# How does the negative feedback of an op-amp control the output?

Discussion in 'General Electronics Chat' started by NichA, Jun 10, 2007.

1. ### NichA Thread Starter Member

Jun 10, 2007
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Ok, so I'm learning about op-amps and I'm just trying to figure out how negative feedback works to control the output.

I'm looking at an op-amp with a voltage Vin at the + terminal and the output voltage Vout is being fed back into the - terminal. I'm looking at an image in the ebook on this site - http://www.allaboutcircuits.com/vol_3/chpt_8/4.html

Here's how I see it, and I know I'm wrong but I just want you guys to help me see what I'm missing here:

SO... lets say that Vin is 1V. If we give the amplifier a huge gain of say.. 10^6, then Vout would be 1,000,000 Volts(assuming we have an unlimited voltage source connected to the amp), right? And this 1,000,000 Volts gets fed back into the - terminal of the amp. The new calculation of our differential amp would be 1V - 1,000,000V. So now -999,999V gets multiplied by the gain of 10^6 and our Vout goes even higher!!!

I know my view on this is all jacked up but could someone explain to me why this logic is wrong?

2. ### hgmjr Moderator

Jan 28, 2005
9,030
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A simple way to look at opamp behavior is that the opamp, in the presence of negative feedback, continuously adjusts its output so that the positive and the negative inputs are maintained at the same voltage.

In the example of the unity gain buffer in which the output is directly connected to the negative input, this simple rule means that whatever voltage appears on the input is almost immediately matched by an adjustment in the output so that the opamp's negative input is equal. Of course the power supply to the opamp places constaints on the mininum and maximum voltage that the opamp's output can swing as well as the acceptable range of input voltages.

hgmjr

3. ### cumesoftware Senior Member

Apr 27, 2007
1,330
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The negative-feedback allows the op-amp to "monitor itself". The output will control itself, so the tendency of the output is to follow acurately the input. Don't forget that the op-amps (and this also includes some audio power amps) have a very high gain, so they will hit the positive or negative rail at the sligthest voltage diference in the inputs.
What negative feedback does is to minimize the voltage diferences in the inputs, so the output will really follow the input (with a much lesser gain). The way this is done is because it the voltage in the negative input exceeds the one in the positive input, the output will go down. Otherwise it will go up. So the output is being constantly adjusted. Think this like a heater thermostat that shuts down when the temperature attains a certain level. This is negative feedback in which you have a constant, controlled temperature.

4. ### NichA Thread Starter Member

Jun 10, 2007
13
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I really appreciate your comments, but you failed to answer my question.

I asked you to explain to me why my logic was wrong in the example I gave. All the stuff you said is something I can read in a book but I don't really understand how it work as I mentioned in my first post.

Is there anyone who can help?

5. ### cumesoftware Senior Member

Apr 27, 2007
1,330
11
This is not exacly how it works. The output voltage cannot go past the positive rail or negative rail. Also, it the voltage in the negative input is bigger than the one in the positive input, it doesn't get highter. Actually it gets lower. But since an op-amp is an active device, compensation will occur until until both inputs have equal voltages (which is almost immediately, immediately to our time scale).

P.S.: -999999V multiplied by the gain is lower thao 1000000V, as negative numbers are lower than positive numbers.

6. ### NichA Thread Starter Member

Jun 10, 2007
13
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Well I mentioned earlier that we have an unlimited power source so it wouldn't come up and make things more confusing so I don't know why you even mentioned the rails...

Anyways, when I said the output would go even higher what I really meant to say was that the absolute value of the output would go higher. So (looking back to the example I gave in my first post) when the difference at the inputs is -999,999 Volts, that difference will get mulitiplied by the gain which is 10^6 and go up to -999,999,000,000 Volts. And then when this output gets fed back into the negative input our new calculation will be (+ terminal minus the - terminal) 1V - (-999,999,000,000V) which equals 999,999,000,001 Volts. This value will get multiplied by our gain of 10^6 and out new output value will be 999,999,000,001,000,000 !!! PLEASE TELL ME WHERE MY CALCULATION GOT OFF TRACK!!!

7. ### hgmjr Moderator

Jan 28, 2005
9,030
214
OK. So far so good.

This is at least what the output of the opamp would tend to do, however it is thwarted from pulling this off due to the negative feedback. In fact, the instant the output exceeds 1 volt (in the case of a unity gain opamp) say 1.000001V for example, this 1.000001 volts is presented to the negative input where it is subtracted from 1 to get -0.000001. This difference then passes through the opamp where it is multiplied by 1,000,000 to give a voltage at the opamp's output of 1 volt. This 1.00000 V is instantly applied to the negative input of the opamp where the process begins again. At this point the opamp has reached a stable condition (as long as the input stays at 1 V) and thus no further adjustment is made to the output of the opamp.
Due to the fact that negative feedback is virtually instantaneous, the opamp's output is never permitted to stray very far from the value of the voltage applied the input. Again this description is in the context of a unity gain buffer configuration. This same process holds true with other opamp configurations.
As I mentioned earlier, the output of the opamp is never permitted to stray very far from the value of the input voltage (again this is based on the unity gain buffer configuration).

hgmjr

8. ### NichA Thread Starter Member

Jun 10, 2007
13
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Ok hgmjr, so let me see if I've figured out where I went wrong...

I was incorrectly assuming that the output of the amp went straight to 1,000,000 Volts, but how it really works is the output gradually increases to its calculated value of 1,000,000... ? But it never makes it to 1,000,000 because of the negative feedback...

9. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Yes you have it now.

I sensed your rising level of frustration with our well-meaning efforts to assist you in understanding this particular behavior of opamps.

As with any complex subject i.e. opamps, one can have an internalized understanding of the subject but have a great deal of difficulty articulating that understanding to others. That is why we are not all teachers.

Out of this execise the best we can hope for is that you have removed one more obstacle to your understanding of the subject. There is much to learn about opamps before you can truely say that you have mastered the subject.

hgmjr

Jun 10, 2007
13
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Thank you!

11. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Please don't hesitate to return with any other question you have about opamps in particular or electronics in general.

Maybe the next time we can get to the answer you seek more rapidly.

Good Luck,
hgmjr

12. ### NichA Thread Starter Member

Jun 10, 2007
13
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Trust me, I'll be back. You were very helpful.

Actually, I've already posted a new question in the homework section...

13. ### cumesoftware Senior Member

Apr 27, 2007
1,330
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Don't need to be rude. I am not obliged to answer questions, much less the way you expect. You are lucky is someones answers you. Nevertheless, be assure that I won't answer next time.
And I think this also applies here.

14. ### NichA Thread Starter Member

Jun 10, 2007
13
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I apologize for being rude. I let my frustration get the best of me and I guess I just took it out on you. I will try to not make this mistake again and hope that you will accept my apology.

15. ### cumesoftware Senior Member

Apr 27, 2007
1,330
11
Then, in this case, I accept your apologies.

16. ### xn47 Member

May 22, 2007
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It's because of the sensing resistor. right? typically 27 ohms for the 741 OP AMP.