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Let's solve this **step by step** for the given **difference amplifier** (common-mode case). --- ### **Given:** We are asked to find: * ( i_2 ) (current through ( R_2 )) * ( v_+ ) (voltage at non-inverting input) * ( v_o ) (output) in terms of the **common-mode input voltage ( v_{ICM} )** and resistors ( R_1, R_2, R_3, R_4 ). --- ### **Step 1: Apply common-mode condition** Since both inputs are driven by the same signal: [ v_{i1} = v_{i2} = v_{ICM} ] That means there is **no differential signal** — only common-mode voltage is present. --- ### **Step 2: Analyze the non-inverting terminal (+)** At the **non-inverting terminal**, we have a resistive divider between ( v_{i2} ) and ground (through ( R_3 ) and ( R_4 )). Using the voltage divider rule: [ v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} ] --- ### **Step 3: For an ideal op-amp** For an **ideal op-amp**, the input currents are zero, and the voltage difference between the inputs is zero (virtual short): [ v_- = v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} ] --- ### **Step 4: Analyze the inverting terminal (−)** At node ( v_- ), using KCL: [ \frac{v_{i1} - v_-}{R_1} = \frac{v_- - v_o}{R_2} ] Substitute ( v_{i1} = v_{ICM} ): [ \frac{v_{ICM} - v_-}{R_1} = \frac{v_- - v_o}{R_2} ] --- ### **Step 5: Solve for output ( v_o )** Rearrange the equation: [ R_2 (v_{ICM} - v_-) = R_1 (v_- - v_o) ] [ R_2 v_{ICM} - R_2 v_- = R_1 v_- - R_1 v_o ] [ R_1 v_o = R_1 v_- + R_2 v_- - R_2 v_{ICM} ] [ v_o = v_- \left(1 + \frac{R_2}{R_1}\right) - \frac{R_2}{R_1} v_{ICM} ] Substitute ( v_- = v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} ): [ v_o = v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] ] --- ### **Step 6: Expression for common-mode gain** [ A_{CM} = \frac{v_o}{v_{ICM}} = \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} ] ✅ This is the **common-mode gain**. --- ### **Step 7: Expression for current ( i_2 )** From earlier: [ i_2 = \frac{v_- - v_o}{R_2} ] Substitute ( v_- = v_{ICM}\frac{R_4}{R_3 + R_4} ) and ( v_o ) from Step 5: [ i_2 = \frac{1}{R_2}\left[v_{ICM}\frac{R_4}{R_3 + R_4} - v_{ICM}\left( \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right) \right] ] Simplify: [ i_2 = \frac{v_{ICM}}{R_2}\left[\frac{R_4}{R_3 + R_4} - \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} + \frac{R_2}{R_1}\right] ] [ i_2 = \frac{v_{ICM}}{R_2}\left[-\frac{R_2 R_4}{R_1(R_3 + R_4)} + \frac{R_2}{R_1}\right] ] [ i_2 = \frac{v_{ICM}}{R_1}\left[1 - \frac{R_4}{R_3 + R_4}\right] ] [ \boxed{i_2 = \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)}} ] --- ### ✅ **Final Answers** | Quantity | Expression | Marks (as per question) | | -------- | ---------------------------------------------------------------------------------------------------------------------- | ----------------------- | | ( i_2 ) | ( \displaystyle i_2 = \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} ) | 3 points | | ( v_+ ) | ( \displaystyle v_+ = v_{ICM} \frac{R_4}{R_3 + R_4} ) | 3 points | | ( v_o ) | ( \displaystyle v_o = v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] ) | 5 points | --- ### **Step 8: Special case for perfect common-mode rejection** For ideal common-mode rejection ( ( A_{CM} = 0 ) ): [ \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} = 0 ] [ \Rightarrow \frac{R_1}{R_2} = \frac{R_3}{R_4} ] ✅ This is the **condition for perfect CMRR** (common-mode rejection). --- ✅ **Final Summary** [ \boxed{ \begin{aligned} i_2 &= \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} \ v_+ &= v_{ICM} \frac{R_4}{R_3 + R_4} \ v_o &= v_{ICM}\left[\left(1+\frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1}\right] \end{aligned} } ]

