# Ideal operational amplifier circuit with variable voltage gain, how can I calculate the current?

#### WBahn

Joined Mar 31, 2012
28,510
So the current through the pot is just Vi multiplied by the overall pot resistance ?
Review Ohm's Law. Do you multiply a voltage and a resistance to get a current?

If I am correct, then yes.
How did the value of R1 come into play?

Not really, if I am looking at the overall resistance of the pot, but if it is all the way at the top then it would virtually short circuit R1 meaning that all of the current would flow through it, wouldn't it?
Yes, but does it have any effect on how much current flows in the pot?

Well if we say that the total resistance of the pot is Rpot(1-α)+Rpot(α)=Rpot-Rpotα+Rpotα=Rpot Rpot=R (not Rfred but the value of R1) so the current would then be Ix again?
Even if Rpot = R1, the current Ix depends on the position of the pot -- you yourself just got done pointing out that if the wiper is all the way at the top that there would be no current flowing in R1, meaning that Ix would be zero. But there would still be a current flowing in Rpot, wouldn't there?

#### Peterpan101

Joined Mar 4, 2017
20
Sorry, I just got completely confused. Looking at it from the current flow, the total current is just Ix+Ipot. I pot would then be the voltage across the pot (Vi?) divided by the total resistance of the pot?

#### MrAl

Joined Jun 17, 2014
10,098
Hi,

From your previous result you got Ix so you know the current through THAT resistor, however the total input current is through BOTH that resistor and the 'pot' total resistance. The input voltage is Vin as always, so the input resistance must be:
Rin=Vin/itotal

where itotal is just the sum of Ix and iPot.

That's just Ohm's Law.

In this problem you had an expression for Ix, and so you will end up with an expression for Rin as well.

This is actually much simpler than you are probably seeing right now. For example, if some other circuit like this one had Ix=5+b and the pot was Rp, the total current would be:
iTotal=Ix+Vin/Rp=5+b+Vin/Rp

and so the input resistance is:
Rin=Vin/(5+b+Vin/Rp)

That's just a fictitious example and may simplify.

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#### WBahn

Joined Mar 31, 2012
28,510
Sorry, I just got completely confused. Looking at it from the current flow, the total current is just Ix+Ipot. I pot would then be the voltage across the pot (Vi?) divided by the total resistance of the pot?
YES. This is because no current (or negligible current) flows in the wiper. If the wiper were connected to anything other than an op-amp input, that might not be the case.

The current Ix, however, does depend on the wiper position (but not the value of the pot).

So now you should be in a position to express the input resistance in terms of the three resistances and the wiper position. Do it first for unrelated values for the three resistances, R1, R2, and Rpot. Then show what it reduces to in the special case that they are all equal to R.

#### Peterpan101

Joined Mar 4, 2017
20
This is actually much simpler than you are probably seeing right now.
I think so too..just need to wrap my head around it

where itotal is just the sum of Ix and iPot.
so then Ri=Vi/Itotal

where Itotal=Ix+Ipot
thus
$$\small I_total = \frac{V_i-\alpha V_i}{R_1}+\frac{V_i}{R_pot}$$
I'm not sure how simplifiable this is when the R-values are not the same but when they are the two fractions have the same denominator so we then have:
$$\small I_total = \frac{2*V_i-\alpha V_i}{R}$$

putting that into the Rin equation gives us:
$$\small R_in =V_i* \frac{R}{2*V_i-\alpha V_i}$$
=$$\small R_in = \frac{R}{2-\alpha}$$

I don't know where/if R2 comes into play though? Am I overlooking it?

#### WBahn

Joined Mar 31, 2012
28,510
I think so too..just need to wrap my head around it

so then Ri=Vi/Itotal

where Itotal=Ix+Ipot
thus
$$\small I_total = \frac{V_i-\alpha V_i}{R_1}+\frac{V_i}{R_pot}$$
I'm not sure how simplifiable this is when the R-values are not the same but when they are the two fractions have the same denominator so we then have:
$$\small I_total = \frac{2*V_i-\alpha V_i}{R}$$

putting that into the Rin equation gives us:
$$\small R_in =V_i* \frac{R}{2*V_i-\alpha V_i}$$
=$$\small R_in = \frac{R}{2-\alpha}$$

I don't know where/if R2 comes into play though? Am I overlooking it?
Nope, R2 has no effect at all. The opamp isolated the feed back resistance from the input signal source.

Now ask if the answer makes sense at the extreme values of the wiper position? If it does, then you probably have it right.

#### Peterpan101

Joined Mar 4, 2017
20
So with the wiper at the top the input resistance would just be R, because α would be one, making the equation equal to R/1=R
that makes sense as R=Rpot.

