Pi vs T vs L: Which Matching Network

At $2.4\,\mathrm{GHz}$ a real $10\,\Omega$ load into a real $50\,\Omega$ source cannot be matched with nodal Q below $2$. That floor is $\sqrt{50/10 - 1}$. The L-section is the floor: two parts, $\ge 10\,\mathrm{dB}$ from $1.901$ to $2.812\,\mathrm{GHz}$ ($38\%$), and $-9.14\,\mathrm{dB}$ of transducer gain at $4.8\,\mathrm{GHz}$ if you take the low-pass. A Pi or a T does not get wider. At loaded Q $= 5$ the low-pass Pi is three parts, $2.266$–$2.516\,\mathrm{GHz}$ ($10.4\%$), and $-25.51\,\mathrm{dB}$ at $4.8\,\mathrm{GHz}$. Use the L-section when the band is the specification. Use a low-pass Pi when the second harmonic has to be loaded and you can spend one more part. Use a T when the reactance you must absorb is already a series inductor.

The $10\,\Omega$ is a stated optimum load, held constant across frequency. It is not a transistor data sheet.

Why can't an L-section be wider than this?

Two lossless elements stepping between two real resistances have one degree of freedom left after the match is satisfied, and the resistance ratio spends it. The nodal Q is fixed at $\sqrt{R_{\mathrm{high}}/R_{\mathrm{low}} - 1}$. Here that is exactly $2$. $f_0/Q$ is $1.2\,\mathrm{GHz}$. The $\ge 10\,\mathrm{dB}$ span is $911\,\mathrm{MHz}$, which is the same order and not the same definition.

Only topology A exists, because $10\,\Omega$ is below $50\,\Omega$ and the load conductance is too high for topology B. The low-pass solution, A+, is a shunt $2.653\,\mathrm{pF}$ at the source and a series $1.326\,\mathrm{nH}$ at the load. It passes DC: no series capacitor, no shunt inductor. The high-pass, A−, is a series $1.658\,\mathrm{nH}$ and a shunt $3.316\,\mathrm{pF}$. It blocks DC, it is slightly wider ($2.048$–$3.03\,\mathrm{GHz}$, $40.9\%$), and it does almost nothing to the second harmonic.

You cannot type a lower Q into this L-section. A Pi or a T asked for a loaded Q at or below $2$ collapses back to the same L-section. The solver keeps that L-section on the card as the reference and tells you the request was below the minimum.

What does loaded Q buy on a Pi or a T?

A Pi is two L-sections back to back through a virtual resistance lower than both ends. A T uses a virtual resistance higher than both ends. The loaded Q you type is the larger of the two section Qs, so it sits strictly above the L-network minimum. Bandwidth and harmonic rejection are what you are buying. Component count goes from two to three.

• L, Q = 2
• Pi, Q = 5
• Pi, Q = 10

On this real load the Pi and the T at the same Q have the same $\ge 10\,\mathrm{dB}$ span and the same transducer gain at $4.8\,\mathrm{GHz}$. The curves overlay. The parts do not. The low-pass Pi at Q $= 5$ is shunt $6.631\,\mathrm{pF}$, series $899\,\mathrm{pH}$, shunt $13.59\,\mathrm{pF}$. The low-pass T is series $6.795\,\mathrm{nH}$, shunt $1.798\,\mathrm{pF}$, series $3.316\,\mathrm{nH}$. One coil against two.

At Q $= 10$ the span shrinks to $2.341$–$2.455\,\mathrm{GHz}$ ($4.8\%$) and the $4.8\,\mathrm{GHz}$ gain is $-32.46\,\mathrm{dB}$. The Pi’s series arm is now $472.1\,\mathrm{pH}$ and the load-side capacitor is $29.06\,\mathrm{pF}$. The T’s shunt capacitor is $944.3\,\mathrm{fF}$. Those are different layout problems. A half-nanohenry coil and a sub-picofarad shunt are both places where the pad is no longer a footnote.

When is the Pi the one to build?

Build the low-pass Pi when three things are true at once. The L-section’s $38\%$ band is wider than the channel. The stage in front of the match is a power amplifier and the second harmonic should see a poor load. And the parasitics you already have are shunt capacitors: the drain or collector pad, and the antenna pad. A Pi absorbs a shunt C at each end into the two capacitors. You subtract them from $6.631\,\mathrm{pF}$ and $13.59\,\mathrm{pF}$ before you buy parts. This solver does not know your pads, so you do that subtraction yourself.

The low-pass Pi passes DC through the series inductor. A drain supply can ride that inductor if the shunt capacitors are the only path to ground. A shunt inductor would be a DC short. That is the high-pass and the mixed rows, and they are the wrong rows for a bias feed.

