Potentiometer Problems

[RyanN]

Originally Posted by Jake M

Well, firstly, are you sure you have the wires connected correctly. There should be a diagram of some sort on the pot itself, that looks like this.

(1)__________________(3)
(2)_________|

If the diagram isn't on there, the offset pin on the pot will be number 2. Pin 1 goes to white, 2 to red, and 3 to black, on a PWM cable from an Analog Input.

Other than that, I don't use easyc and I couldn't tell you if that's the problem.

Don't you mean 1 to red, 2 to white, and 3 to black, or 1 to black, 2 to white, 3 to red? Or did I confuse something. I think if you do this, you will release some magic smoke. Please correct me if I'm wrong, I have our potentiometer connected this way and I'm about to connect it tomorrow. Please Please reply.

[DonRotolo]

Originally Posted by RyanN

Don't you mean 1 to red, 2 to white, and 3 to black, or 1 to black, 2 to white, 3 to red? Or did I confuse something. I think if you do this, you will release some magic smoke. Please correct me if I'm wrong, I have our potentiometer connected this way and I'm about to connect it tomorrow. Please Please reply.

You ae correct: One "end" of the pot goes to black, the other "end" goes to red - this puts a 5 volt potential across the entire pot. You then connect the Wiper of the pot to the white, so the RC can read the voltage at the wiper contact (which varies from 0 to 5 volts). Using the diagram, black goes to 1 or 3, red goes to 3 or 1, and WHITE goes to TWO.

Don

[Jake M]

Originally Posted by RyanN

Don't you mean 1 to red, 2 to white, and 3 to black, or 1 to black, 2 to white, 3 to red? Or did I confuse something. I think if you do this, you will release some magic smoke. Please correct me if I'm wrong, I have our potentiometer connected this way and I'm about to connect it tomorrow. Please Please reply.

Why do I always screw things up when I'm trying to help someone? Yes, your wiring is right. 1 to red, 2 to white, 3 to black. There won't be any smoke if you do it incorrectly, though.