Tuesday, March 31, 2015

Notes on Antenna Tuners: The L-Network and Impedance Matching

Recently I've become interested in antenna tuners (due to issues I've recently had with them) and I've begun exploring the topic.  Here are some notes on the most basic impedance matching configuration, the L-network, which consists of two reactive elements.

I rely on Smith Charts to help understand how L-networks match an unknown impedance to our "target" impedance (i.e. Zo).  If Smith Charts and Reflection Coefficients are unfamiliar or you are rusty with them, I would recommend taking a quick look at my post:  A Brief Tutorial on Smith Charts. The important points mentioned in it are:
First, a load's Reflection Coefficient, Γ (Gamma) can be plotted on the complex-plane.  This plotted point will have a magnitude (also known as ρ (rho)), when referenced to the center of the plane (0,0), that is between (and inclusive of) 0 and 1.
Second, the Smith Chart itself is simply an overlay over this plot of the Reflection Coefficient, and this overlay identifies the impedance (or admittance) of the load associated with Γ at any point.
Third, if we add capacitance or inductance in series with our load we can visualize this as moving Γ along the Smith-Chart's circles of constant-resistance (the direction will depend on the sign of the reactance: inductive or capacitive).  And in an analogous fashion, if we add capacitance or inductance in parallel with our load, we will move Γ along circles of constant-conductance.
Fourth, when we add capacitance or inductance to our circuit, this movement of  Γ (and thus the change in impedance or admittance) is only along these circles.  It is not along the other lines representing constant-reactance or constant susceptance.
Fifth, circles of constant SWR can be plotted on the Smith Chart.  These circles are centered at (0,0) -- the chart's Zo, and their radius is ρ, the magnitude of Γ.
Finally, before we start, this is a lengthy post and it really consists of two main sections:

The first section describes the eight L-Match networks, how they work (Smith Charts), and how they can be used (and I explain a common misconception regarding when series-parallel networks should be used).

The second section deals with actual simulations of the four most commonly used L-networks.  The simulations cover power loss, peak voltage across components, current through components, and component values.

Section 1:  L-Network Impedance Matching

First, a definition:

And second: I will often refer to the "target" impedance that we want to match to as Zo.  Assume, for this post, that Zo = 50 ohms.

I hope it's obvious from my Smith Chart discussion (A Brief Tutorial on Smith Charts ) that if our load has a reflection coefficient that lies on a Smith Chart's "circle of constant resistance" that intersects the point (0,0) in the complex plane (and only one constant-resistance circle will intersect this point -- it's the circle representing R = Zo), we can easily match the load to our target impedance by adding a single appropriately-valued capacitor or inductor in series with our load.

Similarly, if Zload's Γ lies on a "circle of constant conductance" that intersects the point (0,0) in the complex plane (and only one constant-conductance circle will intersect this point -- it's the circle representing G = 1/Zo), we can easily match the load to our target impedance by adding a single appropriately-valued capacitor or inductor in parallel with our load.

(click on image to enlarge)

And, of course, if the load's reflection coefficient is already at (0,0) on the complex plane, no additional matching is required.

But for all other possible reflection coefficients on the complex plane, we will need more than one reactive component to match our load to our target impedance.

Of course, we can use any number of components to create this match, but, it turns out, a minimum of two reactive components is sufficient to match a load to our target impedance (and for the purposes of this discussion I'm going to avoid components such as transformers and stick with just capacitors and inductors).

There are eight possible networks that use a total of two reactive components, and these configurations all have a circuit topology that looks like an "L" (if turned sideways and sometimes flipped).  Thus, the name "L-network."

The eight networks are shown in the drawing below:

(click on image to enlarge)

In the drawing above I've paired these networks into "duos."  For example, configurations 1 and 2 form a duo because each network consist of a series capacitance and a parallel inductor (i.e. CsLp or LpCs).  Throughout this post I will refer to configurations that share the same type of component for their series element and the same type of component for their parallel element as duos.  From the chart above, there are four "duo" pairings:

CsLp / LpCs
LsCp / CpLs
LsLp / LpLs
CsCp / CpCs

Continuing on...

