# Can I change the effective angle of a potentiometer?

Discussion in 'The Projects Forum' started by belae1ka, Apr 10, 2017.

1. ### belae1ka Thread Starter New Member

Nov 20, 2016
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I am using a 5V voltage input.

Rather than the Vout going from 0V to 5V when you spin it all the way from one end to the other (around 300 degrees I'd say), I only need to the potentiometer to spin roughly 90 degrees for my application.

So to clarify, I was wondering if there is a way of going from 0V to 5V by turing a potentiometer roughly 90 degrees.

I can't seem to find any of these types of potentiometers online.

I would like to know your thoughts.

Thanks.

Jul 18, 2013
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Feed the potentiometer with a larger voltage that will give you 5v at 90°.
Max.

belae1ka likes this.

Nov 20, 2016
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Thank you.

4. ### AlbertHall AAC Fanatic!

Jun 4, 2014
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Or amplify the smaller voltage from the 90 degree movement up to 5V.

5. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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Or use a larger potentiometer, whose resistance over a 90° turn is equal to the total resistance of the existing potentiometer. The value will be about 3.33 times the existing value. Equal to (300°/90°).

GopherT likes this.
6. ### crutschow Expert

Mar 14, 2008
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Don't see how that gets you to 5V output with a 5V source.

7. ### crutschow Expert

Mar 14, 2008
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Here are some that supposedly have a 90° rotation.

8. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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Ok, the existing circuit has a pit such that (for example) when it is at one end, it's output is 0 volt. Now, when rotated, at some point it's output is 5V Lets say for S&G, it's at the other end of the rotation and 330Ω.

So instead of rotating the full rotation the TS wants the 5V at the 90° mark. So, the resistance between 0° and 90° has to be the same as the full rotation of the original pot. That gets you the same output as before since the resistance is the same as before. The rest of the circuit is irrelevant

9. ### crutschow Expert

Mar 14, 2008
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Sorry, that still makes no sense if he has only a 5V supply.
Post a schematic of what you are proposing.

10. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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There is NO difference in the schematic.

11. ### WBahn Moderator

Mar 31, 2012
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Although no circuit diagram was provided by the TS, it is almost certainly a straight up 3-terminal pot with one and tied to 0 V and the other end tied to 5 V and the wiper providing the output voltage. How will changing the overall resistance of the potentiometer change the (unloaded) output at the wiper?

I think you are envisioning the pot being configured as a rheostat and being in series with another resistor to form the voltage divider.

12. ### AlbertHall AAC Fanatic!

Jun 4, 2014
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But then you couldn't get 5V output from a 5V supply.
Or, if you did it the other way up, you couldn't get 0V.

13. ### crutschow Expert

Mar 14, 2008
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You mean the schematic we have not seen?
You must be envisioning a schematic to propose a change.
So what does that straw-man schematic look like?

14. ### WBahn Moderator

Mar 31, 2012
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Agreed. But I'm having to guess at what he has in mind!

15. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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Forget it. It's a mechanical solution, doesn't make any difference what the remainder of the circuit is and since I can't explain it - I'm outta here

16. ### belae1ka Thread Starter New Member

Nov 20, 2016
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haha this is too good.

Apr 11, 2010
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18. ### WBahn Moderator

Mar 31, 2012
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I don't follow this at all.

It looks like you've got two potentiometers. The bottom of the left one is unconnected and the top of the right one is unconnected but the wipers are connected together.

You are claiming that the only difference between the original (never shown) circuit and your proposed circuit is that the overall resistance of (one) potentiometer has been increased.

19. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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Sigh... it tough when you aren't understood. Which is why I gave up initially. Let me try a much more detailed explanation.

First, I agree with many of you that no circuit has been presented. Hence, I took a black box analysis (?) approach.

The TS wants to replace a pot with one that operates over 90°. We can make several assumptions. It has a resistance whose value I notate as R. The second assumption is that it is a standard pot. A wiper and two terminals between which the wiper forms a voltage divider.

Note that there is nothing required from the original circuit to proceed. The diagram present makes two more assumptions. That the ends of the equivalent pot are Vcc and Gnd. These are not critical to a solution, as the equivalent pot's connection inside the black box would still work; we are simply replacing s potentiometer.

The narrow bands of resistance are equal to 90° of rotation. These bands are the only resistance of the pot which are used.

With the way the wipers are wired, it is equivalent to one pot of the original resistance. So far, electrically identical to the original black box (I'd say circuit, but I agree with everybody since there is no circuit.)

But what about those long, unconnected tails of resistance on the pots? They do nothing electrically, but mechanically they allow the two pots to be ganged together, physically aligned so that the two narrow bands also align.

So how long or how much resistance are these "unconnected tails"? Or what is the desired resistance of the pot, so that over 300°, there is enough resistance such that the resistance of a 90° segment is equal to the total pot resistance in the original black box?

Fortunately, this is easy to calculate. The original pot rotated 300° and we desire only 90° The ratio of the two is equal to 300°/90° or 3.33.

You can observe these calculated values in use on the diagram. The narrow bands at the top and bottom have a resistance of R and the middle band has a resistance of 1.3R. Hence, each unconnected tail has a resistance of 2.3R, but we don't care because this part of the pots are never used.

Hence, mechanically, now we have a pot that within a 90° rotation, ranges over the same value as the original pot. The end stops of the two pots, together, mechanically limit the stacked pot to 90°

Electrically, it is the same as the original pot.

Of course, if one could find a pot of the correct value with 90° stops, forget about this.

But this is doable. It is a combination of electronic and mechanical solutions, but I am used to thinking in that space.

Also, I am working with a potentiometer manufacturer that specializes in stackable pots.

@WBahn , better?

20. ### WBahn Moderator

Mar 31, 2012
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Okay, now I see what you are doing. This is quite different that was seemed to be being implied by Posts #5 and #8 where you seemed to be saying that you replace the existing potentiometer with one that has the same total resistance at the 90° mark and that nothing else needed to be changed. There wasn't anything in those posts that hinted that you were replacing the original pot with two pots, positioning them so that when one was at 90° the other was at the opposite extreme end, ganging them together mechanically, and then leaving the other end of each pot disconnected.

I don't know how viable this approach is for the TS -- they have to be able to mechanically gang the two pots together. That may be far more hassle than it is worth. If they have the ability to do that, it is probably easier to gear up the existing pot so that a 90° rotation of the input shaft turns the pot shaft the full amount.

BTW: I'd recommend tying the dangling end of each pot to the wiper so that it isn't sitting there as an antenna picking up noise (although they are now small loop antennas, but I think that is probably still better than dangling free).