I'm currently reading trough an application note from Analog Devices 2, in particular the "A NARROW-BAND LC MATCHING EXAMPLE AT 100 MHz" section. I re-did the calculations by hand and got formulas for the matching capacitance and inductance that are slightly different than the ones reported in the AN.

The circuit is the one reported below, let's assume L2 resonates with Cin at the frequency of interest. One wants the load impedance Z_C1 plus Z_C2 and the parallel between Z_L1 and RIN to be equal to Rs, the 50Ω source impedance.

Circuit details

What I get is: $$L_1=\frac{R_{IN}\sqrt{R_{S}}}{\omega\sqrt{R_{IN}-R_{S}}}$$ $$C_{MATCH}=\frac{1}{\omega\sqrt{R_{S}(R_{IN}-R_{S})}}$$

while the AN gives:

$$L_1=\frac{\sqrt{R_{IN}R_{S}}}{\omega}$$ $$C_{MATCH}=\frac{1}{\omega\sqrt{R_{S}R_{IN}}}$$

The latter equations are a particular case of the ones I've derived where the source impedance is much less than the input one, as it happens in the case analyzed in the AN.

Am I missing something?


1 Answer 1


Look at your formula here: -


Then consider that \$R_S R_{IN}\$ is much more significant than \$R_S^2\$ on it's own hence, they have simplified the formula to this: -

$$C_{MATCH}=\frac{1}{\omega\sqrt{R_{S} R_{IN}}}$$

The simplification is allowed if transforming to a much higher impedance

But, your formula is of course correct and preferable.

Link to my derivation on high pass L matching networks: -

enter image description here

It's worth noting that \$L_1\$ reduces to a simpler formula and that my "C" formula matches your (preferred) derivation.

  • 1
    \$\begingroup\$ Thanks for the confirmation! \$\endgroup\$ Feb 22 at 13:44

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