actual formula for open loop gain in a non-inverting op-amp?
\$A\$ = open loop gain

\$V_{out}=A(V^+-V^-)\$

Let's first assume \$A=\infty\$, we'll cover A=100 later, and we got an op-amp set up as a non-inverting amplifier.

This is the equation in an ideal case: \$V_{out}=(1+\frac{R_2}{R_1})V_{in}\$ where \$R_2\$ is the feedback resistor and \$R_1\$ goes to ground. 



Let's see if we can get that same answer from our first expression. 

\$V^+ = V_{in}\$

\$V^- = V_{out}\frac{R_1}{R_1+R_2}\$, I hope you can see that it is a voltage divider. 

\$
\begin{align}\\
V_{out}=A(V^+-V^-) \rightarrow V_{out}&amp;=A(V_{in}-V_{out}\frac{R_1}{R_1+R_2})\\
V_{out}&amp;=AV_{in}-AV_{out}\frac{R_1}{R_1+R_2}\\
V_{out}+AV_{out}\frac{R_1}{R_1+R_2}&amp;=AV_{in}\\
V_{out}(1+A\frac{R_1}{R_1+R_2})&amp;=AV_{in}\\
\\
V_{out}&amp;=\frac{A}{1+A\frac{R_1}{R_1+R_2}}V_{in}\\
\\
V_{out}&amp;=\frac{A(R_1+R_2)}{AR_1+R_1+R_2}V_{in}\\
\end{align}
\$



"Hmmmm that doesn't look like \$V_{out}=(1+\frac{R_2}{R_1})V_{in}\$ to me", well let's use limit and let \$A \rightarrow \infty\$. In other words, let's make this op-amp ideal. 

\$\lim\limits_{A \to \infty}\frac{A(R_1+R_2)}{AR_1+R_1+R_2}=\frac{A(R_1+R_2)}{AR_1}=\frac{A}{A}\frac{R_1+R_2}{R_1}=\frac{R_1+R_2}{R_1}=1+\frac{R_2}{R_1}\$

"Ahhh! There it is!". This is more of a sanity check for me since I'm so damn rusty. Let's carry on and see what we get if A = 100. The answer to your question.



\$V_{out}=\frac{A(R_1+R_2)}{AR_1+R_1+R_2}V_{in} \rightarrow V_{out}=\frac{100(R_1+R_2)}{100R_1+R_1+R_2}V_{in}\$, hmmm doesn't look like I can make it look better than that to be honest. 



But let's put some numbers to it to see what would happen. Let's say \$R_1 = R_2 = 1000 Ω\$ and that \$V_{in}=1 \$ V. In an ideal case \$V_{out}\$ should be 2 V.

\$V_{out}=\frac{100(1000+1000)}{100×1000+1000+1000}×1 ≃ 1.96\$ V, hmm not too bad for an open loop gain of 100.



If you want it to behave like an ideal amplifier, knowing that \$A=100\$, then you would have to set this equation:

\$\frac{100(R_1+R_2)}{100R_1+R_1+R_2} = 2\$, or whatever gain you wish to have.

Lock one of the resistors to some value, let's lock \$R_1\$ to 1000 Ω.

\$
\begin{align}\\
\frac{100(1000+R_2)}{100×1000+1000+R_2} &amp;= 2\\
\\
100(1000+R_2) &amp;= 2(100×1000+1000+R_2)\\
\\
100×1000+100×R_2 &amp;= 2×100×1000+2×1000+2×R_2\\
\\
100×R_2-2×R_2 &amp;= 2×100×1000+2×1000-100×1000\\
\\
R_2(100-2) &amp;= 2×100×1000+2×1000-100×1000\\
\\
R_2 &amp;= \frac{2×100×1000+2×1000-100×1000}{100-2}\\
\\
R_2 &amp; ≃ 1040.81 Ω
\end{align}\\
\$

And voilla, now it behaves as if you have infinite open loop gain. If you plug in the numbers as I did before, then you will get 2.00 V, instead of 1.96 V.