# Lesson 11: Broadband matching with Constant Q Lines

Smith Charts are a beautiful way of visualizing what's going on with a circuit. The importance of Smith Charts in the design process has (arguably) waned- few professionals still design matching networks using Smith Charts outside of school. The RF design tools (with optimizers) are generally good enough to solve most problems. However, some of the old techniques still provide a lot of value for finding initial designs of circuits.

### Section 1: Constant Q Lines (heavy on the theory)

Neat Smith Chart Fact

For circuits using only passive components (and often even those with active components), curves on the Smith Chart spin clockwise with increasing frequency.

Constant Q Lines

Smith Charts all about lines, curves, or circles of constant... something. Smith Charts themselves are lines of constant resistance and reactance. Designers often use lines of constant SWR, that any point on the Smith Chart the same distance from the center has the same amount of return loss. Transmission lines make circles centered about their impedance. One of the lesser known features of the Smith Chart is the curves of constant Q.

The Q of a circuit can be defined in multiple ways, often as the ratio of the frequency to the bandwidth. High Q translates into a narrow band circuit- in this case a narrow band match. Equation 11.1

Q can also be defined as the ratio of the reactance to the resistance.

Equation 11.2

This simple relationship means that lines of constant Q can be drawn on a Smith Chart. The relationship is simple: the arc starts at R=X=0; ends at R=X=infinity; and is centered at the Cartesian coordinates of X=0, Y= +/-(1/Q).  Figure 11.1: Smith Chart with Constant Q lines

Consider the Smith Chart below with four Q contours, Q equal to 1, 2, 5, and 10. The drawn matching path is meant to illustrate several points:

1. To achieve a particular Q, the match must choose elements for the matching path to stay within curves of constant Q.

1. The green and red curve start and stop at the same points.

2. For the red line below, the path touches on (but stays below) the Q=1 line.

3. For the green line below, the path touches (but stays below) the Q=4 line.

2. Higher Q means staying closer to the real line, using less reactive components.

1. The green curve uses two matching element.

2. The red line uses eight matching elements.

3. More elements are needed to keep under the lower Q curve, providing that greater bandwidth.

1. To achieve lower Q (which corresponds to a broader band match), the curves get exponentially closer to the real line.

1. Technically, the relationship is 1/Q.

2. It gets progressively harder to get broader band designs.

1. The theoretic best bandwidth of the matching network is a function of the starting point on the Smith Chart

1. Point A is on the real line of the Smith Chart, and has a very high bandwidth solution

2. Point B has a strong reactance component, which bounds the broadband of the match

1. Starting SWR is only weakly coupled to the bandwidth of the match (some practical considerations exist)

1. Point B has a lower SWR, but a higher Q.

2. Point A has a broader band solution, if one is willing to use enough matching components

1. Real world lumped element components have losses and tolerances.

1. When trying to achieve a high Q solution, the losses in real components changes the trajectory of the matching path (lowering the Q of the match); this is referred to as the Q of the part.

2. Loss in matching elements reduces the practicality of many broadband solutions. Each added element reduces the SNR that much more.

3. Cascading the tolerances of parts rapidly expands the bounds over which the manufactured solution will fall.

1. Tapers transmission line matching networks work well by hugging a very low-Q curve.

1. Since tapers are printed, the tolerance is not cascaded and so are fairly manufacturable.

2. Tapered transmission lines still have loss issues, particularly at the high frequency end.

As always, certain Caveats exist. For this particular scenario, I have assumed Foster components (no negative capacitors or inductors, just passive components).

### Section 2: Designing Multi-section quarter-wave elements

Lesson 4: Optimization with variables Introduced multi-section quarter-wave elements empirically. In this section we will briefly examine multi-section matching from the perspective of Q-curves. Consider the simple quarter-wave transformer, from 50Ώ source to 25Ώ load. Remembering that:

Then the quarter-wave circuit should look as follows: Equation 11.3

Then the quarter-wave circuit should look as follows: Figure 11.2: Single-section quarter-wave match

If we treat 35.36Ώ as a half-way point, we can design a multi-section quarter-wave network with the first transmission line matching 50Ώ to 35.36Ώ and the second transmission line as 35.36Ώ to 25Ώ, such that the multi-section network looks as follows: Figure 11.3: Two-section quarter-wave match

The matching of the two networks is shown below. Both matching networks transform 25Ώ to 50Ώ at 10GHz, but the two-section matching network keeps below a smaller Q-curve (0.19 versus 0.38). Figure 11.4: Both matching networks with:

(1) blue circle the load impedance;

(2) red curve the single transmission line,

(3) green and cyan curves the first and second transmission line, respectively,

(4) blue ellipse, constant Q curve of 0.19 Figure 11.5: Performance of the matching networks,

(1) blue line is the single-section network;

(2) red line is the two-section network

The two-section network has almost twice the bandwidth (using our usual metric of return less than -20dB), correlating well with our Q-curve being roughly half as much.

Questions:

11.2.1: Create a two-section matching network from 50Ώ to 100Ώ.

11.2.2: Create a three-section matching network from 50Ώ to 100Ώ.

11.2.3: The match goes asymptotically to -9.5dB. Why might that be?

