When Bandwidth Falls Short, Add a Stage — Sharing the Q

In the L-section note I wrote that Q is what sets the bandwidth. The higher the transformation ratio, the higher the Q, and the faster the bandwidth narrows.

So what do you do when the ratio cannot be changed? The basic remedy is to split the transformation into stages that share the Q.

Insert an intermediate impedance

Matching 200 Ω to 50 Ω in one stage is a ratio of 4, giving

Q = √(200/50 − 1) = 1.7321

Insert an intermediate impedance Zm and go 200 → Zm → 50 in two steps. Choosing Zm as the geometric mean makes the two stages' Q equal:

Zm = √(200 × 50) = 100 Ω

Stage 1: Q = √(200/100 − 1) = 1.000
Stage 2: Q = √(100/50  − 1) = 1.000

A ratio of 4 has become 2 × 2. Q drops from 1.73 to 1.00.

The element values come from applying the L-section procedure to each stage as is:

How much wider

Compare the continuous band where |S11| stays below −15 dB:

1 stage  (Q = 1.73)    0.8715 to 1.1140 × f0   →  24.2 %
2 stages (Q = 1.00×2)  0.8295 to 1.2140 × f0   →  38.4 %

For two extra elements, the bandwidth grows by a factor of 1.6. The more stages, the lower each stage's Q and the wider the band. Taken to the limit, you arrive at a continuous transformation like a tapered line.

The catch — a second passband

As you will have noticed in the figure, the two-stage response has two dips. Apart from the main band, there is another point near 1.72 × f0 that drops to −33 dB.

This is not an error but an inherent property of the two-stage network. More stages mean more match points, and unintended passbands appear outside the design band.

In practice this can be a problem:

• Transmitter output stage — a matching network meant to suppress harmonics passes one particular harmonic well
• Receiver input stage — interference at an unintended frequency gets through
• Measurement — sweep only the design band and you will never notice

This is why you check with the sweep extended to two or three times the design band. In this example, looking only up to 1.4 GHz, you would miss the second dip entirely.

This actually tripped me up while checking the numbers. Finding the −15 dB crossings by bisection jumped right over a hump and returned "99 % fractional bandwidth". Only by stepping through the response did I find a hump in between that comes back up to −12 dB. Finding a bandwidth means not just locating two edges, but confirming that everything between them really is continuous.

The same goes for quarter-wave transformers

The thinking is the same with distributed elements. If the load is purely resistive, a single λ/4 line gives a match (the stub matching note):

1 section:  Zc = √(50 × 200) = 100.00 Ω

Make it two sections and each section's ratio drops, widening the band. With a binomial design:

2 sections:  Zc1 = (50³ × 200)^(1/4) =  70.71 Ω
             Zc2 = (50 × 200³)^(1/4) = 141.42 Ω

Whether lines of those two impedances can be built depends on the board and the fabrication process. 141 Ω can be too narrow to make, and that becomes the practical limit. Compared with multi-stage lumped networks, the difficulty with distributed elements is how little freedom you have in the values.

What to check before adding stages

Adding stages works, but there is an order to things:

1. Is Q really the cause? — Maybe the load model is unphysical and the real band is narrower still (the load models note)
2. Can the ratio be reduced? — If the load side can be redesigned, that is often cheaper
3. Can you afford more loss? — More elements mean more loss. Watch |S21| at the same time (why S11 is not enough)

The third one especially. Adding stages improves the match and worsens the transmission. The loss per element may be small, but with four elements it adds up. In circuits where efficiency is king, it becomes a choice between bandwidth and loss.

Try it

Change the number of stages and compare sweeps, and all of this is quick to verify.

Smith Match — Matching Network Designer

Line up four elements as a two-stage network and widen the sweep to about 0.5–2 GHz. You will see the main band widen and the second dip appear higher up, at the same time. With the sweep narrowed to 0.9–1.1 GHz, you would never notice the latter.