Hello,
I have a piece of test gear I built around 30 years ago. I recently started rebuilding this piece of gear and had to stop because of a potentiometer that is no longer manufactured. The part in question is a 10-turn, 2W, 25KΩ potentiometer. In the current market, all that is available are either 20KΩ or 50KΩ. No longer are the 25KΩ pots available.
So far, I tried adding a 50KΩ resistor across the element of a 50KΩ, 10-turn, 2W pot. It turned out to be close in its span, but not the same. Is this all I can expect or is there something or some way better?
Is there a way to perhaps "pad" a 50KΩ, 10-turn pot to act just like the 25KΩ, 2W, 10-turn pot that I have in my original piece of gear?

 

Welcome to AAC!

A schematic showing how the pot is being used would be helpful.

So far, I tried adding a 50KΩ resistor across the element of a 50KΩ, 10-turn, 2W pot. It turned out to be close in its span, but not the same.

That's because the parallel resistance doesn't change linearly.

So, it's only what you wanted at both extremes of the adjustment.

EDIT: I saw some 25K 10 turn pots on eBay if you're okay with paying $75-200 and taking a gamble that they're real.

 

Welcome to AAC!

A schematic showing how the pot is being used would be helpful.

That's because the parallel resistance doesn't change linearly.
View attachment 357654
So, it's only what you wanted at both extremes of the adjustment.

EDIT: I saw some 25K 10 turn pots on eBay if you're okay with paying $75-200 and taking a gamble that they're real.

Here is a copy of a portion of the schematics. I'm sorry that I can't include more of the drawing, but my customer will not allow it.

 

Hello,

R6 is parallel with the potmeter.
To have a closer parallel value with an 50 k pot, lower R6 to 15 k.
To have a closer parallel value with an 20 k pot, raise R6 to 25 k.
The schematic does not show the load of the potmeter.
I think the 20 k solution will likely work.

Bertus

 

Hello,

R6 is parallel with the potmeter.
To have a closer parallel value with an 50 k pot, lower R6 to 15 k.
To have a closer parallel value with an 20 k pot, raise R6 to 25 k.
The schematic does not show the load of the potmeter.
I think the 20 k solution will likely work.

Bertus

Thanks for the suggestion. I'll be trying this, later this week when I am able to test under load.

 

The wiper is connected to R7 going left, and something unknown going to the right. What is the impedance of the stuff on the right. This determines the total load impedance on the wiper, which is necessary to evaluate the relative merits and issues with a 20K versus 50K replacement. I'm leaning toward 20K, but need more data.

ak

 

The wiper is connected to R7 going left, and something unknown going to the right. What is the impedance of the stuff on the right. This determines the total load impedance on the wiper, which is necessary to evaluate the relative merits and issues with a 20K versus 50K replacement. I'm leaning toward 20K, but need more data.

ak

Hello,
Thanks for replying.
The 25KΩ pot is a remote location on the machine. I show it on the schematics for my convenience during testing.
With that said, Yes, the wiper connects to R7 at "G". "162" terminates to the grid of a 2500 volt Thyratron (#: 5557).
I hope this helps. I apologize for being vague, but I have an 'NDA' regarding the unit in its entirety. Which is why I could only show a small portion of the schematics.