Half-Wave Rectifier:

If a diode and load resistor are connected in series with an AC source as illustreated in Figure, the resultant voltage v₀ = i₀ R_L and current i₀ through the load resistor R_L 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

                                      -------- (1)

where the applied voltage is v = V_p sin wt, period T = 1/f, and w = 2π f. As an instance, the average or DC voltage, if the applied voltage is 120 V_ac, would be 120 × (√2 / π) or 54 V. The DC current passing through the load resistor R_L is I _DC  = ( V_p  /π) R_L .

Figure: Half Wave Rectifier and Its Waveforms