Consider a differential amplifier shown in the figure below:

$If the two input terminals are connected together to an input common mode signal and R_{2} R_{3} = R_{1} R_{4} then the input common mode resistance is equal to$

R_{1} + R_{3} | | R_{2} + R_{4}

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R_{1} | | R_{3} + R_{2} | | R_{4}

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(R_{1} + R_{2}) | | (R_{3} + R_{4})

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(R_{1} + R_{3}) | | (R_{2} + R_{4})

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Solution

The correct option is C $(R_{1} + R_{2}) | | (R_{3} + R_{4})$

$V_{+} = V_{cm} \frac{R_{4}}{R_{3} + R_{4}}$

$V_{+} = V_{-}$

$i_{2} = \frac{V_{cm}}{R_{3} + R_{4}}$

$i_{1} = \frac{V_{cm}}{R_{1}} - \frac{V_{cm} R_{4}}{R_{3} + R_{4}} \times \frac{1}{R_{1}} = \frac{V_{cm}}{R_{1}} \times \frac{R_{3}}{R_{3} + R_{4}}$

$i_{1} = i_{1} + i_{2} = \frac{V_{cm}}{R_{1}} \times \frac{R_{3}}{R_{3} + R_{4}} + \frac{V_{cm}}{R_{3} + R_{4}}$

$\frac{1}{R_{1}} = \frac{i}{V_{cm}} = \frac{1}{R_{1}} \times \frac{1}{\frac{R_{4}}{R_{3}} + 1} + \frac{1}{R_{3} + R_{4}}$

$If we replace \frac{R_{4}}{R_{3}} with \frac{R_{2}}{R_{1}}, then$

$\frac{1}{R_{1}} = \frac{1}{R_{1} + R_{2}} + \frac{1}{R_{3} + R_{4}}$

$R_{1} = (R_{1} + R_{2}) | | (R_{3} + R_{4})$

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Similar questions

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