In what way does current flow through a potentiometer?

[rtareen]

TL;DR

Self-studying electronics and I'm confused about the current flow through a potentiometer.

Above is a picture of a potentiometer with terminals A, B, and W. Does current always from from A to W or can current flow directly from A to B? Is there usually both? And what do they mean when they say a potentiometer is an effective "voltage divider"?

 

[phinds]

If you USE only A and B then you just have a resistor. Normally, there would not BE a "B".

 

[rtareen]

If you USE only A and B then you just have a resistor. Normally, there would not BE a "B".

Lets say we have wires connected to all three terminals and current is flowing into terminal A. Would current flow to both W and B?

 

[phinds]

Lets say we have wires connected to all three terminals and current is flowing into terminal A. Would current flow to both W and B?

Generally, yes, but really it depends on what you have them hooked to. You COULD create a circuit where current flowed INTO B

 

[Tom.G]

Hi @rtareen,

The drawing seems to show a view from the back without a cover, the shaft and knob would be on the 'far' side of the potentiometer.

The most common use of a potentiometer is as a volume control on a device.

The full audio signal would be connected to the 'A' terminal, and the 'B' terminal would be connected to circuit Ground.

The 'W' terminal would connect to the audio amplifier that drives the speaker/earphones.

The Black strip is a resistive material, often measuring 5000 or 10 000 Ohms between the 'A' and 'B' ends if used as a volume control.

With the wiper positioned at the 'A' contact (full clockwise from the front), it would be getting the full audio signal; when fully counter-clockwise, the wiper connects to the 'B' signal, which is Ground (no signal).

As you may have figured out by now, when the wiper is at some position other than the ends, it gets only a proportion of the audio signal.

If you have studied voltage dividers (two or more resistors in series across a voltage source), you have learned that the voltage at the junction of the two resistors depends on the ratio of the resistor values (and of course the supplied voltage).

As the potentiometer Wiper is rotated, there is a varying resistance between 'A' and wiper and a varying resistance between wiper and 'B'; as one increases the other decreases. This changes the ratio of the resistance values to the wiper, making the potentiometer effectively a variable voltage divider.

Hope this helps!

Cheers,
Tom

 

[LvW]

You can see the whole device as a series connection of two resistors: One resistor between A and W and the other one between W and B.
When the wiper is moved the one resistor increases its value and the other one descreases by the same amount (and the sum remains unchanged).
Because of the series connection the same current does exist in both parts (both resistors). And because the voltage across each resistor is directly proportional to the resistor value (constant current when the connected voltage source between A and B remains fixed) the voltage is "divided" depending on the wiper position.

 

[darth boozer]

If you imagine the track straightened out, as shown in the attached picture, you can see that the voltage across A and B is the sum of the voltage across A and W and the voltage across W and B. Depending on the position of the wiper, these voltages will change but the sum will remain the same. Thus the voltage is divided.

 

[Averagesupernova]

My first thought on this was if the OP needs to ask about voltage division, a potentiometer is not the place to start. No disrespect intended. After all, at one time all of us didn't know 'where to start'.

 

[LvW]

My first thought on this was if the OP needs to ask about voltage division, a potentiometer is not the place to start. No disrespect intended. After all, at one time all of us didn't know 'where to start'.

I rather think a potentiometer would be a very good start. Let me explain:

During circuit analysis we very often apply Ohms law in the form V=I*R. That means: We treat the voltage V as it would be "created" (developed) across R by the current I. This works well and all the calculations are correct. However - physically spoken - it is wrong. Vice versa - it is the voltage (and the associated E-field within the conducting material) that allows the movement of charges (current I).
A current through the resistor R can never produce the corresponding voltage V across the resistor R.

This can be best explained (visualized) by one single resistance Rtot (like the total potentiometer path) with a wiper which divides the total resistance Rtot into two parts.
It is clear that the voltage between A and B (see the figure in the original question) creates a corresponding E-field within the material - and allows the current I=V_AB/Rtot. Now - it is obvious that between A and W (and also between W and B) we have smaller voltages because of V=E*L (E-field E constant, L=length between A and W or W and B, respectively). And - of course - the sum of both again is V_AB.
For my opinion, this explains best the principle of voltage division.
If we would start with the other alternative - two discrete resistors - the explanation based on the E-field would be not so clear and we would use instead the relation V=I*R.
As mentioned above, this would work - however, on the basis of a physically wrong assumption.
Not to forget - here we are in a "Physics Forum" and each good electronic engineer should know the physical background of the equations he is using.

(Remark: Some similar considerations apply in connection with the use of the relation Ic=beta*Ib, and some people derive from this equation that Ib would control Ic - but this is not the case. But according to my experience - a very "critical" question).

Remark: Rather often I can read that the voltage-current relation would be a chicken-egg problem. I think, this is not the case and such a wrong view results probably from a wrong interpretation of Ohms law (as explained above). Therefore: No current without a driving voltage - voltage first and current as a result. As a consequence - in reality, there are no real current sources.