Each of the eight L-network configurations can be used to transform an unmatched impedance to a matched impedance, and each do this either by first transforming Zload's impedance into an intermediary impedance whose Γ lies on the Smith Chart "circle of constant-resistance" representing R = Zo (e.g. 50 ohms), or by first transforming Zload's impedance into an intermediary impedance which lies on the Smith Chart "circle of constant-conductance" representing G = 1/Zo (e.g. 0.02 Siemens).

The L-network component nearest the load performs this transformation to an impedance on either of these two circles.  And, if we were to move from the load towards these two "target" circles by incrementally changing this component, we would find that the changing intermediary impedance moves along paths consisting of other constant-resistance or constant-conductance circles.

The next component (the one closest to the source), then transforms the intermediary impedance (or admittance -- you can think of it as either) to our destination Zo.  This transformation essentially moves the impedance along either the R=Zo circle or along the G=1/Zo circle until it reaches the center of our Smith Chart (coordinate (0,0) on Γ's complex plane) -- our match point!

The diagram below shows how each of the eight L-network configurations can transform Zload first along a circle to one of two circles (constant-resistance or constant conductance) that intersects the center of the chart, and then finish the transformation by moving the impedance from that intermediary point, along this circle (either the R=Zo or G=1/Zo circle), to the center point (0,0) of the chart.  The image below illustrates this:

(click on image to enlarge)

Note that the impedance (and thus Γ) always moves along one circle or another (were we to incrementally change its reactance or susceptance).  It does not move along the other (non-circle) lines that represent constant-reactance or constant-susceptance.  (Although in real-life there might be some small amount of movement along these other lines, due to resistive loss in real-life inductors and capacitors).

Although the diagram above shows the Smith Chart divided into eight "regions" and a different L-Network in each, L-Networks can actually be used across multiple regions.  But some are more useful than others, in that some can match impedances over a larger Smith Chart area than can others.

The dual-inductor and dual-capacitor networks are limited in the regions they cover.  For example, let's look at the LsLp / LpLs duo.  First, matching with LsLp:

(click on image to enlarge)

And matching with LpLs:

(click on image to enlarge)

These two configurations can only match impedances within the bottom area of the Smith Chart (shown in yellow, below) and no where else:

(click on image to enlarge)

In an analogous fashion, dual-capacitor networks also have limited usefulness because they, too, cannot cover the entire Smith Chart.  Here's the region spanned by the CsCp and CpCs duo:

(click on image to enlarge)

On the other hand, the LsCp / CpLs duo of  networks can cover impedances on the entire Smith Chart, as shown below, where one network covers the yellow region and the other network covers the unshaded region:

(Click on image to enlarge)

(Note that I call the LsCp / CpLs duo "Low-pass" L-networks because, ostensibly, they provide a low-pass function while also impedance matching, although the amount of filtering they actually provide depends upon Zload's impedance).

Similarly, the CsLp / LpCs duo of "High-pass" L-networks can also cover impedances on the entire Smith Chart:

(click on image to enlarge)

Clearly, if we're going to design an antenna tuner based upon L-networks, then using either the LsCp / CpLs combination or the Cs/Lp / LpCs combo is the way to go -- we can cover the entire Smith Chart with an inductor and a capacitor and a switch (to flip the chosen configuration between series-parallel and parallel-series).

Calculating Actual Component Values

How can we calculate L-Network component values?

One way is to use a Smith Chart.  If we know the load's complex impedance, we can quickly come up with an L-network to match it to our target impedance.  Programs such as Smith V3.10 are a great tool for doing this and will give us actual component values.

But if we're trying to discover a range of values to match a corresponding range of  load impedances and SWRs, determining component values via a Smith Chart quickly becomes tedious.  Can we instead use a formula to calculate values?

If you take a look on the internet, you'll come across at least two different ways to calculate L-network component values:

The first method uses Q.  Here's an example:  http://www.ece.ucsb.edu/~long/ece145a/Notes5_Matching_networks.pdf

Personally, I don't use this method and I'm also not sure that it is an all-encompassing technique.  Note that the load is assumed resistive.  What happens if the load is complex, which it could easily be?

