Let's solve the problem step-by-step. --- ## **Given Data and Assumptions** - The is a basic difference amplifier. - OP-AMPs are ideal. The question involves connecting \( v_{1} \) and \( v_{in2} \) to the amplifier. - The goal: Find the common-mode gain \(A_{}\) and output \(v\) in terms of \(v_{cm}\) and resistor values. - The circuit diagram shows: - \( R_1 \) connected to \( v_{in1} \) - \( R_3 \) connected to \( v_{in2} \) and ground through \( R_4 \) - Feedback resistor \( R_2 \) - Non-inverting input (+) connected to \( v_{in2} \) and grounded via \( R_4 \). --- ## **Step 1: Understanding the circuit** - The circuit is a difference amplifier, which ideally subtracts the two input voltages \( v_{in1} \) and \( v_{in2} \). - For the common-mode analysis, assume both inputs have the same voltage \( v_{cm} \). --- ## **Step 2: Express the voltages at the non-inverting terminal (+)** - The non-inverting terminal is connected to \( v_{in2} \) through \( R_3 \) and to ground through \( R_4 \). - When both \( v_{in1} \) and \( v_{in2} \) are equal to \( v_{cm} \), the voltage at the non-inverting terminal \( v_+ \) is: \[ v_+ = v_{in2} \times \frac{R_4}{R_3 + R_4} \] - This is a voltage divider between \( v_{in2} \) and ground. --- ## **Step 3: Find the voltage at the inverting terminal (-)** - Since the op-amp is ideal, the voltage at the inverting terminal \( v_- \) is virtually equal to \( v_+ \): \[ v_- \approx v_+ \] - The inverting terminal is connected to \( R_1 \) (connected to \( v_{in1} \)) and \( R_2 \) (feedback resistor connected to \( v_o \)). - The node at \( v_- \) forms a voltage divider between \( v_{in1} \) and \( v_o \): \[ v_- = \frac{ R_2 \times v_{in1} + R_1 \times v_o }{ R_1 + R_2 } \] But since \( v_- \approx v_+ \), and \( v_+ \) is determined as above, for the common-mode case: \[ v_{in1} = v_{in2} = v_{cm} \] --- ## **Step 4: Derive the common-mode gain \(A_{cm}\)** - The output voltage \( v_o \) in response to common-mode input \( v_{cm} \): \[ v_o = A_{cm} \times v_{cm} \] - To find \( A_{cm} \), write the expression for \( v_o \) when both inputs are equal: \[ v_{-} = v_+ = v_{cm} \times \frac{ R_4 }{ R_3 + R_4 } \] - The inverting input voltage: \[ v_- = \frac{ R_2 \times v_{in1} + R_1 \times v_o }{ R_1 + R_2 } \] - Since \( v_{in1} = v_{cm} \), and \( v_- \approx v_+ \): \[ v_{cm} \times \frac{ R_4 }{ R_3 + R_4 } = \frac{ R_2 \times v_{cm} + R_1 \times v_o }{ R_1 + R_2 } \] --- ## **Step 5: Solve for \( v_o \) in terms of \( v_{cm} \)** Multiply both sides by \( R_1 + R_2 \): \[ \left( R_1 + R_2 \right) \times v_{cm} \times \frac{ R_4 }{ R_3 + R_4 } = R_2 \times v_{cm} + R_1 \times v_o \] Rearranged: \[ \left( R_1 + R_2 \right) \times v_{cm} \times \frac{ R_4 }{ R_3 + R_4 } - R_2 v_{cm} = R_1 v_o \] Factor \( v_{cm} \): \[ v_{cm} \left[ \left( R_1 + R_2 \right) \times \frac{ R_4 }{ R_3 + R_4 } - R_2 \right] = R_1 v_o \] Finally, solve for \( v_o \): \[ v_o = \frac{1}{ R_1 } \times v_{cm} \times \left[ \left( R_1 + R_2 \right) \times \frac{ R_4 }{ R_3 + R_4 } - R_2 \right] \] --- ## **Step 6: Expression for common-mode gain \(A_{cm}\)** \[ A_{cm} = \frac{v_o}{v_{cm}} = \frac{1}{ R_1 } \times \left[ \left( R_1 + R_2 \right) \times \frac{ R_4 }{ R_3 + R_4 } - R_2 \right] \] --- ## **Summary:** - **Common-mode gain:** \[ \boxed{ A_{cm} = \frac{1}{ R_1 } \left[ \left( R_1 + R_2 \right) \frac{ R_4 }{ R_3 + R_4 } - R_2 \right] } \] - **Output in terms of common-mode voltage:** \[ v_o = A_{cm} \times v_{cm} \] This completes the derivation based on the circuit and ideal op-amp assumptions. --- If you have specific resistor values, you can substitute them into the formula to compute the actual common-mode gain.

