Rapid design procedure for matching resistive and complex impedance terminations using Lsection circuit
There are numerous tutorials on the literature discussing RF matching networks, yet a full understanding of the design methodology is sometimes confusing. This tutorial will try to present a practical and quick method for designing an Lsection RF matching network between a source (Zs) and a load (Zl), when both impedances are real, i.e. resistive with Zs = Rs and Zl=Rl or complex, i.e. Zs=Rs+jXs, and Zl=Rl+jXl.
Page Contents 

Part 1: Matching Resistive Terminations (Z=R) 
Part 2: Matching Complex Impedance Terminations (Z=R+jX) 
Summary 
References 
Part 1: Matching Resistive Terminations (Z=R)
Let’s suppose that we want to match two resistive terminations, Rhigh (Rh) and Rlow (Rl), with Rh>Rl. The matching network has to be inserted between Rh and Rl to ‘match’ the impedances (or resistances in this case, as reactances of terminations are zero, i.e. Z = R + j0 = R) and eliminate reflections (and hence maximise power transfer) between these two components.
An Lsection matching network consists of two reactances. As Rh>Rh the reactances of the Lsection have to be connected so that one of them is in parallel with the higher value Rh, namely Xp and the other in series with the lower value Rl, namely Xs. This will ensure that there is a proper stepdown resistance change as we view the circuit from Rh to Rl. This is shown schematically below:
Xp and Xs are the reactances of the Lsection network. These reactances need to have opposite signs, that is: if Xp is a capacitor then Xs is an inductor and vice versa. This observation results in a matching network with two possible solutions. We will see later some reasons of why one solution might be preferred over the other.
From the reactance values we can extract the component values, using the following popular relations:
Where f is the frequency where matching will take place, L is the inductance in Henry (H) and C is the capacitance in Farad (F). Notice that reactances are frequency dependent, which means that the Lsection network will only provide optimum matching at a specific frequency – unfortunately, this is something that cannot be avoided, so expect that the further away you operate the matching network from the design frequency the more the mismatch will be introduced between the two resistive terminations.
The reactance values of the matching network components are related with the Qfactor as follows:
While the values of the two resistive terminations are interrelated with the Qfactor by the following equation:
Based on the above equations, given Rh and Rl someone is now able to calculate the element values of the reactive components of the Lmatching network. Let’s proceed with an example to understand this better.
Example: Match a 20 Ohm resistor with a 75 Ohm resistor at 1 GHz.
Let’s assign Rl= 20 Ohm and Rh= 75 Ohms.
From the Q factor equation (5), Q=1.658 (note: Qfactor does not have a unit)
From equation (3), Xs=33.16 Ohms and from equation (4), Xp=45.23 Ohms. These are the reactive values (in Ohms) of the components to be calculated.
As mentioned before, one reactance of the Lsection matching network has to be a capacitor and the other an inductor. This results in a matching network circuit with two possible solutions.
Solution 1: Let’s assume that Xs is an inductor and so Xp is a capacitor. From equations (1) and (2) this means that the values are found to be L=5.28nH and C=3.52pF, respectively. The complete circuit with the matching network values is shown below:
So, effectively, we have a matching network with a shuntC and a seriesL elements, reassembling a low pass filter.
Solution 2: Take Xs as a capacitor and so Xp must be an inductor. From equations (1) and (2) Xs=1/(2πfC) and Xp=2πfL. Doing the calculation results in C=4.8pF and L=7.2nH, with the complete circuit shown below:
In this case, there is a shuntL followed by a seriesC, forming a high pass filter.
We can verify the correctness of our solutions by going through a quick simulation of the
S parameters of the above 2 matching networks. The results for both circuits are shown below:
On the figure above, we investigate the Sparameters of both solutions when viewing the circuits from the end of the high resistance Rh termination towards the Rl (left to right as we view the circuits – ignore labelling of terminations, just observe the values shown if you are not familiar with circuit simulation). We can notice that there is a nearly perfect match at 1GHz (this is shown from the ‘dip’ of the S11 and S33 parameters at 1GHz frequency). Additionally, the transmission Sparameters (S21 and S43) for each of those two circuits show that there is no attenuation at 1GHz, hence allowing all the power to flow from Rh to Rl without any attenuation or loss of power at 1GHz. This is desirable and illustrates the zero insertion loss of the matching network. Also, from the graphs, notice the lowpass filter behaviour of the shuntC, seriesL matching network and the highpass behaviour of the seriesC, shuntL matching network.