Also, this technique requires that the source resistance, Rs, be known (e.g. the 2011 edition of the ARRL Handbook states that "To design an L-Network, both source and load impedances must be known.")

Unfortunately, the source impedance is not always known.

For these reasons, I prefer the second technique for calculation L-network component values.  It can be used with any complex load impedance, and it is independent of the source impedance.

This second technique uses two sets of equations.  You can find the derivation of these equations here, for example:   http://www.ece.msstate.edu/~donohoe/ece4333notes5.pdf.

I'll get to these equations in just a bit, but first...

A slight digression into a common misconception...

Practically every reference that I've come across states that one should select series-parallel versus parallel-series networks per the following conditions:

If Rload > Zo:  Use series-parallel configurations such as LsCp or CsLp.

If Rload < Zo:  Use parallel-series configurations such as CpLs or LpCs.

For example, here's an image from the ARRL Handbook for Radio Communications, 2011 (figure 20.9, page 20.10) showing just that (although they use Rs in lieu of Zo):

(click on image to enlarge)

But there's a problem...

The Rload > Zo constraint is incorrect.  In fact, it is too constraining.  Series-parallel networks (e.g. LsCp or CsLp) can be (and are) used in other regions in the Smith Chart and not just when Rload > Zo.

Let me repeat that:

(This misconception seems to be very common, and I'm just using the page from the ARRL Handbook (2011) as an example.  One can easily find many other examples which repeat this same misconception).

The Correct Constraint:

The correct constraint can be expressed two ways:

or

The two are equivalent:

Update
7 April 15
I've given this topic its own blog post.
For a more detailed explanation, go here:

Okay, on to the simulations!

Section 2:  L-Network Simulations

Simulation Methodology:

I've created an Excel spreadsheet to let me analyze and compare the various L-Network configurations.  (Excel handles complex number calculations, but those of you who are more comfortable with, say, Matlab, could use it to do the same thing).

The Excel spreadsheet calculates the following:
1. Component values of the L-network, based upon reactances calculated by the spreadsheet given Operating Frequency, SWR, and L-network configuration.
2. Equivalent-Series-Resistance (ESR) in each matching-network inductor and capacitor, given pre-defined Q's and the spreadsheet's calculated component reactances.
3. Reflection Coefficient, as seen by the source (and modified by component ESR).
4. Power dissipated (as a percent of total power) by all ESRs and by the load.
5. Peak-voltage-across and current-through each L-Network Component (ditto for the load).
For a given SWR, Excel calculates these values as it steps around the appropriate "circle of constant SWR" (on the Smith Chart) in two degree increments (180 points, total).

Power Dissipation in LsCp / CpLs and CsLp / LpCs Matching Networks:
Let's calculate  inductor power dissipation for the CsLp / LpCs duo and for the LsCp / CpLs duo.  Is there any difference in power dissipation between the inductor being placed in series versus in parallel?  (I'll use a high SWR to accentuate any differences).

The simulation conditions are:
1. Frequency  = 3.5 MHz
2. Zo = 50 Ohms
3. SWR = 32:1
4. Capacitor Q (Qc) = 2000
5. Inductor Q (Ql) = 100
6. Power = 1000 watts
(click on image to enlarge)

First, some notes regarding the plot above:

The X-Axis corresponds to the angle of Zload's Γ (Reflection Coefficient) when expressed in Polar Coordinates (ρ, theta).  To create these plots Excel increments this angle in 2 degree increments around a Smith Chart's "circle of constant SWR" (thus keeping ρ fixed,) starting on the Smith Chart's right-hand x-axis (i.e. theta = 0 degrees) and rotating through a full 360 degrees.

And as we go around the circle, Excel's equations will be switching from one configuration of a duo to the duo's other configuration as we pass into the Smith Chart area for that configuration (e.g. LsCp will flip to its corresponding CpLs configuration and then back again later).

