Avizor
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I have a doubt about wire losses in transformers. In the magnetic design cookbook of Texas Instruments, copper losses are estimated with Rac and Rdc in a flyback converter (https://www.ti.com/lit/ml/slup127/slup127.pdf page 12), and I agree with this method because a flyback transformer isn't a true transformer, it's coupled inductors, so they have a dc level.
In a true transformer there's no dc component, but in the cookbook (https://www.ti.com/lit/ml/slup126/slup126.pdf page 11) they still consider dc losses.
How can an ideal transformer have a dc current? Considering it's a real transformer with a sinusoidal waveform, a dc level would saturate the transformer.
In my opinion, the copper losses should be:
Pdc = Rdc*Idc2 , with Idc = 0, then Pdc = 0
Pac = Rac * Iac2 , with Rac = AC/DC ratio * Rdc, and Iac = Irms of the sinusoidal waveform
I can't find any clear info about this, can anyone help? Thanks!
In a true transformer there's no dc component, but in the cookbook (https://www.ti.com/lit/ml/slup126/slup126.pdf page 11) they still consider dc losses.
How can an ideal transformer have a dc current? Considering it's a real transformer with a sinusoidal waveform, a dc level would saturate the transformer.
In my opinion, the copper losses should be:
Pdc = Rdc*Idc2 , with Idc = 0, then Pdc = 0
Pac = Rac * Iac2 , with Rac = AC/DC ratio * Rdc, and Iac = Irms of the sinusoidal waveform
I can't find any clear info about this, can anyone help? Thanks!