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Can I reverse the wiring on a reverse log pot to convert it to a log/audio pot taper? Basically can I convert a "G" type taper pot to a "D" type taper -- http://www.bourns.com/docs/Product-Datasheets/91_95.pdf -- if I reverse the wiring.

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You can add a resistor between wiper and common end of the pot to get these modifications to the response versus angular position of an otherwise linear pot: -

enter image description here

Next, here is the difference between a D type and G type response (top left): -

enter image description here

So, as an approximation, you can convert (somewhat) a D response to a G response (by adding a resistor as per the top diagram) but you can't go the other way without compromizing the pot's ability to turn the sound off completely (adding a fixed resistor to the common end of the pot to somehow raise the G response towards a D response).

So, depending on how accurate you want to make the conversion, on the face of it the answer is unfortunately no.

However, you could use a small micro driving a digipot circuit and use the micro to convert the G pot position to the equivalent D pot position (on the digipot) but, I suspect that is going to far.

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  • \$\begingroup\$ Thanks for the linear-to-log pot approximation method. Will use this if I cannot make the "G" taper work. I think the "G" taper in the bourns datasheet means something different from the plots you have shared. It says "G" is "CCW audio taper", which I believe means its LOG taper between 1-2 when turning in the CCW direction. DOes this mean it will be LOG taper between 2-3 when turning in the CW direction (is my main question)? \$\endgroup\$ – O.K. Sep 18 '16 at 17:02
  • \$\begingroup\$ Yes you can reverse the wiring and it will effectively swap response between the two top graphs in my 2nd picture. \$\endgroup\$ – Andy aka Sep 18 '16 at 17:22
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Can I reverse the wiring on a reverse log pot to convert it to a log/audio pot taper ... if I reverse the wiring?

Yes. It will work in reverse rotation. e.g., Turn knob clockwise to turn the volume down.

It could be novel. Break out of the convention. At last - a volume control for left-handed people!

[OP's comment:] What if I want it to be right handed -- i.e. turn knob clockwise to increase volume. Can I achieve this by wiring between terminal 2-3 (instead of 1-2, which will be antilog)?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Standard, left-handed and OP's idea.

Swapping 2 and 3 doesn't work as you think. You will be shorting the input to ground when you turn fully clockwise. If your source can tolerate this you may find some combination that is acceptable in use but it will be a strange setup.

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  • \$\begingroup\$ Thanks Transistor. What if I want it to be right handed -- i.e. turn knob clockwise to increase volume. Can I achieve this by wiring between terminal 2-3 (instead of 1-2, which will be antilog)? I think it can be, but want some confirmation before ordering it. \$\endgroup\$ – O.K. Sep 18 '16 at 17:08
  • \$\begingroup\$ See the update. Why are you messing about with anti-log pots if you haven't ordered yet? I assumed you had purchased the wrong part and were stuck. Order the right part! \$\endgroup\$ – Transistor Sep 18 '16 at 17:48
  • \$\begingroup\$ The reason I am messing with this is becasuse I am unable to find the right part (CW log) in the required ohm value; but I can get a CCW log part. I am sorry if I wasnt clear, but (c) is not what I was suggesting. Your (b) actually answers my question. You have not really used a CW anti-log pot in (b), you have used a CW log pot with reversed/swapped wiring. Essentially you are using the CW log pot in "antilog" configuration. If you had used a CW anti-log pot, '1' should have still been ground and '3' still IN (assuming that the convention is that '3' is always the CW terminal). Thanks! \$\endgroup\$ – O.K. Sep 18 '16 at 18:12
  • \$\begingroup\$ I will be doing similar to what you have done from (a) to (b) -- just that you are using a CW log pot, while I will be using a CCW anti-log pot (because thats whats available) \$\endgroup\$ – O.K. Sep 18 '16 at 18:13

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