In the bottom position α would be zero, so the total input Resistance Rin=R/2
That also makes sense because the two resistors are connected in parallel so:

$$\small \frac{1}{Rtot} = \frac{1}{Rpot}+\frac{1}{R1}=\frac{2}{R}$$

#### WBahn

Joined Mar 31, 2012
28,510
So with the wiper at the top the input resistance would just be R, because α would be one, making the equation equal to R/1=R
that makes sense as R=Rpot.

In the bottom position α would be zero, so the total input Resistance Rin=R/2
That also makes sense because the two resistors are connected in parallel so:

$$\small \frac{1}{Rtot} = \frac{1}{Rpot}+\frac{1}{R1}=\frac{2}{R}$$

Yes!

If you routinely add stuff like this to the work you turn in on every problem (or every problem in which it is possible, and it is usually possible), then a couple of things will happen. First and foremost, you will learn a lot more from the problem then you otherwise would have. Second, you will catch most of the mistakes that you might have made. Third, you will almost certainly bias the grader in your favor so that even when a mistake gets through (or perhaps you mess up on another problem) you will likely get significantly more partial credit -- or possibly extra credit outright. Remember, graders are human -- they see so much sloppy, poorly reasoned crap that they really like to reward good, solid work. I remember one time getting a 100% on a Statics assignment that I had made a pretty big error on one problem. The grader pointed it out and added the comment (paraphrasing after 30 years), "You should have lost 10 pts for this, but the work is so beautiful that I just can't bring myself to deduct any points. I'm confident you will learn from the error and not make it again." Oddly enough, that was before I was in the habit of doing sanity checks and, if I had, I would definitely have caught that error because the total forces would not have summed to zero. I have no recollection what the problem was, but I remember that much.

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#### Peterpan101

Joined Mar 4, 2017
20
Yes!!

Wow, I'll definitely keep that in mind for future exercises. The whole process here really made me feel like I gained a way deeper understanding (knowing where the formula in the textbook actually comes from, rethinking my answer as to if/why it makes sense and structuring and labeling my work correctly..) thank you all!

#### MrAl

Joined Jun 17, 2014
10,098
I think so too..just need to wrap my head around it

so then Ri=Vi/Itotal

where Itotal=Ix+Ipot
thus
$$\small I_total = \frac{V_i-\alpha V_i}{R_1}+\frac{V_i}{R_pot}$$
I'm not sure how simplifiable this is when the R-values are not the same but when they are the two fractions have the same denominator so we then have:
$$\small I_total = \frac{2*V_i-\alpha V_i}{R}$$

putting that into the Rin equation gives us:
$$\small R_in =V_i* \frac{R}{2*V_i-\alpha V_i}$$
=$$\small R_in = \frac{R}{2-\alpha}$$

I don't know where/if R2 comes into play though? Am I overlooking it?

Hi again,

Very good question. To find out the right answer you would just include R2 in ALL of the calculations and then see if it drops out in the final equation for Rin or whatever else you are solving for.
If R2 drops out then it's value does not matter, for the most part, unless of course it is some extreme value which interferes with the standard operation of the op amp itself which would be a practical concern not so much theoretical.

Before you do that however, you also have to define what "R" is for the pot itself. On the original schematic you do not show if R=R1 or R=R2 or if R just has it's own value (like 10k when R1=1k and R2=1k for example). So you have to define what R is for the pot first.
If we guess that the value is R1, then you just use R1 in it's place and calculate the Rin that way, and whatever drops out completely is eliminated from the result for Rin.

We also have been assuming that the open loop gain for the op amp is infinite, which is usually acceptable for theory. If that changes then we really have to start all over again because R2 may not cancel out anymore or other changes may happen, most notable that the open loop gain AOL will probably also then appear in the expression for Rin. Note it would not take that much more to calculate all this, and then plug in the known values later and perhaps try a few different values to get a feel for how these other things change the input resistance. Some things will have a large effect and other things may have a much smaller effect.

I dont know if you are doing the calculations by hand on paper or using an automatic algebraic system but the software that helps do the calculations helps a lot, as long as you can use it also on a test you might get.

Example:
Vout=Vin*((a-1)*Aol*R2+a*Aol*R1)/(R2+(Aol+1)*R1)

note this includes alpha 'a', R2, and Aol the open loop op amp gain. Note the pot resistance which we might have called Rp is not in that expression not because it was made equal to R1 but because it dropped out with only alpha 'a' remaining. That's because the output voltage of a voltage divider with no load does not depend on the divider resistance (in theory). That does not mean it will not show up in the expression for Rin however.

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