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\section{MATLAB code}   

\subsection*{Inductor design}

\begin{par}
N87 material is chosen. Air gap of 0.5 mm is chosen as the A\_L value matches well Value coming from the datasheet in [H]
\end{par} \vspace{1em}
\begin{verbatim}
A_L=201/10^9;
% Air gap length in [m]
g=0.5/1000;
% Core volume in [m³]
V_core=5350/10^9;
% Number of turns. Equation comes from application note
N=round(sqrt(L_sn/A_L));
% With rounded number of turns new value for snubber inductance in [H]
L_sn_new=A_L*N^2;
% Diameter of round middle leg of core in [m]
d_round=9.8/1000;
% Gap area
A_gap=pi*(d_round/2)^2;
% Max flux density
B_max=u0*N*30/g;

% Core loss calculation when line is extrapolated towards 10 kHz
% Relative core losses depending on volume in [W/m³]
P_core_rel=30000;
% Core losses in [W]
P_core=P_core_rel*V_core;

% Copper losses
% Skin depth in [m]
skd=7.5/sqrt(f_s)/1000;
% Rule of thumb: 10 A [rms] per mm². We have 3 A [rms].
A_copper=0.3/10^6;
% Next standard area of copper in [m²]
A_copper_new=0.5/10^6;
% Length of copper needed in [m]
l_copper=N*pi*d_round;
% DC resistance of copper depending on A_copper in [Ohm]
R_DC=rho*l_copper/A_copper_new;
% Diameter of copper in [m]
d_copper=2*sqrt(A_copper_new/pi);
% AC resistance of copper in [Ohm] which current do I need?
R_AC=d_copper/skd*R_DC;
% DC copper losses in wire
P_copper=R_DC*I_IGBTrms^2;
% Total losses in the inductor in [W]
P_L=P_core+P_copper;
\end{verbatim}