# Understanding Negative feedback loop circuit with an Ideal Op-Amp

The following is given as an example of a negative feedback loop in my textbook, note that the op-amp is assumed to be ideal.

However I do not seem to understand the logic behind this acting as a negative feedback loop.

Suppose if $$\V_{in}\$$ is $$\0.01V\$$ after passing through the op-amp. It should be amplified to $$\+6V\$$.

Now $$\V_{out}\$$ is $$\+6V\$$ hence the inverting input will also become $$\+6V\$$ thus the difference between the two inputs becomes $$\-5.99V\$$ thus $$\V_{out}= -6V\$$ and this process shall repeat over and over again. Would the circuit then just cycle between being $$\+6V\$$ and $$\-6V\$$ until the power supply is disconnected?

How is this considered a negative feedback, am I missing something crucial?

• Would Electrical Engineering be a better home for this question? Apr 10, 2021 at 15:54

## 2 Answers

The op amp generates a voltage equal to the difference of the two inputs. Hence, suppose we start with $$\V_{out} = 0V = V_{-} = V_{+}\$$. Let's consider two cases.

1. The input becomes positive, $$\V_{+}=0.1V\$$ . The output is equal to the difference of the two inputs. Thus, it is equal to $$\V_+\$$, and we end up with $$\V_+ = 0.1V = V_{out} = V_-\$$.
2. The input becomes negative, $$\V_{+}=-0.1V\$$. Again, the output of the op amp is equal to the difference of the two inputs. Thus, it is equal to $$\V_+\$$, and we end up with $$\V_+ = -0.1V = V_{out} = V_-\$$.

this process shall repeat over and over again. Would the circuit then just cycle between being +6V and −6V until the power supply is disconnected?

No. As you noted, each time around the loop the input difference is less. Eventually the output is essentially equal to the input and the circuit is stable. What you are describing actually happens. If you look at opamp datasheets, particularly "high speed" ones, there usually is a scope shot showing the "output settling time". Because it takes a finite length of time for changes at the input to appear at the output, you can see the loop settle in real time.

The output never truly equals the non-inverting input voltage. There is an input offset voltage error. Think of this as a microvolt-to-millivolt sized battery in series with the input. The output goes to wherever it needs to go such that the input differential pair is happy, but the two base-emitter junction forward voltages never are exactly equal, and this shows up as an error.