Operational Amplifier(Op-Amp)

Problem-Solving Strategy for solving the Op-Amp Circuits

STEP 1. Use the ideal op-amp model: $A_o=\infty$, $R_i=\infty$, $R_o=0$

$\Rightarrow$ $i_+=i_-=0$, $v_+=v_-$

STEP 2. Apply nodal analysis to the resulting circuit.

STEP 3. Solve nodal equations to express the output voltage in terms of the op-amp input signals.

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed.

If there are any errors in solution, please let me know in the comments

If we assume an ideal op-amp, its condition is as follows:

$i_+=i_-=0$, $v_+=v_-$

The node equations for the network are

$\frac{v_3-v_1}{R_2}=\frac{v_1-v_2}{R_1}=\frac{v_2-v_4}{R_2}$

$\frac{v_4-v_5}{R_3}=\frac{v_5}{R_4}$

$\frac{v_3-v_5}{R_3}=\frac{v_5-v_o}{R_4}$

To find $v_o$, $v_3$ and $v_5$ should be expressed in terms of $v_1$ and $v_2$, which are

$v_3=(1+\frac{R_2}{R_1})v_1-\frac{R_2}{R_1}v_2$

$v_5=-\frac{R_2R_4}{R_1(R_3+R_4)}v_1+(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})v_2$

Since $v_o=-\frac{R_4}{R_3}v_3+(1+\frac{R_4}{R_3})v_5$

$v_o=\frac{R_4}{R_3}(1+\frac{2R_2}{R_1})(v_2-v_1)$