I have following power electronic converter enter image description here

and I need to determine inductance of the filtering chokes. I have deduced following formula $$L=\frac{V_{dc}\cdot(1-d)}{f_s\cdot\Delta i_L}$$ (based on \$v=L\frac{di}{dt}\$), where \$V_{dc}\$ is dc link voltage, \$f_s\$ is switching frequency (=12 kHz), \$\Delta i_L\$ is desired current ripple and \$d\$ is duty ratio (\$0\leq d\leq1\$). My problem is that I would like to use Sinewave Pulse Width Modulation so the duty ratio is changing sinusoidally. I don't know what value to substitute. The solution could be to choose the worst case but I don't know what value of \$d\$ coresponds to the worst case. Can anybody give me an advice how to solve that? Thanks in advance.

  • \$\begingroup\$ Possible duplicate of Filtering inductance calculation \$\endgroup\$ – JRE Oct 18 '16 at 10:07
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    \$\begingroup\$ Why did you delete the original question? It had some comments that may have been of use in answering your question. \$\endgroup\$ – JRE Oct 18 '16 at 10:16
  • \$\begingroup\$ I used the comments in the new question. \$\endgroup\$ – Steve Oct 18 '16 at 10:18
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    \$\begingroup\$ Still, though, it wasn't necessary to create the new question at all. \$\endgroup\$ – JRE Oct 18 '16 at 10:20
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    \$\begingroup\$ That's what editing is for. \$\endgroup\$ – JRE Oct 18 '16 at 10:35

The solution could be to choose the worst case but I don't know what value of d coresponds to the worst case

You have derived a formula for L and the worst case scenario is when L is maximum. With inductors, a smaller value is usually preferred for technical reasons so, choose a value of d that makes L large.

However, I doubt that your formula is correct because if I choose d to be zero then the value of L isn't infinite as I would expect it to be.

Alternatively go and down load LTSpice (free from LT) and simulate solutions to double check your formulas.

  • \$\begingroup\$ I have started with this assumption \$v_L=+V_{dc}\$ for \$0\leq t \leq d\cdot T_s\$ and \$v_L=-V_{dc}\$ for \$d\cdot T_s\leq t \leq T_s\$. The inductor current at the end of the first interval is \$i_L(d\cdot T_s=\frac{1}{L}V_{dc}\cdot d\cdot T_s\$. This value of inductor current is initial condition for integration the voltage over the other interval. The inductor current at the of the other interval is \$i_L(T_s)=-\frac{1}{L}V_{dc}\cdot T_s(1-d)+\frac{1}{L}V_{dc}\cdot d\cdot T_s\$. \$\endgroup\$ – Steve Oct 18 '16 at 11:24

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