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Rectifier

Curricula: Sek II, electronics courses

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Aim

Study the full-wave bridge rectifier and the smoothing of the ripple by a capacitor.

The bench

Source: sine, amplitude 10 V, 50 Hz. A bridge of four diodes, a 1 kΩ load. The 100 µF smoothing capacitor is connected to the load through a switch, which is open. Oscilloscope across the load: CH1, 5 V/div, 5 ms/div.

Procedure

  1. Look at the waveform across the load and find the period of the ripple.
  2. Close the capacitor's switch. To see the ripple, in the oscilloscope menu switch CH1 to the AC input, use the knobs to set 0.2 V/div and a trigger level of 0; estimate the peak-to-peak ripple.
  3. In the capacitor's settings set 1000 µF and switch CH1 to 20 mV/div.
  4. Open the switch and remove the leftmost diode — you get a half-wave rectifier.
Expected result

Across the load there are positive half-waves with no dips below zero. The peak is about 8.9 V: about 1.1 V is lost across the two conducting diodes (real silicon rectifier diodes lose about 1.3 V, and the peak is lower, about 8.7 V). The ripple frequency is 100 Hz — 2 divisions. With the 100 µF capacitor the voltage is almost constant, about 8.5 V; the ripple is about 0.67 V — 3.4 divisions on the AC input; with 1000 µF — about 0.07 V. With one diode removed, one half-wave per period remains, 50 Hz.

Questions

  1. Why is the ripple frequency of a full-wave rectifier twice the mains frequency?
  2. How does the peak-to-peak ripple depend on the load current?
Answers
  1. The bridge turns the negative half-wave over, so the load receives both half-waves of each period in the same direction. One period of the source, 20 ms at 50 Hz, gives two identical humps, and the ripple repeats every 10 ms: 100 Hz.
  2. The ripple is roughly proportional to the load current. Between the peaks the capacitor alone feeds the load and loses a charge of \(I \cdot \Delta t\), so its voltage drops by \(\Delta U \approx I \cdot \Delta t / C\), where \(\Delta t\) is at most the 10 ms between peaks. For 8.5 V / 1 kΩ = 8.5 mA and 100 µF this gives at most 8.5 mA · 10 ms / 100 µF = 0.85 V; the bench shows 0.67 V, because the capacitor starts recharging before the next peak.