OP-AMP simple problem

Thread Starter

OggiSerbia

Joined Dec 28, 2017
34
I am new in all this so I would be grateful if somebody with more experience can check my solution of this electronics problem.

Thanks in advance!

Problem: Screenshot_2.jpg
Vout (Vo) should be solved for R.

*OP-AMP is ideal!

My solution: IMG_20171228_122217.jpg
 

WBahn

Joined Mar 31, 2012
29,976
Thanks for showing your work. That's refreshing.

Before I point anything out in your work, what have you don't to check the validity of it? That should become an automatic part of solving any problem, both now in school and after you get out into the "real world" (where there usually won't be anyone to check it for you).

Pick some limiting cases, such as ΔR = 0 and/or R1=R2 and see if everything works out for, say, Vi = 1 V.

If those work, then pick something slightly more complex, such as ΔR = R and check that.

If that also works, then it is highly likely that you got it right.

Do all of those work?
 

Thread Starter

OggiSerbia

Joined Dec 28, 2017
34
Thank you very much.

Firstly I had tried with ΔR = 0 and it worked. After that, R1=R2 and ΔR=R were giving me bad results. So it was enough to check my solution again and I noticed a mistake in my calculation in row 4 - my equation was bad.
I've corrected it and now it works perfectly for any case.

Here's the correction: IMG_20171228_122217.jpg

Thanks again. See you soon.
 

WBahn

Joined Mar 31, 2012
29,976
So what was you new solution?

I can't quite read the changes you made, but if they say what I think they do, then they aren't correct.
 

WBahn

Joined Mar 31, 2012
29,976
It says: (Vi - V2)/R = (V2 - V1)/R

And from there I got: Vi + V1 = V2

So it is not correct?
So (Vi - V2) / R would appear to be the current flowing downward in the righthand R, while (V2 - V1) / R is the current flowing downward in the R just below it. That seems just fine.

But I'm a bit puzzled how you got from there to Vi + V1 = V2.

You might ask if this makes sense. What if Vi = 2 V and V1 = 1 V. Would it make sense at a point midway between them would be 3 V?

You might check your math a bit more closely.
 

Thread Starter

OggiSerbia

Joined Dec 28, 2017
34
I am sorry, I didn't write it well. In the picture it is fine.

I forgot to multiply V2 by two.
So, the equation should be:
Vi + V1 = 2*V2

Here's the calculation:
(Vi - V2) / R = (V2 - V1) / R
Remove R from both sides and we get
Vi - V2 = V2 - V1
Now we add V1 and V2 to both sides and get the final equation:

Vi + V1 = 2 * V2
 

MrAl

Joined Jun 17, 2014
11,389
I got:

Vo = (R1+R2)/(2*R2) * [1-(R+ΔR)/R] * Vi
Hi,

Happy to say, that looks correct.

A few tricks you can use based on these individual 'theories'::
1. Inverting op amp circuit.
2. Resistive voltage divider.
3. Non inverting op amp circuit.

The three 'tricks' above help to solve the network or at least just check it.
 
Last edited:

MrAl

Joined Jun 17, 2014
11,389
Hi,

Yes that is obviously correct. I must have copied his equation incorrectly. What i think happened was the two square brackets did not get interpreted correctly with the software. In any case though, i will correct my previous post a little later today as his solution looks correct now.
Thanks for posting.

Also that's not a bad idea posting the image of the equation rather than the typed out equation or the Tex equation or whatever is being used now.
 

Thread Starter

OggiSerbia

Joined Dec 28, 2017
34
Hi,

Yes that is obviously correct. I must have copied his equation incorrectly. What i think happened was the two square brackets did not get interpreted correctly with the software. In any case though, i will correct my previous post a little later today as his solution looks correct now.
Thanks for posting.

Also that's not a bad idea posting the image of the equation rather than the typed out equation or the Tex equation or whatever is being used now.
 

Thread Starter

OggiSerbia

Joined Dec 28, 2017
34
Hi,

Yes that is obviously correct. I must have copied his equation incorrectly. What i think happened was the two square brackets did not get interpreted correctly with the software. In any case though, i will correct my previous post a little later today as his solution looks correct now.
Thanks for posting.

Also that's not a bad idea posting the image of the equation rather than the typed out equation or the Tex equation or whatever is being used now.
Yes, I know. I was writing through the phone so I couldn't upload the picture. When I am on the PC I upload it, of course.

Thanks everyone!
 

MrAl

Joined Jun 17, 2014
11,389
Yes, I know. I was writing through the phone so I couldn't upload the picture. When I am on the PC I upload it, of course.

Thanks everyone!
Hi again,

Also note you can use those three 'tricks' to do this circuit rather quickly.

1. Inverting op amp gain: G=-Rfb/Rin.
2. Non inverting op amp gain: G=Rfb/Rin+1
3. Voltage divider: Vo=(v1-v2)*RL/(RU+RL)+v2 where RU is the upper resistor and RL the lower resistor. When v2=0 this defaults to v1*RL/(RU+RL).
 
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