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So the solution to a homework problem has this:

Voltage ripple for 3 phase HALF wave rectified 240√2 - 240√2 sin30

and Voltage ripple for 3 phase FULL wave rectified

240√6 - 240√6 cos30

I can understand that 240√2 would be peak phase voltage and at 30º is the lowest point of output voltage – this is how I think equation 1 works

But for equation 2 there's and extra √3 thrown in (assuming cos30, being 1/2, can be replaced with sin30 due to the same reasoning about 30º that I stated above)

Why the difference in these two?

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1 Answer 1

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Formulas's for the rectifier polyphased systems are:

Circuit P3 Half wave give Umean = Uco = q/pi * Vm * sin (pi/q) with q=3

Voltage max-min = Vm - Vm* sin (pi/2 + pi/q)

Ondulation factor = Ko = (Vmax-Vmin)/(2*Uco) = 0.302

For Veff= 220 V rms -> Uco = 260 V

enter image description here

And

Circuit PD3 Full wave gives Umean = Uco = 2*q/pi * Vm * sin (pi/q) with q =3

Voltage max-min = 2* Vm -2* Vm* sin(pi/2 + pi/(2*q))

Ondulation factor = Ko = (Vcmax-Vcmin)/ (2*Uco) = 0.070

For Veff = 220 V rms -> Uco = 520 V

enter image description here

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  • \$\begingroup\$ thank you, but i dont understand how these are derived and I think that could be really helpful to know \$\endgroup\$ Commented Jun 23 at 16:47
  • \$\begingroup\$ These formulas are derived from "inspection" of the curves. Uco is integrated over a 1/3 (for P3) or a 1/6 (for PD3) interval of 20 ms (50 Hz european means). These can be found in a course over "Rectified Power Supplies" (Pn, PDn or Sn supplies) on a "infinite" inductive load (to simplify calculus where currents are "rectangular" waves). Any number of phases can be used 3,4,6,9,12 24 ...). For the two phases systems ( P2, PD2 or bridge), some formulas must be changed. \$\endgroup\$ Commented Jun 23 at 17:23

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