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Let's solve this **step by step** for the given **difference amplifier** (common-mode case). --- ### **Given:** We are asked to find: * ( i_2 ) (current through ( R_2 )) * ( v_+ ) (voltage at non-inverting input) * ( v_o ) (output) in terms of the **common-mode input voltage ( v_{ICM} )** and resistors ( R_1, R_2, R_3, R_4 ). --- ### **Step 1: Apply common-mode condition** Since both inputs are driven by the same signal: [ v_{i1} = v_{i2} = v_{ICM} ] That means there is **no differential signal** — only common-mode voltage is present. --- ### **Step 2: Analyze the non-inverting terminal (+)** At the **non-inverting terminal**, we have a resistive divider between ( v_{i2} ) and ground (through ( R_3 ) and ( R_4 )). Using the voltage divider rule: [ v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} ] --- ### **Step 3: For an ideal op-amp** For an **ideal op-amp**, the input currents are zero, and the voltage difference between the inputs is zero (virtual short): [ v_- = v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} ] --- ### **Step 4: Analyze the inverting terminal (−)** At node ( v_- ), using KCL: [ \frac{v_{i1} - v_-}{R_1} = \frac{v_- - v_o}{R_2} ] Substitute ( v_{i1} = v_{ICM} ): [ \frac{v_{ICM} - v_-}{R_1} = \frac{v_- - v_o}{R_2} ] --- ### **Step 5: Solve for output ( v_o )** Rearrange the equation: [ R_2 (v_{ICM} - v_-) = R_1 (v_- - v_o) ] [ R_2 v_{ICM} - R_2 v_- = R_1 v_- - R_1 v_o ] [ R_1 v_o = R_1 v_- + R_2 v_- - R_2 v_{ICM} ] [ v_o = v_- \left(1 + \frac{R_2}{R_1}\right) - \frac{R_2}{R_1} v_{ICM} ] Substitute ( v_- = v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} ): [ v_o = v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] ] --- ### **Step 6: Expression for common-mode gain** [ A_{CM} = \frac{v_o}{v_{ICM}} = \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} ] ✅ This is the **common-mode gain**. --- ### **Step 7: Expression for current ( i_2 )** From earlier: [ i_2 = \frac{v_- - v_o}{R_2} ] Substitute ( v_- = v_{ICM}\frac{R_4}{R_3 + R_4} ) and ( v_o ) from Step 5: [ i_2 = \frac{1}{R_2}\left[v_{ICM}\frac{R_4}{R_3 + R_4} - v_{ICM}\left( \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right) \right] ] Simplify: [ i_2 = \frac{v_{ICM}}{R_2}\left[\frac{R_4}{R_3 + R_4} - \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} + \frac{R_2}{R_1}\right] ] [ i_2 = \frac{v_{ICM}}{R_2}\left[-\frac{R_2 R_4}{R_1(R_3 + R_4)} + \frac{R_2}{R_1}\right] ] [ i_2 = \frac{v_{ICM}}{R_1}\left[1 - \frac{R_4}{R_3 + R_4}\right] ] [ \boxed{i_2 = \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)}} ] --- ### ✅ **Final Answers** | Quantity | Expression | Marks (as per question) | | -------- | ---------------------------------------------------------------------------------------------------------------------- | ----------------------- | | ( i_2 ) | ( \displaystyle i_2 = \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} ) | 3 points | | ( v_+ ) | ( \displaystyle v_+ = v_{ICM} \frac{R_4}{R_3 + R_4} ) | 3 points | | ( v_o ) | ( \displaystyle v_o = v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] ) | 5 points | --- ### **Step 8: Special case for perfect common-mode rejection** For ideal common-mode rejection ( ( A_{CM} = 0 ) ): [ \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} = 0 ] [ \Rightarrow \frac{R_1}{R_2} = \frac{R_3}{R_4} ] ✅ This is the **condition for perfect CMRR** (common-mode rejection). --- ✅ **Final Summary** [ \boxed{ \begin{aligned} i_2 &= \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} \ v_+ &= v_{ICM} \frac{R_4}{R_3 + R_4} \ v_o &= v_{ICM}\left[\left(1+\frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1}\right] \end{aligned} } ]

