Designing an L-Section Match — Taking 200 Ω Down to 50 Ω

The L-section is where matching networks begin. It takes two elements and the values can be worked out by hand. Here we follow the numbers all the way through for one case: matching a 200 Ω load to 50 Ω at 1 GHz.

Why two elements are enough

As the previous note explained, each element can only move along certain paths:

• Series elements — only along constant-resistance circles
• Shunt elements — only along constant-conductance circles

These two families of circles intersect. So ride one circle to the intersection, switch to the other there, and go into the center. That is why two elements are enough: as long as an intersection exists, an L-section can solve it.

Which side the shunt element goes on

An L-section can have its shunt element on the load side or on the source side. The rule is simple: put the shunt element on the higher-resistance side.

Adding an element in shunt always lowers the apparent resistance. Here the load is 200 Ω and the target is 50 Ω, so we are going down. The shunt element therefore sits on the load side. If the load were the lower of the two, the shunt element would go on the source side.

Get this wrong and there is no intersection — no solution.

Once Q is fixed, the values follow

The transformation ratio sets the Q of the network.

Q = √(R_high / R_low − 1)

Here R_high = 200 and R_low = 50, so

Q = √(200/50 − 1) = √3 ≈ 1.7321

Both element values follow from this Q. An easy way to remember it: the shunt element works against the higher resistance, and the series element against the lower one.

Shunt susceptance  B = Q / R_high = 1.7321 / 200 = 8.6603 mS
Series reactance   X = Q × R_low  = 1.7321 × 50  = 86.603 Ω

Converting to element values at 1 GHz, with ω = 2π × 10⁹:

C = B / ω = 8.6603e-3 / 6.2832e9 = 1.3783 pF
L = X / ω = 86.603   / 6.2832e9 = 13.783 nH

Shunt C of 1.3783 pF, series L of 13.783 nH. These are the textbook values.

Checking the intermediate point

The intermediate point tells you whether the values are right.

With the shunt C added across the 200 Ω load, the impedance is

Z = 1 / (1/200 + j·8.6603e-3) = 50.000 − j86.603 Ω

The real part is exactly 50 Ω. This is the "carried to the intersection" state. The shunt element's job is done; all that is left is a capacitive reactance of −j86.603.

Add the series L's +j86.603:

Z = 50.000 − j86.603 + j86.603 = 50.000 + j0 Ω

The imaginary part cancels and we land exactly on the center. The shunt element sets the real part; the series element cancels the imaginary part — that is all an L-section does.

Q also sets the bandwidth

This is where it starts to matter in practice. Above, Q was just a number for computing element values — but the same Q sets the bandwidth.

The element values can be made exact, but only at the single design frequency. Move off frequency and the reactances of C and L change, and the match falls apart. How quickly it falls apart is Q.

For the network above, the range where |S11| stays below −15 dB is

0.871 GHz to 1.114 GHz   →  24.3 % fractional bandwidth

Now vary the load resistance and compare under the same conditions:

The higher the transformation ratio, the faster the bandwidth collapses. An L-section matching 800 Ω to 50 Ω gives you a little under 100 MHz at 1 GHz.

So when the bandwidth falls short, fine-tuning element values will not fix it. The cause is that Q is too high, and the remedy lies elsewhere:

• Split the L-section into two stages that share the ratio (each stage has a lower Q)
• Bring in transmission lines or stubs
• Ask whether the transformation ratio on the load side can be reduced in the first place

"Trimming element values" and "changing the topology" are different remedies for different problems. Look at Q first, and work out which problem you have — that is the order to do things in.

The other solution

Here we chose shunt C and series L, which is a low-pass arrangement. The high-pass arrangement, shunt L and series C, matches the same 200 Ω to 50 Ω. The Q is the same, so the element values come out symmetrically.

Which one to use is not about how good the match is. It depends on what you want the transmission response to do:

• Want to suppress harmonics → low-pass (shunt C, series L)
• Want to block DC or avoid low-frequency noise → high-pass (shunt L, series C)

Two networks that look equivalent if you only watch the match become different things once you look at |S21|. This is where matching is not just about S11 comes into play.

Try it

The procedure above can be followed by eye on the chart.

Smith Match — Matching Network Designer

Set the load to 200 Ω and add a shunt C: the point moves along a constant-conductance circle and reaches 50 − j86.6 where it meets the 50 Ω real-part circle. Add a series L there and you are at the center. Switch to the sweep view and you can check the bandwidths in the table above directly.