In analog PID controller implementation, how is the integral (I) mode realized using op-amp networks in practical hardware?

Introduction / Context:
Analog PID controllers implement proportional, integral, and derivative actions using op-amp circuits. This question focuses on the integral action and what components are actually necessary for a true integrator.

Given Data / Assumptions:

• The controller is implemented with operational amplifiers and passive components.
• The error signal is conditioned to avoid saturation and noise where possible.
• We aim for a practical, realizable integrator (not an ideal mathematical block).

Concept / Approach:

A practical integrator requires both a resistor and a capacitor in a topology that causes the output to be proportional to the time integral of the input. Simply having only resistors cannot integrate; using an op-amp and only a capacitor without the proper RC placement is incomplete and non-robust. The standard inverting integrator uses a resistor at the input and a capacitor in the feedback path, creating Vout proportional to the integral of Vin.

Step-by-Step Solution:

Verification / Alternative Check:

A Bode plot of the circuit shows a -20 dB/decade slope and -90 degrees phase shift over the integration range, matching integral behavior.

Why Other Options Are Wrong:

Option a (op-amp + resistors) implements proportional or summing, not integration. Option b (op-amp + capacitor only) lacks the necessary RC relationship and biasing. Option d is incorrect because a and b are not sufficient by themselves.

Common Pitfalls:

Ignoring saturation due to input offset; forgetting to include a high-value resistor in parallel with the capacitor for DC stability; omitting input protection and anti-alias networks.

Final Answer:

Op-amp with an RC network that places the error across a capacitor (true integrator)