And now, a couple of observations on the plot above:
1. Maximum inductor power dissipation (as a percentage of total power) is the same irrespective of configuration.  That is, matching networks with their inductor-in-series and matching networks with their inductor-in-parallel both have the same max dissipation when measured around the entire constant-SWR circle!
2. Maximum inductor power dissipation peaks at roughly 270 degrees (i.e. -90 degrees) and is high from (very roughly) 180 degrees to 360 degrees.This is the lower region of the Smith Chart:  Rload < Zo and Xload has capacitive reactance.
3. If designing to match a known load impedance in a particular "Γ angle" pie-slice region of the Smith Chart, note that one configuration might have lower power dissipation than the other.

Here's the same plot, but at an SWR of 4:1.  Note the interesting symmetry of the power-dissipation curves around 90 degrees and 270 degrees.

(click on image to enlarge)

Now let's now take a look at capacitor power dissipation:

Again, the simulation conditions are:
1. Frequency  = 3.5 MHz
2. Zo = 50 Ohms
3. SWR = 32:1
4. Capacitor Q (Qc) = 2000
5. Inductor Q (Ql) = 100
6. Power = 1000 watts
(click on image to enlarge)

Observations:
1. Peak power dissipation is the same regardless of the capacitor being a shunt element or a series element.
2. Again, note the symmetries around 90 and 270 degrees.
3. And note how the capacitor power dissipation has the same shape as the inductor power dissipation for the same SWR (32:1).  It's just that the peak power is scaled down and the maximum is around 90 degrees, rather than around 270 degrees.
4. If designing to match a known load impedance in a particular "Γ angle" pie-slice region of the Smith Chart, note that one configuration might have lower power dissipation than the other.
For reference (and comparison), here is a side-by-side plot of power dissipation for the inductor and the  capacitor used in the LsCp / CpLs configuration duo.

(click on image to enlarge)

If Qc equals Ql, capacitor power loss and inductor power loss have the same peak value:

(click on image to enlarge)

Conclusion 1:
When considering all loads possible for a particular SWR, there is no difference in inductor maximum power dissipation between the CsLp / LpCs configuration duo and the LsCp / CpLs configuration duo, nor in capacitor power dissipation between these two duos, given equivalent component Q's.  Both result in the same component power dissipation, but at different Reflection Coefficient angles (relative to the positive x-axis) when comparing one duo to the other.

So power-dissipation is not a differentiating factor when designing a general-purpose tuner.

Let's now look at...

Component Current and Peak-Voltage in LsCp / CpLs and CsLp / LpCs L-Networks:

Using the same Excel spreadsheet I've calculated the peak-voltage across each component as well as the current through them.  Again, the simulation conditions are:
1. Frequency  = 3.5 MHz
2. Zo = 50 Ohms
3. SWR = 32:1
4. Capacitor Q (Qc) = 2000
5. Inductor Q (Ql) = 100
6. Power = 1000 watts
First, a plot of Peak-Voltage:

(click on image to enlarge)

And here, Current:

(click on image to enlarge)

They might be a bit difficult to figure out.  So here's a table listing the maximum values for each:

The plots above were for an SWR of 32:1.  What happens to them if the SWR is lower, say, 4:1?  Let's take a look

First, a plot of Peak-Voltage:

(click on image to enlarge)

And here, Current:

(click on image to enlarge)

And again, the maximums in tabular form:

Comments on the Voltage and Current Simulations:
1. The plots are frequency-independent.  This is because component Q, as defined for the simulations, is also frequency independent.  In actual life, Q would probably vary with frequency (e.g. coil Q degrading at higher frequencies).
2. If inductor-Q and capacitor-Q are identical, the peak power dissipated in each will be identical.
3. If designing to match a known load impedance in a particular "Γ angle" pie-slice region of the Smith Chart, note that one configuration might have lower peak voltage or current than another.
Conclusion 2:
When considering all loads for a particular SWR, there is little difference in Peak-Voltages or in Current between the CsLp / LpCs configuration duo (i.e. High-Pass network) and the LsCp / CpLs configuration duo (i.e. Low-Pass network).  (What differences there are become more noticeable at high SWRs.)

