trying to accomplish.  Might even be worth trying B5k, as some people prefer the "feel" of linear taper.  
It is far better to grasp the universe as it really is than to persist in delusion, however satisfying and reassuring. - Carl Sagan

spargo

The B5k by itself does not work well in the circuit.  If I use the B25k with a resistor to make it a reverse log, how do I wire the resistor?

soggybag

I think you can put a 10K resistor across pins 1 and 3, to make a C5K.

12Bass

6.8 k will get closer (5.34 k).  Try it across 1 and 3.
It is far better to grasp the universe as it really is than to persist in delusion, however satisfying and reassuring. - Carl Sagan

spargo

I went with the 6.8k resistor on the B25k pot, and it sounds and acts exactly like the B5k does - was hoping to not have 90% of the volume in the last 5% of the knob.

jasperoosthoek

That's because of the 2.7k resistor in the circuit.  It can also be considered to be parallel to the wiper. So in other words, the 6.8k resistor doesn't do much as there is already a 2.7k resistor connected the wiper. If you look at this picture http://www.geofex.com/article_folders/potsecrets/logpot.gif you can see that the first part (10%) of the characteristic is actually very similar to the pot without resistor at all. And it is exactly that part of the characteristic that you are most interested in.
Simply said: I don't think anything other than a real reverse log pot will work in this schematic. Well, not exactly anything, an audio taper and two gears to reverse direction will work too  :icon_biggrin:
[DIYStompbox user name]@hotmail.com

zombiwoof

How about just putting a 10k resistor across the outer lugs to make it a 5k audio pot, and wire it backwards.  I think it would work in reverse then, right, but have the taper you want (if I'm figuring this correctly)?.  Not perfect, but it would get you through until you get the correct Rev Audio pot.

Al