Why is the following true?
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.
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Sign up to join this communityWhy is the following true?
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.
Just follow the following approach:
$$\begin{align*} &= \lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}}\\\\ &= \lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}}\cdot\frac{\frac{1}{A}}{\frac{1}{A}}\\\\ &= \lim_{A\rightarrow\infty}\frac{1}{\frac{1}{A}+\frac{R_1}{R_1+R_F}}\\\\ &= \frac{1}{\frac{R_1}{R_1+R_F}}\\\\ &= \frac{R_1+R_F}{R_1}\\\\ &= 1+\frac{R_F}{R_1} \end{align*}$$
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.
$$\lim_{A\to \infty} \frac{1}{\frac{1}{A}+\frac {R_i}{R_i+R_f}} = \frac{1}{0+\frac {R_i}{R_i+R_f}} =\frac {R_i+R_f}{R_i} $$
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}\$
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.
\$\infty/\infty\$ is an indeterminate form, so use L'Hôpital's rule
$$\lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}} = \frac{\frac{d}{dA}(A)}{\frac{d}{dA}(1+A\frac{R_1}{R_1+R_F})} = \frac{1}{\frac{R_1}{R_1+R_F}} = 1 + \frac{R_F}{R_1}$$