5
\$\begingroup\$

Why is the following true?

Enter image description here

I can't seem to understand why. Shouldn't it just be infinity/1+infinity which is approximately 1?

It's comparing the vo/vin of a non-inverting amplifier with finite gain with that of one with infinite gain.

\$\endgroup\$
  • 1
    \$\begingroup\$ infinity/(1+infinity) is an indeterminate form, it isn't equal to 1. If you get an indeterminate form it means you haven't found a solution and you need to try some other approach. \$\endgroup\$ – user253751 Mar 25 '18 at 23:06
17
\$\begingroup\$

Just follow the following approach:

$$\begin{align*} &= \lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}}\\\\ &= \lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}}\cdot\frac{\frac{1}{A}}{\frac{1}{A}}\\\\ &= \lim_{A\rightarrow\infty}\frac{1}{\frac{1}{A}+\frac{R_1}{R_1+R_F}}\\\\ &= \frac{1}{\frac{R_1}{R_1+R_F}}\\\\ &= \frac{R_1+R_F}{R_1}\\\\ &= 1+\frac{R_F}{R_1} \end{align*}$$

\$\endgroup\$
  • \$\begingroup\$ Perfect. Thank you. \$\endgroup\$ – AlfroJang80 Mar 24 '18 at 21:07
8
\$\begingroup\$

This is of infinity/infinity form. So we will do a work around to calculate this limit. $$\frac{A}{1+\frac {AR_i}{R_i+R_f}} = \frac{A}{A(\frac{1}{A}+\frac {R_i}{R_i+R_f})} = \frac{1}{\frac{1}{A}+\frac {R_i}{R_i+R_f}} $$

Now you can apply limits.

$$\lim_{A\to \infty} \frac{1}{\frac{1}{A}+\frac {R_i}{R_i+R_f}} = \frac{1}{0+\frac {R_i}{R_i+R_f}} =\frac {R_i+R_f}{R_i} $$

\$\endgroup\$
  • \$\begingroup\$ Perfect. Thank you. I think this is the same approach as jonk's. \$\endgroup\$ – AlfroJang80 Mar 24 '18 at 21:07
5
\$\begingroup\$

We can write your equation in a slightly different way $$A_{CL} = \frac{A}{1 + Aβ}$$

Where \$β = \frac{R_1}{R_1 + R_F}\$

And now if we divide this by \$A\$ we are going to get this:

$$A_{CL} = \frac{1}{(1/A) + β}$$

So, now \$A\$ is approaching the infinity \$(1/A = 0)\$

we can see that the closed loop gain is equal to:

\$\Large \frac{1}{\beta} =\frac{R_1+ R_F}{R_1} =\frac{R_1}{R_1}+\frac{R_F}{R_1}= 1 +\frac{R_F}{R_1}\$

\$\endgroup\$
5
\$\begingroup\$

A less rigorous method is to look at the denominator, and notice that, as A gets very large, A(R1/(R1 + RF)) gets much larger than 1, so the 1 can be discarded.

Then the ratio is easily evaluated and the A's drop out.

\$\endgroup\$
5
\$\begingroup\$

\$\infty/\infty\$ is an indeterminate form, so use L'Hôpital's rule

$$\lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}} = \frac{\frac{d}{dA}(A)}{\frac{d}{dA}(1+A\frac{R_1}{R_1+R_F})} = \frac{1}{\frac{R_1}{R_1+R_F}} = 1 + \frac{R_F}{R_1}$$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.