getting rid of trim pots, how?

[stopstopsmile]

if you have a circuit that contains a trim pot how do you get rid of the trim pot?

I am working on a project, its simple but it has a trim pot, I have tweaked around with it and I like one position on the pot and don't plan on moving it again.

If I wanted to remove this pot, how would I do this.  I understand pots are really just resistors.  So.  How would I find the resistor value of the trim so I can just dump it. 

[aron]

Take the trim pot out of the circuit. Measure from one outer lug to the wiper with your meter set on resistance, write the value down. Now measure from the other outer lug to the wiper with your meter set on resistance, write that down.

Now replace the trim pot with those two resistors with the values that you wrote down. The junction of the two resistors is now your middle lug (wiper). The other two ends are the outer lugs of the trim pot. Take care to put the correct value resistors in the same locations as the measurements of the trim pots.

[axg20202]

...and if the trim pot is being used as a simple variable resistor with only two lugs connected, just remove the trimmer, measure the resistance between these two lugs and replace with a single resistor of the same resistance...

[jakenold]

... and if the trimpot is not being used at all, just remove it!

[gez]

Bear in mind that with large value trimmers, the input resistance of the meter used for measuring can load whatever resistance is on the pot, giving inaccurate readings.

One way round this is to wire up a FET input op-amp (preferably one that doesn't mind its inputs being pulled to ground) as a follower on a breadboard and connect its + input to the wiper of the trimmer via croc-clips (or probe) with (same) power on to both circuits.  Take measurements from the output of the follower.

Above trick only works for circuits that run from low voltage supplies (you'll fry the op-amp if you try this in a tube amp!)

Alternatively, if you know the input impedance of the meter, you can calculate the actual resistance by accounting for the meter's loading.

[stopstopsmile]

on a schematic how would i know if just 2 or 3 lugs are being connected?  would it look on a schematic like its just a resistor with an arrow?  so that would mean two lugs?

[stopstopsmile]

I wish I could post a schematic example,

but I see on schematics a trim that looks like a resistor but it has an arrow that goes through the path. Is this 2 lug?

I have then seen a trim that looks like a resistor but the arrows back end connects into the circuit, which to me looks like all 3 lugs are connected. 

correct?  or am I completely off?

[gez]

One way round this is to wire up a FET input op-amp (preferably one that doesn't mind its inputs being pulled to ground) as a follower on a breadboard and connect its + input to the wiper of the trimmer via croc-clips (or probe) with (same) power on to both circuits.  Take measurements from the output of the follower.

Forgot to point out that the above method is a way to measure resistance of 'in-circuit' trimmers, though what you actually measure is the voltage at the wiper.  Once you have this voltage, you need to rearrange the voltage-divider formula to calculate actual resistances. 

Re lugs in schematics.  Pots/trimmers have three terminals.  In a schematic, there will be 3 lines connected to these terminals (often they're labelled).  If one of the lines is short and connects to nothing/thin air, only 2 lugs are connected.