OpAmp problem

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Hi...

I was here trying to solve an OpAmp problem, but I'm stuck.

I can't find a way of writing an equation for v_R1 in function of va and vb, which I know and confirmed to be equal to v1 (measured with the voltmeter).

I found v_R1 = (R1*vo)/(R1 + R2) and that vo = ((R1 + R2)*v_R1)/R1
By the OpAmp thumbs-up rules, I know that the OpAmp will try everything it cans to keep v+ and v- at the same potential, so v1 = v_R1.

Now, I need to get rid of v1 from the vo equation, but I can't find the way!
Any help would be appreciated!
 

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Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Your circuit shows positive feedback. Me thinks this is not your intention.

Oh, my mistake... But now, if I try to turn it upside down on LTSpice and try to run the simulation again, LTSpice keeps simulaitong for ever, and if I cancel the simulation it returns this error:

Singular matrix:check nodes u1:11 and v1
Iteraction Nº. 1


I don't know what this is!
 

crutschow

Joined Mar 14, 2008
25,247
Oh, my mistake... But now, if I try to turn it upside down on LTSpice and try to run the simulation again, LTSpice keeps simulaitong for ever, and if I cancel the simulation it returns this error:

Singular matrix:check nodes u1:11 and v1
Iteraction Nº. 1


I don't know what this is!
It simulates fine for me.
Did you reverse the power leads when you inverted the op amp?
 

WBahn

Joined Mar 31, 2012
26,046
Hi...

I was here trying to solve an OpAmp problem, but I'm stuck.

I can't find a way of writing an equation for v_R1 in function of va and vb, which I know and confirmed to be equal to v1 (measured with the voltmeter).

I found v_R1 = (R1*vo)/(R1 + R2) and that vo = ((R1 + R2)*v_R1)/R1
By the OpAmp thumbs-up rules, I know that the OpAmp will try everything it cans to keep v+ and v- at the same potential, so v1 = v_R1.

Now, I need to get rid of v1 from the vo equation, but I can't find the way!
Any help would be appreciated!
Can you write v1 in terms of va and vb?

Can you write v_R1 in terms of vo?

Now set v1 equal to v_R1 and, viola, both v1 and v_R1 are gone.
 

WBahn

Joined Mar 31, 2012
26,046
Forget about the opamp part of the circuit completely. The opamp has no effect on v1.

You have two voltage sources, va and vb, connected by two resistors. What is the voltage at the junction of the two resistors?

How much current is flowing in the two resistors? What is the voltage drop across one of the resistors? What is the voltage at the center node give the voltage drop across one of the resistors and the voltage on the other side of the resistor?
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
You have two voltage sources, va and vb, connected by two resistors. What is the voltage at the junction of the two resistors?
It will be v_R1, because the OpAmp will always try to set the same voltage in both it's inputs so that the voltage drop between them is approximately 0 V. v_R1 = vo *R1(R1 + R2).

How much current is flowing in the two resistors?
Here I'm not sure if I can simply divide va/Ra and vb/Rb to find those 2 currents. By inspection I can tell the ia = -ib, because at v1 cannot be any current flowing, so ia and ib must cancel each other so that no current gets out of that junction into the non-inverting input of OpAmp.

What is the voltage drop across one of the resistors?
I don't know how to answer this with a formula. I can only say that as these 2 branches are in a parallel setup, the voltage drop of each branc must be equal!

What is the voltage at the center node give the voltage drop across one of the resistors and the voltage on the other side of the resistor?
I'm not sure but I think I didn't understood the question!
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
I tried this:

v_R1 = R1*vo/(R1+R2)

Then for the non-inverting input I wrote:
va - vb + Rb*I1 + Ra*I1 = 0
I1 = (vb - va)/(Ra + Rb)

v1 = va + Ra*I1 or v1 = vb - Rb*I1

replacing I1 in the above equation by (vb - va)/(Ra + Rb), I get:
Imagem1.png

But is this the easiest way? Jesus, I struggled a lot!
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Was it really THAT much of a struggle? Especially now that you know how to go about solving it?
It was a big struggle for me, not for the math, but for the time I took to understand how to take v1 in terms of va and vb... And of course that I need to come up with some mnemonic to shorten the time I take in the analysis itself! I can't take this long on an exam!

I can yet, give it another look:

ex_1a.png
 
Last edited:

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
But now, my question is:

Could I write this equation only from knowing the basic non-inverting amplifier gain equation, which is:

non_inv.png
 

Jony130

Joined Feb 17, 2009
5,180
Well, yes. All you have to do is to solve for V1 in terms of Va and Vb.

V1 = Va - I1*Ra and I1 :

I1 = (Va - Vb)/(Ra + Rb)

So we have

V1 = Va - (Va - Vb)/(Ra + Rb) *Ra = Va - (Va - Vb)*Ra/(Ra + Rb)

Vo = (1 + R2/R1) * [Va - (Va - Vb)*Ra/(Ra + Rb)] And notice that my solution is slightly different than yours.

Or we can use a superposition and solve for Vo directly.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Well, yes. All you have to do is to solve for V1 in terms of Va and Vb.

V1 = Va - I1*Ra and I1 :

I1 = (Va - Vb)/(Ra + Rb)

So we have

V1 = Va - (Va - Vb)/(Ra + Rb) *Ra = Va - (Va - Vb)*Ra/(Ra + Rb)

Vo = (1 + R2/R1) * [Va - (Va - Vb)*Ra/(Ra + Rb)] And notice that my solution is slightly different than yours.

Or we can use a superposition and solve for Vo directly.

Oh yes, but that is roughly what I did, right?
You just took out the minus sign inside the vb-va parenthesis and therefore it comes -(va-vb).

I'm asking this because I always hear our teacher just throwing out the vo expressions with just quick analysis by inspection and I have the idea that I'm really missing something here!
 
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