GATE 2024 EE Q.61: solution and answer

GATE 2024 · EEQ.61 · 2 marks · NAT

A difference amplifier is shown in the figure. Assume the op-amp to be ideal. The CMRR (in dB) of the difference amplifier is ________ (rounded off to 2 decimal places).

GATE 2024 EE question 61: Single op-amp difference amplifier. — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

ANSWER

The CMRR of the difference amplifier is about 40.51 dB (ratio about 106.02).

Official answer: 39.50 to 41.50Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

Assume an ideal op-amp: virtual short, so $v_{-} = v_{+}$ , and no current into either input. By superposition, $v_{o} = - \frac{R _{f}}{R _{1}} v_{in 1} + (1 + \frac{R _{f}}{R _{1}}) \frac{R _{4}}{R _{3} + R _{4}} v_{in 2}$ where $R_{1} = 10.5 k Ω$ , $R_{f} = 101 k Ω$ , $R_{3} = 9.5 k Ω$ and $R_{4} = 101.5 k Ω$ .
Gain from $v_{in 1}$ (inverting path): $A_{1} = - \frac{101}{10.5} = - \frac{101 \times 111}{10.5 \times 111} = - \frac{11211}{1165.5} = - 9.61905$ .
Gain from $v_{in 2}$ (non-inverting path): $A_{2} = \frac{111.5}{10.5} \cdot \frac{101.5}{111} = \frac{11317.25}{1165.5} = 9.71021$ .
Split the inputs into differential and common-mode parts: $v_{in 1} = v_{c} - \frac{v _{d}}{2}$ and $v_{in 2} = v_{c} + \frac{v _{d}}{2}$ . Substituting into $v_{o} = A_{1} v_{in 1} + A_{2} v_{in 2}$ gives $v_{o} = (A_{1} + A_{2}) v_{c} + \frac{A _{2} - A _{1}}{2} v_{d}$ .
Common-mode gain: $A_{c} = A_{1} + A_{2} = \frac{11317.25 - 11211}{1165.5} = \frac{106.25}{1165.5} = 0.091163$ . It is not zero because the resistor ratios are mismatched: $R_{f} / R_{1} = 9.619$ while $R_{4} / R_{3} = 10.684$ .
Differential gain: $A_{d} = \frac{A _{2} - A _{1}}{2} = \frac{11317.25 + 11211}{2 \times 1165.5} = \frac{22528.25}{2331} = 9.66463$ .
Ratio (the common denominator $1165.5$ cancels): $CMRR = \frac{A _{d}}{A _{c}} = \frac{22528.25}{2 \times 106.25} = \frac{22528.25}{212.5} = 106.0153$ .
Convert to decibels: $CMRR_{dB} = 20 lo g_{10} (106.0153) = 40.507 \approx 40.51 dB$

KEY IDEA

With an ideal op-amp, the output is $A_{1} v_{in 1} + A_{2} v_{in 2}$ , where $A_{1} = - R_{f} / R_{1}$ and $A_{2} = (1 + R_{f} / R_{1}) R_{4} / (R_{3} + R_{4})$ . The two gains are equal and opposite only when $R_{f} / R_{1} = R_{4} / R_{3}$ . Any mismatch leaves a common-mode gain $A_{c} = A_{1} + A_{2}$ alongside the differential gain $A_{d} = (A_{2} - A_{1}) /2$ , and $CMRR = ∣ A_{d} / A_{c} ∣$ , expressed in dB as $20 lo g_{10}$ of that ratio. Here the op-amp is ideal, so the finite CMRR comes entirely from the mismatched resistors.

WATCH OUT FOR

Three common slips. (1) Taking $A_{d}$ as just $R_{f} / R_{1} = 9.619$ (or $R_{4} / R_{3}$ ), which only holds for a matched bridge, and so getting a wrong ratio. (2) Leaving out the $(1 + R_{f} / R_{1})$ factor in the non-inverting path. (3) Rounding $A_{1}$ and $A_{2}$ to two or three digits before subtracting them. $A_{c}$ is the small difference of two nearly equal numbers, so early rounding throws the CMRR off badly. Keep exact fractions over the common denominator $1165.5$ until the last step. Using $10 lo g_{10}$ instead of $20 lo g_{10}$ for this voltage-gain ratio is another way to land outside the key.

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GATE 2024 EE Q.61: solution and answer

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