2
\$\begingroup\$

I‘m trying to understand exactly what an L-section Network does. I understand it tries to eliminate the imaginary part of the impedance and match the real part. However I do not understand what excatly both components do and why the inductor(in my example, more specifically in a Highpass Downward L-match) has to be a shunt inductor and the capacitor has to be in series? Thank you for any help.

\$\endgroup\$
3
  • \$\begingroup\$ Simply put, the impedance ratio creates an attenuator by rising f causing Xc to decrease while XL increases passing high f voltage to a damping resistor load \$\endgroup\$ Jan 15, 2021 at 14:12
  • \$\begingroup\$ Thanks, what is meant by Xc and XL? \$\endgroup\$ Jan 15, 2021 at 15:12
  • \$\begingroup\$ Reactive impedance, look it up \$\endgroup\$ Jan 15, 2021 at 21:02

1 Answer 1

3
\$\begingroup\$

I‘m trying to understand exactly what an L-section Network does.

What you are describing in your question is a loss-less, high-pass, L-pad impedance matching circuit as per this example: -

enter image description here

The L-pad is used to match a lower impedance on the left with a higher impedance on the right. Either end can be source of course. In other words; you can match a higher source impedance to a lower load impedance or vice versa.

The formulas for L and C are dependant on the operating frequency of interest (\$\omega\$), the input impedance (\$R_{IN}\$) and the output impedance (\$R_L\$): -

$$C = \dfrac{1}{\omega\cdot R_{IN}}\cdot\sqrt{\dfrac{1}{\frac{R_L}{R_{IN}}+1}}$$

$$L = R_{IN}\cdot R_L\cdot C$$

Here you can find a calculator that saves you crunching the numbers by hand: -

enter image description here

That website also provides proofs for the formulas.

why the inductor(in my example, more specifically in a Highpass Downward L-match) has to be a shunt inductor and the capacitor has to be in series?

Well, you can make a low-pass version like this: -

enter image description here

Both work the same; at the frequency of interest, you can provide loss-less impedance matching as opposed to a wideband lossy impedance matcher like this: -

enter image description here

I understand it tries to eliminate the imaginary part of the impedance and match the real part.

It provides loss-less impedance matching i.e. it makes the input impedance looks resistive and the output impedance looks resistive at the desired operating frequency. Either side of the mid-operating frequency, it's not quote perfect, but, it's good enough for most signals just like an antenna is not quite perfect either side of it's target mid-point frequency.

\$\endgroup\$
4
  • \$\begingroup\$ Thank you very much. I see that a different arrangements have a different effect. Do you know why that is? Like what the singular component does to achieve these different results? \$\endgroup\$ Jan 15, 2021 at 12:44
  • \$\begingroup\$ @SimonGerber there is no superficial or obvious explanation to this other than following the theory of examining the input and output impedances mathematically and finding the value of \$\omega\$ where both are resistive. If you find one, you find the other of course. But, the different effects can only be realized by following the math (as detailed in the websites I linked). \$\endgroup\$
    – Andy aka
    Jan 15, 2021 at 12:47
  • \$\begingroup\$ @SimonGerber if you have your answer you should take the 2 minute tour to understand what motivates people to give free answers then, when you understand that, you can formally accept my answer unless you have something you still don't understand and need clarification via a comment. \$\endgroup\$
    – Andy aka
    Jan 16, 2021 at 14:28
  • 1
    \$\begingroup\$ Ah I am new here and didn't understand one could/should do that. i'm sorry \$\endgroup\$ Jan 16, 2021 at 14:49

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.