TIDA-010054: RMS Current wtih Leakage Inductance

Hi Michael,

Yes, the circulating RMS currents increase when the leakage inductance is increased. This is because the phase changes with the associated leakage inductance to get the desired power transfer and hence the circulating increases with increased phase.

I have attached a MATLAB script which you can run to see the above statement.

Please let me know if you require any further clarifications.

P=10000;
L=[30e-6 40e-6 50e-6 60e-6 70e-6 80e-6];
F=100000;
n=1.6;
V1=800;
V2=500;
I_prms = zeros(length(L));
I_srms = zeros(length(L));
m = 1.6*(V2/V1);
for i=1:length(L)
    XL= 2*pi*F*L(i);
    I_base= V1/XL;
    phi = pi/2*(1-sqrt(1-((8*F*L(i)*P)/(n*V1*V2))));
    i_0 = 0.5*(2*m*phi + (1-m)*pi)*I_base;
    i_1 = 0.5*(2*phi - (1-m)*pi)*I_base;
    I_prms(i) = sqrt(i_0^2 + i_1^2 +i_0*i_1*(1-2*phi/pi)/3);
    I_srms(i) = n*I_prms(i);
end
plot(L,I_srms)

P=10000;

L=[30e-6 40e-6 50e-6 60e-6 70e-6 80e-6];

F=100000;

n=1.6;

V1=800;

V2=500;

I_prms = zeros(length(L));

I_srms = zeros(length(L));

m = 1.6*(V2/V1);

for i=1:length(L)

XL= 2*pi*F*L(i);

I_base= V1/XL;

phi = pi/2*(1-sqrt(1-((8*F*L(i)*P)/(n*V1*V2))));

i_0 = 0.5*(2*m*phi + (1-m)*pi)*I_base;

i_1 = 0.5*(2*phi - (1-m)*pi)*I_base;

I_prms(i) = sqrt(i_0^2 + i_1^2 +i_0*i_1*(1-2*phi/pi)/3);

I_srms(i) = n*I_prms(i);

end

plot(L,I_srms)

Thanks for the explanation.

One more question to your voltage transfer ratio in your matlab script.

It is for every Input equal to 1.

n=Vin/Vout = 1,6

m= n*(Vout/Vin) = Vin/Vout * Vout/Vin = 1

So for every Voltages you get the same Transfer Ratio?

Will be awesome if you can share your MathCad Script and / or Matlab Script (figure 17) as well.

Michael

Hi, 

have another question to your equation (6)

Is compared it to the ones you referred in your Documents [3 Texas Instruments, Modeling and Optimization of Bidirectional Dual Active Bridge DC–DC Converter Topologies] on page 107 and  [2. Texas Instruments, Performance of a 25kW 700V Galvanically Isolated Bidirectional DC-DC Converter Using 1.2kV Silicon Carbide MOSFETs and Schottky Diodes] on page 107.

And there is a big difference. 

When i set the phase shift in the equations to pi/2 (90°) i get different results.

In your equation i get: (n*V1*V2)/(8*pi*fs*L)

and in the other ones: (n*V1*V2)/(8*fs*L)

So there is a difference of 1/pi from both equations. 

Witch one is the rigt one?

Hi Michael,

The second one is correct. The equation 6 in the design guide has a pi^2 which should actually be pi.Sorry for the typo, I will make the change.

Thank you