![[differential-amp.png]] ``` Consider the op-amp circuit shown in the figure, where $v_{1}$ is the input connected to the inverting input of the op-amp and $v_{2}$ is the input connected to the non-inverting input of the op-amp. Use double dollar signs for LaTeX equation formatting. a) Find an expression for the output voltage $v_0$. b) Consider that $R_4=R_2$ and $R_3=R_1$. Find the new expression for the output voltage $v_0$. ``` (Basic Engineering Circuit Analysis (11th Edition), Example 4.5, page 158) > [!Solution]- > a) > $ > v_{0} = \frac{R_{2}}{R_{1}} \left( 1+\frac{R_{1}}{R_{2}} \right) \frac{R_{4}}{R_{3}+R_{4}}v_{2}-\frac{R_{2}}{R_{1}}v_{1} > $ > > b) > $ > v_{0} = \frac{R_{2}}{R_{1}}(v_{2}-v_{1}) > $ > [!Claude 3.5]- > Claude 3.5 fails: > > Let's approach this problem step-by-step: > > a) To find an expression for the output voltage $v_0$: > > 1. This is a differential amplifier circuit with feedback. > 2. For an ideal op-amp, we can assume that $v_- = v_+$ (virtual short) and that no current flows into the op-amp inputs. > 3. Let's write equations for $v_-$ and $v_+$: > > $v_- = v_1 - i_-R_1$ <--- WRONG! > $v_+ = v_2\frac{R_4}{R_3 + R_4}$ > > 4. Since $v_- = v_+$, we can equate these: > > $v_1 - i_-R_1 = v_2\frac{R_4}{R_3 + R_4}$ > > 5. The current $i_-$ also flows through $R_2$, so: > > $v_0 = v_- + i_-R_2 = v_2\frac{R_4}{R_3 + R_4} + i_-R_2$ > > 6. Solving for $i_-$ from step 4: > > $i_- = \frac{v_1 - v_2\frac{R_4}{R_3 + R_4}}{R_1}$ > > 7. Substituting this into the equation from step 5: > > $v_0 = v_2\frac{R_4}{R_3 + R_4} + \frac{R_2}{R_1}(v_1 - v_2\frac{R_4}{R_3 + R_4})$ > > 8. Simplifying: > > $v_0 = \frac{R_2}{R_1}v_1 + (1 - \frac{R_2}{R_1})v_2\frac{R_4}{R_3 + R_4}$ > > This is the general expression for $v_0$. > > b) If $R_4 = R_2$ and $R_3 = R_1$: > > 1. Substituting these into our expression: > > $v_0 = \frac{R_2}{R_1}v_1 + (1 - \frac{R_2}{R_1})v_2\frac{R_2}{R_1 + R_2}$ > > 2. Simplifying: > > $v_0 = \frac{R_2}{R_1}v_1 + (1 - \frac{R_2}{R_1})\frac{R_2}{R_1 + R_2}v_2$ > > 3. This can be further simplified to: > > $v_0 = \frac{R_2}{R_1}(v_1 - v_2) + v_2$ > > This is the expression for $v_0$ when $R_4 = R_2$ and $R_3 = R_1$. > [!GPT-4o after some corrections]- > > You're right, my earlier analysis contained an error. Let's correct the approach and derive the accurate expressions. > > ### Part (a): General Expression for the Output Voltage \(v_0\) > > 1. **Op-Amp Characteristics:** > - The voltage at the inverting input \(v_-\) and the non-inverting input \(v_+\) will be equal due to the virtual short concept in an ideal op-amp. > - There is no current into the op-amp inputs, i.e., \(i_+ = 0\) and \(i_- = 0\). > > 2. **Determine \(v_+\) and \(v_-\):** > > - For the non-inverting input, \(v_+\): > $ > v_+ = v_2 \frac{R_4}{R_3 + R_4} > $ > This is derived from the voltage divider formed by \(R_3\) and \(R_4\). > > - For the inverting input, we use the voltage at the inverting input \(v_-\): > $ > v_- = v_1 - \left(\frac{v_1 - v_o}{R_1 + R_2}\right)R_1 > $ > > Since \(v_+ = v_-\): > $ > v_2 \frac{R_4}{R_3 + R_4} = v_1 - \frac{v_1 - v_o}{R_1 + R_2} R_1 > $ > > $ > v_2 \frac{R_4}{R_3 + R_4} = v_1 \left(1 - \frac{R_1}{R_1 + R_2}\right) + v_o \frac{R_1}{R_1 + R_2} > $ > > Simplify the terms: > $ > v_2 \frac{R_4}{R_3 + R_4} = \frac{v_1 R_2}{R_1 + R_2} + v_o \frac{R_1}{R_1 + R_2} > $ > > Solve for \(v_o\): > $ > v_o \frac{R_1}{R_1 + R_2} = v_2 \frac{R_4}{R_3 + R_4} - \frac{v_1 R_2}{R_1 + R_2} > $ > > $ > v_o = \frac{(R_1 + R_2)}{R_1} \left( v_2 \frac{R_4}{R_3 + R_4} - \frac{v_1 R_2}{R_1 + R_2} \right) > $ > > Thus, the general expression for the output voltage \(v_0\) is: > $ > v_o = \frac{R_1 + R_2}{R_1} \left( v_2 \frac{R_4}{R_3 + R_4} - \frac{v_1 R_2}{R_1 + R_2} \right) > $ > > ### Part (b): Simplified Expression with \(R_4 = R_2\) and \(R_3 = R_1\) > > Given: > $ > R_4 = R_2 \quad \text{and} \quad R_3 = R_1 > $ > > Plug these values into the expression found in part (a): > $ > v_o = \frac{R_1 + R_2}{R_1} \left( v_2 \frac{R_2}{R_1 + R_2} - \frac{v_1 R_2}{R_1 + R_2} \right) > $ > > Simplify: > $ > v_o = \frac{R_1 + R_2}{R_1} \cdot \frac{R_2}{R_1 + R_2} \left( v_2 - v_1 \right) > $ > > $ > v_o = \frac{R_2}{R_1} \left( v_2 - v_1 \right) > $ > > Thus, the simplified expression for the output voltage \(v_o\) is: > $ > v_o = \frac{R_2}{R_1} (v_2 - v_1) > $