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In an RC differentiator, which statement best describes the capacitor's charging behavior?

Difficulty: Easy

Correct Answer: The capacitor charges exponentially at a rate depending on the RC time constant

Explanation:


Introduction / Context:
An RC differentiator produces an output proportional to the rate of change of the input when the RC time constant is much smaller than the signal's time scale. Understanding how the capacitor charges in response to a step or edge is key to predicting the waveform spikes.


Given Data / Assumptions:

  • RC differentiator: series capacitor, shunt resistor, output taken across the resistor.
  • Consider response to a step/rising edge to discuss charging behavior.
  • Linear components; no saturation.


Concept / Approach:
For a step change, the current through the capacitor is i_C = C * dV_in/dt. Immediately after the step, a large current flows, and the node voltages adjust per KVL/KCL. The capacitor voltage changes exponentially toward its new value with time constant tau = R * C, even in the differentiator topology (because the network remains first order).


Step-by-Step Solution:

For a step input from 0 to V_s at t = 0+, the initial capacitor voltage cannot change instantaneously, so V_C(0+) = 0.The subsequent evolution follows V_C(t) = V_s * (1 - exp(-t / tau)) in the canonical charging scenario.Thus, the charge trajectory is exponential with time constant tau = R * C, not purely dependent on input amplitude alone.Therefore, the correct qualitative statement is that the capacitor charges exponentially with a rate determined by RC.


Verification / Alternative check:
Laplace-domain analysis of the differentiator transfer function H(s) = sRC / (1 + sRC) shows a first-order pole at s = -1/(RC). Time-domain responses to steps necessarily follow exponential laws governed by that pole.


Why Other Options Are Wrong:

  • Only input amplitude: Ignores the RC time constant that sets the rate.
  • Charges when input is decreasing: For a falling edge, the capacitor discharges (or charges to opposite polarity); the statement is misleading without context.
  • Charges to approximately one time constant: Time constant has units of time, not voltage.


Common Pitfalls:

  • Confusing differentiator small-signal behavior with exact edge dynamics.
  • Interpreting 'time constant' as a voltage level instead of a time scale.


Final Answer:
The capacitor charges exponentially at a rate depending on the RC time constant

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