Calculating V0 on a Complex Op Amp Circuit

[dbakg00]

Homework Statement

See attached picture. My objective is to calculate V0 on the 2k resistor. I'm used to seeing op amps that are more simple looking. I'm trying to reduce this op amp circuit to a simpler one (my attempt is the bottom picture). My question is: did I simplify the circuit correctly? I subtracted the 1v from the 2v source since they are working against each other and I added the resistors since they are in series. Thanks in advance for any help you can give.

Homework Equations

The Attempt at a Solution

 

[mfb]

Those two sources do not work against each other, they both have the same direction.

 

[dbakg00]

How could they both have the same direction? The negative terminal of the 1v touches the negative terminal of the 2v. Sorry if this is a dumb question...I'm new to circuits.

 

[rude man]

How could they both have the same direction? The negative terminal of the 1v touches the negative terminal of the 2v. Sorry if this is a dumb question...I'm new to circuits.

Use superposition principle.

1. let V1 = 0 and solve for Vout.
2. let V2 = 0 and solve for Vout.
3. add the two Vout s.

Your "simplified" circuit is invalid, sorry.

 

[skeptic2]

I agree, your simplified circuit is not the same. Try calculating the voltage at the inverting input of the op amp of the top circuit and show us what you get.

 

[dbakg00]

Use superposition principle.

1. let V1 = 0 and solve for Vout.
2. let V2 = 0 and solve for Vout.
3. add the two Vout s.

Your "simplified" circuit is invalid, sorry.

So if I let the 1V source = 0, do I still consider both the 5k and the 2.5k resistors? I know that in an inverting amplifier, vo = (-R2/R1)vi. R2 is obviously 10k. What is confusing me is what value I would use for R1. If I did the superposition as you suggested, when I killed the 1V source, would R1 = 5k? And then when I killed the 2V source, R1 = 2.5k? Am I on the right track?

 

[dbakg00]

When I do superposition, I get the following:

(short circuiting the 1V source)
V01 = Vs1(-R2/R1)
V01 = 2(-10/2)
v01 = -4 V


(short circuiting the 2V source)
V02 = Vs2(-r2/R1)
V02 = 1(-10/2.5)
V02 = -4 V

V0 = V01 + V02
V0 = (-4) + (-4)
V0 = -8

how does that look?

 

[rude man]

So if I let the 1V source = 0, do I still consider both the 5k and the 2.5k resistors? I know that in an inverting amplifier, vo = (-R2/R1)vi. R2 is obviously 10k. What is confusing me is what value I would use for R1. If I did the superposition as you suggested, when I killed the 1V source, would R1 = 5k? And then when I killed the 2V source, R1 = 2.5k? Am I on the right track?

Yes.

 

[rude man]

When I do superposition, I get the following:

(short circuiting the 1V source)
V01 = Vs1(-R2/R1)
V01 = 2(-10/2)
v01 = -4 V


(short circuiting the 2V source)
V02 = Vs2(-r2/R1)
V02 = 1(-10/2.5)
V02 = -4 V

V0 = V01 + V02
V0 = (-4) + (-4)
V0 = -8

how does that look?

Very good. You made a typo on V02 (2nd line) but still came up with the right answer.

 

[dbakg00]

Thanks to all for the help!

 

[mfb]

How could they both have the same direction? The negative terminal of the 1v touches the negative terminal of the 2v. Sorry if this is a dumb question...I'm new to circuits.

They are "in parallel" - not exactly due to the resistors, but they share the same direction of voltage between ground and the negative input of the op amp.