Hello! Consider this operational amplifier circuit,a subtractor to be more precise


I need to find the Ua (output voltage) using superposition; Now I dont have any values given for the resistors but that they are all the same (R1=R2=R3=R4=R5) so I just need to find the expressions.I've tried it like this. I let U1 be active and all of the other voltages get short circuited. When I simplify the circuit in that way I get a simple inverting operational amplifier that looks like this;

By no means a pretty picture but I think it will suffice. So I get that Ua is simply UA=−R5/R1 * U1 And according to the solutions this is correct. Now I did the same thing for U2 and I get the same circuit (simplified) but the first resistor is R2 and not R2 so the voltage is UA=−R5/R2 *U2 But according to the solutions it shouldnt be the resistor R2 but rather the resistor R1

It states "Same circuit as for U1" and gives the same equation but with voltage U2. I do not understand this; how can it be the same resistor? If we put it on ground and give voltage to R2 how can we get the same circuit if we do it vice versa? I'd assume I am simplifying the circuit wrong,that is why I posted the second picture,is the simplified circuit correct?

Thanks in advance!

EDIT: Professor said mistake in phrasing of resistor values; it should be

R1=R2=R3=R4 and R5=R6; so the resistors 1-4 are the same and 5-6 as well.

 

With the given circuit, the output will assume a value that makes both inputs assume values that make the difference very close to zero. So close that they are for all practical purposes equal to each other. Now you can use KCL to write equations for each of the inputs.

 

how can it be the same resistor?

It's not.
It's R2, the resistor connected to U2.

 

It's not.
It's R2, the resistor connected to U2.

Sooo I was not wrong but rather the solution? Or is it since all the resistors are equall it does not matter? Couls you please elaborate more. Thank you!

 

With the given circuit, the output will assume a value that makes both inputs assume values that make the difference very close to zero. So close that they are for all practical purposes equal to each other. Now you can use KCL to write equations for each of the inputs.

I am not sure what you are trying to tell me,I assume you mean that having U1 active or U2 it does not matter since they are pretty much the same, hence they can be described with the same equation. Is this what you meant?

 

The gist is that you calculate each input separately (U1, U2, etc.) with the other inputs grounded, whether the input resistors are the same or different.
The you use superposition to combine them all together at the output.

 

The gist is that you calculate each input separately (U1, U2, etc.) with the other inputs grounded, whether the input resistors are the same or different.
The you use superposition to combine them all together at the output.

That is what I wanted to do, and when I do that for U2 I dont see how you get -R5/R1* U2. I know what I need to do I just dont see how they get this when U2 is active.

 

If all of the resistors have the same value then:

\( -\cfrac{R5}{R1}(U_2)\;=\;-\cfrac{R5}{R2}(U_2)\;=\;U_A \)

KCL (Kirchoff's current Law) says the sum of the currents at a node must add up to zero. For all the inputs that are zero, the current through the input resistor MUST be 0!! This is entirely consistent with Ohm's Law.

 

If all of the resistors have the same value then:

\( -\cfrac{R5}{R1}(U_2)\;=\;-\cfrac{R5}{R2}(U_2)\;=\;U_A \)

Okay that would make sense, but this poses a question, why make it confusing for the student? Why not just make it intuitive and put R2. I will try to complete the whole problem, hopefully I do not stumble anywhere.

Thank you

 

It states "Same circuit as for U1" and gives the same equation but with voltage U2. I do not understand this; how can it be the same resistor?

It's a bit of sloppiness on the writer's part. They are relying on the assumption that all of the resistors are the same value, which is why they can claim that it is the same circuit as for U1. They really should just state something like

R = R1 = R2 = R3 = R4 = R5

and then use R and not one of the explicit resistance designators.

 

Hello! Consider this operational amplifier circuit,a subtractor to be more precise
View attachment 281514

I need to find the Ua (output voltage) using superposition; Now I dont have any values given for the resistors but that they are all the same (R1=R2=R3=R4=R5) so I just need to find the expressions.I've tried it like this. I let U1 be active and all of the other voltages get short circuited. When I simplify the circuit in that way I get a simple inverting operational amplifier that looks like this;
View attachment 281515
By no means a pretty picture but I think it will suffice. So I get that Ua is simply UA=−R5/R1 * U1 And according to the solutions this is correct. Now I did the same thing for U2 and I get the same circuit (simplified) but the first resistor is R2 and not R2 so the voltage is UA=−R5/R2 *U2 But according to the solutions it shouldnt be the resistor R2 but rather the resistor R1

It states "Same circuit as for U1" and gives the same equation but with voltage U2. I do not understand this; how can it be the same resistor? If we put it on ground and give voltage to R2 how can we get the same circuit if we do it vice versa? I'd assume I am simplifying the circuit wrong,that is why I posted the second picture,is the simplified circuit correct?