As expected, the response of the two circuits remains identical when viewing Sparameters from the Rl side. For the interested reader, this is shown below:
Now, someone may ask: Which solution between these two matching networks is the best? Well, there is no general answer and it depends on the application or the nature of the terminations used. For example, if a DC current blocking between the two terminations is desired, then a shuntL and a seriesC solution is preferred, as the seriesC behaves like an open circuit for DC currents (useful hint for RF amplifier design). Or alternatively, for microstrip PCB realisation, it might be easier to consider a shuntC and a seriesL as a solution, as a shuntC can be implemented using an open circuit microstrip transmission line and the seriesL as a short section of a transmission line between the two resistive terminations – this avoids the need to insert vias between the top and bottom layer of a microwave board when matching the two resistors. On a different case, if the high frequency components need to be attenuated or filtered out, the shuntC and seriesL matching network would be preferred as it behaves like a low pass filter (it’s a 2nd order filter, so you don’t expect much outofband attenuation). So, it can be seen that both matching network solutions are ‘good’ as long as someone match it with the system application under consideration.
Smith Chart Solution:
For brevity, the above solutions can equally be calculated using a Smith Chart too. It is not the purpose of this article to analyze the solution using a Smith Chart, so it left to the reader as an exercise! The two possible solutions using the same example values above, are shown below when viewing the circuit from Rl=20 Ohms (it does not matter which termination you take as a start to design the matching network on the Smith Chart – starting from Rl or Rh and going to the other side of the circuit will yield in the same solution).
Solution 1: SeriesL, shuntC (I purposely inverted the order of the circuit, that is I put Rl=20 Ohms to the left side, to familiarise you with the order of the topologies – notice that the component values remain the same  allow some small deviation in calculated numbers due to the error introduced during the drawing of the lines on the Smith Chart, otherwise the exact same solutions for the component values result). Note: Zin=Rh=75 Ohms .Zin is the impedance looking towards the 20 Ohm termination starting from the end of the Lsection, as shown on the schematic below. The Smith Chart line plot aims to create a conjugate matching between a termination and the rest of the circuit. If there is a load of Z=R+jX to be matched with the circuit, the impedance looking from the Z load into the circuit must be Z’=R=jX. This ensures that the imaginary parts when added in series cancel each other leaving equal values of resistive ports to be matched, providing maximum power transfer. That is, in our example, the 75 Ohms load (see circuit below) is looking towards a 75 Ohm circuit, providing the ideal match to it.
Solution 2: (Note: Zin=Rh=75 Ohms.)
[For resistive terminations, it reminded that a quarter wavelength transformer can also be used to provide a necessary matching between resistive ports. In this case only one resistor needs to be inserted between the two terminations to be matched. The value of this resistor is the geometrical mean of the resistances to be matched, that is: . Pluggingin the numbers of our example, results in R=38.7 Ohms. Then, based on the transmission lines properties (coaxial, microstrip PCB, etc)and effective dielectric constant, we calculate the quarter wave length from the operating frequency.]
It can be seen that it is generally extremely easy to design an Lsection matching network for resistive terminations using standard equations. This tutorial will now be used to proceed to the matching of two terminations when either or both are not real, that is, they include some reactance value too. This will be shown later. Although it requires some additional steps compared to the matching of purely resistive terminations, you will see that it is generally easy to grasp the concept.
Resistive matching L network calculator v1.0  Excel file (20KB)

Part 2: Matching complex impedance terminations (Z=R+jX)
In this part, we will investigate how an Lmatching network can be designed to match complex terminations that either or both contain a complex (imaginary) impedance. The situation is presented schematically in the next figure, with Rh>Rl as before and Zh=Rh+jXh, Zl=Rl+jXl:
To design the Lnetwork in this case, we follow these next steps:
 Generate the L matching network circuit layout with Xp positioned towards Rh and Xp positioned towards Rl , with Rh>Rl (as explained in the previous part).
 (If necessary) Transform each impedance termination into a circuit such that there is a parallel equivalent port circuit in parallel with Xp and a series equivalent port circuit in series with Xs. This will simplify the impedance calculations later on (the figure above has Xl and Xl in series with with Xs so no such transformation is required. On other other hand the series Rh and Xh need to be converted into a parallel equivalent circuit to become in parallel with Xp).
