# Current rating of center-tapped full wave rectifier vs full wave bridge rectifier

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?

• Keeping reading Art of Elec. it's good medicine for an EE student. There's a lab workbook that can also teach you a lot. Makes you build stuff rather than just reading about it. – George Herold Jul 20 '14 at 22:25
• I believe I do have access to the lab workbook so I'll start using it. Thanks! I've been told that I should only read the analog portions of the book though, as the digital parts are outdated. – Ammar Jul 21 '14 at 5:38

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.