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RC circuit on the oscilloscope

Curricula: Tle, APC, Sek II

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Aim

Observe the charging and discharging of a capacitor on the oscilloscope and measure the time constant.

The bench

AC voltage source — a square wave from 0 to 5 V (amplitude 2.5 V, DC offset 2.5 V), 10 Hz. A 1 kΩ resistor and a 10 µF capacitor in series (\(\tau\) = 10 ms). Oscilloscope: CH1 — the source, CH2 — capacitor, DUAL mode, 2 V/div, 20 ms/div, triggered on CH1 at 2.5 V.

Procedure

  1. Examine both waveforms.
  2. Use the TIME/DIV knob to set 5 ms/div.
  3. Measure the time it takes the voltage across the capacitor to rise from 0 to 3.16 V (63% of 5 V).
  4. In the capacitor's settings, change 10 µF to 22 µF and measure the time to 63% again.
  5. Return to 10 µF and raise the source frequency to 50 Hz.
Expected result

On each rising edge of the square wave the capacitor charges, and on the falling edge it discharges, exponentially. The voltage reaches 63% in 10 ms — 2 divisions at 5 ms/div. With 22 µF, \(\tau\) grows to 22 ms, and within a half-period the capacitor no longer discharges to zero — the minimum is about 0.5 V. At 50 Hz the half-period (10 ms) is shorter than several \(\tau\), and the capacitor does not have time to charge fully.

Questions

  1. Why can't the voltage across a capacitor change instantly?
  2. What square-wave frequency should you choose so that the capacitor has time to charge almost fully?
Answers
  1. The voltage across a capacitor is set by its charge, \(U = Q / C\), and the charge changes only as fast as the current brings it to the plates. The resistor limits that current to at most 5 V / 1 kΩ = 5 mA, so the voltage changes gradually; a jump would need an infinitely large current.
  2. The half-period should last at least \(5\tau\); after that the capacitor is charged to more than 99%. With \(\tau\) = 10 ms that is 50 ms, a period of 100 ms, so \(f \le 1 / (10\tau)\) = 10 Hz: the 10 Hz set on the bench is the highest such frequency.