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Potentiometer question

R

Ramma2

Platinum Member
If I remove a potentiometer from a circuit board (its a switch that would be used for like a joystick throttle) would that tell the system that the switch is all the way off 0% or all the way on 100%?

I need to remove 2 from a circuit board to get it to fit in a project box.

Thanks!
 
You would break the connection and it just wouldn't work. Basically it allows from 0V-12V for example...you remove it and there will be no voltage going through the circuit.
 
replace it with a resistor that is in the middle of the range of the pot
or at the high or low end , depending on where the pot is normally set
 
measure the resistance of the pot (after you remove it) in this "off" state and replace it with a resistor of similar value and the circuit will retain its characheristics of when the pot was in the circuit

you are replacing a variable resistor with a static resistor
 
Originally posted by: FoBoT
measure the resistance of the pot (after you remove it) in this "off" state and replace it with a resistor of similar value and the circuit will retain its characheristics of when the pot was in the circuit

you are replacing a variable resistor with a static resistor
Click to expand...

qft. ftw.
 
If the Pot controls the flow of the electricity through the circuit by regulating it, then removing it and making it a straight-through connection means it will allow all electricity to flow through, meaning it will stay at 100%. If it's not there, there is nothing to offer resistance.


Maybe it would be better if you just told us the dilema at hand?
 
The pot is usually there for adjustment or calibration.
If it's a two legged setup, you could replace with a fixed resistor (good luck finding 23.1 kohm resistor by the way), but if it's three legged setup w/ either ends hooked up across a voltage source and a center going to a circuit, you'll need two resistors of exact values.
 
Here is what I'm working on:

http://webpages.charter.net/paul21/5.jpg

Its an Xbox controller circuit board. I'm going to be putting 2 of them in a project box, so I just need to remove the pots from the back of the board so the 2 boards will fit. I don't need the circuit to be completed as I'm not using them one they're in the box.

PS the boxes here on the front of the board are from the analog sticks. The pots for the triggers are on the back, they are the ones I want to remove. I'm leaving the joystick pots on there because they look like a pita to desolder and remove.
 
Originally posted by: IAteYourMother
jumper it if you want 100%
Click to expand...

have to be careful on that one. having an open circuit won't hurt anything as most pots have a pretty high resistance (i.e. 1M ohms) at the upper end, so an open circuit won't look much different to the system as the pot turned all the way to max resistance. however, most don't go down to 0 ohms (if you used a jumper) but 300 or so. if you use 0 ohms you stand a good chance of frying something.
 
Originally posted by: Cattlegod
Originally posted by: IAteYourMother
jumper it if you want 100%
Click to expand...

have to be careful on that one. having an open circuit won't hurt anything as most pots have a pretty high resistance (i.e. 1M ohms) at the upper end, so an open circuit won't look much different to the system as the pot turned all the way to max resistance. however, most don't go down to 0 ohms (if you used a jumper) but 300 or so. if you use 0 ohms you stand a good chance of frying something.
Click to expand...

well fine, jumper it if you want 200% 😛
 
Originally posted by: Ramma2
If I remove a potentiometer from a circuit board (its a switch that would be used for like a joystick throttle) would that tell the system that the switch is all the way off 0% or all the way on 100%?

I need to remove 2 from a circuit board to get it to fit in a project box.

Thanks!
Click to expand...
A potentiometer is a VARIABLE RESISTOR, not a switch.
It's value is zero ohms at one end of travel and whatever the rated value is at the other. Removing it from the circuit OPENS the signal path, rendering it inoperable.



 
Originally posted by: tikwanleap
Originally posted by: FoBoT
measure the resistance of the pot (after you remove it) in this "off" state and replace it with a resistor of similar value and the circuit will retain its characheristics of when the pot was in the circuit

you are replacing a variable resistor with a static resistor
Click to expand...

qft. ftw.
Click to expand...

Quoting it again. This is the best advice in the thread because it assumes nothing about the circuit, and also answers your question.
 
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