I was going through the Art of Electronics (Horowitz & Hill) and in the section on center-tapped full wave rectifiers (pg.47) it says:
"The output voltage is half what you get if you use a bridge
rectifier. It is not the most efficient circuit in terms of
transformer design, because each half of the secondary is used only
half the time. Thus the current through the winding during that time
is twice what it would be for a true full-wave circuit. Heating in the
windings, calculated from Ohm's law, is I^2R, so you have four times
the heating half the time, or twice the average heating of an
equivalent full-wave bridge circuit. You would have to choose a transformer with a current rating 1.4 (square root of 2) times as
large, as compared with the (better) bridge circuit; besides costing
more, the resulting supply would be bulkier and heavier. "
Why would the current rating be ~1.4 times more than the bridge circuit?
Also, 'Equivalent full-wave bridge circuit' refers to one with the same output characteristics as the center-tapped circuit right?
Best Answer
Well, at a current \$\frac{1}{\sqrt2}\$ lower your \$I^2R\$ heating will be half, so the overall copper heating would be the same as with the bridge.
Win Hill et al. are simplifying things a bit here by ignoring the diode drops. In the case of a requirement to get a very low voltage where the diode drops make significant difference, and/or if the granularity of the lamination size selection leads to a transformer that has more capability than required, the centre-tapped configuration may make a lot of sense.
Also, not all the transformer losses are copper losses, so doubling copper losses does not increase the total transformer losses by 2:1, especially with cheap lams.