## Peak current of single phase half wave rectifie

Assuming the voltage across the transformer secondary be sinusoidal of peak value V_{SM}, instantaneous value of the voltage applied to the rectifier is given as

**V _{S} = V_{SM }sin ωt**

If the diode is assumed to have a forward resistance of R_{F} ohms and reverse resistance equal to infinity, then the current flowing through the diode (or load resistance R_{L}) is given as

**i = I _{MAX} sin ωt for 0 ≤ ωt ≤ π**

**i = 0 for π ≤ ωt ≤ 2π**

Where peak value of current flowing through diode (or load resistance R_{L}) is given as

**I _{MAX } = V_{SM}/(R_{F} + R_{L})**

### Related Topic

**Single-phase half-wave rectifier****AVERAGE AND RMS VALUE OF SINGLE-PHASE HALF-WAVE RECTIFIER****Efficiency of single-phase half-wave rectifier****Frequency Component of Half-Wave Rectifier Voltage and Current****Ripple Factor of single phase Half-Wave rectifier****Peak Inverse Voltage (PIV) of single phase half wave rectifier****Transformer Utilization Factor (TUF) of single phase half wave rectifier****Advantage and Disadvantage of single-phase half-wave rectifier**