Can I convert a potentiometer to 2 resistors?

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Discussion Overview

The discussion revolves around the concept of converting a potentiometer into two resistors and understanding the implications of this conversion in a circuit. Participants explore the theoretical and practical aspects of potentiometers, including their behavior in voltage divider configurations and their application in op-amp circuits.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the potentiometer effectively splits its total resistance into two parts based on the position of the slider.
  • Others argue that the total resistance remains constant regardless of how the potentiometer is divided, similar to cutting a string into two pieces.
  • A later reply questions the appropriateness of using the potentiometer in the given circuit, suggesting potential reliability issues.
  • Participants discuss the calculation of voltage at specific points when the potentiometer is set to a certain percentage of its total value, with some confusion about the correct interpretation of these values.
  • There are challenges regarding the calculations and assumptions made about the voltage divider rule as applied to the potentiometer.
  • Some participants clarify that the resistance values derived from the potentiometer depend on the slider's position, which can create confusion in calculations.
  • There is a mention of the importance of not directly connecting the wiper of a potentiometer to an op-amp input due to design concerns.

Areas of Agreement / Disagreement

Participants generally agree on the basic principle that a potentiometer divides its resistance, but there is significant disagreement and confusion regarding the calculations and implications of this division in specific circuit contexts. The discussion remains unresolved with multiple competing views on the correct approach.

Contextual Notes

Some limitations include unclear definitions of terms and assumptions about the circuit configuration that have not been explicitly stated, leading to confusion in calculations and interpretations.

Who May Find This Useful

This discussion may be useful for individuals interested in circuit design, particularly those working with potentiometers and op-amps, as well as those looking to deepen their understanding of voltage dividers and resistance calculations.

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I think that is correct. Wherever the slide is splits the total resistance in two
 
Femme_physics said:
Is that how it works basically?

That is exactly how it works. The two resistor values sum to the total resistance of the potentiometer.

It's like cutting a piece of string: You can cut it anywhere but the lengths of the two pieces still add up to the length of the whole string.
 
Berkeman -- is it just because I didn't do p/2 ? Why would 2 people said I split it correctly, though?If I may carry on with the exercise...I need to find the value at point a IF the potentiometer is at 25% of its valuehttp://img31.imageshack.us/img31/2733/nyetvibor.jpg

Did I get it right?
 
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I need to find the value at point a IF the potentiometer is at 25% of its value
Hello again Femme_physics! :smile: You say 25%, but in your working you've set it at the 50% mark. Which should it be?
 
Hey NasOx!

Oops, I forgot to divide by two. But the result of Va = 2 [V] still ends up the same :)

And then Va = Vb due to the fact it's a feedback op-amp

Now I imagine this is an inverter op-amp, ergo

5d482bf78fb686a8ed1e8ebc894fdf5a.png


Vc = Vout = -8 [V]

Who's the queen? :)
 
When a is 25%, the 8K potentiometer is divided into two sections. One is 25% of 8K and the other is 75% of 8K. End-to-end the potentiometer still totals 8K.
 
So you have, for example, RP1=0.75P and RP2=0.25P, and Vi=4v, and you still think Va=2V?? What happened to our "potential divider" expert from two weeks ago?
Who's the queen?
The crown has yet to be claimed. ♔[/size]
 
  • #10
gneill said:
When a is 25%, the 8K potentiometer is divided into two sections. One is 25% of 8K and the other is 75% of 8K. End-to-end the potentiometer still totals 8K.
Wait, gneill, I'll reiterate the question word by word

Potentiometer P is set at 25% of its value (the resistance between point a and Earth is 25% of P value).

I'm just all the more confused now. If I split the potentiometer to 2 resistors, don't they both must equal to the value of I set the potentiometer at? According to the question IMO it's 2 volts TOTAL. i.e. their sum is 2 volts and they split equally...the way I read it

NasOx-- just saw ur reply give me a moment :)
 
  • #11
Femme_physics said:
Wait, gneill, I'll reiterate the question word by word

Potentiometer P is set at 25% of its value (the resistance between point a and Earth is 25% of P value).
...so the portion above point a must be 75% of P value.
I'm just all the more confused now. If I split the potentiometer to 2 resistors, don't they both must equal to the value of I set the potentiometer at? According to the question IMO it's 2 volts TOTAL. i.e. their sum is 2 volts and they split equally...the way I read it
The physical implementation of a potentiometer has a resistive strip whose total end-to-end resistance is the "value" of the potentiometer. Each end of the strip has a contact. There is a third contact that can slide from end to end along the surface of the strip. This contact "divides" the strip into two portions according to the slider location. The resistance between the sliding contact and an end of the strip is proportional to the length of strip lying between the contact point and that end. The portions can be any ratio of lengths (and thus resistances) but their sum must still add up to the net length (resistance) of the strip.

attachment.php?attachmentid=46153&stc=1&d=1334241513.gif


From the original circuit diagram it looks like the potentiometer can "tap" any value between 0 and 4V.
 

