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Below two paragraph source from The Art of Electronics third edition at page 33.

The circuit in Figure 1.63 is called a center-tapped fullwave rectifier. 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. To develop some intuition on this subtle point, consider two different configurations that produce the same rectified dc output voltage: (a) the circuit of Figure 1.63, and (b) the same transformer, this time with its secondary cut at the center tap and rewired with the two halves in parallel, the resultant combined secondary winding connected to a full-wave bridge. Now, to deliver the same output power, each half winding in (a), during its conduction cycle, must supply the same current as the parallel pair in (b). But the power dissipated in the winding resistances goes like I²R , so the power lost to heating in the transformer secondary windings reduced by a factor of 2 for the bridge configuration (b).

Here’s another way to see the problem: imagine we use the same transformer as in (a), but for our comparison circuit we replace the pair of diodes with a bridge, as in Figure 1.62, and we leave the center tap unconnected. Now, to deliver the same output power, the current through the winding during that time is twice what it would be for a true full-wave circuit. To expand on this subtle point: heating in the windings, calculated from Ohm’s law, is I²R , so you have four times the heating for 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 compared with the (better) bridge circuit; besides costing more, the resulting supply would be bulkier and heavier.

Figure 1.63

I’m not sure how the circuit is exactly connected about and (b) the same transformer, this time with its secondary cut at the center tap and rewired with the two halves in parallel, the resultant combined secondary winding connected to a full-wave bridge. I think cut that in cut at the center tap is like remove earth, so is it like below?

guess circuit

About the power lost to heating in the transformer secondary windings reduced by a factor of 2 for the bridge configuration (b). I feel that the author didn't explain it in detail. I find a solution that is easy to understand, but I not sure it is correct.

The wave of output voltage of configuration (a) (Figure 1.63) is:

The wave of output voltage of configuration (a)

The wave of output voltage of configuration (b) is

The wave of output voltage of configuration (b)

The frequency and the period are related by the equation $$T = \frac{1}{f}.$$

$$P_a = I^2 R = P_b$$

Energy that lost to heating in the transformer secondary windings is equal to power multiply time.

$${W_a \over W_b} = {P_a \over P_b} {t_a \over t_b} = \frac{1}{f_a}{f_b} = 2$$

Is it correct?

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    \$\begingroup\$ Your first graph, in blue, is incorrect. The output waveform of both circuits is the same (save for the minor additional voltage drop of circuit B's extra diodes). In fact, the second graph is also incorrect, unless you intend to ignore the capacitor; the capacitor acts to maintain the output voltage near its peak. \$\endgroup\$ Commented Jun 17 at 14:08
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    \$\begingroup\$ As alluded to by @periblepsis, transformer current only flows around the voltage peaks, so a train of narrow pulses. Copper losses (pri and sec) are proportional to RMS I squared. Hence transformers are rated in VA not W. I don't think it's feasible to calculate RMS currents directly because the system is so non-linear. The way it used to be done was to refer to graphs in eg. ‘Analysis of Rectifier Operation’ by O. H. Schade (1943). Condensers and vacuum tubes but same principles! These days you might estimate a solution using rules of thumb and fine tune using SPICE in time domain (.tran). \$\endgroup\$ Commented Jun 18 at 1:00
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    \$\begingroup\$ As @SteveH indicates, Schade's work is seminal and is still being referenced in far more modern books on the topic. For example, OnSemi's Rectifier Application Handbook writes, "The best practical procedure for the design of single–phase capacitor–input filters still remains based on the graphical data presented by Schade in 1943." A crappy copy of Schade's paper can be found here. Better copy is here. \$\endgroup\$ Commented Jun 18 at 4:10
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    \$\begingroup\$ @Tom Most of these things were important when I was growing up, all so many years ago. We didn't have access to sophisticated switching power supply methods, except one: the push-pull primary version, which was widely used in the 1950's and 1960's for mobile HAM radio transceivers placed in the trunk of a car. These used what we called vibrators which looked like this to perform the switching operation for the primary. However, with access to modern ICs today, the weight, bulk, and expense isn't competitive. So the knowledge is fading away rapidly. \$\endgroup\$ Commented Jun 18 at 4:35
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    \$\begingroup\$ @Tom If a very serious hobbyist who may be interested in repairing old equipment, or otherwise interested in learning about important historical circuits, or otherwise just like learning for its own sake then all this will be useful and fun. If you are looking to get hired as a modern electronics guru for a company, then wasting a lot of time here won't help all that much (except with older interviewers, I suppose.) Just the transformer will be about 1 kg per 50 W out. So 100 W output will be 2 kg. It won't be cheap (all measures), as efficient, and the power factor will be less optimal. \$\endgroup\$ Commented Jun 18 at 4:48

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The thing with mains frequency power transformers is that you typically want to fill the winding window.

When you cut the winding at the center tap and connect them in parallel you get the equivalent of one of the half-windings but with half the resistance. Since power loss is proportional to winding resistance for the same current the power loss is halved. They did not state this explicitly, but if you imagine it, it should be clear.

You have not labeled the graphs completely but I think you have gotten something wrong. The power loss of your top and bottom graphs is identical if the peak currents and resistances are the same. That’s because current is squared, so polarity doesn’t matter. Frequency is unimportant as long as you calculate over a complete cycle and divide by that time. The energy over a half cycle if each is half, but the time is half so then power is the same.

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  • \$\begingroup\$ Thank you for your answer. I feel that your answer has been very helpful in helping me understand this question. \$\endgroup\$ Commented Jun 17 at 7:29

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