Doing so just reaffirms the power is not balanced. I see power distributed among windings... but nowhere close to being balanced. A single-phase load sees the direct winding as 1z and the indirect windings as 2z. Where's the balance?
Leyton transformer connections
- Thread starter winnie
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wwhitney
Senior Member
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OK, so the Leyton configuration uses the magnetic field of the core like an inductor to store energy during part of the cycle, thereby transforming the constant power delivery of the balanced 3 phase Pin into the double frequency sinusoidal (offset, with non-zero average) single phase power Pout?
Cheers, Wayne
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The description notes that if you use identical transformers (which you do not have to do) some paths will have twice the coil resistance. But this affects only the magnitude of the power losses, which are in the 1-5% range, not the load current and power.
To my mind a current distribution of 100:98:98 is close enough to balanced to keep everybody happy.
If not, just double the wire size for the indirect windings. Then totally balanced.
mobile
:blink: The 2 circuit wires for the indirect windings' current are the same 2 circuit wires that carry the direct winding's current.
When the current carried by those 2 wires reaches node "a" on one side of the transformer secondary it sees the indirect windings as a 2z impedance and the direct winding as a 1z impedance, divides accordingly then through the parallel pathways provided by the windings only to recombine at node "c", then back out to and through the load where the loop is thus completed.
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