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Inverting and Non-Inverting Amplifiers Questions and Answers

1.

The transfer function of the following differentiating circuit is :

(Figure: an inverting op-amp circuit. The input \(V_{1}(S)\) reaches the inverting input through R1 in parallel with a capacitor C; R2 is the feedback resistor from the output to the inverting input; the non-inverting input is grounded; the output is \(V_{2}(S)\).)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The transfer function is \(-\left[\frac{R_{2}(R_{1}C_{s} + 1)}{R_{1}}\right]\), where \(C_{s}\) means C times s.

For an inverting amplifier the gain is \(-Z_{f}/Z_{in}\). Here \(Z_{f} = R_{2}\), and the input element is \(R_{1}\) in parallel with C.

\[ \begin{aligned} Z_{in} &= \frac{R_{1}}{1 + sR_{1}C} \\ \frac{V_{2}}{V_{1}} &= -\frac{R_{2}}{Z_{in}} \\ &= -\frac{R_{2}(sR_{1}C + 1)}{R_{1}} \end{aligned} \]

Option B swaps \(R_{1}\) and \(R_{2}\). Options C and D drop the minus sign of the inverting amplifier, and D also changes the sign inside the bracket.

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2.

An integrator contains 100 kΩ and 1 μF capacitor. If the voltage applied to the integrator is 1 Volt, what is the output of the integrator after 1 sec ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The output is 10 V in magnitude.

An op-amp integrator gives \(v_{o} = -\frac{1}{RC}\int v_{i}\, dt\). With a constant input the output ramps linearly:

\[ \begin{aligned} RC &= 100 \times 10^{3} \times 1 \times 10^{-6} = 0.1\ \text{s} \\ |v_{o}| &= \frac{1 \times 1}{0.1} = 10\ \text{V} \end{aligned} \]

For the usual inverting integrator the output is -10 V, which is within the swing of a typical supply. 5 V and 15 V come from wrong time constants, and 0 V would mean no integration at all.

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