external circuit stranded loss calculation
Hi,
I designed a circuit like below in external circuit excitation.When I calculate according to the datas I got from the analysis, I find the stranded loss value different from the stranded loss value shown by the program.
The calculation formula I made with the data I obtained as a result of the analysis is as follows.
rmscurrent(winding1)^2*R8+rmscurrent(winding2)^2*R9+rmscurrent(winding3)^2*R10+rmscurrent(winding4)^2*(R23+R20)+rmscurrent(winding5)^2*(R21+R24)+rmscurrent(winding6)^2*(R22+R25)
My calculation result = 6000W stranded loss value provided by the program = 5000W.
What is the reason for it to come out different? Is there a mistake in my calculations? or will the error in the winding dimensions drawn on the model affect these losses?
Thanks in advance.
Best Answer

icellb1 Forum Coordinator
Hi, @johnfides ,
The strandedloss shown in Maxwell result is calculated based on the winding in the Maxwell side, the resistances added in external circuit are extra resistances which would not be included in the R for loss calculation in Maxwell. Since you are using current source in external circuit, the resistances value would not even affect loss calculation. If you use voltage source in external circuit, the resistance would change the current input in the Maxwell project and thus affecting the loss.
Answers
Hi, @johnfides ,
Is there a Maxwell project to which this external circuit is linked? If so, could you share some info about the Maxwell project and maybe some screenshots about the 6 windings in Maxwell? Thank you.
Hi,@icellb1
My model is a typical transformer. I share pictures below.
Thanks in advance.
Hi, @johnfides ,
The strandedloss shown in Maxwell result is calculated based on the winding in the Maxwell side, the resistances added in external circuit are extra resistances which would not be included in the R for loss calculation in Maxwell. Since you are using current source in external circuit, the resistances value would not even affect loss calculation. If you use voltage source in external circuit, the resistance would change the current input in the Maxwell project and thus affecting the loss.
Thank you so much @icellb1.