So, component peak-voltage or current is not a deciding factor in choosing between the CsLp / LpCs and the LsCp / CpLs configurations for a general-purpose tuner.

So far the performance of the two configuration-duos is essentially equivalent.  So let's look now at component values...

Component Values in LsCp/CpLs and CsLp/LpCs L-Network Tuners:

Again, using the EXCEL spreadsheet to calculate L-Network component values.

And again, the simulation conditions are:
1. Frequency  = 3.5 MHz
2. Zo = 50 Ohms
3. SWR = 32:1
4. Capacitor Q (Qc) = 2000
5. Inductor Q (Ql) = 100
6. Power = 1000 watts
Let's look at Capacitor values:

(click on image to enlarge)

That's interesting, there's an outlier!

Let's look at Inductor values:

(click on image to enlarge)

Another outlier!

If I calculate the maximum and minimum inductance and capacitance for these two networks, they are:

Note the huge maximum values required for the CsLp / LpCs High-Pass networks!

(Also note:  Excel runs its calculations in discrete steps 2 degree increments, so the actual maxima and minima are probably a bit different from those listed in the table above, unless they happen to lie exactly on a 2-degree increment).

Let's lower the SWR to 4:1 and look at these component values again:

Inductance:

(click on image to enlarge)

Capacitance:

(click on image to enlarge)

At a 4:1 SWR, inductance for the High-Pass networks peaks at a Γ angle of about 52 degrees.  And capacitance peaks at an angle of about 232 degrees.  Let's take a look at what's going on...

First, let's look at what's happening at 52 degrees.  Given the SWR of 4:1, let's plot this point on a Smith Chart:

(click on image to enlarge)

And now, plotting the path to a match:

(click on image to enlarge)

So what's going on?

Because this is the CsLp (High-Pass) network, we only need to move a small distance along a "circle of constant conductance" (conductance circle because we are adding a shunt component across the load -- in this case, inductance) to reach the R=50 "circle of constant resistance" (from which it will be a straight shot down to our match point).

A very small movement along a "circle of constant conductance" is equivalent to adding a very large reactance in parallel with the load. In the limit, if we added a parallel element of infinite ohms of reactance (i.e. an open-circuit) across Zload, there would be no movement along the constant-conductance circle.  So, if we require only a very small movement along the constant-conductance circle, we need to add a relatively large reactance in parallel with Zload.  For the CsLp network, this means a large inductor.

Let's do the same thing at the 232 degree Γ angle:

(click on image to enlarge)

Now we're looking at an LpCs (again, High-Pass) network.  Given the position of the load on the Smith Chart, we only need to move a small distance along a "circle of constant resistance" (because we are adding a component in series with the load -- in this case, a capacitor) to reach the G=0.02 "circle of constant conductance" (from which it will be a straight shot up to our match point).

A very small movement along a constant-resistance circle is equivalent to adding a very small reactance in series.  In the case of the LpCs network, the series element is a capacitor (and the movement is counter-clockwise), and for there to be a low reactance the capacitance needs to be relatively large.

(One way to look at this:  in the limit, if our series element were a short circuit (zero ohms reactance), Γ would not move at all.  That is, if we were already on the G=0.02 "circle of constant conductance" (so that no movement was first required along a "circle of constant resistance", there would be no series-capacitor at all -- it would be replaced with a short.)

This implies: if using the LpCs - CsLp High-Pass L-Network networks in an antenna tuner, one might want to add a switch to short out the capacitor (i.e. minimize series reactance) and change the inductor switch from single-pole double-throw to single-pole triple-throw (with the middle position disconnecting the inductor from the matching network -- maximizing parallel reactance).
Something like this:

(Note that vacuum-variable caps will often short at the "maximum capacitance" end of their travel, so no parallel-switch would be needed in that case).

So, we've determined that the LpCs / CsLp High-Pass networks can sometimes require very larger inductors and capacitors compared to the LsCp / CpLs Low-Pass networks, depending upon Zload.

What component-value drawbacks do the LsCp / CpLs Low-Pass networks have?

