[SOLVED] Stupidly back to basics question

[prestonee]

this is about op amp closed loop gain.

I always considered the gain to be A0/(1+A0*B)

i have never questioned this and really havent solved for it until today... dun dun dunnn

If i look at just a simple Rf/Ri feedback wrapped around a fully diff amp, and apply the beta to the above equation i get the wrong gain.
I also thought Beta was Ri/Rf, which gives the right gain of about 1/B. however this was corrected by a coworker to be Ri/(Ri+Rf)
so please correct me with as much force as you see fit.
Beta= Ri/(Ri+Rf), example: Ri=5k, Rf=10k
B=1/3
if A0 is large, say 10e3, I get a Gain of approximately 3, not 2. I have found online an equation where this number is multiplied by (1-B) which then results in 2!
I guess I have along the years lost sense of this. Please help.

 

[LvW]

this is about op amp closed loop gain.
I always considered the gain to be A0/(1+A0*B)

The answer to your problem is simple:
The above formula applies for an input signal that is applied to the opamp input terminal directly.
Example: Non-inverting gain.

However, in case of an inverter only a part of the input signal is applied to the (inverting) input terminal.
Using your notation this part is H1=Rf/(Ri+Rf).

Using the above feedback formula and neglecting 1/Ao (very large open-loop gain) you arrive at

Gain=-H1/B=-Rf/Ri.

 

[prestonee]

Aw thank you, you are right I overlooked the fact that Beta looking into the sj of amp = beta looking back into vsig for non inverting , while i have to voltage divide for the inverting.
and since the full diff is treated as inverting .. yep i got it, thanks for the missing link.
-Pb