1

As the theory goes, following are the 2 cases of op-amp operation:

  1. If there is a valid -ve feedback, inverting terminal volt = non inverting terminal voltage
  2. If there is no -ve feedback, o/p acts like a comparator and its output will be saturated to either +VCC or VSS

Original question: Why does the below op-amp circuit output not getting into saturation (+Vcc or VSS) even though it is open-loop?

enter image description here

In the above diagram V(VCC)=10V and I don't see any straight-forward feedback path, but the o/p is settled to 9V instead of 10V. Do i need to dig further on the op-amp model to resolve this?

Edit: LT1001 opamp datasheet

Add on question: How can we confirm whether the op-amp has -ve feedback?

In the below circuit, the simulation results show op-amp is in linear mode of operation (V+=V-).

enter image description here

How do I prove the below circuit has -ve feedback?

  1. The red path indicates a potential path from o/p to -ve input terminal. But since V- is permanently tied to GND thru R18- i dont see any -ve feedback action coming into play.

  2. The blue path indicates possible -ve feedback action. If output increases, the gate voltage of M2 increases, current increases, R5 voltage becomes more -ve -> eventually bringing the output lower.

Is this analysis correct?

Divya K.S
  • 2,489
  • 16
  • 31
  • Many op-amps cannot drive their output to the positive or negative supplies - check the datasheet for the one you are using. – Peter Bennett Jan 24 '22 at 07:10
  • Please note I was forced to merge these 2 questions in single post since moderator had tagged the 2nd one as duplicate. In my opinion, these are two different questions altogether. Duplicate question: https://electronics.stackexchange.com/questions/605628/how-can-we-confirm-whether-the-op-amp-has-ve-feedback – Divya K.S Jan 24 '22 at 11:16
  • 2
    When drawing schematics, try to keep the circuit much more compact than you did - we can't magnifiy the drawing enough to read the text. – Peter Bennett Jan 24 '22 at 16:42

2 Answers2

4

The op-amp will strongly amplify any difference on its inputs. So, in open-loop the output should swing to one of the rails. If you see that happening, there is either no or not enough negative feedback.

Whether it actually does make it to the rail is limited by the op-amp output stage voltage swing. For common types like the LM741 is only able to swing within 1.5V of the rail. The model you’re using reflects that limitation (I can’t read it from your schematic - too fuzzy.) Check the op amp datasheet.

If you need the full swing choose a ‘rail-to-rail’ output op-amp that supports it.

MORE: ok, so you added details about first sim’s op-amp, the LT1001, which indeed has swing limited to within 1V of the rails. You've changed to a rail-to-rail type for the second sim.

Now, to your question. If the op-amp feedback is actually closing the loop, you will see (almost) no voltage difference on the inputs. The ‘almost’ depends on the op-amp open-loop gain. If there’s more than that tiny difference, the loop isn’t closed and the op-amp is probably hitting a rail.

So, what's going in your circuit? The tiny scale doesn't help, but with some magnification I finally noticed that the (-) input is simply tied to ground through R18. There is no negative feedback happening, so it's definitely running open-loop. It's functioning as a comparator against ground.

My suggestion? Try using Falstad (https://www.falstad.com/circuit/). It's not as accurate as LTSpice, but it is much better at showing how the circuit is behaving in a dynamic, interactive way.

hacktastical
  • 49,832
  • 2
  • 47
  • 138
  • 2
    To add: Full swing output is often called "rail to rail out" when looking for components with the supplier component search for example. – Ralph Jan 24 '22 at 07:30
0

You write in your question:

But since V- is permanently tied to GND thru R18- i dont see any -ve feedback action coming into play.

GND is essentially an arbitrary label. We typically label one of the 'main' power supply nodes as 'GND' because this usually makes interpreting the circuit easier - but there's no hard requirement to do that - so if your initial choice of GND makes it difficult to interpret the circuit then choose a different node as your GND reference. Just make sure that you're consistent - don't start your analysis with GND at one node and them move it to a different node half-way through!

So for your circuit, consider labelling the other side of R5, where it connects to V3 and R19, as GND instead.
Doing this makes it easier to see that an increasing current through the MOSFET and R5 will cause the voltage at the opamp's (-) inverting input to rise with respect to to the voltage at its (+) non-inverting input.

So since the rest of the circuit after the opamp's output does not invert the signal (an increasing voltage at the output will increase the MOSFET's Vgs, causing it o conduct more current through itself & R5), your circuit therefore does give the opamp negative feedback.

brhans
  • 14,373
  • 3
  • 34
  • 49