The spreadsheet calculations show that no large capacitor is required -- in fact, we have the opposite problem -- the closer Zload is to the R=Zo "circle of constant resistance", the smaller the capacitor must be  to move the impedance along a "circle of constant conductance" to the R=Zo "circle of constant resistance (recall that for the CsLp High-Pass network the inductor had to be larger the closer Zload was to the circle).

Again, this is because we want the shunt element to have a very small effect on Zload; therefore, we don't want to change it much.  And so the shunt element has to be a large impedance.  If the shunt element is a capacitor, its value must be small for it to have a large impedance.

Similarly, the CpLs Low-Pass network has the opposite problem of the LpCs High-Pass network -- the closer Zload is to the G=1/Zo "circle of constant conductance", the smaller the inductor must be to move Zload onto that circle.

This is because we want the series element to have a very small effect on Zload -- we don't need to change it much.  So the series element must have a very small impedance.  In this case, the series element is an inductor, so it must have a very small inductance if the series-element's impedance is to be small.

So, whereas LpCs / CsLp High-Pass networks can sometimes require very large inductors and capacitors compared to the LsCp / CpLs Low-Pass networks (depending upon Zload), LsCp / CpLs networks can have very small inductors and capacitors compared to the LpCs / CsLp networks.

Here's a table of the maximum and minimum component values required to match an SWR of 4:1.  Notice how the low-pass networks (LsCp/CpLs) have smaller minimum values than the high-pass networks.

At minimum inductance Γ's angle is around 128 degrees.

And at mimimum capacitance Γ's angle is around 308 degrees.

Below are the Smith Chart locations for Low-pass network minimum component values and High-pass network maximum component values (for SWR = 4:1):

(click on image to enlarge)

Let's continue the analogy with the High-Pass networks to its logical conclusion...

If the value of Ls and/or Cp of our components cannot be made small enough, we could add switches to 1) short out the inductor (i.e. replacing it with a short  circuit which has minimal inductance), and 2) open the connection to the capacitor, making the resulting Cp capacitance very small, too.

Just like this:

Looks like the same switch configuration that I drew for the High-Pass duo, doesn't it?

Summarizing:  overall, the High-Pass networks can have the largest value components if used as a general-purpose tuner.  But -- when matching a specific impedance, sometimes a High-Pass network will have smaller component values than the Low-Pass network.  As an example, consider the following Zload:

Zload = 5 - j400 @ 3.75 MHz

Either of these two networks can match this load to 50 ohms:

CpLs (Low-pass):  C = 2550 pf, L = 17.6 uH

CsLp (High-pass):  C = 33.4 pf, L = 12.9 uH

Note that the High-Pass network uses smaller components.  So, although High-Pass networks can result in much larger component values at certain load impedances than Low-Pass networks, at other times their component values will be smaller than the equivalent Low-Pass network.

Jack Belrose, in his article "On the Quest for an Ideal Antenna Tuner," QST, Oct. 2004, describes a matching network that let's the user select any of the four networks (CsLp, LpCs, LsCp, CpLs) to get around this problem of network-specific "out-of-range" component values that can occur in different areas of the Smith Chart.

With that thought in mind, I created the following plot showing "allowed" networks if we limit their values to lie between maximum and minimum values.

(click on image to enlarge)
Note that in two regions the High-Pass network's values are so large that only the Low-Pass network has values in range.  And even then, the capacitance of the low-pass network can become quite large.

If we create the same plot at an SWR of 32:1 (in lieu of 4:1) to accentuate the differences between networks, we see something interesting:

(click on image to enlarge)

The component values of the High-Pass network are almost always lower than those of the Low-Pass network, except in those regions where the High-Pass network's capacitance or inductance values skyrocket.

On the other hand, for low SWR loads, the Low-Pass network uses smaller component values than do High-Pass networks.  Here's how they compare at an SWR of 2:1

(click on image to enlarge)

So what does all of this mean?

To me it means that, if working with loads that have a high SWR,  you're probably better off going with High-Pass network to keep component values down (except in the regions of skyrocketing values).