 Cancel the imaginary parts of each termination (Xh and Xl, or their equivalent found in step 2) by adding virtual reactances of opposite sign in series or in parallel with each termination, depending on the topology of the termination. This will leave each termination with the real part needed to be matched.
 Proceed with the design of the Lnetwork by matching resistances Rh and Rl, as presented in the previous part.
 Check if component simplification can apply further, i.e. two components in parallel or in series. Series impedances are added (Xtot=X1+X2), parallel impedances are paralleled (Xtot=(X1*X2)/(X1+X2) ). Sign of X1 and X2 can be different. If resulting impedance is Xtot>0, then it’s an equivalent inductor and if Xtot<0 then it’s an equivalent capacitor.
Let’s illustrate the design procedure with an example.
Example: Match a Z1 = 20j30 Ohms termination with a Z2 = 75+j10 Ohms termination at 1 GHz.
Since Re{Z2}>Re{Z1}, let’s assign Z2=Zh and Z1=Zl. The appropriate L matching network circuit has been shown on the figure above with Xp placed in parallel with Zh and Xs placed in series with Xl. As explained before, there are two possible circuits that can provide the required matching at the frequency of interest. Let’s first proceed with the solution when
Xp=jX (capacitor) and Xs=+jX (inductor).
Solution 1: Xp=jX (capacitor) and Xs=+jX (inductor)
The topology of the complete circuit is the following:
Following the procedure according to step 2 (from above), it is convenient to convert Zh(right port as we look the circuit) into an equivalent parallel combination of an inductor and a resistor. This will simplify element calculations with Xp later on, as these will be in connected in parallel. The transformation is presented below:
Xh=+j10 Ohms and Rh=75 Ohms, so we get Xh_p=+j572.5 Ohms. Doing the calculations for Rh_p we get Rh_p=76.3 Ohms. So, effectively our circuit has now the following form:
By doing this transformation the electrical properties of the circuit remain unchanged and no harm has been done to the equivalency of the circuit.
Now, the next step would be to insert appropriate reactances next to each port in order to cancel Xl and Xh_p. In this case, we do this by inserting in series an inductor of inductance Xind=Xl between Xl and Xs, and a capacitor of capacitance Xcap=Xh_p in parallel with Xp and Xh_p. This takes us to the following circuit:
We will make a short pause at this point. The circuit might have a complicated appearance, but in reality all we did is to leave the two ports with the resistive part only. The L matching network consists of Xs and Xp and sees terminations with only Rl and Rh present. Xl and Xind, together with Xcap and Xh_p, although present in the circuit, are in effect invisible. Note that Xl and Xh_p (or the equivalent Xh) are characteristics of the port and are internally contained within the terminations  they cannot be removed or replaced, but their complex part is now ‘cancelled’ with the addition of the external reactances. So, at this stage we summarize the following values for the reactances for the above circuit:
Xl=j30 Ohms (the imaginary part of Zl)
Xind=+j30 Ohms (the inductor to cancel the imaginary part of Zl)
Xs and Xp to be calculated later on
Xcap=j572.5 Ohms (the capacitor to cancel the imaginary part of Zh_p)
Xh_p=+572.5 Ohms (the imaginary part of Zh after transformation)
So, we now proceed with the design of the L matching network between 2 resistive ports with Rl=20 Ohms and Rh=76.3 Ohms, as described at the beginning of this tutorial. Going through the calculations we find that Q= 1.678, Xs=33.56 Ohms, Xp=45.49 Ohms.
Now, we can check for grouping of elements.
Xp is in parallel with Xcap, so their equivalent impedance is (Xp*Xcap)/(Xp+Xcap)=[(45.49)*(572.5)]/(45.5772.5)=42.18 Ohms (negative sign indicates capacitance). Xs and Xind are connected in series so they can add up Xs+Xind=33.56+30=63.56 Ohms. Converting Zh back to its series equivalent and reactance values into element values at 1GHz, this simplifies the overall circuit as follows:
Final Circuit (solution 1): SeriesL, parallelC
This completes the design of the L matching network.
We can confirm the validity of our calculations by going through a circuit simulation:
It can be seen that the circuit has an excellent matching characteristic at 1GHz, verifying the accuracy of our calculations. Note that although the L matching network provides a low pass filtering response, near the 0 Hz the series capacitor within the Zl port blocks any DC from passing through – this is indicated by the S21 value which becomes very small at DC (and S11 where any signal from port 1(Zl) is reflected and can’t appear at the side of port 2 (Zh) ).
