## 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

#### = V_{L(ac)}/V_{L(dc)} = V_{r(rms)}/V_{L(dc)}

In term of current, we have

#### γ = I_{L(ac)}/I_{L(dc)}

We see from above,

#### I_{L(ac) }= √I_{L}^{2 }– I_{L(dc)}^{2}

#### γ = I_{L(ac)}/I_{L(dc) }= √I_{L}^{2} – I_{L(dc)}^{2}/I_{L(dc) }= √(I_{L}/I_{L(dc)})^{2} – 1

Now,

#### I_{L}/I_{L(dc)} = form factor K_{f}

#### γ = √K_{f}^{2}–1

In this case of Half-Wave rectifies with resistive load but no filter K_{f }= π/2 = 1.57

#### γ = √1.57^{2} -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

#### I_{L(ac)} = √I_{L1}^{2} + I_{L2}^{2} + I_{L3}^{2} +………

#### = √(I_{LM}/2√2)^{2} + (√2I_{LM}/3 π)^{2} + (√2I_{LM}/15 π)^{2}+ ………..

#### = 0.358I_{LM}

#### γ = I_{L(ac)}/I_{L(dc) }

_{ }= 0.385 I_{LM}/(I_{LM}/π)

#### = 0.358I_{LM}/0.318I_{LM }

_{ }= 1.21

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