I don't really need anyone to solve this for me all the way, but I just need help. My issue is that I don't understand how to analyze an opamp circuit when there are two voltage sources connected to inputs in the way they are above.

I mean, the voltage at one input node is supposed to equal the voltage at another input node, but this would make no sense in this case since it seems I would have 3V = 1V..

 

Hint: The reason it makes no sense is because the circuit has positive feedback.

 

Hmm, sorry but I don't understand the hint. I'm reading this from Sadiku and it seems all they cover is negative feedback.

Is it possible one of the voltage sources is subtracted from the other and then one becomes a short?

 

Hello Mawangs1
op-amp works in two mode
1-amplifing
2-switch-and saturated between two statues(output shoud be 0,+VCC,-VCC)

if you don't see a feedback between input and output it is look like a switch and do this simple job
1-if v(+)>v(-) output=+VCC
2-if V(+)=V(-) output=0
3-if v(+)<V(-) output=-VCC

if it has a feedback resistor its better to upload your circuit schematic here

 

Hmm, sorry but I don't understand the hint. I'm reading this from Sadiku and it seems all they cover is negative feedback.

Is it possible one of the voltage sources is subtracted from the other and then one becomes a short?

With positive feedback the op amp no longer operates as a linear amplifier. It is thus saturated at either the maximum or minimum output voltage it can deliver.

The voltages do subtract from each other but that still leaves the positive feedback.

 

What you're saying makes sense. What would I do if I'm using nodal analysis though? The output voltage is stated to be 2V but I'd just be guessing at how that answer works out.....

 

Hint: The reason it makes no sense is because the circuit has positive feedback.

I'm betting it's unstable with the positive feedback. If you took a snapshot, you think the output was 2 volts, but it won't stay that way for long.

 

What you're saying makes sense. What would I do if I'm using nodal analysis though? The output voltage is stated to be 2V but I'd just be guessing at how that answer works out.....

There is a quasi-stable solution for the output at 2V. But in reality any small noise at the positive input will be amplified as the same polarity noise at the output which is fed back to the positive input for further amplification and so on, which means the output will rapidly end up at one of the rails.

For the output to be a stable at 2V output, the +3V would need to be connected to the plus input and the 1V would need to be connected between the output and the minus input for proper negative feedback.

The point is, the circuit shown is incorrect for a stable output voltage at other than one of the rails.

 

Does it change anything if I said this was assumed to be an ideal amplifier? All the problems that this book currently lists deal with an ideal op amp, so I guess what I'm trying to say is that it may be a little to in depth to consider saturation or stability for this analysis right now.

 

Does it change anything if I said this was assumed to be an ideal amplifier?

No, not at all. The problem is the positive feedback and its lack of stability. It's eventual output is unpredictable.

 

I don't really need anyone to solve this for me all the way, but I just need help. My issue is that I don't understand how to analyze an opamp circuit when there are two voltage sources connected to inputs in the way they are above.

I mean, the voltage at one input node is supposed to equal the voltage at another input node, but this would make no sense in this case since it seems I would have 3V = 1V..

No, remember that voltages mean a DIFFERENCE in potential...... the 3V source is referenced to ground but the 1V source is NOT (so it can move with respect to ground).

BTW, I'd wager the person who made up this problem got the op amp inputs reversed accidentally, as the problem makes no sense this way as shown. Assume the feedback path through 1V is to the negative input:

The positive input would be held at 3V (and so would the inverting input) but there would be a 1V drop going to the output. So that means Vo would be 2V across a 2k resistor for a 1mA output load current.