Smith Chart solution for matching Zl=20j30 Ohms with Zh=75+j10 Ohms using a series L and a shunt C matching network:
Starting from Zl (could be Zh too), the Smith Chart solution involves the generation of a L matching network which will result in establishing a conjugate matching between Zh and the input of the remaining circuit looking towards Zl. This is shown in the circuit below. On this circuit, Zin is the impedance looking into the Lsection towards Zl when viewed from Zh and it is equal to 75j10 Ohms. This means that when the output of the circuit is connected with Zh=75+j10 Ohms the imaginary parts cancel, leaving only equal resistance values presented resulting in the desired match. Below, is also shown the Smith Chart plot of the matching procedure.
Solution 2: Xp=+jX (inductor) and Xs=jX (capacitor)
The solution in this case is extracted following similar steps as those in solution 1, noting the sign inversion. We summarize the reactance values of our circuit:
Xl=j30 Ohms (the imaginary part of Zl)
Xind=+30 Ohms (the inductor to cancel the imaginary part of Zl)
Xs=j33.56 Ohms
Xp=+j45.5 Ohms
Xcap=j572.5 Ohms (the capacitor to cancel the imaginary part of Zh_p)
Xh_p=+572.5 Ohms (the imaginary part of Zh)
Now we check if simplification of the elements can be applied. Note, that in this case we have inductance in series with capacitance on the side of Zl, and inductance in parallel with capacitance on the side of Zh. It is crucial to consider carefully the signs of the reactances when doing impedance simplification:
For the series case (side of Zl), we simply add the values of the reactances: Xind+Xs=+3033.56=3.56 Ohms (capacitor). For the shunt case, (side of Zh), we have that: (45.5)*(572.5)/(45.5572.5)=+49.43 Ohms (the resulting plus sign indicates that the equivalent impedance is inductive). Converting reactances into element values and transforming Zh back into its initial form this takes us to the final solution:
Final Circuit (solution 2): SeriesC, parallelL
Checking this with the Sparameter simulation we get:
From the above graph, we notice that the accuracy of our calculations is confirmed.
Smith Chart solution for matching Zl=20j30 Ohms with Zh=75+j10 Ohms using a series C and a shunt L matching network:
Note the minor discrepancy between the calculated value elements and those extracted from the Smith Chart. It is evident that Smith Chart solution provides a very quick calculation method, but since this is done by hand and drawing, we get approximate results (although still pretty accurate). On the other hand, the circuit solution based on equations provides a nearly perfect solution (taking away the rounding of numbers). So, depending on the application, the RF circuit designer chooses a solution accordingly.
This concludes the design of the matching network with complex terminations.
Summary:
On this tutorial we investigated the design of a L matching network when the ports to be matched are real or complex. In either case, the resulting matching network has two solutions forming a lowpass or a highpass network, depending on how an inductor and a capacitor are placed.
When the ports to be matched are resistive we get a straightforward solution using a small number of simple equations.
In the case of complex terminations, the situation on the first instance requires the insertion of appropriate inductors or capacitors in series or in shunt in order to eliminate the complex part of the terminations. At this point it might be necessary to transform a termination into its shunt equivalent circuit to simplify the calculation of element grouping – it is also possible that a termination can be given as a shunt circuit e.g. say you want to match a 50 Ohms source with the input shunt circuit of a transistor as given on a manufacturer’s datasheet. In this case, you might have to convert the termination into a series equivalent or leave it as it is depending on the circuit. After doing this step, we check for grouping of elements noting the sign of the reactances to give the solution of the final circuit.
We have also seen that a Smith Chart solution provides an extremely quick solution to the same problem, although it can be less accurate in terms of component value calculation due to the drawing on the chart. For highQ circuits where the width of the passband is narrow, the accuracy of the element values is more important in order to provide the necessary match at the frequency of interest.
Nowadays, there are dozen of computer programs available on the market, which solve matching network problems automatically, but going through the entire process stepbystep, as described on this tutorial, gives a clear insight and understanding of the intermediate steps followed.
References
[1] Christopher Bowick ‘RF Circuit Design’ 1982, Newnes.
[2] Computer Software Smith Chart program ‘Smith V3.10’ (Demo version), Bern University of Applied Sciences (www.fritz.dellsperger.net).