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Operational Amplifier Fundamentals Questions and Answers

1.

The output voltage (\(V_{o}\)) of the following Op-amp circuit is :

In the circuit, \(V_{2}\) = 4 V is applied through \(R_{3}\) = 1 kΩ to the inverting (-) input, with a feedback resistor \(R_{4}\) = 1 kΩ from the output \(V_{o}\) to that input; \(V_{1}\) = 2 V is applied through \(R_{1}\) = 1 kΩ to the non-inverting (+) input, which is connected to ground through \(R_{2}\) = 1 kΩ.

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The output voltage is - 2 V.

This is a difference amplifier. The non-inverting input sees \(V_{1}\) through the divider \(R_{1}, R_{2}\), and the gain to the output is set by \(R_{3}\) and \(R_{4}\):

\[ \begin{aligned} V_{+} &= V_{1}\frac{R_{2}}{R_{1} + R_{2}} = 1\ \text{V} \\ V_{o} &= \left(1 + \frac{R_{4}}{R_{3}}\right)V_{+} - \frac{R_{4}}{R_{3}}V_{2} \\ &= 2(1) - 4 = -2\ \text{V} \end{aligned} \]

With all resistors equal, the circuit simply gives \(V_{1} - V_{2}\). The option +2 reverses the subtraction, and \(\pm 4\) uses only one input.

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Answer: B

Negative feedback leads to an increase in the input impedance and the bandwidth.

With series (voltage) negative feedback, as in the non-inverting amplifier, the input impedance is multiplied by the factor (1 + Aβ), the output impedance is divided by the same factor, and the bandwidth is extended by it while the gain falls by it, keeping the gain-bandwidth product constant.

So the output impedance decreases rather than increases, which rules out option A; option C wrongly says the bandwidth decreases; and option D ignores the effect altogether.

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