2kB Presence Pots

[Darkbluemurder]

... are a value that seems hard to find. I really want them because I much prefer the 5kB value over the 25kB value. With 25kB the useable part of the knob seems to be between 9 and 10, with the 5kB between 8 and 10. 2kB should spread the useable range over a wider travel.

Does anybody know who carries them?

Thanks a lot.

[glasman]

... are a value that seems hard to find. I really want them because I much prefer the 5kB value over the 25kB value. With 25kB the useable part of the knob seems to be between 9 and 10, with the 5kB between 8 and 10. 2kB should spread the useable range over a wider travel.

Does anybody know who carries them?

Thanks a lot.

The ones that I buy are an Alpha part that I pick up at a local store. They do have a website but not sure if they ship internationally. They are also a GREAT place for mini switches.

www.circuitspecialists.com

Gary

PS> I am not associated with Circuit Specialists other than being a regular customer.


Gary

[dogears]

Why not adjust the 5k so it is closer to 2K? The presence pot is just a variable resistor.

... are a value that seems hard to find. I really want them because I much prefer the 5kB value over the 25kB value. With 25kB the useable part of the knob seems to be between 9 and 10, with the 5kB between 8 and 10. 2kB should spread the useable range over a wider travel.

Does anybody know who carries them?

Thanks a lot.

[Darkbluemurder]

Of course I could solder a 4k7 resistor from CW to the wiper and one from the wiper to ground to simulate a 2kB pot if I cannot find a 2kB. I used a similar technique to bring down 250kA pots to approx. 120kA.

[dogears]

DBM,

Only 1 resistor is required. The pot has the input and wiper connected already. I'd go ahead and try it.... Neat and effective.

Of course I could solder a 4k7 resistor from CW to the wiper and one from the wiper to ground to simulate a 2kB pot if I cannot find a 2kB. I used a similar technique to bring down 250kA pots to approx. 120kA.

[glasman]

DBM,

Only 1 resistor is required. The pot has the input and wiper connected already. I'd go ahead and try it.... Neat and effective.

Of course I could solder a 4k7 resistor from CW to the wiper and one from the wiper to ground to simulate a 2kB pot if I cannot find a 2kB. I used a similar technique to bring down 250kA pots to approx. 120kA.

Actually I was working on this issue today. Mr. Ayan and I were discussing the merits of using a reverse log pot for the presence control. A 10K pot and 2500 ohm (or a standard value that is close) would make a dandy 2K rev log pot.

Gary

[jaysg]

Mr. Ayan and I were discussing the merits of using a reverse log pot for the presence control. A 10K pot and 2500 ohm (or a standard value that is close) would make a dandy 2K rev log pot.

Good company -- Ed Jahns spec'd a 25K rev. log pot for the Series II Fenders. fwiw, Paul Rivera doesn't like to look back, but he did tell me once that the idea for the Series II amps was to go after the Robben Ford/Dumble tone of that day. It somewhat explains the changes to the clean tone on those amps. No cigar, unfortunately.

[Tdale]

What exactely would you do to make a 5K (or 10K for that matter) into a 2K, using one resistor on the precence pot? On the hybrid-a schem, the input is on one side, while the other side is connected to the wiper.

Tommy

[mlp-mx6]

You would put a single 3.3K (or slightly larger) resistor either...

1. From wiper to ground.
2. From wiper to "input".
3. Across the end lugs like the .001 on the Bass pot for Skyliner.

Option 1 will approximate a log pot. Option 2 will approximate a reverse log pot. Option 3 will be closer to linear, IIRC. Try it for yourself and see which way you like - several lately have been suggesting #2.

[glasman]

Here is a quick graph and circuit showing how to make the rev log pot from a 10K and 2490 ohm resistor.

[img:911:623]http://glaswerks.com/images/revlog.jpg[/img]

[llemtt]

Hi Gary

I can't understand how it works, shouldn't it be "reversed"? I expect 0 resitance at 100% (ten) and 2k at 0% (zero), it actually seems to be log instead of anti-log.

btw Tommy you can even go with a 1k pot!

teo

[glasman]

Hi Gary

I can't understand how it works, shouldn't it be "reversed"? I expect 0 resitance at 100% (ten) and 2k at 0% (zero), it actually seems to be log instead of anti-log.

btw Tommy you can even go with a 1k pot!

teo

Teo, you know you are right, that will not work as drawn. Thats was happens when you start thinking without the mandatory 3 cups of coffee in the morning....

The cap needs to go the top of the pot, the resistor from the Top of the pot to the wiper. The PI ground connection to the Wiper. No Connection on the bottom of the pot.

I think that fixes it .



Gary

[llemtt]

Thanks Gary

I'll try again but lot of time ago I almost went to a conclusion that it's impossible to make a reverse log out of a lin pot I'll redo the homework

Another solution would be to use a log pot and to reverse panel graphics -> 0-10 but counterclockwise

In the meantime I go with 1k lin.

teo

[dogears]

Why go 1k lin?? Just make a 2k. You could use a 5k..... Or, use the 1k, but stick a 470ohm resistor in front. I really think having the presence 1/2 on all time at a minimum is not good.... Sometimes setting it at 10 oclock is a good tone. 1/2 on is noticeable.

Thanks Gary

I'll try again but lot of time ago I almost went to a conclusion that it's impossible to make a reverse log out of a lin pot I'll redo the homework

Another solution would be to use a log pot and to reverse panel graphics -> 0-10 but counterclockwise

In the meantime I go with 1k lin.

teo

[llemtt]

My favorites suppliers have plenty of 5k and 1k lin but they don't carry 2k lin

5k lin with a parallel 3.3k isn't ok because it becomes log-like

now with 1k I usually run between 2-5 so it's almost ok to me, I definitely have to search for a good reverse-log supplier

cheers
teo