# Q: How to use circuit analysis on a circuit ideal diodes with ideal op-amps

I have been practicing how to analyze op-amps with diodes as I never really learnt them and thought it would be a handy skill to finally learn.

I have been attempting a question on here that I found, but I am getting stuck.

Here's is the schematic

simulate this circuit – Schematic created using CircuitLab

I am doing a case where the diode is off (assuming ideal). Where I need to find the drop of the diode to be < 0, however I have no idea what is the output voltage of the op-amp. Any hints or suggestions will be helpful!

EDIT: I tried doing the negative input and it looks like the diode is both on and off?

simulate this circuit

• This unity gain voltage follower only works for positive inputs. Diode drop is irrelevant since the differential input must be 0V Commented Jul 15, 2018 at 2:06
• Ill re-edit the question to a more difficult one then. Edited.
– Leoc
Commented Jul 15, 2018 at 2:07
• Now you have the opposite inverting negative peak detector. But because there is no load to gnd it is less reliable. Commented Jul 15, 2018 at 2:52
• Hint: consider the op-amp is also a non-linear circuit with more than one operating mode. Commented Jul 15, 2018 at 3:04
• @copper.hat, yes but when I suggested OP should consider that, they told me they want to use an ideal op-amp model. Commented Jul 15, 2018 at 3:28

In comments you said you want to consider an ideal op-amp model.

For your second diagram, that means the op-amp's inverting input is at +1 V. It isn't at 0 V as you indicated because there's no feedback from the op-amp output to drive it to 0 V.

The node marked "-1V or 1V?" is also at +1 V by using Ohm's law on R3.

The node marked "????" is at $-\infty$ V because the ideal op-amp has infinite gain.

What this really means is that an ideal op-amp model is not useful for analyzing this circuit. You should consider a model that includes saturation behavior so that the op-amp output can not be driven outside the power supply rails.

• Yeah you're right, you helped me solve it by saying the output will be driven to -infinity. Essentially the negative feedback is non operative then :) Which is the correct answer
– Leoc
Commented Jul 15, 2018 at 3:44