Skin effect for a rectangular wave

I am going to connect a 200kHz rectangular voltage wave to the primary winding of a transformer. It is made of 0.2 mm diameter copper wire. Presumably the current wave will also be rectangular (when load is connected to the secondary winding). I would like to know the ratio of resistance of the primary winding to the 200 kHz square wave to its resistance to DC.

Hello,

This is a difficult one, as it is not only skin effect that counts, but also "proximity effect". The proximity effect is that the field from one turn induces eddy current in adjacent turns. So the overall AC loss resistance will be above the value based on skin depth calculation.

The additional loss due to proximity depends on how the windings are built-up and can be significant. All turns close together will increase proximity effect loss, turns in the fringe field of an air-gap will also increase the loss.

As an indication for just the skin depth I would use the fundamental frequency and ignore the rest. The first harmonic has 1.27*(amplitude of square wave with 50% d.c.).

You are right that when the magnetizing current is far below primary loaded current, wave shape is also square wave.

I am going to use a square wave without DC component.
Can you give me any idea how big will be the losses due to proximity effect?
I am going to have 1000 turns of the 0.2 mm diameter copper wire on 1/2 of the circumference of a toroid, OD=32mm, ID=19mm, thickness=9.5mm.

Hello,

This will give you about 150 to 200 turns/layer. So you will have about 4..5 layers on top of each other with virtually no interwinding and interlayer spacing..

I am very sorry, but don't have any experience with such an arrangement at 200 kHz. You may do a search on: Ferreira Dowell proximity loss factor (without quotes). This will guide you further.

You can be lucky that your D/skin depth ratio is not that large. So expect the proximity losses to be not 10 times higher then the losses based on skin depth only

I found something, but I do not know the definition of this loss factor. What does it mean that the loss factor value is e.g. = 5?
By the way, I also have a different problem: I want to place many (20 to 70) transmitting antennas - either short dipoles or half-wave dipoles close together, e.g. on the surfaces of small-diameter concentric cylinders (say 10 - 12 cm diameters, 2m wavelength, 150 MHz). Will the proximity effect lead to big losses? (Skin depth is 5.4 micron, I want to use 2mm diameter wire).

Hello,

If you can get "TRANSFORMER AND INDUCTOR DESIGN HANDBOOK" (McLyman), check chapter 4. It discusses proximity also without elaborate math (based on Dowell method graph).

You have to find the definition of the proximity effect factor in the document you have. The one I have uses a normalized factor (G with ^ on top of it). This link may be useful: http://engineering.dartmouth.edu/inductor/intro.shtml.

Note that your transformer setup doesn't use primary/secundary interleave, so you have relative high proximity loss. Also don't forget the core loss, it may be significant (depending on core material).

Added after 3 minutes:

Regarding the dipoles. When the seperation between the dipoles is very large with respect to diamater of the dipoles, I would ignore the proximity loss.

OT: what is the purpose of putting many dipoles close to each other with respect to wavelength? If the dipoles are very small w.r.t. wavelength, your main loss may be in the matching circuits (if requried).

Regarding dipoles: currently I cannot disclose the purpose. If the distance between neighboring dipoles is 5 times as large as the diameter of the dipoles, is this sufficient to ignore proximity losses?

Hello

With distance > 5*diameter, you can assume a circular current distribution in each dipole, so you can model the loss in each dipole via skin depth approximation. Off course there will be interaction (mutual coupling), so when you feed all dipoles from same source(s), the current in all dipoles will not be equal.

Are you using an EM field solver to do the math? If so, just divide one of the dipoles in many segments perpendicular to the dipole axis (segments in horizontal direction for a vertically oriented dipole). The current density in each horizontal segment will show you if the current distribution assumption is valid.