# Op-amps arrrgh

1. May 27, 2007

### samski

everyone loves these things but i cant get my head around it... any explanations seem to go from blindingly simple to technical very quickly... i understand logic comparators but as soon as negative feedback is included i get all confuzzled.

OK so i understand that the output will become G((V+)-(V-)). So if you use 100% negative feedback, i don't undertand why you will get the same output out as you put in... this is what i am refering to - http://web.telia.com/~u85920178/begin/opamp00.htm [Broken] in the first part of the section "The Op-Amp As A Comparator - Analogue"

it says: "The only stable condition that can exist is if the + and - inputs are the same voltage."

however i was under the impression that if the + and - input are the same, the output is 0v because G*(x-x)=G*0=0. therefore this wouldn't be a stable input...

im also quite confused about limiting gain by using resistors and things but i guess ill understand that if i understand the basics...

i hope somebody can help...
thanks!

sam

Last edited by a moderator: May 2, 2017
2. May 27, 2007

### waht

Op-amps have a very high gain 100,000x so even a small voltage across the inputs will bring the output to the highest available voltage from the power supply. Doesn't sound that useful unless using window comparators

With a gain of 100,000 what would have to be the voltage across inputs to produce output of 5 volts? , which is well below the power voltage for the IC.

Well simple, divide that and you get .000005 Volt across inputs to make 5V

Think about that? This is unrealistic.

So if some output is sampled back to the input, it's possible it can make a voltage difference between inputs 0.000005 volts or just call it 0, that depends on hundreds of different feedback configurations, you could study, but the general quick rule with feedbacks is

1. The output will do whatever is necessary to make the voltage across inputs zero. If it can't achieve that, the output will swing to positive or negative power supply.

This is from Horowithz Art of Electronics book, The Classic.

As an example, look at the simplest op-amp feedback circuit "the buffer"
You see output is directly hooked up to one input.

So if you apply voltage to the other input, output will do anything it can to to make the voltage difference between inputs zero. The only way it can do that is to swing to the actual input voltage.

Last edited: May 27, 2007
3. May 27, 2007

### samski

hmm thanks thats a much better way of thinking of it. the thing that is still bugging me is that that doesnt follow output v = G(vplus-vneg)

o well... i'll have another look and see if it makes more sense now... :D

4. May 27, 2007

### abdo375

If you have a couple of hours on your hands, may I suggest that you take a look at these videos, the explanation is simply beautiful

PS there are two lecture about op-amps, the First one is about the basic theory and application, and the other one is about Positive feedback.

5. May 28, 2007

### AlephZero

You are right, in theory.

But suppose G = 100000 and V1 is at 10V.

Then V2 = 100,000(V1-V2)
V2 = V1 x (100000/100001) = 9.9999V

For most practical purposes, assuming that V2 = V1 is near enough.

6. May 28, 2007

### samski

thanks aleph.... thats a good way of working it out :D

sorry abdo i couldnt see any links on your post so i couldnt watch the videos you were referring to...

anyway, thats cleared things up a bit...

thanks guys!

sam