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Two different sites with L-match calculators show different LC schematics as shown below. Note that Schematic 1 shows C1 on the source-side and Schematic 2 shows C1 on the load side. Both sites calculated the L and C values to be the same, and input impedance for both is 50-ohms.

Question: Does L/C order matter in this context?

Schematic 1

L-match with C on the source-side

Schematic 2

L-match with L on the source-side

This answer is about component ordering in a loop, but LC vs. CL ordering in an L-match circuit doesn't appear to be a loop in the same sense since the load is parallel with the shunt component (but maybe it is?).

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  • \$\begingroup\$ The resistor connected to the LC-junction has a higher value than the other (series) resistor. \$\endgroup\$
    – glen_geek
    Sep 27, 2022 at 0:29
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    \$\begingroup\$ You are not showing the calculated values from both sites so, how can this be properly answered? \$\endgroup\$
    – Andy aka
    Sep 27, 2022 at 7:49
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    \$\begingroup\$ @Andyaka, good point, but both sites present the same values as noted in the first paragraph and the 1st schematic shows the values. \$\endgroup\$
    – KJ7LNW
    Sep 27, 2022 at 20:20

3 Answers 3

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At one fixed frequency, either topology can be used, and it doesn't really matter which. However, you may want your network to provide matching over a range of frequencies. If that is the case, then it become necessary to know the frequency-impedance characteristics of your source, and the item that will be driven.

For example if you are driving a half-wave antenna with a broad-band rf amplifier, you would probably choose the topology that best performs a match between the shape of the transmitter's complex impedance graph w.r.t. frequency, to the shape of the antenna's complex impedance graph w.r.t frequency. At least the best match for the frequency range of interest.

Edit:

The user a_concerned_citizen performed an analysis using a 50 ohm resistance as a load impedance of 50 ohms, rather than the 10 + j10 ohms specified in the original question. Using the 50 ohm load, and different capacitor and inductor values, the analysis seems to show that the two configurations perform identically over all frequencies.

However, the original question specified a complex load impedance, not a purely resistive one, and used different capacitor and inductor values. I have performed my own analysis below using a complex impedance for a load. Note that the figures for 15.915 nH and 63.622 pF provided by the original poster were obtained by using 100 MHz in the L-pad calculators. The original poster's schematic shows a 10 + j10 ohm load. At 100 MHz, this would be equivalent to a 10 ohm resistor in series with a 15.915 nH inductor. So, that is what I have used for the load.

schematic

simulate this circuit – Schematic created using CircuitLab

Here is the frequency responses I get.

enter image description here

enter image description here

EDIT 2:

Observing the frequency responses of the two circuits, I note that at the 100 MHz point, which is what was used in the calculators, the two configurations have different outputs. This, and working the problem by hand, leads me to believe that the All about circuits calculator computes the value for the shunt capacitor first configuration, and not the shunt capacitor second configuration shown in the image following the calculator. That is, both calculators calculate the same values because they correctly calculate the value for the configuration shown below by the Analog Devices Calculator.

enter image description here

and NOT the configuration shown by All About Circuits.

enter image description here

enter image description here

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  • \$\begingroup\$ @aconcernedcitizen's answer has a graph showing they are identical from 0-1GHz, though you hint at (nonlinear?) behavior that could mean placing the shunt on one side or the other would be more desired. Can you elaborate on what you mean by "shape" and how the topoligies could or would present different behaviors? (Lets assume ideal components, so no coupling or ESL, unless those are what you are getting at) \$\endgroup\$
    – KJ7LNW
    Sep 27, 2022 at 20:42
  • \$\begingroup\$ @KJ7LNW Please read the edit I have made to my answer. I have performed a simulation that shows that with the complex impedance of 10 + j10 ohms specified in your question, the two configurations do not perform the same over all frequencies. \$\endgroup\$ Sep 28, 2022 at 15:12
  • \$\begingroup\$ Just to reiterate the point: an impedance Z = a+jb is just that, an impedance; it cannot possibly tell you anything about the equivalent circuit [beyond an RL or RC at exactly that frequency], which also depends on how impedance changes with frequency. So, always seek out enough information -- measure Z at multiple frequencies -- to determine a useful equivalent over the range of interest. And then model to that. And the above shows one consequence of this difference. If the load were instead like a ferrite bead (X ~= R), yet another response would be seen, and so on. \$\endgroup\$ Sep 28, 2022 at 15:31
  • \$\begingroup\$ @KJ7LNW Math Keeps Me Busy has a point (+1) so I've updated my answer. The part about series or shunt first elements remains and, by the looks of it, the site uses 1st element as shunt C because of that (50 > 10+10j). \$\endgroup\$ Sep 28, 2022 at 18:28
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\$\color{red}{\text{There was a(n unintended) mistake, please see the edit!}}\$ Feel free to downvote.

In the first picture you're not showing the input impedance but, if it is accounted for then the two circuits have an identical transfer function:

$$H(s)=\dfrac{Z_L}{Z_S}\dfrac{\dfrac{1}{LC}}{s^2+\left(\dfrac{1}{Z_SC}+\dfrac{Z_L}{L}\right)s+\dfrac{1}{LC}\left(\dfrac{Z_L}{Z_S}+1\right)} \tag{1}$$

Where \$Z_L\$ and \$Z_S\$ are the load and source imedances, respectively. The identity is only valid for the same imposed conditions: if there's anything different between the two circuits (\$Z_S\$, frequency, etc), all bets are off. A quick SPICE test confirms it:

identical LC circuits

V(a) is slightly offset for better viewing but, the two completely overlap, otherwise, as seen in the phase plot.


