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I have doubt about beta of feedback loop of inverting amplifier.

enter image description here

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

$$A_{cl} =\frac{R_f}{R_in}$$

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  • \$\begingroup\$ Any book on op amps can answer this. Have you done any research? What is your source for these equations? \$\endgroup\$ – Barry Sep 22 '17 at 13:08
  • \$\begingroup\$ missing the sign (-) and GBW of OA. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 22 '17 at 18:12
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The circuit is not exactly equivalent to the two blocks. Some steps must be taken before arriving to that formula.

enter image description here

\$\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:

enter image description here

(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.

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