1
\$\begingroup\$

I have a question concerning the nodal admittance matrix of a winding that has n turns. If each turn of the winding had its own model, is it possible to consider my winding a n-port network?

\$\endgroup\$
6
  • 1
    \$\begingroup\$ You could probably ... if there were no "interaction" between the elementary model "1 turn"... \$\endgroup\$
    – Antonio51
    Aug 1, 2022 at 10:58
  • \$\begingroup\$ @Antonio51 By interaction you mean, couplings rights? -either capacitive ou inductive- \$\endgroup\$
    – Wallflower
    Aug 1, 2022 at 11:13
  • \$\begingroup\$ Yes. But if all these "couplings" are included in your "N-port model", you can define the base model for the "whole windings" (very complicated thing), not for a part (for each turn). \$\endgroup\$
    – Antonio51
    Aug 1, 2022 at 12:55
  • \$\begingroup\$ @Antonio51 I can easily define the base model for the winding (all turns included) but I am more interested in the turn behavior. I am more confused on whether I can define each turn as a port. I mean does is it physically make sense or it nonsense? \$\endgroup\$
    – Wallflower
    Aug 2, 2022 at 7:48
  • 1
    \$\begingroup\$ It does physically make sense ... But probably a "bit" complicated without checking at "lab". See my answer ... A try to modeling... \$\endgroup\$
    – Antonio51
    Aug 2, 2022 at 17:03

1 Answer 1

1
\$\begingroup\$

I tried something like this for a "little" inductor (4 turns).
But I do not have all the measuring instruments to evaluate the "consistency" of the model.
However, it can be a "beginning" ...

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ This is a start for me, thank you! \$\endgroup\$
    – Wallflower
    Aug 3, 2022 at 14:13
  • 1
    \$\begingroup\$ RF Microelectronics (Prentice Hall Communications Engineering and Emerging Technologies Series) Razavi. You should read "chapter 7". \$\endgroup\$
    – Antonio51
    Aug 3, 2022 at 16:00
  • \$\begingroup\$ Will do, thank you very much Antonio !! \$\endgroup\$
    – Wallflower
    Aug 3, 2022 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.