Question

Integrator using op-amp uses

• A. Resistor in feedback
• B. Capacitor in feedback
• C. Inductor in input
• D. No feedback

Hint:

- Integrator: Capacitor placed in the feedback path, resistor placed in the input path. - Differentiator: Resistor placed in the feedback path, capacitor placed in the input path.

Answer 1

The Correct Option is B

Concept: An op-amp integrator is a fundamental operational circuit that performs the mathematical operation of integration over time. It produces an output voltage that is directly proportional to the time-integral of the applied input signal. The physical realization of this mathematical relationship relies on replacing the standard feedback resistor of an inverting amplifier with a reactive element, specifically a capacitor. Let us analyze the circuit using basic electronics relationships:
• The input voltage \(V_{\text{in}}(t)\) is applied through an input resistor \(R\) connected to the inverting terminal.
• A feedback capacitor \(C\) is connected directly between the output terminal and the inverting input terminal.
• The non-inverting terminal is grounded, creating a virtual ground (\(0\text{ V}\)) at the inverting node. Applying Kirchhoff’s Current Law (KCL) at the virtual ground input node: \[ I_{\text{in}}(t) + I_f(t) = 0 \quad \Rightarrow \quad I_{\text{in}}(t) = -I_f(t) \] The current flowing through the input resistor \(R\) is written as: \[ I_{\text{in}}(t) = \frac{V_{\text{in}}(t) - 0}{R} = \frac{V_{\text{in}}(t)}{R} \] The dynamic relationship governing current and voltage across a feedback capacitor is given by: \[ I_f(t) = C \frac{d(V_{\text{out}}(t) - 0)}{dt} = C \frac{dV_{\text{out}}(t)}{dt} \] Equating these two currents according to KCL: \[ \frac{V_{\text{in}}(t)}{R} = -C \frac{dV_{\text{out}}(t)}{dt} \] To isolate the derivative term: \[ \frac{dV_{\text{out}}(t)}{dt} = -\frac{1}{RC} V_{\text{in}}(t) \] Integrating both sides of this expression with respect to time from an initial time \(0\) to a final time \(t\): \[ V_{\text{out}}(t) = -\frac{1}{RC} \int_{0}^{t} V_{\text{in}}(\tau) \, d\tau + V_{\text{out}}(0) \] This mathematically demonstrates that placing a capacitor in the feedback path allows the op-amp circuit to function as a time-domain integrator.