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}
}
]
Please give me this content as it is line by line without any changes
Question:
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}
}
]
Please give me this content as it is line by line without any changes
This is AI generated Response. Asked by: Mohin Created at: 2025-10-20 08:34:44
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Let's solve this **step by step** for the given **difference amplifier** (common-mode).
---
### **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} \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} = ) ):
[
\left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} =
]
[
\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}
}
]
Please give me this content as it is line by line without any changes
Question:
please give me the entire answer in render equations without any errors
This is AI generated Response. Asked by: Mohin Created at: 2025-10-20 08:35:49
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Sure! Here’s the content with rendered equations:
# Difference Amplifier Analysis (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:
\[
\frac{v_{i1} - v_-}{R_1} = \frac{v_- - v_o}{R_2}
\]
Substituting \( v_{i1} = v_{ICM} \):
\[
\frac{v_{ICM} - v_-}{R_1} = \frac{v_- - v_o}{R_2}
\]
---
## Step 5: Solve for Output \( v_o \)
Rearranging the equation:
\[
R_2 (v_{ICM} - v_-) = R_1 (v_- - v_o)
\]
\[
R_2 v_{ICM} - R_2 v_- = R_1 v_- - R_1 v_o
\]
\[
R_1 v_o = R_1 v_- + R_2 v_- - R_2 v_{ICM}
\]
\[
v_o = v_- \left(1 + \frac{R_2}{R_1}\right) - \frac{R_2}{R_1} v_{ICM}
\]
Substituting \( v_- = v_+ = v_{ICM} \frac{R_4}{R_3 + R_4} \):
\[
v_o = v_{ICM} \left[ \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right]
\]
---
## 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:
\[
i_2 = \frac{v_- - v_o}{R_2}
\]
Substituting \( v_- = v_{ICM}\frac{R_4}{R_3 + R_4} \) and \( v_o \) from Step 5:
\[
i_2 = \frac{1}{R_2}\left[v_{ICM}\frac{R_4}{R_3 + R_4} - v_{ICM}\left( \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} - \frac{R_2}{R_1} \right) \right]
\]
Simplifying:
\[
i_2 = \frac{v_{ICM}}{R_2}\left[\frac{R_4}{R_3 + R_4} - \left(1 + \frac{R_2}{R_1}\right)\frac{R_4}{R_3 + R_4} + \frac{R_2}{R_1}\right]
\]
\[
i_2 = \frac{v_{ICM}}{R_2}\left[-\frac{R_2 R_4}{R_1(R_3 + R_4)} + \frac{R_2}{R_1}\right]
\]
\[
i_2 = \frac{v_{ICM}}{R_1}\left[1 - \frac{R_4}{R_3 + R_4}\right]
\]
\[
\boxed{i_2 = \frac{v_{ICM} R_3}{R_1 (R_3 + R_4)}}
\]
---
## 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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