Home Made LPL

sudden_impulse · Post by **sudden_impulse** » Thu Dec 18, 2008 8:22 pm

I'm planning on making my own LPL, and already bought the parts but haven't opened them yet. The potentiometer I want is the 50k-ohm linear-taper, correct? That's the kind I bought, and the only 50k-ohm I saw there. Thanks for the help guys.

lowpoke · Post by **lowpoke** » Thu Dec 18, 2008 10:18 pm

I just picked up th eparts for mine yesterday! Total cost $7.50.
You might find this thread helpful, as did I:
http://phoenixphorum.com/wtb-lpl44-vt65 ... ight=lpl44

sudden_impulse · Post by **sudden_impulse** » Thu Dec 18, 2008 10:22 pm

I actually had him post in my thread in the FS section when I was looking for one, he gave me the idea to do it, but nobody mentions if the pot is linear or not. I'll give it a go when I have a few spare minutes this weekend, the pot was only like $3, so if it doesn't work (hopefully it does) I'm not too bad off.

Grim0013 · Post by **Grim0013** » Sun Dec 21, 2008 2:14 am

sudden_impulse wrote:I actually had him post in my thread in the FS section when I was looking for one, he gave me the idea to do it, but nobody mentions if the pot is linear or not. I'll give it a go when I have a few spare minutes this weekend, the pot was only like $3, so if it doesn't work (hopefully it does) I'm not too bad off.

Now, it's been some time since I've really done much in the way of electronics, but....

When it comes to whether or not to use a linear pot, either should work. Where you will see a difference is in how the control will react to input. A linear model means that turning the nob x degrees will always result in y attenuation regardless of the knob's current position. with a non-linear (what's the right term? progressive? exponential?) pot, as you increase attenuation, each additional degree you rotate the knob results in progressively great increases in attenuation. Like starting from 0, x degrees = y attenuation, but say, x*2 = y^2 and x*3 = y^3, as a rough example. The idea is to use non-linear when you want to be able to have fine control at the lower end of adjustment, but still be able to get to a high attenuation without needing to turn the knob all the way around 13 times to get there.

All that said, based upon what the LPL does (I've never had one though) I would guess that it would work best with a linear pot.

Post by **stipud** » Sun Dec 21, 2008 9:17 am

It is a linear pot.

http://download.phoenixphorum.com/TechS ... L_Info.pdf

sudden_impulse · Post by **sudden_impulse** » Sun Dec 21, 2008 7:58 pm

Thanks guys, I'll get it started tomorrow probably.

Post by **Jacampb2** » Sun Dec 21, 2008 8:12 pm

Grim0013 wrote:
with a non-linear (what's the right term? progressive? exponential?) pot, as you increase attenuation, each additional degree you rotate the knob results in progressively great increases in attenuation. Like starting from 0, x degrees = y attenuation, but say, x*2 = y^2 and x*3 = y^3, as a rough example. The idea is to use non-linear when you want to be able to have fine control at the lower end of adjustment, but still be able to get to a high attenuation without needing to turn the knob all the way around 13 times to get there.

The term is "logarithmic taper" or "Audio taper" and it actually actually follows a logarithmic curve that approximates a humans hearing perception. They are typically used for volume controls. Very basically, when used to control volume, they tend to sound "linear" to us, even though it is not. There are other uses for log taper pots, but not many. There are also "reverse-log" taper pots available commercially, and you can also have virtually any curve made if you want to special order...

As for the LPL pot, yes it is linear, and no it would not matter much if you used a log taper pot, but it would be more difficult to adjust at the upper end.

Later,
Jason