Operational-amplifier integrator behavior — The ramp (output) voltage of an ideal op-amp integrator under a constant input signal will:

Introduction:An op-amp integrator converts an input voltage into the time integral at its output. With a constant (DC) input, the output of an ideal integrator is a linearly changing ramp. Recognizing this linear relationship is essential when designing ramps, function generators, and control systems that rely on integrator dynamics.

Given Data / Assumptions:

• Ideal op-amp with a feedback capacitor Cf and input resistor Rin.
• Constant input voltage (or piecewise constant segments).
• Linear region of operation (no saturation of the output rails).

Concept / Approach:

For an inverting integrator, the fundamental relation is Vout(t) = −(1 / (Rin * Cf)) * ∫ Vin dt + Vout(t0). When Vin is constant, the integral is Vin * t, producing a linear ramp with slope proportional to Vin and inversely proportional to Rin * Cf. The sign is inverted for the classic inverting topology.

Step-by-Step Solution:

Verification / Alternative check:

Bench observation with a square-wave input shows triangular (piecewise linear) output—direct evidence of linear rate change. Simulation (SPICE) reproduces the same slope proportionality to Vin and 1/(Rin * Cf).

Why Other Options Are Wrong:

• exponentially: Exponential arises in RC passive charging, not ideal op-amp integration under DC input.
• always increasing: Sign depends on Vin; negative inputs produce rising ramps.
• constant: Only when Vin = 0, otherwise a slope exists.
• oscillate sinusoidally: Requires additional dynamics; not a property of a simple integrator.

Common Pitfalls:

• Forgetting saturation limits; real outputs clip when rails are reached.
• Ignoring bias currents and offsets that can cause unintended drift (pseudo-ramp) at Vin ≈ 0.

Final Answer:

increase or decrease at a linear rate