I have doubt about beta of feedback loop of inverting amplifier.
The expression for the feedback gain is :
$$A_{cl} =\frac{A}{1 + A \cdot \beta}$$
what is value of beta here? and how do I reach to final expression
$$A_{cl} =\frac{R_f}{R_in}$$
\$\begingroup\$
\$\endgroup\$
2
-
\$\begingroup\$ Any book on op amps can answer this. Have you done any research? What is your source for these equations? \$\endgroup\$– BarrySep 22, 2017 at 13:08
-
\$\begingroup\$ missing the sign (-) and GBW of OA. \$\endgroup\$– Tony Stewart EE75Sep 22, 2017 at 18:12
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
The circuit is not exactly equivalent to the two blocks. Some steps must be taken before arriving to that formula.
\$\beta\$ in fact is not \$R_{in}/R_{f}\$.
It is \$R_{in}/(R_{in}+R_f)\$ instead. Which is the same as the non inverting configuration.
Removing the internal summing node (notice that the signs are brought to the external one!) you get:
(Then if you want you can bring the first block inside the loop, but \$A_{OL}(s) \cdot \beta\$ does not change. And that's the term you'll be considering to analyze the stability of the closed loop system).
If \$A_{OL} = \infty\$, then you get:
$$ A_{CL}=-\frac {R_{f}}{R_{in}+R_{f}} \cdot \beta^{-1} = - \frac {R_{f}}{R_{in}} $$
In general (not in this case, assuming an ideal OA ), you could also have another block (transfer function) at the output.
