# Calculating the feedback factor of the inverting amplifier?

In the feedback chapater of the Razavi book, he introduces another way to calculate the DC transfer gain of amplifiers in feedback loop. This is given by $$\\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} = \frac{A_0}{1 + \beta \cdot A_0}\$$. If $$\\beta \cdot A_0 \gg 1\$$, then $$\\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} \approx \frac{1}{\beta}\$$. In the circuit below, if $$\R_1\$$ and $$\R_2\$$ are very large. Therefore, we do not affect the open-loop performance of the circuit. $$\R_1\$$ and $$\R_2\$$ form a voltage divider at the output. The feedback voltage is the voltage drop across $$\R_2\$$, which is $$\\frac{R_2}{R_1 + R_2} V_{\mathrm{out}}\$$. The feedback factor $$\\beta\$$ is therefore $$\\frac{R_2}{R_1 + R_2}\$$. The closed-loop DC voltage gain is $$\\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} = 1 + \frac{R_1}{R_2}\$$.

The transfer function of the inverting amplifer is well-known. It is $$\\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} = - \frac{R_1}{R_2}\$$. This means that its feedback factor is $$\\frac{R_2}{R_1}\$$. How can we apply the method I used for the derivation of the feedback factor in the non-inverting amplifier to the case of the inverting amplifier?

Here is the derivation for the inverting case. \begin{align*} V_\text{out} &= A_{0}(V_{+} - V_-) \tag1 \\\\ V_\text{out} &= -A_0V_x \tag2 \\\\ V_x &= V_\text{in}\frac{R_1}{R_1+R_2} + V_\text{out} \underbrace{\frac{R_2}{R_1+R_2}}_{\beta} \tag3 \\\\ V_\text{out} &= -A_0V_\text{in}\frac{R_1}{R_1+R_2} - A_0V_\text{out} \underbrace{\frac{R_2}{R_1+R_2}}_{\beta} \tag4 \\\\ V_\text{out} \bigg(1+A_0 \frac{R_2}{R_1+R_2} \bigg) &= -A_0V_\text{in}\frac{R_1}{R_1+R_2} \tag5 \\\\ \frac{V_\text{out}}{V_\text{in}} &= - \frac{A_0 \frac{R_1}{R_1+R_2}}{1+A_0 \frac{R_2}{R_1+R_2}} \tag6 \\\\ \frac{V_\text{out}}{V_\text{in}} &= - \frac{A_0R_1}{R_1+R_2+A_0R_2} \to -\frac{R_1}{R_2} \ \ \ \text{for} \ \ \ A_0 \to \infty \tag7 \end{align*}