Any potentiometer experts out there?

Why does it have to be 42.8k, if you are only using two terminals?
Which two terminals are you using? If you're lucky, the high range of the pot will be on the side of the taper where you reach 25k in about a fifth of the rotation, with an A50k pot. If that's the case, you don't need to worry about the far end of the rotation going over the 42.8k point, because it will only be at the very end.

FWIW, 42.8k is a -14.4% tolerance of 50k. Many pots have 20% tolerances. See if you can't find one that is far under-tolerance.

 

1: If its a simple voltage divider I can't see a reason why it shouldn't work since you are always working with a percentage of the range. You won't get the full range load-vise, though.

2: Should be fine, I think.

The big question is whether or not the design actually need a 42.8k or if it can get away with a 50K? (Or even 47K)


Resistors rarely work alone, so can you change the value of any other components to make the 50K fit?

 

+1

Pots aren't that tight toleranced. Order 5 and use the closest one, or try and bend the tab so the pot stops at 42.8 ( if you can with that type).

 

This is what I'm curious about. 42.8k is a very unusual value for a pot. Since pots control a resistance range, and not one specific resistance, their values tend not to be important. You can often use whatever value is closest to the range you need, given the parts available to you. This is especially relevant here, given that the OP has specified that he will only be using two terminals, so thus, it is a variable resistor, and not a part of any sort of voltage divider circuit that would take into account the total resistance of the pot as the wiper sweeps across it.

An A50k pot should be fine here.

 

My suggestion still stands. Find an under-tolerance A50k pot and be happy with it.

 

As a side note, I'm not sure where you were getting 1.205M from, as a parallel resistance, but to get 42.8k from 50k, you would apply a 297.222k resistance across the 50k.

 

You're going to mess with the taper. In any case, when you know your value, if you want to get ridiculous with an exact 42.8k, you can use this formula:
If R[SUB]1[/SUB] parallel to R[SUB]2[/SUB] equals R[SUB]Total[/SUB], and R[SUB]1[/SUB] and R[SUB]Total[/SUB] are known, then R[SUB]2[/SUB]= 1/([1/R[SUB]Total[/SUB]]-[1/R[SUB]1[/SUB]])

 

Maybe I misunderstood you, but if you put the resistor across the outer lugs (not the wiper) the resistance will go past 48.2 (or whatever) at one point near the top. You'd probably want to put the parallel resistor from the wiper to one side. Or something.

Also, you probably will end up using a trim pot if an exact resistance is required.

Also, resistance varies with temperature. You might want to consider that if you're looking for very high precision.

Luck!

 

The formula is above.

 

Ok, if you haven't figured this out yet, 1/([1/42800]-[1/45800])= 653413.333 Ohms. Since 653k is not a common value for resistors, you will either have to accept the closest substitute, or put a few resistors in series. You should be able to find a 680k resistor. This makes 42.9k, which is close enough. If you're dead set on 42.8k (There are temperature concerns to consider, and such.), you can do a 470k resistor in series with a 180k, or a 620k in series with a 33k, or whatever combination of values are available to you, that add up to ~653k.

 

I wonder if the OP has thought about tolerances on his other components. Maybe he doesn't really want 42.8 after all...

 

Exactly. Usually you either design using commonly available parts, or if you really need the precision, you tune the circuit using a 10-turn pot, or some precision fixed resistors in some other part of the circuit.

 

Dominic, have you tried the 45.8k pot yet? I didn't see which direction the pot was going, but a fifth of turn might take you into the low 20k range. Sweating over the 3k difference is ridiculous.