0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Possible duplicate of Filtering inductance calculation \$\endgroup\$ – JRE Oct 18 '16 at 10:07
  • 2
    \$\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
  • 1
    \$\begingroup\$ Still, though, it wasn't necessary to create the new question at all. \$\endgroup\$ – JRE Oct 18 '16 at 10:20
  • 1
    \$\begingroup\$ That's what editing is for. \$\endgroup\$ – JRE Oct 18 '16 at 10:35
1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.