Help with Double Potentiometer Wiring

Without seeing the rest of the circuit I can’t quite figure out how it’s suppose to work. Assuming the input is hooked to 100nF cap and output taken from the 33nF cap I would guess the following arrangement:

The “horizontal” part of the dual gang pot: The end connecting the series string of 2.2K resistor and 33nF capacitor should be pin #3. The other end is naturally pin #1 since wiper is the centremost pin (pin #2).
The “vertical” pot: The end connecting to common through a 2.2K resistor should be pin #1. The other end (the upper one) is naturally pin #3.

This would seem to create a variable hi-pass filter. If either one of the potentiometers is connected differently the result is just an attenuator circuit. Very interesting circuit, where is it from?

Isn’t a dual gang pot usually just two back-to-back potentiometers sharing the same shaft? (Thus the pinout of the other potentiometer will not be “flipped around"). I could be wrong here but I would check that the pinout shown in the picture you attached is the correct one. Maybe this is your problem.
 

blindsjc

Member
2006-11-28 9:35 pm
Hi teemuk,
I'm not sure about the pot pins,
looking inside (lttle openned windows)
the pot contacts appear like the picture, but
I'm really not sure, and my DMM died today,
I don't have any way to test it. Maybe a led
test, yeah, I will do it right now.

About your instructions, sorry but I don't undertand
the meaning of vertical and horizontal parts.

This is really a variable filter, used like a mid
range control at a heavy metal pedal. I will
test with a led and put the results. Thanks
again master. Wait just a minute.

Blindsjc
 
Ok, I guess my description was a bit confusing: With the terms horizontal and vertical I was referring to how the potentiometers are oriented in the schematic. Anyway, I’m a bit lost with how that circuit works so you better ignore my first post since there’s a good chance that it had some incorrect information.

But don't worry, this should be easy thing to assemble: Each row with three pins is one potentiometer. There are two pots so you have two rows. Once you figure out which pin in each row is the wiper (it’s likely the centremost in both) you can connect at least one of those two potentiometers correctly. (The pot is linear so it doesn’t matter which way you put it in as long as the wiper pin is connected correctly). Regardless of which one of the potentiometers you choose to fit in first you don't have to alter that part of the circuit no more.

You will also know which pin is the wiper pin of the other potentiometer in the dual package. (If you don’t you must measure). Likely the pinout is identical. Connect the two wipers together with a wire. Now, you have basically a 50/50 chance to connect this other pot to correct orientation. You should have two wires coming in and two unused potentiometer pins (likely pins 1 and 3) waiting for connection. Connect them anyway you like and if the circuit doesn’t seem to work just reverse the wiring of those pins. (Remember: You only need to tweak one of the potentiometers since the other one is “automatically” correct.)

That’s all you can do. If this doesn’t work then the error is most likely in some other parts of the circuit.

I hope this post simplified things.
 
Ok, back from the sleep. :)

Two possibilities, only one may be correct. See how the wiring in one of the potentiometers always stays identical. If this doesn't work then the problem is elsewhere.
 

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blindsjc

Member
2006-11-28 9:35 pm
Hi teemuk,
Just tried the two options and there are interesting things.

At the first option the sound comes overloaded like two
mixed signals fighting. Not good.

The second option is fine when set to one side and bad
when set to the other side. So, I think this is the right
option and my pot appears to be with problems in one
of the two parts. I will buy a new pot, a expensive one,
and try again. I revised the other parts of circuit and
appears to be ok.

What do you think about it?