hiya,
i don't have a 5k pot at hand... if i'd want to build the mosfet booster a la jack orman, or the sho, could i use a 10k pot or a 5k neg log?
would i have to change something in the circuit to be able to use the 10k pot? ???
thanks,
dualaud
Quote from: dualaud on May 29, 2007, 12:57:48 PM
i don't have a 5k pot at hand... if i'd want to build the mosfet booster a la jack orman, or the sho, could i use a 10k pot or a 5k neg log?
C5K would function the best. :icon_wink:
Yeah, a C5K is actually better than a B5K for that circuit! Good luck with it, nice sounding little booster.
hiya mark :D
and newbie builder,
would C5K translate to 5k neg log?
because thats what i've found in the drawer. ;)
i used a 5k normal log. its a all bunched up at the loud end but i dont mind as the pot spends most of its time there ;D
Quote from: dualaud on May 29, 2007, 01:11:07 PM
hiya mark :D
and newbie builder,
would C5K translate to 5k neg log?
because thats what i've found in the drawer. ;)
Yes Sir......that'd be it. :)
You can use ANYTHING above 5k, and adjust the taper to your liking with paralleled resistors so that you can go from 0 ohms up to around 5-10k. See The Secret Life of Pots at geofex to find out how.
Good point Mark. Looking at that you could use a 50k lin pot and 10k resistor to get an anti log pot with a max value of about 8.3k. A 25k pot would get closer to 5k total but they are not as common.
In this case dualaud is in luck as he already has a 5k reverse log pot!
Stew.
thanks for your help! :)
i can't believe that i have the right pot. usually thats exactely whats missing... :icon_rolleyes:
Quote from: grapefruit on May 29, 2007, 04:29:14 PM
Good point Mark. Looking at that you could use a 50k lin pot and 10k resistor to get an anti log pot with a max value of about 8.3k. A 25k pot would get closer to 5k total but they are not as common.
Sorry if I post this reply with this delay.
I have a A25K pot and I need a 5K to rebuild an AMZ Mosfet booster: I read the article on geofex.com but still I'm not sure where have I to connect the 10K resistor you were talking about above.
If I put it from one lug to the wiper the total resistance in one external point is still 25K, if I put it between the two external lugs leaving the wiper free the resistance still isn't 5k... where I got wrong in my mind rambling?
Thank you for your help,
Marcello
You'll want to wire the tapering resistor between the wiper and the lug that isn't tied directly to the wiper. The maximum resistance will then be R1 * R2 / (R1 + R2), where R1 is the maximum resistance of the pot, and R2 is the resistance of the fixed resistor.
The 5k reverse-audio taper pot for my Mosfet Boost (http://www.muzique.com/schem/mosfet.htm):
http://www.smallbearelec.com/Detail.bok?no=612
regards, Jack
Thanks for the replies!
Anyway still I have some doubt
Let's suppose connecting a 10K from the wiper lug to one of the external lug of a 25K.
When the wiper is on the oppsite of the lug connected to the resistor the maximum resistance will be the parallel of the to values, 7,14K. When the wiper is on the other side the resistance will be 25K.
...am I right?
bye
...no worry, I've understand how the pot value it works in this particular circuit, so there is no problem, thank you everybody.
Marcello