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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?

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    \$\begingroup\$ 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. \$\endgroup\$ – George Herold Jul 20 '14 at 22:25
  • \$\begingroup\$ 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 I've been told that I should only read the analog portions of the book though, as the digital parts are outdated. \$\endgroup\$ – Ammar Jul 21 '14 at 5:38
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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.

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  • \$\begingroup\$ Thank you for the reply. I'm going off topic here, but how much would lamination size affect the capability of the transformer? Would it not be a minor difference? \$\endgroup\$ – Ammar Jul 20 '14 at 19:41
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    \$\begingroup\$ @Ammar For bespoke transformers, each standard lamination step is something like 30% in watts but you can stack the laminations thicker or thinner if you don't care about filling a bobbin. Here are the steps in off-the-shelf transformers from Tamura (3FD series): 1.1 2.4 6 12 - as you can see they tend to go up in steps of ~2:1 in watts, so a factor of 1.4 is less than the step. \$\endgroup\$ – Spehro Pefhany Jul 20 '14 at 19:53

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