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  • #12
So you have, for example, RP1=0.75P and RP2=0.25P, and Vi=4v, and you still think Va=2V?? What happened to our "potential divider" expert from two weeks ago?

She didn't have to deal with weird potentiometer thingies!

...so the portion above point a must be 75% of P value.
ahh...I see! So...

http://img337.imageshack.us/img337/9110/pottttt.jpg

yes?
 
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  • #13
yes?
No. Let's go back to your diagram in post #5 of this thread. See on RP1 where you wrote "p/2"? Replace that writing with 3p/4. See RP2? Replace "p/2" there with p/4.

That's the only change. Now analyze it. https://www.physicsforums.com/images/icons/icon6.gif
 
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  • #14
NascentOxygen said:
No. Let's go back to your diagram in post #5 of this thread. See on RP1 where you wrote "p/2"? Replace that writing with 3p/4. See RP2? Replace "p/2" there with p/4.

That's the only change. Now analyze it. https://www.physicsforums.com/images/icons/icon6.gif

So P in that case is still its original value. It's only from point a to the ground that the 0.25 is?

Fine, but how was I supposed to have known all that by the question if it hasn't been defined like that?
 
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  • #15
Femme_physics said:
So P in that case is still its original value. It's only from point a to the ground that the 0.25 is?
The connection is 25% along the length of the resistor element (i.e., measured from the earthed end). So now the op-amp's (+) input is being supplied 25% of the 4 volts.
Fine, but how was I supposed to have known all that by the question if it hasn't been defined like that?
<shrugs>
 
  • #16
Femme_physics said:
Berkeman -- is it just because I didn't do p/2 ? Why would 2 people said I split it correctly, though?

The voltage divider nature of the potentiometer is correct. It is bad design practice to connect the wiper of a potentiometer directly to just an opamp input. The reason is described in the post that I linked to.
 
  • #17
gneill said:
...so the portion above point a must be 75% of P value.

The physical implementation of a potentiometer has a resistive strip whose total end-to-end resistance is the "value" of the potentiometer. Each end of the strip has a contact. There is a third contact that can slide from end to end along the surface of the strip. This contact "divides" the strip into two portions according to the slider location. The resistance between the sliding contact and an end of the strip is proportional to the length of strip lying between the contact point and that end. The portions can be any ratio of lengths (and thus resistances) but their sum must still add up to the net length (resistance) of the strip.

attachment.php?attachmentid=46153&stc=1&d=1334241513.gif


From the original circuit diagram it looks like the potentiometer can "tap" any value between 0 and 4V.

Ahhh perfect explanation and diagram. See, I thought that a potentiometer is a device that regulate the amount of resistance of a resistor. But it just regulates the spread of differences...makes sense.


The connection is 25% along the length of the resistor element (i.e., measured from the earthed end). So now the op-amp's (+) input is being supplied 25% of the 4 volts.

Yes sir! I do most certainly can prove it with calculations, too!


http://img513.imageshack.us/img513/4789/page42d.jpg
 
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  • #18
No, not quite there yet. ☹[/size][/color]

Does it not seem strange that you arrived at your answer without involving Va anywhere in the calculation? You went to all the trouble of determining what the potentiometer setting meant, then calculated the voltage at the non-inverting input of the op-amp, then calmly discarded all that work and went back to the Rf/Rin equation you remembered from a simpler circuit previously
UDocr.gif


I'm afraid you don't get out of it that easily. :smile:

So let's go back to your step ➁...

Va=Vb is correct, and as always this relationship applies for any op-amp functioning as an amplifier.
You determined Va = 1V. Therefore, we can say Vb = 1V

Now, let's look at the network connected to the op-amps (–) input. There's Vi at one end of R1 and Vb at its other end. So what is the expression for current through R1? Use Ohm's Law.

Next, look at R2. What is the voltage at each of its ends? So determine an expression for the current through R2.

The op-amp, as always, draws practically no current into either of its input terminals, so we are left with the customary relation, current in R1 = current in R2. So equate your expressions for these currents, and solve for Vout.

See how you go this time. :smile:
 
  • #19
Does it not seem strange that you arrived at your answer without involving Va anywhere in the calculation? You went to all the trouble of determining what the potentiometer setting meant, then calculated the voltage at the non-inverting input of the op-amp, then calmly discarded all that work and went back to the Rf/Rin equation you remembered from a simpler circuit previously

I guess I just didn't think they are related...hmm...

