0
\$\begingroup\$

I'm trying to match a complex source to a complex load with an LC network at 10 kHz only. For instance a 1626 + j2280 Ω source at 10 kHz to a 122k - j3.5*10^6 Ω load. Can someone provide or document the process for going about this?

\$\endgroup\$
3
  • \$\begingroup\$ Since this may be a homework problem, I'll only suggest that the first stem would be to compute the lead or lag required to match source to load. \$\endgroup\$ Feb 12, 2021 at 21:41
  • \$\begingroup\$ Homework problem or not, the Smith chart was designed to solve this. How tho use the chart is something of an art in itself. Well, he said 'only 10kHz' so maybe he's needing the analytical solution. I actually forgot how to do that:P \$\endgroup\$ Feb 13, 2021 at 8:22
  • \$\begingroup\$ It's gonna be a big inductor. \$\endgroup\$
    – Andy aka
    Feb 13, 2021 at 10:33

1 Answer 1

0
\$\begingroup\$

Start by looking at the source resistive component and the load resistive component.

  • Source resistance: 1.626 kΩ
  • Load resistance: 122 kΩ
  • Impedance ratio: step-up 1:75

You can match source and load resistances with a step-up transformer having a ratio of \$\sqrt{75}\$.

Then, you need to counteract the -j3.5 MΩ complex part of the load. Because it's negative we know it's capacitive so, an appropriate value series inductor would do the job. It will have an impedance of 3.5 MΩ at 10 kHz. So, if that inductor were put on the primary side of the transformer, the required impedance drops by 75 to 46.67 kΩ. At 10 kHz, that's an inductance of 742.7 mH.

But, you already have some inductance in the source; +j2280 Ω is an inductance of 36.3 mH so, all you need to do is add 706.4 mH in series with the source and you'll get optimum power transfer from source to load at 10 kHz (via the transformer).

Of course the main tricky part is the transformer but you should be able to engineer this from ferrite cores providing the power transfer isn't too high. Try and simulate what I've described is my advice.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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