# Question for stability of negative feedback OP-AMP

• goodphy
In summary: But one of the most important aspects is: The closed-loop gain Acl will be defined by external resistors only and NOT (!) by the intrinsic properties of a (real) opamp. This gives as the possibility to design amplifiers with very low offset, very low THD, very high input impedance, very low output impedance etc etc Yes - that is the secret of stabilization caused by negative feedback. And this is only ONE benefit of negative feedback. There are some more. But one of the most important aspects is: The closed-loop gain Acl will be defined by external resistors only and NOT (!) by the intrinsic properties of a (real) opamp. This gives as the possibility to design amplifiers with very low offset,
goodphy
Hello.

I've studied the golden rules of the feedback OP-AMP and applying this to voltage follower shows that voltage gain (Vout/Vin) is 1.

Thus, Vout should eventually follows Vin when Vin suddenly changed. I've tried to follow this process for clear feeling by drawing pictures as shown in the attached image.

Since AV is very large for typical OP-AMP, It looks Vout becomes divergent! (or becomes saturation value.) This divergence is even more clear when feedback equation is approximates to (V+ - V-)AVn where n is number of looping. (You can get this approximation by looking last shown Vout in the image.)

What was I wrong? Why does feedback make the system crazy?

#### Attachments

• OP-AMP feedback process in step by step.jpg
20.7 KB · Views: 651
I am afraid, you are mixing open-loop gain Aol with closed-loop gain Acl=1.
Hence, the third figure in your sequence is wrong:
Here is the correct sequence assuming Acl=1(because of full negative feedback):
1.)Vin=3V, Vout=3V
2.) Sudden jump from Vin=3V to Vin=5V: In the very first moment we still have Vout=3V (time constant of the device),
3.) Hence Vdiff=5-3=+2V (this is the diff. voltage directly between the input nodes)
4.) Therefore, Vout tends to increase (direction to pos. supply) and passes a value of slightly below 5V, which causes Vdiff~some µV
5.) This is a new equilibrium because the tiny Vdiff (µV range) multiplied with the (large) open-loop gain "holds" the value of Vout at a value very close to 5V (perhaps 4.99995, depending on the open-loop gain Aol).

Dschumanji and goodphy
LvW said:
I am afraid, you are mixing open-loop gain Aol with closed-loop gain Acl=1.
Hence, the third figure in your sequence is wrong:
Here is the correct sequence assuming Acl=1(because of full negative feedback):
1.)Vin=3V, Vout=3V
2.) Sudden jump from Vin=3V to Vin=5V: In the very first moment we still have Vout=3V (time constant of the device),
3.) Hence Vdiff=5-3=+2V (this is the diff. voltage directly between the input nodes)
4.) Therefore, Vout tends to increase (direction to pos. supply) and passes a value of slightly below 5V, which causes Vdiff~some µV
5.) This is a new equilibrium because the tiny Vdiff (µV range) multiplied with the (large) open-loop gain "holds" the value of Vout at a value very close to 5V (perhaps 4.99995, depending on the open-loop gain Aol).
Thanks for replying! But I think I need a little more help for the detailed process.

Could you tell me more about details of 4th process in your reply that how Vout suddenly gets the value slightly below 5V at 1st forward feeding? I thought it should be 2Aol in volt since open loop gain Aol is intrinsic property of OP-AMP so Vout = VdiffAol always holds from inputs to output regardless of feedback loop.

goodphy said:
... open loop gain Aol is intrinsic property of OP-AMP so Vout = VdiffAol always holds from inputs to output regardless of feedback loop.
.
Yes - that`s correct. Let me try to explain in more detail.
....4.) Because Vdiff is heavily enlarged (by 2 V) the opamp "feels" that the input is overdriven and the output tends to go into positive saturation. However, it does not "jump" to the supply rail (because of internal time constants) but there will be a finite rise time.
And - as I have mentioned - on the way to the pos. supply voltage the output crosses a voltage which again (as bevor in the 3V case) exactly fulfills the condition mentioned by you: Vout = Vdiff*Aol . At this point (in this moment) the rising is stopped (equilibrium) because a further increase of the feedback signal would again decrease the diff voltage Vdiff.
Example: Vout=5.1 V. That means: Feedback voltage (at the inv. terminal) is also 5.1V and Vdiff=5-5.1=-0.1 V.
Hence, the opamp sees a negative diff. voltage and tends to go back to negative output voltages (and will stop again at an output which fulfills Vout=Vdiff*Aol).

There is only one stable condition (equilibrium) caused by negative feedback: In my example:
Vin=5V, Vout=+4.99995V, Vdiff=0.00005V.
From this, we can calculate back:
Aol=Vout/Vdiff=99999.
OK?

Sergei Gorbikov, Dschumanji and goodphy
Yes thanks!

So what my picture does not show is finite response time OP-AMP. My picture implicitly assumed that response time of OP-AMP is a brink of instance, which is impossible in real world! Thus as on the way of Vout increase triggered by sudden input voltage change, It finds equilibrium point where OP-AMP gain equation Vout = AOLVdiff holds and stay there. It looks like OP-AMP finds comfort "home" during journey at which it intended to go somewhere at first time.

Yes - that is the secret of stabilization caused by negative feedback. And this is only ONE benefit of negative feedback. There are some more.

Dschumanji and goodphy

## 1. What is negative feedback in an OP-AMP?

Negative feedback in an OP-AMP (operational amplifier) is a technique used to stabilize the output voltage by feeding a portion of the output signal back to the input. This helps to reduce any distortion or changes in the output caused by variations in the input signal or other external factors.

## 2. How does negative feedback affect the stability of an OP-AMP?

Negative feedback improves the stability of an OP-AMP by reducing the gain of the amplifier at high frequencies. This prevents the amplifier from oscillating or producing unwanted noise, making it more reliable and accurate.

## 3. What are the potential issues with negative feedback in an OP-AMP?

If the amount of negative feedback is too high, it can cause the amplifier to become slower and less responsive. On the other hand, if there is too little negative feedback, the amplifier may become unstable and produce oscillations or noise. Finding the right balance is crucial for optimal performance.

## 4. How can the stability of negative feedback in an OP-AMP be tested?

The stability of negative feedback in an OP-AMP can be tested by applying different input signals and observing the output. If the output remains stable and the gain is consistent, then the negative feedback is working effectively. Additionally, simulations and mathematical calculations can also be used to analyze the stability of an OP-AMP.

## 5. What are some common applications of negative feedback in OP-AMPS?

Negative feedback is commonly used in OP-AMPS for various applications such as amplification, filtering, and signal conditioning. It is also used in control systems, audio equipment, and communication devices to improve stability and accuracy.

Replies
50
Views
7K
Replies
3
Views
1K
Replies
10
Views
3K
Replies
12
Views
2K
Replies
33
Views
5K
Replies
7
Views
1K
Replies
8
Views
2K
Replies
6
Views
2K
Replies
2
Views
1K
Replies
6
Views
9K