I'm afraid you don't get out of it that easily.

That's usually how it turns out :)

So let's go back to your step ➁...

Va=Vb is correct, and as always this relationship applies for any op-amp functioning as an amplifier.
You determined Va = 1V. Therefore, we can say Vb = 1V

Agreed

Now, let's look at the network connected to the op-amps (–) input. There's Vi at one end of R1 and Vb at its other end. So what is the expression for current through R1? Use Ohm's Law.

Ir1 = Va/R1 = 1/3000 = 0.333 [mA]

Next, look at R2. What is the voltage at each of its ends? So determine an expression for the current through R2.

The voltage at R2 must be 3 [V]...as 4-1 = 3[V]

Ir2 = Vr2 / R2 = 0.5 [mA]

The op-amp, as always, draws practically no current into either of its input terminals, so we are left with the customary relation, current in R1 = current in R2. So equate your expressions for these currents, and solve for Vout.

http://img59.imageshack.us/img59/1320/thingyhere.jpg

Same result for when using the formula!
 
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  • #20
Femme_physics said:
Ir1 = Va/R1 = 1/3000 = 0.333 [mA]
Nope. What's the potential drop across R1? It's not Va --- What happened to Vin? Vin is at one end of R1 and Va at the other, so the potential difference is...
 
  • #21
Ah, ah...my bad

Ir1 = Vin - Va / R1

Ir1 = 3 / 3000

Ir1 = 1 [mA]
 
  • #22
Femme_physics said:
Ah, ah...my bad

Ir1 = Vin - Va / R1

Ir1 = 3 / 3000

Ir1 = 1 [mA]

Correct. Now, since the input to the op-amp does not draw any current, what happens to that 1mA?
 
  • #23
Now, let's look at the network connected to the op-amps (–) input. There's Vi at one end of R1 and Vb at its other end. So what is the expression for current through R1? Use Ohm's Law.

Ir1 = Va/R1 = 1/3000 = 0.333 [mA]

That's not the Ohm's Law I know. :rolleyes:

My Ohm's Law says that the current through a resistor is 1/R times the voltage difference across that resistor. You need to establish the voltage difference across R1. Re-read what I have quoted at the top here.
 
  • #24
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  • #25
It's -5 V actually :smile: A small algebra sign issue.

Note that you don't always have to trace out a complete loop to apply voltage sums. If you know the potential at one location in a circuit and have all the potential changes between that point and another point, you can simply add up all the changes to determine the net change from start to finish.

In this case you know the potential at b is Va, and have determined the potential drop across R2 to be VR2 = -I*R2. So Vout = Va + VR2.

A quick way to analyze this type of op-amp arrangement is to proceed as follows:
1. Determine the "reference" potential at the +input of the op-amp. This will also be the potential at the -input.
2. Determine the current from the input that flows through the input resistor due to the potential difference across it.
3. Determine the potential drop across the feedback resistor due to this same current.
4. Add this drop to the known potential at the -input to find Vout.

attachment.php?attachmentid=46220&stc=1&d=1334416965.gif
 

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  • #26
It's -5 V actually A small algebra sign issue.

Right! I see it.
Note that you don't always have to trace out a complete loop to apply voltage sums. If you know the potential at one location in a circuit and have all the potential changes between that point and another point, you can simply add up all the changes to determine the net change from start to finish.

In this case you know the potential at b is Va, and have determined the potential drop across R2 to be VR2 = -I*R2. So Vout = Va + VR2

Hm, still, not that much faster, so I rather be fundamental :smile:

1. Determine the "reference" potential at the +input of the op-amp. This will also be the potential at the -input.

Yes, made sure to realize that a few days ago

. Determine the current from the input that flows through the input resistor due to the potential difference across it.
3. Determine the potential drop across the feedback resistor due to this same current.
4. Add this drop to the known potential at the -input to find Vout.

Good system. I'll try to get more practice applying KVL until I get those op-amps right

BTW the next question ask me what happens if I set the potentiometer to zero...I think I got it, but just for verification..:

http://img809.imageshack.us/img809/8698/rollinglikethis.jpg
 
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  • #27
Yup. That's it. Or more succinctly:

##V^- = V^+ = 0V##
##Vout = V^- - \frac{Vin}{R1}R2##
 
  • #29
Yes again :smile:
 
  • #30
Yes!

Just to check, I'm asked if there's a linear connection between how you set "P" in precent, to Vout

Now, I presume that when Vout is -8 volts it's considered a higher voltage than -5 and even 0, since "voltage" by definition is all about potential difference?

So, in my opinion there is a linear connection since Vout increases the lower the potentiometer is!
 

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