Is there a simple method to estimate the copper loss of Litz wire?

Question

I would like to know if it is possible to estimate the loss of Litz wire by using a simple equation.

For a planar transformer, I can estimate the DC loss by calculating the resistance at low frequency.

How can I do that with Litz wire?

Answer 1

You should further explain how the transformer is being used.

I suggest you read the following answers given electronics.stackexchange on the subject of winding losses. If you have a university library nearby, see if they have the books authored by E. C. Snelling.

General discussion of design parameters.
Proximity loss calculations.
Optimal wire diameter and interleaving.

To sum up the above discussions:
Losses in magnetics (transformers and inductors) have the following major loss mechanisms: DC resistance (what you measure with an Ohm meter), AC resistance (eddy losses), and additional AC losses when you near the resonant frequency of the magnetics.

DC loss:
The following formula is for round conductors, either single strand or muli-stranded like Litz or bunched conductors.
$$ R_{dc} = {{4 \rho_c N l_w} \over {\pi s d^2}}$$
\$ \rho_c = \$ resistivity of annealed copper = 1.72e−8 Ω⋅m at 20°C
\$ N = \$ number of turns
\$ l_w = \$ mean length of one turn [m] [ferrite data books give this value under bobbin specifications)
\$ s = \$ number of wire strands
\$ d = \$ wire strand diameter [m]

Derivation of this formula:
The standard equation for conductor resistance is given in basic electronic and physics books as:

$$ R = {\rho_c {L \over A}} $$

\$ L \$ = length of the conductor.
For a circular bobbin wound inductor/transformer
\$ L = N \; l_w \$
where \$ N \$ = number of turns, \$ l_w \$ = mean length of one turn around the bobbin.

\$ A \$ = the cross-sectional area of the conductor.
For a circular conductor, \$ A = \pi \; r^2 = \pi {d^2 \over 4}\$
Where \$ r \$ = radius of the conductor, \$ d \$ = diameter of the conductor.

MWS Wire has a table of Litz wire characteristics, including DC resistance.

AC loss:
AC loss is primarily dominated by proximity effect losses, the eddy loss due to the magnetic field of adjacent conductors at frequencies below a couple MHz. Skin effect, the eddy loss due to the magnetic field of the conductor, is present, but usually not significant below a couple MHz. AC loss is is dependent on frequency, wire diameter (larger is not always better as borne out of the proximity loss equation), number of strands (proximity effect losses can be reduced by using bunched or litz wire), interwinding capacitance (affects resonant frequency), number of turns, and eddy losses in the core material which is dependent on flux density and frequency.

Larger diameter wire reduces DC loss.
Smaller diameter wire reduces AC loss.
If operating at a single frequency there is an optimal wire diameter. The graph below shows the calculated loss resistances versus wire diameter at a given frequency (200 kHz in the example plot), number of turns, and bobbin geometry. \$R_{dc}\$ is the DC resistance and \$R_{pe}\$ is the resistance due to proximity effect.

You may ask, is the equation for proximity effect loss accurate? From personal experience, it is. I'm usually within 5% as measured on a HP4195 impedance analyzer.

Litz Wire
The picture you show of stranded wire is not Litz wire. That is a common stranded hookup wire with 7 strands. The picture you show is wrong as the strands should be twisted so they don't splay out when the insulation is stripped. Plus, stranded wire for magnetics normally doesn't use thick insulation. Litz and bunched conductor wire used for magnetics consists of individually insulated strands of wire which reduces eddy losses. Uninsulated strands will not reduce eddy losses as efficiently as insulated strands. The image below is 25/40 HPN-155 Litz wire which has the approximate same cross sectional area of #26 AWG solid wire.
25 = number of strands
40 = 40 AWG diameter wire (0.079 mm dia) which is the individual strand diameter
HPN = heavy polyester-nylon insulation
155 = 155°C insulation thermal class.

Real Life Calculations
In response to the OPs comment about how this is done in real life... Lets say we want to wind magnetics using an RM5 core with 20 turns of 25/40 Litz wire. From an old Ferroxcube handbook (circa 1982), the graph says we can fit about 25 turns of #24 wire (equivalent diameter of 25/40 Litz) on a single section bobbin. BTW, the newer Ferroxcube handbooks removed a lot of design graphs such as the turns per bobbin graph.

The following image is from a modern Ferroxcube ferrite handbook. average length of turn, \$ l_w \$, is 24.9 mm for a winding that occupies all of the winding area. If the windings don't fill up the full area, you'll need to figure out a new average (mean) winding length. From the above graph, we can expect the bobbin to be 80% full with 20 turns of 25/40 Litz wire. In practice, I may choose to ignore the difference in \$ l_w \$ because it will be "close enough".

The total length of the winding is \$ N \times l_w = 20 \times 24.9 = 498 \; mm = 0.498 \; m\$.

The resistance of one strand of #40 AWG (0.0787 mm dia) wire can be figured out from the basic formula. $$ R = {\rho_c {L \over A}} = {1.72 \text{E−8} {0.498 \over 4.87 \text{E-9}} = 1.76 \; \Omega} $$ Since the 25 strands are in parallel, divide the resistance of the single #40 strand by \$ s = 25 \$, thus, \$ R_{dc} = {1.76 \over 25} = 0.070 \Omega \$ .

The \$ R_{dc} \$ equation starts with the basic resistance of a conductor equation which is fit to the case of a practical situation.

If you want to be pedantic, since Litz wire is twisted, the effective length of the wire is increased. The paper I referenced by Tang & Sullivan in the comments discusses this.

The real test is to wind the magnetics and test on an impedance analyzer and a DC ohmmeter. The AC losses increase when you approach resonance. If winding a transformer, you will be interested in the leakage inductance. Usually, one tries to minimize leakage inductance, however, there are cases where some amount of leakage inductance is desirable.