If the radio is a single narrow channel at $2.4\,\mathrm{GHz}$ and you do not care about $4.8\,\mathrm{GHz}$, the L-section is the smaller bill of materials and the more forgiving tune. Loaded Q is not a quality score. It is how hard you are asking the network to work.

When is the T the one to build?

Build the T when the parasitic you must absorb is series inductance: a bond wire, a long package lead, a skinny trace you cannot widen. The low-pass T at Q $= 5$ already wants $6.795\,\mathrm{nH}$ on the source side and $3.316\,\mathrm{nH}$ on the load side. A bond wire of a nanohenry or two disappears into those arms. It does not disappear into the Pi’s $899\,\mathrm{pH}$ series coil. You would be adding inductance the Pi then has to cancel.

The cost is two inductors instead of one, and a shunt capacitor that shrinks as Q rises. At Q $= 10$ that capacitor is $944.3\,\mathrm{fF}$. I would not plan on a discrete $1\,\mathrm{pF}$ being that accurate once it is soldered. If you do not have a series parasitic to swallow, the Pi is the same bandwidth with an easier set of parts.

What about the second harmonic?

Hold the load at $10\,\Omega$ and look at $4.8\,\mathrm{GHz}$. The low-pass L-section delivers $-9.14\,\mathrm{dB}$. The low-pass Pi or T at Q $= 5$ delivers $-25.51\,\mathrm{dB}$. At Q $= 10$ that is $-32.46\,\mathrm{dB}$. The high-pass Pi at the same Q $= 5$ delivers only $-4.48\,\mathrm{dB}$, worse than the low-pass L-section. “Pi” is not a harmonic filter. Low-pass is the harmonic filter. The extra Q makes it deeper.

This is not a filter specification. There is no layout, the load was not allowed to move, and there is no transmission zero placed on $4.8\,\mathrm{GHz}$. A real collector waveform will not present $10\,\Omega$ at the harmonic. The comparison is still the right one between these networks, because the load model is held still while the topology changes. The same discipline, on an L-section whose load does move, is why an L-match is narrow.

What this comparison is not

It is not a stability analysis. A conjugate match of a transistor is not a $K$-factor. It is not a vendor load-pull file. It is not component Q: every row above is lossless, and loss is applied after you pick the topology. It is not a claim that a Pi can beat Fano. A finite lossless ladder does not match an arbitrary impedance over an arbitrary band, and on a real step the L-section is already the widest two-element match. More sections, or a different load, are what “wider” requires. A Pi and a T only go narrower.

References

• C. Bowick, RF Circuit Design, the chapter on Pi and T matching. Loaded Q is the larger of the two L-section Qs through a virtual resistance.
• D. M. Pozar, Microwave Engineering, the chapter on impedance matching and tuning.
• R. M. Fano, “Theoretical limitations on the broadband matching of arbitrary impedances,” Journal of the Franklin Institute, 1950.

Related

• How to design an L-network and choose a solution
• Why an L-match is narrow
• How to match a chip antenna with a Pi network
• Smith chart calculator

FAQ

When should I use a Pi network instead of an L-network?

Use the L-section when its bandwidth is the specification. At 2.4 GHz a real 10 Ω load into 50 Ω has L-network Q of exactly 2, two parts, and a ≥10 dB span of 1.901–2.812 GHz. A Pi cannot be wider than that. Use a low-pass Pi when you want a higher loaded Q: at Q = 5 it is three parts, 2.266–2.516 GHz, and −25.51 dB at 4.8 GHz against the L-section’s −9.14 dB. The Pi also absorbs a shunt capacitor at each end.

What is the difference between a Pi and a T matching network?

On a real 10 Ω to 50 Ω step they have the same loaded Q, the same ≥10 dB bandwidth, and the same transducer gain at the second harmonic. The parts differ. The low-pass Pi at Q = 5 is one 899 pH inductor and two capacitors. The low-pass T is two inductors, 6.795 nH and 3.316 nH, and a 1.798 pF shunt. Use the T when a series parasitic, such as a bond wire, has to be absorbed. Use the Pi when the parasitics are shunt capacitors.

Can a Pi network be more broadband than an L-network?

Not on a real-to-real step. The L-section Q is already the minimum, sqrt(Rhigh/Rlow − 1). A Pi or a T asked for a loaded Q at or below that minimum collapses to the L-section. Higher loaded Q only gets narrower. At Q = 10 the same 10 Ω load is ≥10 dB from 2.341 to 2.455 GHz.

Does a Pi network reject harmonics?

The low-pass ones do, relative to the L-section, on a load that does not move. At 4.8 GHz with 10 Ω held constant, the low-pass L-section is −9.14 dB, the low-pass Pi at Q = 5 is −25.51 dB, and the high-pass Pi at Q = 5 is only −4.48 dB. Low-pass is what rejects the harmonic. The word Pi is not.