# How loop gain works...??

Discussion in 'General Electronics Chat' started by Himanshoo, Feb 9, 2016.

1. ### Himanshoo Thread Starter Member

Apr 3, 2015
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In the following article I didn't got this phrase....

If is less than unity, the signals will be gradually attenuated into insignificance despite the fact that they are reinforcing each other at the input"

My reason...
If the open loop gain (A) depends upon the difference of the signal at the input, loop gain depends upon A since loop gain = , thus its clear that loop gain also depends upon the difference of the signal at the input..so how can we say that signals will be attenuated despite they reinforce each other at input..??

Thanks!

Last edited: Feb 9, 2016
2. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Say you borrow a dollar from me. It's my very last dollar but I'm a good guy.

Next week you pay me back, but you're short so you can only give me 90 cents. "Fine" I say and off we go.

The following week you again borrow my last dollar, but I can only give you 90 cents. And again you are short so you pay me back 81 cents.

If this keeps going on, how much can you borrow from me a year from now?

(If you link to where you found that article perhaps someone can make more sensible comments on your question.)

Himanshoo likes this.
3. ### AnalogKid Distinguished Member

Aug 1, 2013
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1,295
But it doesn't. The open loop gain is a function of the circuit design, not any input conditions. The *output* depends on the input, but that is a different thing. My guess is that the article is about oscillations, either intentional or unintentional. What you are describing is the Barkhausen stability criterion. The point is that while an opamp might have near-infinite open loop gain and near-zero phase shift and group delay at DC, everything changes at high frequencies, so much so that an amplifier can oscillate when it shouldn't, and an oscillator won't oscillate when it should.

https://en.wikipedia.org/wiki/Barkhausen_stability_criterion

ak

4. ### Himanshoo Thread Starter Member

Apr 3, 2015
236
6
Nice sort of explanation....but I would appreciate if you would link your example with the topic in discussion....

5. ### hp1729 Well-Known Member

Nov 23, 2015
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How is the input "re-applied and degraded"?

I can see the point about oscillations. Less signal is fed back in each interval.

6. ### ErnieM AAC Fanatic!

Apr 24, 2011
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When A times beta is less that one it is analogous to paying me back less that you borrowed.

You need unity or greater than unity to oscillate.

7. ### Himanshoo Thread Starter Member

Apr 3, 2015
236
6
Exactly...
The \$ 100 input is degrading to 90 cents and so on...
Its the output voltage that is generally attenuated....and in case of oscillation the feedback signal is reinforced (added) to the input and in case of stability the feedback is subtracted...so in both the cases what alters is the feedback voltage not the input voltage....