# Bought a board that has an adjustable pot - how to measure ohm vals

Discussion in 'General Electronics Discussion' started by pityocamptes, Sep 25, 2012.

1. ### pityocamptes

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0
Jul 26, 2012
I bought a cheap booster board that has an adj pot for voltage output. I want to remove the pot and instead run it to a resistor box for hardwired (switched) voltage outputs. Can I measure the pot resistance for a specific voltage while the pot is on the board or does the pot have to be removed, once voltage output value is reached and then measure the ohm value on the pot for a comparable resistor value? Oh, and one more thing, the pot has three legs, what should I do with the 3rd leg when I hard wire in a resistor selection box? Thanks!

2. ### BobK

7,682
1,688
Jan 5, 2010
Most likely you can measure it in circuit, since it would likely just be setting a voltage for a comparator, which should not draw much current. If the 3 legs all have separate connections you would need 2 resistors to replace the pot. If one of the outer legs is connected to the middle leg (or not connected) you could do it with one.

Bob

3. ### pityocamptes

79
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Jul 26, 2012

So how would that be in a circuit - where I need two resistors? Like this:

http://diyaudioprojects.com/Technical/Voltage-Regulator/

When I get the board in the mail I can take some pics...

4. ### BobK

7,682
1,688
Jan 5, 2010
No, in that circuit, the pot is used as a variable resistor, i.e. only two legs are used.

Let's say you have a 10K pot. And at some position, the center leg to the left leg reads 3K. The the center to the right leg would read 7K, and you could replace the pot with those two resistors wiring one end together with that junction replacing the center leg on the pot.

Bob

5. ### pityocamptes

79
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Jul 26, 2012
How would that work though if I want a resistor box that allows me to switch to multiple FIXED voltage outputs?

6. ### BobK

7,682
1,688
Jan 5, 2010
You're getting ahead of yourself. It would depend on how the pot is connected, which you don't seem to know yet. It is most likely used as a variable resistor (like the LM317) example you showed, and you would only need one resistor.

If it is used a a potentiometer, then most likely one leg is to ground. For each switched pair of resistors you would connect the ground sides together and then uses a 2-pole switch to swith the top of the dividers to the the where the other outer leg is connected and the middle of the dividers to where the center lug was connected.'

Do you have a schematic for the converter board?

Bob

7. ### pityocamptes

79
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Jul 26, 2012
No I do not have a schematic of the board, I will wait until I receive it... thanks.

8. ### pityocamptes

79
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Jul 26, 2012
OK, just got the board. The side of the adjustable pot says:

BAOTER
3296
xchc electron
_________1__________
3 V^V^V^V^V^V^V^V^V^V 1
_________2__________

CW

The top of the pot says:

194 T
W 103

The bottom the pot has all three legs in line if that makes a difference. Thanks!

79
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Jul 26, 2012

10. ### CocaCola

3,635
5
Apr 7, 2012
It's a 10K pot, but you still need to figure out how it's hooked aka what legs go where in the circuit...

11. ### pityocamptes

79
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Jul 26, 2012
So then I need to remove it?

12. ### CocaCola

3,635
5
Apr 7, 2012
Maybe or take an real educated guess based on what you can see and test with a continuity tester...

79
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Jul 26, 2012

14. ### CocaCola

3,635
5
Apr 7, 2012
Ok it's being used as a variable resistor, you can even see where they left a place for a fixed resistor if you wanted to go that route...

Measure between these two highlighted points...

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15. ### pityocamptes

79
0
Jul 26, 2012
Ok, so I just need to come off those two pads to my adjustable resistor box?

16. ### pityocamptes

79
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Jul 26, 2012
Also, how would be the best way to test the var pot resistor so that I can match the ohm value to a fixed resistor, etc. so that I can make a resistor box that will have a switch for different fixed outputs? Thanks!

17. ### CocaCola

3,635
5
Apr 7, 2012
Best way socket that pot or make it removable, but keep the two pins on the right that are shorted when in circuit also shorted when removed, dial it in on the circuit remove from circuit and simply measure resistance between the two pins that are not shorted...