Derive Gain from open loop mode

Question

I was given in classroom some theory on operational amplifiers, how to apply Kirchoff's Laws and apply the ideal conditions to understand the general working principles of the circuits using it,

^{simulate this circuit – Schematic created using CircuitLab}

I understand the next set of equations are valid for the design given:
\begin{cases} \frac{V-V_{in}}{R_{1}}= & \frac{V_{out}-V_{in}}{R_{1}+R_{f}} \end{cases}

\begin{cases} V_{out}= & -A_{VOL}V \end{cases}

But I don't get how the next equations is justified, I was told only it comes from the previous ones, and after a while I can't derive how to obtain the next arrangement:
\$A_{V}=\frac{V_{out}}{V_{in}}=\frac{A_{VOL}R_{f}}{-R_{1}-R_{f}-A_{VOL}R_{1}}\$

so I try to get the transfer function first (taking into account this is a inverter opamp configuration) \$\frac{V-V_{in}}{R_{1}}=\frac{V_{out}}{R_{1}+R_{f}}-\frac{V_{in}}{R_{1}+R_{f}}\$ \$\frac{V}{R_{1}}-\frac{-V_{in}}{R_{1}}=\frac{V_{out}}{R_{1}+R_{f}}-\frac{V_{in}}{R_{1}+R_{f}}\$

\$\frac{V}{R_{1}}-\frac{-V_{in}}{R_{1}}+\frac{V_{in}}{R_{1}+R_{f}}=\frac{V_{out}}{R_{1}+R_{f}}\$

\$V_{in}(\frac{1}{R_{1}+R_{f}}-\frac{1}{R_{1}})+\frac{V}{R_{1}}=\frac{V_{out}}{R_{1}+R_{f}}\$

\$v_{+}=v_{-}=V=0\ V \$

\$V_{in}(\frac{1}{R_{1}+R_{f}}-\frac{1}{R_{1}})=\frac{V_{out}}{R_{1}+R_{f}}\$

\$V_{in}(\frac{R_{1}-(R_{1}+R_{f})}{R_{1}(R_{1}+R_{f})})=\frac{V_{out}}{R_{1}+R_{f}}\$

\$(R_{1}+R_{f})(\frac{R_{1}-(R_{1}+R_{f})}{R_{1}(R_{1}+R_{f})})=\frac{V_{out}}{V_{in}}(\frac{R_{1}-(R_{1}+R_{f})}{R_{1}})=\frac{V_{out}}{V_{in}}\$

\$(\frac{R_{1}-R_{1}-R_{f})}{R_{1}})=\frac{V_{out}}{V_{in}}\$

\$H(s)=(\frac{-R_{f})}{R_{1}})=\frac{V_{out}}{V_{in}}\$

Indeed \$H(s)\$ looks like the expected form, but I have no idea how to to get the former expression of A_VOL or what it means.
So, what do these equations mean and how can I get it:
\$A_{V}=\frac{V_{out}}{V_{in}}=\frac{A_{VOL}R_{f}}{-R_{1}-R_{f}-A_{VOL}R_{1}}\$

Answer 1

I usually use superposition for this kind of approach, it is quite straightforward: determine the voltage at pin (-) when \$V_{in}=0\;V\$ while \$V_{out}\$ is alive and vice versa. Sum both results and you have the voltage at the negative pin. This voltage is minus the output voltage divided by the open-loop gain of the op-amp. Write the equation, solve for \$V_{out}\$ and rearrange:

Run a quick SPICE sim to check the dc operating point and compare it with the result you obtain from the solver:

In your equations, you write V(-) = V(+) and this only true if the open-loop gain is conveniently considered as infinite, not in this case.