### Section 3: Designing a Simple Matching Network

Consider the schematic below, a simple matching circuit. SER3 and SEC6 represent a complex load of 100Ώ and 5pF capacitance in series. We would like the impedance at Port 1 to be matched to 50Ώ. This is a common matching problem. Figure 11.6: Simple matching circuit with SER3 and SEC6 as the load

The impedance of L1 and C2 follow the convention of: Equation 11.4

Therefore the starting impedance of L1 at 1GHz is 1256Ώ (large enough for our purposes to be an open and therefore not affect the circuit) and the starting impedance for C2 at 1GHz is 8Ώ (small enough for our purposes to be a short and therefore not affect the circuit). In effect, this circuit starts out as looking at the load formed by SER3 and SEC6. The impedance looking into Port 1 to begin with is shown below. Figure 11.7: Impedance looking into the load formed by SER3 and SEC6

In order to match the impedance at Port 1, we will start by decreasing the value of L1 until the center of the curve intersects (approximately) with the 50Ώ circle, as shown below: Figure 11.8: L1 has been adjusted to shift the curve to the 50Ώ circle

To finish or match at Port 1, we will decrease the value of C2 until the center of the curve intersects the center of the Smith Chart (which is “matched” point). Figure 11.9: C2 has been adjusted to make the circuit appear at Port 1 to be "matched"

The log view of the impedance gives us our bandwidth. Your plot may look a little different but should have the same basic shape (the null might be in a different place). Assuming that we treat the return loss being less than -20dB for our metric of bandwidth, the bandwidth below is approximately 175MHz centered at 750 MHz, dividing these values gives a fractional bandwidth of 23%. If we cared to center this, we could perform that match over the actual bandwidth of interest, but for instructional purposes having long curves is useful (easier to see on the page). Figure 11.10: Plot of the bandwidth for the simple match

Questions:

11.3.1: What value of L1 was needed to shift the impedance looking into Port 1 to center on the 50Ώ circle?

11.3.2: What value of C2 was needed to shift the impedance looking into Port 1 to center on the 50Ώ point?

11.3.3: What was the bandwidth and center of your circuit? What was the fractional bandwidth?

Main Point

It is easy to design a matching network by hand using the Smith Chart. For a simple matching network like this, it may be more efficient to use an optimizer to perform a gradient search.

### Section 4: Designing a Matching Network with Constant Q Lines

Given the introduction to this lesson, one suspects that the constant Q curves could be used to improve our matching network. Excellent observation!

If we go back to C2 being 200pF, L1 being a matched condition, and in the Plot window find the “Q curve” box and enter a value of 1.3, we should see the plot below. The match is centered approximately with a Q of 1.23. We will use this as a relative measure for now, a quality factor and not a quantitative factor, if you will. Figure 11.11: Partial match of the circuit with a Q of 1.23

We will add additional elements and reset the variables L1 = 200nH and C2 = 200pF. Figure 11.12: More complicated matching network

Starting from L1, and working to the left (C2, L3, C4, L5, and C6 in that order), begin to adjust the variables but this time make the “Q Curve” equal to 1 and bring the center of the impedance curve to the line of constant Q, as shown below (6 elements, 6 curves). Figure 11.13: Matching stages from adding the first element to the last element

In the above plot:

S_2_2 represents the impedance looking into L1

S_3_3 represents the impedance looking into C2

S_4_4 represents the impedance looking into L3

S_5_5 represents the impedance looking into C4

S_6_6 represents the impedance looking into L5

S_7_7 represents the impedance looking into C6

The log view of the impedance gives us our bandwidth. Your plot may look a little different but should have the same basic shape (the null might be in a different place). Assuming that we treat the return loss being less than -20dB for our metric of bandwidth, the bandwidth below is approximately 525MHz centered at 1075 MHz, dividing these values gives a fractional bandwidth of 49%. Figure 11.14: Return loss of the larger matching network

Questions:

11.4.1: What value of inductors and capacitors were used to create the matching network?

11.4.2: What was the bandwidth and center of your circuit? What was the fractional bandwidth?

11.4.3: Try adding an optimizer block with a simplex optimizer, and (remembering the semicolon between initial value and the limits) add the following optimization limits to the variables L1, C2, L3, C4, L5, and C6:

1. capacitors to have the following optimization: ; <opt min="0.1 pF" max="200 pF" step = "0.1 pF" />

2. inductors to have the following optimization: ; <opt min="1 nH" max="200 nH" step="0.1nH" />

What is the new bandwidth? What is the new fractional bandwidth?

11.4.4: Keep the optimizer block using the simplex optimizer, this time set the initial values of all of the capacitors to “200pF” and the initial values of the inductors to “200nH”. Allow the optimizer to start from scratch. After running the optimizer, what is the new bandwidth? What is the new fractional bandwidth?

Main Point

While it requires more resources (time, elements, space), it is possible to improve the bandwidth of a match by adding more elements. The improvement follows a qualitative relationship with the curves of constant Q. In the example above the bandwidth was tripled (assuming you ran the optimizer correctly in Question 3) at the cost of tripling the components. The bandwidth was not improved when the optimizer started from scratch (unless you added a random stage and got exceedingly lucky). Starting from a good design and using the optimizer properly provides a much better solution than not using the optimizer at all, or if we only used the optimizer.