[edit]

I forgot to add that the two topologies differ for a reason: using a shunt element as the first one is useful when \$Z_S>>Z_L\$, and series 1st element for \$Z_s<<Z_L\$. Or when there is significant current through the respective element. This is not a rule, it's a convenience but, slightly more than useful for current source input, for example (e.g. DACs). Or, if you feel that your output may end up to be disconnected or high impedance then your 2nd picture may be a better choice.


[edit 2]

Given Math Keeps Me Busy's edit there are two points I need to (re-)address.

I recalculated the transfer function and I realize I made a simplification that resulted in the two transfer functions to be the same only if \$Z_S==Z_L\$! Otherwise, the two transfer functions are these:

$$\begin{align} A(s)&=\dfrac{\dfrac{1}{LC}}{s^2+\left(\dfrac{1}{Z_LC}+\dfrac{Z_S}{L}\right)s+\dfrac{1}{LC}\left(\dfrac{Z_S}{Z_L}+1\right)} \tag{2} \\ B(s)&=\dfrac{Z_L}{Z_S}\dfrac{\dfrac{1}{LC}}{s^2+\left(\dfrac{1}{Z_SC}+\dfrac{Z_L}{L}\right)s+\dfrac{1}{LC}\left(\dfrac{Z_L}{Z_S}+1\right)} \tag{3} \end{align}$$

(1) is the same as (3) but, (2) has \$Z_L\$ and \$Z_S\$ reversed. When they are equal, the two transfer functions are the same. For the particular case of \$Z_S=R_i\$ amd \$Z_L=R_x+jL_x\$ (@100 MHz), the two transfer functions become:

$$\begin{align} A(s)&=\dfrac{1}{LC}\dfrac{s+\dfrac{R_x}{L_x}}{s^3+\left(\dfrac{R_x}{L_x}+\dfrac{R_i}{L}\right)s^2+\left(\dfrac{R_iR_x}{LL_x}+\dfrac{1}{L_xC}+\dfrac{1}{LC}\right)s+\dfrac{R_i+R_x}{LL_xC}} \tag{4} \\ B(s)&=\dfrac{1}{R_iL_x(L+L_x)C}\dfrac{s+\dfrac{R_x}{L_x}}{s^2+\dfrac{R_iR_xC+L+L_x}{R_i(L+L_x)C}s+\dfrac{R_i+R_x}{R_i(L+L_x)C}} \tag{5} \end{align}$$

So now, (4) is a 3rd order, while (5) is a 2nd order, both having a simple zero due to the load:

2nd and 3rd order

Now, the OP is about the order of the LC elements and if it matters. Yes, it does, for cases where \$Z_s\neq Z_L\$ -- such as this case. I did say that the two circuits are the same "for the same imposed conditions", but I did not specify enough differences. Mea culpa.

Still, I have to note that, even with the calculated values, the maximum power transfer does not occur at 100 MHz: the peaks are slightly higher, though not by much.

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  • \$\begingroup\$ Minor update about the differences between the two circuits. \$\endgroup\$ Sep 27, 2022 at 18:00
  • \$\begingroup\$ Ultimately this is FM input from an antenna into to an LNA that I'm trying to match, so not a DAC. Can you extrapolate on why it is "useful" when Zs>>ZL or vice-versa? \$\endgroup\$
    – KJ7LNW
    Sep 27, 2022 at 20:44
  • \$\begingroup\$ @KJ7LNW If ZS=1m then a shunt C as the 1st element will have very little influence, since the source acts as a voltage source. It's not that it doesn't have influence, it's that it will have very little, and the value will need to compensate for that. So a series 1st element will make more sense. But if ZS~ZL, or even an order of magnitude similar, it's GoodEnough®. \$\endgroup\$ Sep 28, 2022 at 5:56
  • \$\begingroup\$ Math Keeps Me Busy's answer deserves the check mark. \$\endgroup\$ Sep 28, 2022 at 18:30
  • \$\begingroup\$ Thanks for the edit! Willingness to edit a post like this deserves an upvote! I moved the check mark as you suggested, thanks for the help on this! \$\endgroup\$
    – KJ7LNW
    Sep 28, 2022 at 22:27
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In Schematic 1, L1 and Z1 can be interchanged.

In Schematic 2, L and C cannot be interchanged.

The labeling in Schematic 2 is incorrect. See below:

schematic

simulate this circuit – Schematic created using CircuitLab

Components in a loop may be interchanged (ordered) if and only if they are in series.

The rules for interchanging components are:

  1. Elements in series may be interchanged, but their orientation must be preserved.
  2. Elements in parallel may be interchanged, but their orientation must be preserved.

In the diagram above:

\$Z_{P}\$ may be interchanged with \$R_{1}, L\$ or \$Z_{S}\$.

\$C\$ may be interchanged with \$R_{2}\$.

The rules are consistent whether or not the components form an L-match.

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  • \$\begingroup\$ R1 isn't really a resistor: its that site's representation of the source impedance, which is 50-ohms provided by the stimulus down a feedline. Thus, R1 would never be interchanged...Does that change your answer? \$\endgroup\$
    – KJ7LNW
    Sep 27, 2022 at 20:27

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