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I'm trying to derive the ripple factor for a full wave rectifier with a filter.

This parameter is defined as the ratio of the root mean square (rms) value of the ripple voltage (Vr_rms) to the absolute value of the DC component of the output voltage:

γ = Vr_rms / Vdc

I use the following plot with an assumption of |AB| = T/2.

enter image description here

First I want to write an expression for ripple rms voltage Vr_rms.

The ripple amplitude as seen in the plot is Vr. This can be approximated(using Taylor series expansion's first term for exponential decay) and given for full wave rectifier with a filter as:

Vr = |XY| = t/τ * Vm, and in this case t = T/2 and τ = R*C

Vr = Vm / (2*CRf) and rms value of a triangle wave is given as Vp/sqrt(3) so:

Vr_rms = Vm / (2*sqrt(3)xCRf)

Now I want to find the average DC voltage Vdc. This can be approximated as:

Vdc = Vm - (Vr/2)

This yields the ripple factor:

γ = Vr_rms / Vdc

γ = [Vm / (2*sqrt(3)xCRf) ] / [Vm - (Vr/2)]

γ = [1 / (2*sqrt(3)xCRf) ] / [1 - (Vr/(2*Vm))] and since Vr/Vm = 1/(2*CRf)

γ = [1 / (2*sqrt(3)xCRf) ] / [1 - (1/(4*CRf))]

I'm stuck at this point I cannot progress and the texts give this factor as:

enter image description here

How is this derived? I cannot reach this..

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  • \$\begingroup\$ If T=period of the rectified sine wave, then it seems clear from your plot that AB is not equal to T/2. Note that T is a fixed value while Vr is a function of T and RC. Thus as RC varies, Vr will vary, and hence AB will also vary. \$\endgroup\$ – Barry Oct 18 '17 at 2:24
  • \$\begingroup\$ Noo AB is taken T/2 because ripple is assumed very small. Read the texts. It is an approximation assumes RC>>T \$\endgroup\$ – HelpMee Oct 18 '17 at 2:34
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The RMS value of a triangle wave is its peak value/sqrt(3). In your case the peak value of the triangle wave is Vr/2, not Vr which is the peak-to-peak value. This correction will produce a factor of 4 in your final result instead of 2. If you then make the approximation that RC>>T, then CRf>>1. With that approximation, the denominator in your final equation can be approximated as 1 since 1/(4CRf)<<1. The result is that your final equation reduces to the desired result.

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  • \$\begingroup\$ Your argument seems very correct. I didn't notice that. Please see my previous question I had a similar issue of reaching an equation: electronics.stackexchange.com/questions/334871/… \$\endgroup\$ – HelpMee Oct 18 '17 at 2:54
  • \$\begingroup\$ It's the same idea. In general, 1/(1+x) can be approximated by 1-x if x<<1. In your case, x=1/4CRf. Since CRf>>1, x<<1. Thus the 2 results are the same. \$\endgroup\$ – Barry Oct 18 '17 at 3:01

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