 

[rtareen]

Hey everybody thanks for all the responses. For some reason I never got any email notifications so I thought this thread was dead. Sorry for not replying. I will take eveything into account

 

[rtareen]

Hello all,

For the sake of not starting a new thread, can somebody tell me if my drawing would be a correct way to set up a potentiometer as a voltage divider? What I'm trying to figure out is how you would physically lay out the voltage divider. This is just a guess but I hope its right.

 

[rtareen]

You can see the whole device as a series connection of two resistors: One resistor between A and W and the other one between W and B.
When the wiper is moved the one resistor increases its value and the other one descreases by the same amount (and the sum remains unchanged).
Because of the series connection the same current does exist in both parts (both resistors). And because the voltage across each resistor is directly proportional to the resistor value (constant current when the connected voltage source between A and B remains fixed) the voltage is "divided" depending on the wiper position.

How can it be the same current if some current from terminal A ends up in W and some goes to B? Or maybe if Current from both A and W both sum to the current in B? I don't understand what you mean.

 

[Tom.G]

For the sake of not starting a new thread, can somebody tell me if my drawing would be a correct way to set up a potentiometer as a voltage divider? What I'm trying to figure out is how you would physically lay out the voltage divider. This is just a guess but I hope its right.

Yes, that would work.

Often in a schematic diagram, the terminal at far clockwise rotation (as viewed by the person operating the device) would be labeled with "CW".

Now can you specify which direction you are assuming would increase/decrease Vout?

Which terminal would be considered the "Common" or Ground terminal?

Cheers,
Tom

 

[rtareen]

Yes, that would work.

Often in a schematic diagram, the terminal at far clockwise rotation (as viewed by the person operating the device) would be labeled with "CW".

So you're saying terminal A would be labeled CW? I'm guessing that's just another notation for labeling the terminals, right? What about in the counter-clockwise direction? What would we call that one if we stick to this convention?

Now can you specify which direction you are assuming would increase/decrease Vout?

I'm not 100% confident in this answer, but here it goes. $V_{out}$ would be in parallel with whatever we have as the right-hand side resistance. Moving the wiper clockwise decreases this arc. So it would decrease $V_{out}$. Moving it CCW would increase the length and thus increase $V_{out}$.

Which terminal would be considered the "Common" or Ground terminal?

If I had to choose I would say terminal A because the point is to have $V_{out}$ at some specific voltage, and that takes up both W and B?

 

[Tom.G]

So you're saying terminal A would be labeled CW? I'm guessing that's just another notation for labeling the terminals, right? What about in the counter-clockwise direction? What would we call that one if we stick to this convention?

Yes, that is the usual convention for labeling. It helps the person that is building the thing to know which wire goes to which end. If they get it backwards, the radio they are building will have full volume when the user thinks they are turning it down! (Ouch if you are wearing earphones.)

(Exception: Sometimes the connection labeling reflects the manufacturers labeling, such as 1, 2, 3, rather than CW. The 1, 2, 3, being what is shown on the manufacturers mechanical drawing for the part. This makes it a little easier for the person designing a PC board but a little tougher for someone troubleshooting a problem with it.)

The other two terminals are not generally labeled as they are obvious once one end is labeled; the wiper is drawn with an arrow on its end at some position along the resistor, leaving only the CCW end. This image shows how a potentiometer is shown on a schematic diagram.

The "Common" terminal name is used generically for a connection or terminal that is used for both Input and Output (along with some other more confusing definitions!) Which of the teminals in your original image meets this definition?

Your answer to the which direction increases/decreases the output confuses me a little; as in the CW you answered, does that refer to the original drawing as shown, or to the definition in post#13?

Cheers,
Tom

 

[LvW]

How can it be the same current if some current from terminal A ends up in W and some goes to B? Or maybe if Current from both A and W both sum to the current in B? I don't understand what you mean.

What do you mean with "ends up in W" ? A current cannot "end up". When "W" is an open terminal (no load connected) there will be no current through this terminal. The current will go from A to B and then back to the driving voltage source. But the VOLTAGE between A and W will be smaller than between A and B because the length between A and W (and with it the voltage producing E-field) is just a part of the total length (from A to B).

 

[berkeman]

So you're saying terminal A would be labeled CW? I'm guessing that's just another notation for labeling the terminals, right? What about in the counter-clockwise direction? What would we call that one if we stick to this convention?

One of the reasons it's important to know which end is CW on a potentiometer is so that you can design your circuit to be "intuitive" when the user turns the potentiometer. For example, we are all used to the volume going up when you turn the volume knob CW. And lights get brighter and numbers increase with CW motion of controls.

