# pot value in voltage divider?

Discussion in 'General Electronics Chat' started by tracecom, Nov 22, 2013.

1. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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I am having trouble posting this, so I'll try again. If it's a duplicate, I'll try to remove it.

When using a pot as a voltage divider, does it make any difference what the resistance value of the pot is? I know that the voltage out is just a ratio and can be attained with any pot value, but are there other reasons for choosing a particular resistance value. Is there any advantage to using a 1k versus a 100k, or vice versa, and if so why?

Thanks.

2. ### MrChips Moderator

Oct 2, 2009
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Yes it does matter on two points:

1) input impedance
2) output impedance

The resistance value of the pot will determine the current draw - the lower the resistance the higher the current.

1) Rule of thumb

You want the resistance value of the pot to be at least 10 times greater than the resistance of the voltage source. This will limit the effect of the pot to 10% on the source.

2) Rule of thumb

You want the load connected to the voltage divider (center terminal of the pot) to be at least 10 times greater than the resistance of the pot. Again this will limit the load effect on the voltage output of the divider to 10%.

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3. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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Mr Chips,

Thanks for the reply. I will let that bounce around in my head in the hopes I can fully understand. I want to ask a question, but I am not sure what to ask. Maybe if you gave an example for each rule of thumb?

What has prompted my interest is that I needed easy access to variable reference voltages for comparator circuits. So, I assembled an array of three 1k pots, each arranged as a voltage divider. It seems to be doing what I need, but I wonder if I should have chosen different resistance pots.

I know that's a rambling response, but maybe you can understand what I am trying to get at.

Thanks.

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4. ### Bernard AAC Fanatic!

Aug 7, 2008
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Do not exceed wattage rating of pot, I assume that most small trim type pots are about 1/8W to 1/4W, & should not be more than slightly warm.
For small battery applications keep pot as high resistance as load can tolerate.
With 12V supply,1k pot, power = 144 mW & output relativley low impedance; sounds like a good choice.

Last edited: Nov 22, 2013
5. ### MrChips Moderator

Oct 2, 2009
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To clarify, the resistance of the pot will depend on the application, situation and circuit topology.

Here is a simple example to illustrate the 10x rule.

Suppose the voltage source to be divided has a resistance of 1kΩ.
Then I could select a 10kΩ pot.
I would have to make sure that the load on the center wiper of the voltage divider is greater than 100kΩ.

But note that these are not absolute rules. It just means that when I adjust the pot I will not get a perfectly linear relationship, i.e. I wouldn't get 50% reduction when the pot is set to exactly 50%. If this is simply a trimmer adjustment then the non-linear relationship is not critical.

In your particular situation, three 1kΩ pots in parallel is equivalent to one 333Ω load on your source. Can the source handle this extra current? Do you really need to waste that much current?

The input bias current of an LM339 comparator (a commonly use IC) is less that 1μA. Hence the input resistance is greater than 1MΩ.

You could very well increase the resistance of the pots to 10k-100kΩ and still satisfy the 10x rule of thumb while reducing the current drain and power dissipation of the pot.

Edit: BTW, don't forget to put 0.1-10μF smoothing capacitors on the wiper terminal of your pot in order to stabilize the voltage reference.

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6. ### MrChips Moderator

Oct 2, 2009
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Here is another way of putting it.

Suppose your voltage divider is being fed from the 5V supply.
A 1kΩ pot is going to take 5mA of which one LM339 input will take 25nA.

Instead, a 50kΩ pot will take 100μA which is still 4000 times 25nA, way more than the 10x rule.

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7. ### #12 Expert

Nov 30, 2010
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All this proportionality complicates things, doesn't it?
I use a rule I learned from a real engineer, "If you design for (at least) 1 ma you will usually avoid noise problems". That's for precision analog design. It works for circuit practice, most of the time. You will know if you're using an LED that needs 10 ma or a buzzer that needs 50 ma. For just experimenting with chips, 1 ma works most of the time, and it won't burn up your pots.

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8. ### wayneh Expert

Sep 9, 2010
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[referring to #6, not #12 who snuck in] That's the way I look at the choice. Calculate the current needed from the wiper. Multiply by 10. Choose a pot that gives at least that much current. If that required pot value is greater than, say, 10K, I just use a 10K and forget about the tiny amount of power I might be wasting by not using a 50k or a 100k instead, even though those values would be OK. Of course I'd use the 50k if that's what I have on hand. I don't use pots over 500k.

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9. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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Do you mean a .1uF and a 10uF or just one cap per pot that is in the range of .1uF to 10uF?

And exactly how should they be connected: from wiper to ground?

Thanks again.

Last edited: Nov 24, 2013
10. ### MrChips Moderator

Oct 2, 2009
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I mean any thing in that range, from wiper to ground.

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11. ### MikeML AAC Fanatic!

Oct 2, 2009
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Stare at this. It shows the deviation from the expected 5V output as a function of the load resistance for a 10K pot. It is just a consequence of the Thevenin Equiv. of the pot, i.e. 5V in series with 5K (worst case when the pot is centered).

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12. ### WBahn Moderator

Mar 31, 2012
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Whether you go wiper to ground or wiper to the other terminal depends on what your noise sensitivities are. That can usually be answered by asking what you want the output of the pot to do if the supply voltage were to go up a bit. Would you want the voltage at the wiper to stay the same, or would you want it to track the supply voltage? Different circuits lead to different answers to that question. You put the cap across whichever terminals will lead to the answer you want being the answer in practice.

Mar 31, 2012
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And also

3) noise

14. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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I plan to use the output as a voltage reference, and may provide the input voltage in either of two ways: from a separate regulated supply, or from the power supply of the circuit under test. In the first case, I can control the input voltage, and in the second case, I would want the output to track the supply voltage. Therefore, I want to put the properly sized cap in the place that facilitates noise reduction and lets the output voltage vary with the input. The input voltage will most often be 5 to 12 VDC.

So, what value cap would be best, and from wiper to what?

Thanks

15. ### #12 Expert

Nov 30, 2010
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Just throw in any old cap in the range of 10 uf to 100 uf.

5K and 10 uf gives a time constant of .05 seconds which translates to 20 HZ. That will keep your filter under the power line frequency.

16. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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Here's the finished product. I changed to 10k pots and used 10uF caps.

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17. ### #12 Expert

Nov 30, 2010
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You really are pretty neat about physical products. I wouldn't show you a photo of the last potentiometer board I made! What I have laying around is sheet metal. Drill some holes, attach some pots, and it looks like something I found in a junk yard.

18. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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Thanks...I think.

My electronic theory knowledge is so limited that I depend quite a lot on experiments, and the simpler I can keep the wiring, the better, so I do tend to try to keep things neat, when possible.

19. ### wayneh Expert

Sep 9, 2010
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Looks nice. I'd add labels while you still remember what it is. Voice of my own experience.

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20. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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Right now, I think I will remember.

What was it again?

You are correct; I will label it. I already find little perfboard assemblies that I don't remember building.