BUT: if you need to match Zloads where High-Pass network component values blow up, you will still need to use relatively large values of capacitance and inductance if implementing a Low-Pass network, but at least the component values are now realizable (to some extent).  So, is there any advantage to a High-Pass L given that the Low-Pass network covers all the bases and doesn't blow up?

I don't know.

But this makes me wonder...can a High-Pass T-Network (which is essentially a High-Pass L-Network with an additional series-element) cover the full 360 degrees of an SWR circle with lower component values (in the range of the Low-Pass network or lower) and (roughly) the same amount of loss?  In other words, can the High-Pass T-network cover the cases where the High-Pass L-network blows up?

Hmmm...something to explore!

Power Loss versus Inductance Value

In Conclusion:

If designing a general-purpose antenna tuner, I see little difference in power dissipation and component current and peak-voltages between the CsLp/LpCs High-Pass networks and the LsCp/CpLs Low-Pass networks.  Instead, the main difference I see is with component values.

Note though, that at certain load impedances CsLp/LpCs High-Pass networks require much higher "maximum" values of inductance and capacitance compared to the LsCp/CpLs Low-Pass networks.  But outside of the "blow up" impedances, the High-Pass network will often have lower component values than the Low-Pass networks for loads with high SWRs.

L-Network Extras...

1.  I was wondering what the maximum component values would need to be for Low-Pass L-Networks if they were to be guaranteed to convert certain SWRs down to a 1:1 SWR match.  Here's a table:

2. [Added 13 June 2015]  I was recently wondering if there was a correlation between inductance value and power loss when using a Lowpass L-Network.  That is, did power-loss increase as inductance increased (as we move around a Smith Chart's "Circle of Constant SWR" (and assuming inductance Q was far worse than capacitance Q)?

Well, not really, as you can see from the plot below:

(click on image to enlarge)

Max power loss occurs while inductance is increasing, rather than aligning with the inductance curves.  So power loss does not correlate well with inductance.

BUT, there is useful information here, never the less...

Given a Lowpass L-network, maximum power loss occurs in the CpLs configuration (inductance in series with load).

And if in the LsCp configuration (which usually has lower power loss than the CpLs configuration), power loss does correlate with inductance value.

So, if you can adjust the length of your feedline, my opinion is: try to adjust it so that...
1. The L-Network is in its LsCp configuration for as broad a frequency swath as possible (rather than the CpLs configuration), and
2. The value of L is minimized.
This should minimize power loss due to current through the inductor's parasitic resistance.

(If you cannot do both of these, then I would shoot for one or the other.  But it's difficult to say if you should minimize L in the CpLs configuration over choosing the LsCp configuration with a higher value of L.  This decision really depends upon the Q (parasitic resistance) of the components being used)

Other Stuff...

1.  A Note on my EXCEL Simulations:

It's quite possible that there are errors in my spreadsheet.  However, I've compared my results to the L-Network simulations displayed by W8JI on his Antenna Tuners webpage, and they concur.   (Although, please note: there is a mistake in W8JI's first L-network simulation (1.8MHz, load = 15+j0) -- it states that the coil Q is 200.  However, the results are actually for a coil Q of 250).

So I feel fairly confident in my simulations, but there could always be a mistake -- it is a very large spreadsheet, after all, so the possibility for error is there.  Please let me know if you come across one!

2.  Resources:

You can download the Smith-chart software that I've used to make the Smith-Chart plots in this blog post from this site:  Smith V3.10

W9CF T-Network Tuner Simulation.  This simulator can also be used to calculate power-loss in High-Pass L-Networks.  To do this, you can manually force the high-pass T-Network into a high-pass L-Network by setting one of the capacitors to a very large capacitance (i.e. low-reactance) value.  Find the component values with, say, a Smith Chart.  Then you can enter these values into the simulator by turning the simulator's knobs.  Time consuming if you want to test a lot of values, but a good check if, say, you want to verify that your simulations are in the ballpark.

