Question

Calculate the output voltage of an op-amp for input voltages of \(V_{i1}=150\mu V\) and \(V_{i2}=140\mu V\). The amplifier has a differential gain of \(A_d=4000\) and the value of CMRR is \(100\):

• A. \(145 \text{ mV}\)
• B. \(14.5 \text{ mV}\)
• C. \(45.8 \text{ mV}\)
• D. \(40.006 \text{ mV}\)

Hint:

For op-amp problems involving CMRR: \[ \text{CMRR}=\frac{A_d}{A_c} \] Always calculate:
• Differential voltage \(V_d=V_1-V_2\)
• Common-mode voltage \(V_c=\frac{V_1+V_2}{2}\) Then use: \[ V_o=A_dV_d+A_cV_c \] to obtain the final output voltage.

Answer 1

The Correct Option is A

Concept: An operational amplifier responds to two types of input signals:
• Differential input voltage \[ V_d = V_{i1} - V_{i2} \]
• Common-mode input voltage \[ V_c = \frac{V_{i1}+V_{i2}}{2} \] The output voltage of a practical op-amp is given by: \[ V_o = A_dV_d + A_cV_c \] where:
• \(A_d\) is the differential gain
• \(A_c\) is the common-mode gain The relation between CMRR and gains is: \[ \text{CMRR}=\frac{A_d}{A_c} \] Thus, \[ A_c=\frac{A_d}{\text{CMRR}} \] A high value of CMRR indicates that the amplifier strongly amplifies differential signals while rejecting common-mode signals.

Step 1: Calculate the differential input voltage. Given: \[ V_{i1}=150\mu V \] and \[ V_{i2}=140\mu V \] Therefore, \[ V_d=V_{i1}-V_{i2} \] Substituting the values: \[ V_d=150\mu V-140\mu V \] \[ V_d=10\mu V \] Converting into volts: \[ V_d=10\times10^{-6}V \] \[ V_d=1\times10^{-5}V \]

Step 2: Calculate the common-mode input voltage. The common-mode voltage is: \[ V_c=\frac{V_{i1}+V_{i2}}{2} \] Substituting the values: \[ V_c=\frac{150\mu V+140\mu V}{2} \] \[ V_c=\frac{290\mu V}{2} \] \[ V_c=145\mu V \] Converting into volts: \[ V_c=145\times10^{-6}V \] \[ V_c=1.45\times10^{-4}V \]

Step 3: Determine the common-mode gain using CMRR. Given: \[ A_d=4000 \] and \[ \text{CMRR}=100 \] Using: \[ A_c=\frac{A_d}{\text{CMRR}} \] Substituting the values: \[ A_c=\frac{4000}{100} \] \[ A_c=40 \]

Step 4: Calculate the output voltage due to differential input. The differential output component is: \[ V_{od}=A_dV_d \] Substituting the values: \[ V_{od}=4000\times10\mu V \] \[ V_{od}=40000\mu V \] \[ V_{od}=40\text{ mV} \]

Step 5: Calculate the output voltage due to common-mode input. The common-mode output component is: \[ V_{oc}=A_cV_c \] Substituting the values: \[ V_{oc}=40\times145\mu V \] \[ V_{oc}=5800\mu V \] \[ V_{oc}=5.8\text{ mV} \]

Step 6: Calculate the total output voltage. The total output voltage is: \[ V_o=V_{od}+V_{oc} \] Substituting the obtained values: \[ V_o=40\text{ mV}+5.8\text{ mV} \] \[ V_o=45.8\text{ mV} \] Hence, the output voltage of the op-amp is: \[ \boxed{45.8\text{ mV}} \] Therefore, the correct option is: \[ \boxed{(C)\ 45.8\text{ mV}} \]