Thanks in advance!

Yes they must be assuming all the resistors are the same value.
Whatever input you allow to be 'active' the resistor associated with that input will be a determining factor in the solution.
For your cases, R1 and R2 appear in the denominator, and if you do input 3 and input 4 instead you get a different solution but they are the same except one has R3 in the numerator and one has R4 in the numerator. If you can assume that R4=R3 then those two could be called the same also.

The more general case though is with different value resistors and if they were all different there would be no two solutions the same.

Next i guess you are going to try to actually solve this circuit entirely?

 

Yes they must be assuming all the resistors are the same value.
Whatever input you allow to be 'active' the resistor associated with that input will be a determining factor in the solution.
For your cases, R1 and R2 appear in the denominator, and if you do input 3 and input 4 instead you get a different solution but they are the same except one has R3 in the numerator and one has R4 in the numerator. If you can assume that R4=R3 then those two could be called the same also.

The more general case though is with different value resistors and if they were all different there would be no two solutions the same.

Next i guess you are going to try to actually solve this circuit entirely?

Yea I've also stated that all the resistors are the same (look at my initial post) but no values but rather a general solution.

Yes I will try to solve the circuit entirely,meaning calculate the output voltage for U3 active,and U4 active,than superpone the solutions. Hopefully no trouble,if so than I will seek for help (yet again).

EDIT: I've written to my professor about these resistor business and it turns out there is a fault in the problem itself,mainly how it is stated; it should state like this;

R1=R2=R3=R4 AND R5=R6 so the resistors 1-4 are the same and 5 and 6 are same value as well.

 

Yea I've also stated that all the resistors are the same (look at my initial post) but no values but rather a general solution.

Yes I will try to solve the circuit entirely,meaning calculate the output voltage for U3 active,and U4 active,than superpone the solutions. Hopefully no trouble,if so than I will seek for help (yet again).

EDIT: I've written to my professor about these resistor business and it turns out there is a fault in the problem itself,mainly how it is stated; it should state like this;

R1=R2=R3=R4 AND R5=R6 so the resistors 1-4 are the same and 5 and 6 are same value as well.

Oh ok great. Then the solution will simplify into a sum of some inputs and difference of some other inputs (i wont say which yet) multiplied by a gain which is a ratio of two resistors. See if you can get that kind of solution. So it turns out it is an adder and subtractor with a gain.

It is also interesting to not assume any values and see if you can come up with that most general solution. Each resistor value R1 to R6 remain completely different from each other in this case. You'll get extra credit for this solution

 

Oh ok great. Then the solution will simplify into a sum of some inputs and difference of some other inputs (i wont say which yet) multiplied by a gain which is a ratio of two resistors. See if you can get that kind of solution. So it turns out it is an adder and subtractor with a gain.

It is also interesting to not assume any values and see if you can come up with that most general solution. Each resistor value R1 to R6 remain completely different from each other in this case. You'll get extra credit for this solution

I was able to solve this. With the confusion of the resistors gone i was able to find that when U3 is active I get R5/R1* U3 the same is when U4 is active.

Now I can superpone them all together and I get Ua = R5/R1× (U3+U4-U1-U2)
Which is whar they have in the solutions as well.

 

I was able to solve this. With the confusion of the resistors gone i was able to find that when U3 is active I get R5/R1* U3 the same is when U4 is active.

Now I can superpone them all together and I get Ua = R5/R1× (U3+U4-U1-U2)
Which is whar they have in the solutions as well.

Yes that's good that's the right solution as you noted already.

So are you going for the extra credit?
Seriously it is interesting but more difficult.

 

Yes that's good that's the right solution as you noted already.

So are you going for the extra credit?
Seriously it is interesting but more difficult.

I would but schedule is quite tight right now,have a lot do to before christmas break. Might come back to it.

 

I would but schedule is quite tight right now,have a lot do to before christmas break. Might come back to it.

Ok just thought you might like to try it. I am sure you can do it just takes a little time.