If a diode and load resistor are connected in series with an AC source as illustreated in Figure, the resultant voltage v0 = i0 RL and current i0 through the load resistor RL shall be as shown. In drawing this waveform, we have supposed that the diode is ideal and therefore, have avoidd the 0.7 V drop while the diode conducts, and have supposed that the diode can be treated as an open-circuit when it is not conducting. Therefore, the circuit of Figure changes AC into unidirectional DC. The equivalent DC value of this voltage is given by
where the applied voltage is v = Vp sin wt, period T = 1/f, and w = 2π f. As an instance, the average or DC voltage, if the applied voltage is 120 Vac, would be 120 × (√2 / π) or 54 V. The DC current passing through the load resistor RL is I DC = ( Vp /π) RL .
Figure: Half Wave Rectifier and Its Waveforms