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## Experiment: The Full Wave Rectifier

We observed that in the half-wave rectifier, we lost half of our signal. To take advantage of the entire signal we use the full-wave rectifier which is shown in Fig. 1.2.

If you follow the current path through a full wave rectifier you will notice that when Vin goes positive, D2 and D4 are on (conducting), while D1 and D3 are off, and current flows in the direction through the resistor as indicated by the arrow. When Vin goes negative, D1 and D3 are on (conducting), while D2 and D4 are off, and current flows in the same direction through the resistor as indicated by the arrow. So, during both halves of the cycle current flows through the load resistor in the same direction, and the entire signal is used.

1.
Set up the circuit in Fig. 1.2, with RL=10K. Set the signal generator to 1kHz with and amplitude of approximately 5V. (Make sure the DC offset on your signal generator is zero.) Sketch the input versus the output as a function of time. Remember, the output is the voltage across the 10K resistor. You may find it easiest to measure the drop across the 10K resistor with two probes.
2.
Calculate the average power dissipated in the resistor. Compare this value to that of the half wave rectifier. Which circuit can provide more power to a load?

Next: DC Power Supplies Up: Diodes and Rectifier Circuits Previous: Experiment: Simple Half-Wave Rectifier:
Neil Goldsman
10/23/1998