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Let N1 turns in the primary winding, N2 turns in the secondary winding.After centre tap, the turns ratio becomes half of secondary winding.In textbooks given that the voltages of the upper winding and lower winding are equal with that of the input with 180 phase diff. but my doubt is that If the turns ratio becomes half, Voltage at upper, lower part of centre tap should be half that of the input. how it possible to have the same voltage as the input?

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  • \$\begingroup\$ You get half the overall secondary voltage across each half of the secondary winding. \$\endgroup\$
    – Chu
    Jul 1, 2017 at 12:11

2 Answers 2

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How it possible to have the same voltage as the input [on only half of the secondary]?

To achieve this the manufacturer would put twice as many turns on the secondary.

Since each half of the transformer secondary is only used on alternate half-cycles the wire rating only needs to be half the capacity of a full-wave rectified single-coil secondary. In effect the weight of the copper would be the same for a single-coil full-wave PSU or a split-coil double half-wave PSU. Similarly the same laminations could be used for either as the power requirement is the same.

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If N2=N1 on the secondary, with a center tap, you get only 1/2 the voltage peak from each leg on each half cycle but for a fixed VA core rating you get twice the current compared to a full-wave bridge which acts as voltage doubler.

Therefore the Voltage output of a half-wave bridge split winding is \$N_s/2N_p\$ with peak voltage , \$Vp=\sqrt{2}~V_{RMS}\$ with a load cap and no load and often 10% conduction loss at rated VA load so the no load voltage is ~ 50% more than Vs(rms).

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