The analysis begins at the inverting input terminal of the op amp, where the voltage is

$$
v_1=\frac{-v_O}{A}=\frac{-v_O}{\infty}=0
$$


Here we have assumed that the circuit is "working" and producing a finite output voltage $v_0$. Knowing $v_1$, we can deternune the current $i_1$ as follows:

$$
i_1=\frac{v_I-v_1}{R_1}=\frac{v_I-0}{R_1}=\frac{v_I}{R_1}
$$


Since zero current flows into the inverting input terminal, all of $i_1$ will flow through $R_2$, and thus

$$
i_2=i_1=\frac{v_I}{R_1}
$$


Now we can determine the voltage at node $x$ :

$$
v_x=v_1-i_2 R_2=0-\frac{v_1}{R_1} R_2=-\frac{R_2}{R_1} v_1
$$


This in turn enables us to find the current $i_3$ :

$$
i_3=\frac{0-v_x}{R_3}=\frac{R_2}{R_1 R_3} v_I
$$

Next, a node equation at $x$ yields $i_4$ :

$$
i_4=i_2+i_3=\frac{v_I}{R_1}+\frac{R_2}{R_1 R_3} v_I
$$


Finally, we can determine $v_0$ from

$$
\begin{aligned}
v_O & =v_x-i_4 R_4 \\
& =-\frac{R_2}{R_1} v_I-\left(\frac{v_I}{R_1}+\frac{R_2}{R_1 R_3} v_I\right) R_4
\end{aligned}
$$


Thus the voltage gain is given by

$$
\frac{v_O}{v_I}=-\left[\frac{R_2}{R_1}+\frac{R_4}{R_1}\left(1+\frac{R_2}{R_3}\right)\right]
$$

which can be written in the form

$$
\frac{v_O}{v_I}=-\frac{R_2}{R_1}\left(1+\frac{R_4}{R_2}+\frac{R_4}{R_3}\right)
$$


Now, since an input resistance of $1 \mathrm{M} \Omega$ is required, we select $R_1=1 \mathrm{M} \Omega$. Then, with the limitation of using resistors no greater than I M $\Omega$, the maximum value possible for the first factor in the gain expression is 1 and is obtained by selecting $R_2=1 \mathrm{M} \Omega$. To obtain a gain of $-100, R_3$ and $R_4$ must be selected so that the second factor in the gain expression is 100 . If we select the maximum allowed (in this example) value of $1 \mathrm{M} \Omega$ for $R_4$, then the required value of $R_3$ can be calculated to be $10.2 \mathrm{k} \Omega$. Thus this circuit utilizes three $1-\mathrm{M} \Omega$ resistors and a $10.2-\mathrm{k} \Omega$ resistor. In comparison, if the inverting configuration were used with $R_1=1 \mathrm{M} \Omega$ we would have required a feedback resistor of 100 MS 2 , an impractically large value!

Before leaving this example it is insightful to enquire into the mechanism by which the circuit is able to realize a large voltage gain without using large resistances in the fcedback path. Toward that end, observe that because of the virtual ground at the inverting input terminal of the op amp, $R_2$ and $R_3$ are in effect in parallel. Thus, by making $R_3$ lower than $R_2$ by, say, a factor $k$ (i.c., $R_3=R_2 / k$ where $k>1$ ), $R_3$ is forced to carry a current $k$-times that in $R_2$. Thus, while $i_2=i_5$, $i_3=k i_1$ and $i_4=(k+1) i_1$. It is the current multiplication by a factor of $(k+1)$ that enables a large voltage drop to develop across $R_4$ and hence a large $v_O$ without using a large value for $R_4$. Notice also that the current through $R_4$ is independent of the value of $R_4$.