How to connect this potentiometer?

Ron H

Joined Apr 14, 2005
7,063
Most single-turn pots are made like this.
One of the diodes is backward in the schematic you posted. I posted a correction below. If the schematic is tiny when you open it, click on it to enlarge it.
Connect the anode of one diode to A and the cathode of the other one to B. The wiper (w) connects to pins 2 and 6 of the 555.
If you still don't know how to connect your pot, post a picture of it.
 

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Thread Starter

deki

Joined Sep 6, 2011
23
Actually just another question: How would I go about to measure the resistance of the potentiometer?

Would I just place the negative probe on either A or B, and the positive one on the wiper? I tried that, and I get either 0k or 10k, depending on whether the negative probe is on A or B.
 

BMorse

Joined Sep 26, 2009
2,675
it really does not matter which side the red or black probe is on, just connect one probe to one of the terminals, and connect the other to the wiper, now slowly turn the pot one way or the other and you should see the value change, it will get lower when wiper is turned towards the terminal with the probe, and higher as you move it closer to the other terminal which should have nothing connected to it. and also, you must do this while pot is out of the circuit and disconnected.
 

Thread Starter

deki

Joined Sep 6, 2011
23
^^Actually thing is, when I put the red probe on the wiper, and the black on say 'A', I get ~9k, yet when the black is on 'B' I get ~1k? (with the dial turned as in the image, more to the left).

Also when I just want to use it as a simple variable resistor in a circuit: do the A and B connections go to Vs and Ground? And the wiper just goes into wherever I'd like the resistance to be?
 
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BMorse

Joined Sep 26, 2009
2,675
^^Actually thing is, when I put the red probe on the wiper, and the black on say 'A', I get ~9k, yet when the black is on 'B' I get ~1k? (with the dial turned as in the image, more to the left).

I take it that you are not moving the wiper while probes are in place, and since you took measurements from both sides and they equal out to 10K, I would suspect that the pot is a 10K pot....
 

Thread Starter

deki

Joined Sep 6, 2011
23
Yes it is a 10k pot just that I don't understand why I get the two different readings depending on where I set the ground probe.
 

Adjuster

Joined Dec 26, 2010
2,148
Yes it is a 10k pot just that I don't understand why I get the two different readings depending on where I set the ground probe.
That is simple enough to explain. If we ignore any contact resistance, which is normally pretty negligible, a potentiometer behaves like two resistors in series joined at the wiper, where the total is the full pot resistance.

If the whole resistance of the pot end-to-end is R, and we define the wiper position as "A", a percentage of its maximum travel*, then one resistance is given by R times A/100, the other is R times (100-A)/100.

Thus if you have a 10kΩ pot, and the wiper is at 10% of its travel, you get one resistance of 10kΩx10/100 = 1kΩ, and the other is 10kΩx(100-10)/100 = 9kΩ.

*Electrical travel along the track resistance that is. The mechanical position may be different, allowing for various errors, and the fact that some potentiometers are made with a deliberately non-linear characteristic, such as "logarithmic " or audio tapers.
 
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