BTW, there are a number of design tips for using potentiometers. For example, there is a Minimum Wiper Current required for most potentiometers to avoid corrosion and changing resistance at the wiper contact. The "Bourns Trimmer Primer" is a great tutorial for how to design reliable circuits using potentiometers:

https://www.bourns.com/docs/default-document-library/bourns_trimmer_primer.pdf

 

[rtareen]

The "Common" terminal name is used generically for a connection or terminal that is used for both Input and Output (along with some other more confusing definitions!) Which of the teminals in your original image meets this definition?

In that case I believe it would be terminal B since its directly wired to both $V_{in}$ and $V_{out}$. However, what I understand of the common is that we usually take it to be at zero voltage relative to all other points in the circuit. Furthermore I'm somewhat aware that in complex circuits there can be many points that are "connected to ground", meaning they all have the same voltage, which is the reference voltage. Can you explain a little more about this?

Your answer to the which direction increases/decreases the output confuses me a little; as in the CW you answered, does that refer to the original drawing as shown, or to the definition in post#13?

I was referring to the second picture I drew, since that one shows the entire layout including $V_{out}$

 

[rtareen]

What do you mean with "ends up in W" ? A current cannot "end up". When "W" is an open terminal (no load connected) there will be no current through this terminal. The current will go from A to B and then back to the driving voltage source. But the VOLTAGE between A and W will be smaller than between A and B because the length between A and W (and with it the voltage producing E-field) is just a part of the total length (from A to B).

Now I think I understand. So you were just saying that the current through A and B are the same when W is open?

 

[rtareen]

One of the reasons it's important to know which end is CW on a potentiometer is so that you can design your circuit to be "intuitive" when the user turns the potentiometer. For example, we are all used to the volume going up when you turn the volume knob CW. And lights get brighter and numbers increase with CW motion of controls.

BTW, there are a number of design tips for using potentiometers. For example, there is a Minimum Wiper Current required for most potentiometers to avoid corrosion and changing resistance at the wiper contact. The "Bourns Trimmer Primer" is a great tutorial for how to design reliable circuits using potentiometers:

https://www.bourns.com/docs/default-document-library/bourns_trimmer_primer.pdf

Thank you so much! I will take a look at this.

 

[Tom.G]

Which terminal would be considered the "Common" or Ground terminal?

If I had to choose I would say terminal A because the point is to have at some specific voltage, and that takes up both W and B?

But in post #11, you show V_out measured between "W" and "B"

In that case I believe it would be terminal B since its directly wired to both V_in and V_out. However, what I understand of the common is that we usually take it to be at zero voltage relative to all other points in the circuit. Furthermore I'm somewhat aware that in complex circuits there can be many points that are "connected to ground", meaning they all have the same voltage, which is the reference voltage. Can you explain a little more about this?

Well, let's see... If you measure V_in, you could connect a meter to terminals "A" and "B".

Then to measure V_out, there is already a meter lead on terminal "B" from the V_in measurement you just made; no need to move that meter lead. Move the other meter lead from "A" to "W" to measure V_out.

As you can see, "B" is used for both measurements, therefore it is called the Common Reference point FOR THOSE MEASUREMENTS.

 

[hutchphd]

Forgive me if this has been covered (I can't find it): For the resistor to act as a perfectly linear potentiometer,, the current allowed to flow on the wiper leg must be effectively zero. This means in practice that the resistance value is chosen low enough that bias current from A to B is much bigger than any current allowed to flow out of V at any configuration of the wiper (when the rest of the circuit is attached to V) .

 

[rtareen]

Forgive me if this has been covered (I can't find it): For the resistor to act as a perfectly linear potentiometer,, the current allowed to flow on the wiper leg must be effectively zero. This means in practice that the resistance value is chosen low enough that bias current from A to B is much bigger than any current allowed to flow out of V at any configuration of the wiper (when the rest of the circuit is attached to V) .

I don't think that was covered here either. At least the part about it being linear. What do you mean by wiper "leg". Is that the same thing as terminal W? Or do you mean both terminals W and B (referring to my second drawing where I drew the voltage divider layout).

 

[rtareen]

But in post #11, you show V_out measured between "W" and "B"

I corrected my answer in post #18 where I said the common for $V_{in}$ and $V_{out}$ is probably terminal B.

By the way, thanks a lot to you and everyone else. This has been a pleasant experience getting so much input from experts. This is by far the longest thread on any post I ever made. I wouldn't have gotten all this by just reading the book and doing nothing else. This is all actually for a class next semester that I am trying to get a head start on. I think I'll be pretty well prepared for this specific topic. I'll definitely be coming back with more questions soon. You will probably see me here frequently for the next few months.

 

[hutchphd]

Is that the same thing as terminal W

Yes That is what I meant (I just called it out wrong). But in order to measure (or use) that voltage something (even just a voltmeter) will attach there and complete a circuit to ether A or B, and use (hopefully a small amount) of current. But that will affect the nice linear voltage ramp ( divider) from A to B. It is easy enough to calculate, if you know the external resistance, but it is always a design consideration when choosing the part.