How to Solve an Inverting Operational Amplifier Problem?

[Katfazack]

Homework Statement

Given the attached figure (with values as shown on the op-amp) find Vo and io.

Homework Equations

None

The Attempt at a Solution

I figure that we're looking at an inverting operational amplifier and I used a voltage divider to find the input voltage [(90/90+10)]*1V = .9V. But I'm stuck on the rest.

I don't think it was very well explained by my professor, and since no two op-amps look alike I'm just wondering about how to solve this problem but also how to approach these problems in general? Any help would be greatly appreciated!

 

[skeptic2]

Are you sure the diagram is correct? As it is, there is DC positive feedback which will result in the op amp output at either the positive rail or negative rail. Could the inputs be reversed?

 

[Katfazack]

Well that's the diagram she gave us... So I'm not sure. I don't really understand that much about op-amps in particular. She just introduced the concept, but the homework is due today!

 

[berkeman]

Well that's the diagram she gave us... So I'm not sure. I don't really understand that much about op-amps in particular. She just introduced the concept, but the homework is due today!

Well then solve it both ways. Solve it as drawn with positive feedback (the output will do as skeptic says, either pinned high or low), and go for the extra credit by reversing the inputs of the opamp to configure it for negative feedback, and solve for the output voltage and current.

 

[Katfazack]

Ok, but I've only solved before with negative feedback. How do I do it with positive feedback?

 

[skeptic2]

Because the supply voltage isn't given it is impossible to calculate what the output voltage and current would be. Since the op amp isn't identified I'd have to assume it was an ideal op amp so that when the output is low, it is zero.

You might mention these objections to solving the circuit with positive feedback and then solve it with the inputs reversed. You can do that, right?

 

[Katfazack]

Oh, yes we are only dealing with ideal op amps. But how do I know that the output is low?

 

[berkeman]

Oh, yes we are only dealing with ideal op amps. But how do I know that the output is low?

You don't, because the circuit is bistable. Because of the + feedback, the output will peg either high or low. Generally with a comparator circuit (which is what is drawn), the input varies about some reference point (in this circuit it would be ground), and the output trips high and low when the input crosses the reference point (plus or minus half of the hysteresis voltage).

In the circuit shown, the input is steady, so you have no way of knowing how the output initialized (which of the bistable states it started in). If this were a real circuit, the output state would be determined at power-up, and might even be different for multiple power-ups.

BTW, the output of the opamp will pin near the rails, so if you have split power supplies for the opamp, the output will pin near those rails and not ground. If it's a single supply opamp, then the low output state will be near ground.