When designing a matching network, how you model the load affects the result more than you might expect. In particular, treating the load as "fixed R + jX" is convenient but not physical. For a single-frequency design it does no harm, but the moment you look at bandwidth, it lies.
What "fixed R + jX" assumes
Say the load is 50 − j50 Ω. The fixed R + jX model treats it as "50 − j50 at every frequency".
But in reality, what produces that −j50 of reactance is a capacitor or something like one. A capacitor's reactance is inversely proportional to frequency:
X_C = −1 / (ωC)Halve the frequency and the reactance doubles; double it and the reactance halves. There is no pure reactance that stays constant with frequency.
So fixed R + jX is "a snapshot at that frequency" — as a frequency response, it is a model that cannot physically exist.
How big is the difference
Compare under the same conditions. Model a load that is 50 − j50 Ω at 1 GHz in two ways:
- Model A — fixed R + jX.
50 − j50at every frequency - Model B — R and C in series.
R = 50 Ω,C = 3.1831 pF
At 1 GHz the two have exactly the same impedance. They cannot be told apart.
Match each with a series L. To cancel −j50, L = 7.9577 nH. Both match perfectly at the design frequency.
Sweep the frequency, and this is what happens (|S11|):
| Frequency | Model A (fixed R + jX) | Model B (series R-C) |
|---|---|---|
| 0.70 × f0 | 0.1483 | 0.3423 |
| 0.85 × f0 | 0.0748 | 0.1611 |
| 1.00 × f0 | 0.0000 | 0.0000 |
| 1.15 × f0 | 0.0748 | 0.1389 |
| 1.30 × f0 | 0.1483 | 0.2565 |
In terms of the fractional bandwidth where |S11| stays below −15 dB:
Model A (fixed R + jX) 72.3 %
Model B (series R-C) 36.1 %A factor of two. With the unphysical model, the bandwidth looks twice what it really is.
Why a factor of two
The reason is the direction in which the reactance moves. Writing out the total reactance after matching makes the difference clear:
Model A: X(f) = ωL − 50 = 50 (f/f0 − 1)
Model B: X(f) = ωL − 1/(ωC) = 50 (f/f0 − f0/f)In Model A, only the inductor moves. The −50 on the load side is fixed, so the error grows only by the inductor's share.
In Model B, the inductor moves one way and the capacitor moves the other. The two that were canceling pull apart from both sides, so the error opens up twice as fast. At 0.7 × f0:
Model A: 50 × (0.7 − 1) = −15.0 Ω
Model B: 50 × (0.7 − 1.4286) = −36.4 ΩThat is what shows up as the bandwidth difference. With a physical load, the match falls apart sooner.
Which one to use
It splits by use:
- Single-frequency design — fixed R + jX is fine. As long as the value at that frequency is right, the result is the same
- Evaluating bandwidth — do not use it. As shown above, the result comes out too optimistic
- When measured data exists — using it as is, is best. The whole modeling question goes away
In practice the third is ideal. If the device maker publishes S-parameters, there is no need to guess a model. It is healthier to think of modeling as a stand-in for when there is no data.
Choosing a model
When there is no data and you have to build a model, choose by where the reactance comes from:
| Source of the reactance | Suitable model |
|---|---|
| Junction capacitance, pad capacitance | Series R-C or parallel R-C |
| Bond wires, leads | Series R-L |
| A capacitive load seen in parallel with a resistance | Parallel R-C |
Series or parallel can be told apart by which way it heads as frequency rises. Series R-C loses its reactance at high frequency and approaches R. Parallel R-C shrinks as a whole at high frequency and heads for the origin. With two or more measured points, you can tell which model fits.
Optimistic estimates come back to bite
What makes this kind of error troublesome is that nothing looks wrong at the design stage. The value at the design frequency is correct and the match is perfect. The problem appears when you build a prototype and measure the bandwidth — only then do you notice it is "narrower than the simulation".
When the bandwidth is narrower than expected, there is an order in which to suspect things:
- Is the load model unphysical? (this note)
- Is Q too high? (the L-section note)
- Are element parasitics at work?
The first is a problem in the calculation, so it can be eliminated before you build anything. Leave it in, and you end up suspecting the second and third and fiddling with hardware.
Try it
When you can switch load models and compare, this difference is easy to see.
Smith Match — Matching Network Designer
Besides fixed R + jX, the load model can be series or parallel R-L / R-C. The note that appears when you choose fixed R + jX — that its frequency response is not physical — is there for exactly the reason described here. Keep the same design, switch only the model, and you can watch the shape of the sweep change.