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Can't make heads or tails out of this potentiometer pinout

post #1 of 4
6/10/11 at 4:18pm
Thread Starter 

PRESENTS ARRIVED TODAY!!!!

 

1004049h.jpg

 

http://www.alps.com/WebObjects/catalog.woa/E/HTML/Potentiometer/RotaryPotentiometers/RK501/RK50114A0001.html

 

Can anyone help me out here? 

 

 

My best guess is that the first and second columns of pins are both inputs, the third and fourth column act like a standard pot, and the fifth column is a pass-through... am I anywhere close? 


Edited by El_Doug - 6/10/11 at 4:20pm
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post #2 of 4
6/10/11 at 6:51pm

How bout 1, 1 circled, and 4 are ground.

3 in, 2 out.

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post #3 of 4
6/10/11 at 6:54pm

Looks like 1 & 3 are connected together to make pin 1

Pin 3 is the other end of the resistor.

 

So, pin 1 & pin 3 connected as in and out will act as a constant resistor value

 

Pin 2 is the tuneable pin, and pin 4 is a for bypass.

 

To hook it up, ground pin 1, apply input to pin 3 and take pin 2 as the output.

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post #4 of 4
6/10/11 at 6:59pm

As it sits, facing us, the pins are from left to right; 1, (circle 1), 2, 3, 4.

The two #1 pins seem connected together. These are one end of the element (the ground end).

Pin 2 (the middle pin) is the wiper.

Pin 3 is the other end of the element. The signal end.

Pin 4 is a shield.

 

The pattern repeats front to rear for four channels.

 

You should double check this with a meter.

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