# Solution to the Difference Amplifier Problem This solution provides a detailed step-by-step approach to determine the common-mode gain and the output voltage for the given difference amplifier circuit. ## **Given Data and Assumptions** - The circuit is a basic difference amplifier. - All operational amplifiers (OP-AMPs) are ideal. - Input voltages \( v_{in1} \) and \( v_{in2} \) are equal to a common-mode signal \( v_{cm} \). - Resistors are denoted as \( R_1, R_2, R_3, R_4 \). --- ## **Step 1: Understanding the Circuit Configuration** - The non-inverting terminal (+) connects to \( v_{in2} \) via \( R_3 \) and to ground through \( R_4 \). - The inverting terminal (-) connects to \( v_{in1} \) through \( R_1 \) and feedback through \( R_2 \). --- ## **Step 2: Voltage at the Non-Inverting Terminal (+)** The voltage at the non-inverting terminal \( v_+ \) can be expressed using a voltage divider formula: \[ v_+ = v_{in2} \times \frac{R_4}{R_3 + R_4} \] Since \( v_{in2} = v_{cm} \): \[ v_+ = v_{cm} \times \frac{R_4}{R_3 + R_4} \] --- ## **Step 3: Voltage at the Inverting Terminal (-)** For ideal OP-AMPs, the voltage at the inverting terminal \( v_- \) is equal to \( v_+ \): \[ v_- \approx v_+ = v_{cm} \times \frac{R_4}{R_3 + R_4} \] Using the node equation for \( v_- \): \[ v_- = \frac{R_2 \cdot v_{in1} + R_1 \cdot v_o}{R_1 + R_2} \] Substituting \( v_{in1} = v_{cm} \): \[ v_{cm} \times \frac{R_4}{R_3 + R_4} = \frac{R_2 \cdot v_{cm} + R_1 \cdot v_o}{R_1 + R_2} \] --- ## **Step 4: Deriving the Output Voltage \( v_o \)** Multiply by \( R_1 + R_2 \): \[ v_{cm} \times \frac{R_4 (R_1 + R_2)}{R_3 + R_4} = R_2 v_{cm} + R_1 v_o \] Rearranging gives: \[ v_{cm} \left[ \frac{R_4 (R_1 + R_2)}{R_3 + R_4} - R_2 \right] = R_1 v_o \] Solving for \( v_o \): \[ v_o = \frac{1}{R_1} \left[ v_{cm} \left( \frac{R_4 (R_1 + R_2)}{R_3 + R_4} - R_2 \right) \right] \] --- ## **Step 5: Common-Mode Gain \( A_{cm} \)** The common-mode gain \( A_{cm} \) is defined as: \[ A_{cm} = \frac{v_o}{v_{cm}} = \frac{1}{R_1} \left( \frac{R_4 (R_1 + R_2)}{R_3 + R_4} - R_2 \right) \] --- ## **Summary of Results** - **Voltage at the Non-Inverting Terminal (+)**: \[ v_+ = v_{cm} \times \frac{R_4}{R_3 + R_4} \] - **Voltage at the Inverting Terminal (-)**: \[ v_- \approx v_+ = v_{cm} \times \frac{R_4}{R_3 + R_4} \] - **Output Voltage**: \[ v_o = \frac{1}{R_1} \left[ v_{cm} \left( \frac{R_4 (R_1 + R_2)}{R_3 + R_4} - R_2 \right) \right] \] - **Common-Mode Gain**: \[ A_{cm} = \frac{1}{R_1} \left( \frac{R_4 (R_1 + R_2)}{R_3 + R_4} - R_2 \right) \] This completes the derivation and provides all required expressions as specified in the question.