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Asked by: Mohin
Created at: 2025-10-20 08:36:56

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# Step-by-Step Solution for the Common-Mode Case of a Difference Amplifier --- ## Given: - Both are driven by the same common-mode voltage: \[ _{i1} = v_{2} = v_{ICM} \] - Resistors: R_1 R_2, R_3 R_4) --- ## 1. Non-inverting terminal voltage \( v_+ \) Using the voltage divider formed by \( R_3 \) and \( R_4 \): \[ v_+ = v_{i2} \times \frac{R_4}{R_3 + R_4} = v_{ICM} \frac{R_4}{R_3 + R_4} \] --- ## 2. Virtual short condition For an ideal op-amp: \[ v_- \approx v_+ = v_{ICM} \frac{R_4}{R_3 + R_4} \] --- ## 3. Inverting node analysis Apply Kirchhoff's Current Law (KCL) at \( v_- \): \[ \frac{v_{i1} - v_-}{R_1} = \frac{v_- - v_o}{R_2} \] Substitute \( v_{i1} = v_{ICM} \): \[ \frac{v_{ICM} - v_-}{R_1} = \frac{v_- - v_o}{R_2} \] Solve for \( v_o \): \[ R_2 (v_{ICM} - v_-) = R_1 (v_- - v_o) \] \[ R_2 v_{ICM} - R_2 v_- = R_1 v_- - R_1 v_o \] \[ R_1 v_o = R_1 v_- + R_2 v_- - R_2 v_{ICM} \] \[ v_o = v_- \left( 1 + \frac{R_2}{R_1} \right) - \frac{R_2}{R_1} v_{ICM} \] Substitute \( v_- = v_+ \): \[ v_- = v_{ICM} \frac{R_4}{R_3 + R_4} \] \[ \boxed{ v_o = v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] } \] --- ## 4. Current \( i_2 \) through \( R_2 \) Recall: \[ i_2 = \frac{v_- - v_o}{R_2} \] Substitute \( v_- \) and \( v_o \): \[ i_2 = \frac{v_{ICM} \frac{R_4}{R_3 + R_4} - v_o}{R_2} \] Plug in \( v_o \): \[ i_2 = \frac{v_{ICM} \frac{R_4}{R_3 + R_4} - v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right]}{R_2} \] Factor \( v_{ICM} \): \[ i_2 = \frac{v_{ICM}}{R_2} \left[ \frac{R_4}{R_3 + R_4} - \left( \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right) \right] \] Simplify numerator: \[ i_2 = \frac{v_{ICM}}{R_2} \left[ \frac{R_4}{R_3 + R_4} - \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} + \frac{R_2}{R_1} \right] \] \[ i_2 = \frac{v_{ICM}}{R_2} \left[ - \frac{R_2 R_4}{R_1 (R_3 + R_4)} + \frac{R_2}{R_1} \right] \] Factor further: \[ i_2 = \frac{v_{ICM}}{R_1} \left[ 1 - \frac{R_4}{R_3 + R_4} \right] \] Recognize: \[ 1 - \frac{R_4}{R_3 + R_4} = \frac{R_3 + R_4 - R_4}{R_3 + R_4} = \frac{R_3}{R_3 + R_4} \] **Final expression:** \[ \boxed{ i_2 = \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} } \] --- ## **Summary of Results** | Quantity | Expression | Points | |------------|--------------------------------------------------------------|---------| | \( i_2 \) | \(\displaystyle \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} \) | 3 pts | | \( v_+ \) | \(\displaystyle v_{ICM} \frac{R_4}{R_3 + R_4} \) | 3 pts | | \( v_o \) | \(\displaystyle v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] \) | 5 pts | --- ## **Condition for Perfect Common-Mode Rejection** To make the common-mode gain zero (\( A_{CM} = \)), set: \[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} = \] which simplifies to: \[ \boxed{ \frac{R_1}{R_2} = \frac{R_3}{R_4} } \] This ensures ideal common-mode rejection. --- **Note:** These expressions are valid under the assumption of an ideal op-amp with infinite open-loop gain and zero input bias current.

