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So I thought the ripple voltage was approximated by the formula

Vr,pp = Vp / fRC

for a half-wave rectifier, and Vp/2fRC for a full-wave.

Yet simulation shows about 55% of this value.

So is the formula wrong? Just very inaccurate? Or under what conditions is it otherwise more accurate?

enter image description here

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    \$\begingroup\$ It only really applies to ripple that is small relative to the DC voltage. The assumption is that the conduction period of the diode is small and the current through the load is essentially constant. Your example does not meet either of those constraints. Try increasing the capacitor to 10uF and look at it then. \$\endgroup\$ – Kevin White Nov 9 '16 at 23:41
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There is a mass of online help to aid calculating this. This one at Hyperphysics is particularly complete with a calculator provided.

enter image description here

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Remember this Rule of Thumb for % voltage ripple, using rectified frequency out f=1/τ

  • If RC = 8 τ then Vpp/Vavg ≈ 10%

  • If RC = 10τ then Vpp/Vavg ≈ 8%

  • Then \$ \frac{Vpp_{ripple}}{V_{avg}} = \frac{ I_{avg}}{I_{pk}}\$= \$\frac{discharge}{charge}\$

i.e. \${\%Vpp_{ripple}}\$ is inverse to peak diode and cap current ratio.

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