Full-Wave Bridge Rectifier
Four diodes arranged as a bridge send both halves of the AC cycle through the load in the same direction. A filter capacitor then smooths the pulses into DC with a little ripple.
Simulation result
The simulator computed these waveforms from the circuit above, over the first 100 ms.
Numbers from the simulation
Computed by the simulator from the part values in the schematic.
- Average DC voltage on the load
- 15.4 V
- Peak-to-peak ripple
- 282 mV
The circuit is copied into a new project in your workspace. If you are not signed in, you sign in with Google first.
How it works
On the positive half cycle, current flows through D2, the load, then D3. On the negative half, through D1, the load, then D4. The current through the load always flows the same way.
The current always passes two diodes, so the output peak sits two diode forward voltages below the source peak (12 V × √2, about 17 V).
Between peaks the capacitor feeds the load and its voltage sags a little. That sag is the ripple, roughly I / (2 × f × C). A bigger capacitor or a lighter load gives less ripple.
Try changing
- Change C1 to 1000 µF. The ripple drops to about half.
- Lower RL to 220 Ω. The load draws more current and the ripple grows.
- Delete C1. The output pulses 100 times a second, twice the source frequency.