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Asked by: Mohin
Created at: 2025-10-20 08:38:21

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# Step-by-Step Solution for the Common-Mode Case of a Difference Amplifier ## Given: - Both inputs are driven by the same common-mode voltage: \[ v_{i1} = v_{i2} = v_{ICM} \] - Resistors: \( R_1, R_2, R_3, R_4 \) --- ## 1. Non-inverting Terminal Voltage \( v_+ \) Using the voltage divider formed by \( R_3 \) and \( R_4 \): \[ v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} \] --- ## 2. Virtual Short Condition For an ideal op-amp: \[ v_- \approx v_+ = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} \] --- ## 3. Inverting Node Analysis Apply Kirchhoff's Current Law (KCL) at \( v_- \): \[ \frac{v_{i1} - v_-}{R_1} = \frac{v_- - v_o}{R_2} \] Substituting \( v_{i1} = v_{ICM} \): \[ \frac{v_{ICM} - v_-}{R_1} = \frac{v_- - v_o}{R_2} \] ### Solve for \( v_o \): \[ R_2(v_{ICM} - v_-) = R_1(v_- - v_o) \] \[ R_2 v_{ICM} - R_2 v_- = R_1 v_- - R_1 v_o \] \[ R_1 v_o = R_1 v_- + R_2 v_- - R_2 v_{ICM} \] \[ v_o = v_- \left( 1 + \frac{R_2}{R_1} \right) - \frac{R_2}{R_1} v_{ICM} \] Substituting \( v_- = v_{ICM} \cdot \frac{R_4}{R_3 + R_4} \): \[ \boxed{ v_o = v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] } \] --- ## 4. Current \( i_2 \) Through \( R_2 \) Recall: \[ i_2 = \frac{v_- - v_o}{R_2} \] Substituting \( v_- \) and \( v_o \): \[ i_2 = \frac{v_{ICM} \cdot \frac{R_4}{R_3 + R_4} - v_o}{R_2} \] Plugging in \( v_o \): \[ i_2 = \frac{v_{ICM} \cdot \frac{R_4}{R_3 + R_4} - v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right]}{R_2} \] Factor out \( v_{ICM} \): \[ i_2 = \frac{v_{ICM}}{R_2} \left[ \frac{R_4}{R_3 + R_4} - \left( \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right) \right] \] ### Simplify: \[ i_2 = \frac{v_{ICM}}{R_2} \left[ - \frac{R_2 R_4}{R_1 (R_3 + R_4)} + \frac{R_2}{R_1} \right] \] Final expression: \[ i_2 = \frac{v_{ICM}}{R_1} \left[ 1 - \frac{R_4}{R_3 + R_4} \right] \] Recognizing: \[ 1 - \frac{R_4}{R_3 + R_4} = \frac{R_3}{R_3 + R_4} \] **Final expression:** \[ \boxed{ i_2 = \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} } \] --- ## **Summary of Results** | Quantity | Expression | Points | |----------|--------------------------------------------------------------|---------| | \( i_2 \) | \(\displaystyle \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)} \) | 3 pts | | \( v_+ \) | \(\displaystyle v_{ICM} \cdot \frac{R_4}{R_3 + R_4} \) | 3 pts | | \( v_o \) | \(\displaystyle v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right] \) | 5 pts | --- ## **Condition for Perfect Common-Mode Rejection** To achieve ideal common-mode rejection (\( A_{CM} = 0 \)): \[ \left(1 + \frac{R_2}{R_1}\right) \frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} = 0 \] This simplifies to: \[ \boxed{ \frac{R_1}{R_2} = \frac{R_3}{R_4} } \] This condition ensures ideal common-mode rejection. --- **Note:** These expressions are valid under the assumption of an ideal op-amp with infinite open-loop gain and zero input bias current.

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