Ripple Factor of single phase Half-Wave rectifier

When define in term of voltage, it is given by

 γ = rms value of ac component/dc value of load voltage

   = VL(ac)/VL(dc) = Vr(rms)/VL(dc)

In term of current, we have

 γ = IL(ac)/IL(dc)

We see from above,

IL(ac) = √IL2 – IL(dc)2

 γ = IL(ac)/IL(dc) =  √IL2 – IL(dc)2/IL(dc) = √(IL/IL(dc))2 – 1


IL/IL(dc) = form factor Kf

γ = √Kf2–1

In this case of Half-Wave rectifies with resistive load but no filter Kf = π/2 = 1.57

γ = √1.572 -1 = 1.21

Alternatively, the value of γ could be found as under:

If we neglect fourth and higher harmonics in the load current, then as seen from above

IL(ac) = √IL12 + IL22 + IL32 +………

       = √(ILM/2√2)2 + (√2ILM/3 π)2 + (√2ILM/15 π)2+ ………..

      = 0.358ILM

γ = IL(ac)/IL(dc)

     = 0.385 ILM/(ILM/π)

   = 0.358ILM/0.318ILM 

    = 1.21




Related Topic

  1. Single-phase half-wave rectifier
  3. Efficiency of single-phase half-wave rectifier
  4. Frequency Component of Half-Wave Rectifier Voltage and Current
  5. Peak Inverse Voltage (PIV) of single phase half wave rectifier
  6. Peak current of single phase half wave rectifier
  7. Transformer Utilization Factor (TUF) of single phase half wave rectifier 
  8. Advantage and Disadvantage of single-phase half-wave rectifier
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