# Electronic – Step-by-step explanation of how voltage follower reaches steady state using negative feedback

Just one minute! I am not trying to understand what negative feedback does eventually, or why it should be used. I am trying to understand how the circuit reaches steady state, and how, step by step, the negative feedback causes Vout to be the same as Vin. This has not been addressed adequately in other answers. Let's assume the op-amp has a gain of 10,000, a supply of 15V, and Vin is 5V.

According to my understanding, this is how it goes:

1. \\$V_{in}\\$ is 5V, so \\$V_{out}\\$ should be 50,000V. However, it is limited to 15V by the power supply of the op-amp.
2. \\$V_{out}\\$ is then applied back to \\$V_-\\$, but it is subtracted from \\$V_{in}\\$ due to it being negative feedback
3. So the differential input voltage is now 5V – 15V = -10V
4. This is then amplified to -15V by the op-amp (because of saturation)
5. Now -15V is applied to \\$V_{in}\\$ through negative feedback, but it is added to 5V, due to double negative
6. So now differential input is 20V, and \\$V_{out}\\$ is 15V (due to saturation)
7. It seems that each time the op-amp will reach saturation, but just invert the output

I have obviously done something wrong here. The output is never going to stabilize at 5V in this way. How does it actually work?

Due to the excellent answers, I (think I) have understood the operation of negative feedback. According to my understanding, this is how it goes:

Let's say for simplicity that the input is a perfect step to 5V (otherwise the output would follow the transient input, making everything 'continuous' and difficult to explain in steps).

1. In the beginning, the input is 5V, and right now the output is at
0V, and 0V is being fed back to \\$V_{in}\\$
2. So now the differential voltage \\$(V_+ – V_-)\\$ is 5V. Since the gain of the op-amp is 10,000, it will want to produce an output of 50,000V (practically limited by the supply voltage), thus the output will start to increase rapidly.
3. Let's consider the point in time when this output reaches 1V.
4. Right now the feedback will be 1V as well, and the differential voltage will have fallen to 4V. Now the 'target' voltage of the op-amp will be 40,000V (because of the 10,000 gain, and again, limited to 15V by the power supply). Thus V_out will keep on increasing rapidly.
5. Let's consider the point in time when this output reaches 4V.
6. Now the feedback will be at 4V as well, and the differential voltage
will have fallen to 1V. Now the op-amp 'target' is 10,000V (limited
to 15V by the supply). Thus \\$V_{out}\\$ will still keep on increasing.

The emerging pattern is: the differential input causes increase in V_out, which causes increase in feedback voltage, which causes decrease in differential input, which decreases the op-amp 'target' output voltage. This cycle is continuous, meaning we can split it into even shorter intervals for investigation. Anyhow:

1. Let's consider the point in time when this output reaches 4.9995V. Right now the feedback is 4.9995V, so the differential voltage will fall to 0.0005V \\$(V_{in} – V_- = 5V – 4.9995V = 0.0005V)\\$. Now the target of the op-amp is \\$0.0005V*10,000 = 5V\\$.

However, if the op-amp reaches 4.9998V, now the differential voltage will be only 0.0002V. Thus, the op-amp output should decrease to 2V. Why doesn't this happen?

I believe I have finally understood the process:

The op-amp output cannot reach 4.9998V. Because as soon as \\$V_{out}\\$ increases above 4.9995V, the feedback will also increase, causing differential input to decrease, bringing the op-amp output back to 4.9995V.

And if the op-amp output decreases to below 4.9995V, the feedback will decrease, causing the differential voltage to increase, bringing the op-amp output back to 4.9995V.

The last two points are the essence of negative feedback. \\$V_{out}\\$ has stabilized as close as possible to \\$V_{in}\\$. If the gain were higher, the difference in \\$V_{out}\\$ and \\$V_{in}\\$ would be smaller. If gain reaches infinity, then output voltage is exactly equal to input voltage, and because of feedback being exactly equal to \\$V_{in}\\$, there would be 0 differential voltage, and a virtual ground would be create between the two inputs.