Impedance Matching Network Designer  This is a useful tool for calculating L-network impedances.  Let Rs = Zo and then enter in your load impedance.  Interestingly, it correctly applies series-parallel configurations even when Rload is less than Zo (e.g. a load of 25+j50 at 3.5 MHz returns an LsCp network of 2.8uH and 1.2nF) in addition, of course, to applying a parallel-series LpCs network.

3.  My Related Posts:

http://k6jca.blogspot.com/2015/03/a-brief-tutorial-on-smith-charts.html
A quick tutorial on Smith Chart basics

http://k6jca.blogspot.com/2015/04/revisiting-l-network-equations-and.html
Revisiting a common assumption that, in fact, is in error!

http://k6jca.blogspot.com/2015/04/notes-on-antenna-tuners-t-network-part-1.html
A look at highpass T-Networks (includes the W8ZR EZ-Tuner)

More on the W8ZR EZ-Tuner:  http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-more-on-w8zr-ez.html  (Note that this tuner is also discussed in the highpass T-Network post).

The Nye Viking MB-V-A tuner and the Rohde Coupler: http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-nye-viking-mb-v.html

The Elecraft KAT-500:  http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-elecraft-kat500.html

4.  Other Interesting Posts:

http://www.w8ji.com/antenna_tuners.htm  (Good information on tuners.  Check out the Q info for roller inductors)

http://www.ece.msstate.edu/~donohoe/ece4333notes5.pdf  (Excellent derivation of equations for synthesizing L networks.  But note the repetition of the error of when to use series-parallel networks.  It's not just when Rload > Zo).

http://www.g3ynh.info/zdocs/z_matcing/part_1.html   (Lots of interesting stuff on this site.  Well worth the read.  And it was this series of posts that got me thinking of impedance matching in terms of conductance, G.).

http://www.nonstopsystems.com/radio/pdf-ant/article-antenna-ideal-ATU.pdf  (Belrose article from October, 2004 QST)

W0QE: Antenna Tunera Paper: http://www.w0qe.com/Papers/Antenna_Tuners.pdf

Standard Caveat:

I could have easily made a mistake anywhere in this blog post.  If something looks wrong, is unclear, or doesn't make sense, please feel free to contact me!

Sunday, March 29, 2015

A Brief Tutorial on Smith Charts

Recently I've become interested in Antenna Tuners and I want to post some notes on them to this blog.  I thought that I'd start off these posts with a quick Smith Chart review/tutorial, given their helpfulness in understanding and analyzing impedance-matching.  So here we go...

A Brief Tutorial on Smith Charts

The foundation of the Smith Chart is a calculated quantity known as the "Reflection Coefficient."  The Reflection Coefficient (usually symbolized with Γ, the Greek "Gamma"), is a measure of impedance mismatch between a load impedance (Zload) and a desired target impedance (the target impedance is usually the Characteristic Impedance, Zo, of a transmission line, but it could also be the Source Impedance of a signal generator, such as a transmitter).  For this discussion I will use 50 ohms as our target impedance.

The Reflection Coefficient can be calculated with this formula:

Note that Zload is a complex quantity with both real and imaginary parts:

Therefore, Γ is also a complex quantity.  Let's look at an example.  Consider the following load impedance:

Its impedance mismatch (as represented with its Reflection Coefficient) from our target impedance of Zo = 50 ohms calculates to be:

Γ = 0.8 + j0.4

And we can plot its position (using Cartesian coordinates) on the complex plane:

(click on image to enlarge)

In addition to expressing Γ's position on the complex-plane using Cartesian coordinates, we can also express its position in terms of Polar-coordinates using rho (symbol: ρ), representing the magnitude of Γ, and theta (symbol: θ), the angle it subtends from the positive x-axis.

Again using our example of the Reflection Coefficient for Zload = 50 + j200 (i.e. Γ = 0.8 + j0.4), this Γ, in Polar coordinates (rho, theta), would be:

(0.894, 26.6°)

where ρ = 0.894 and θ = 26.6 degrees:

(click on image to enlarge)

Let's plot some more Reflection Coefficients on our Complex plane.  I'm going to chose eight more values for Zload, all with a resistance (i.e. "real") value of 50 ohms, and various reactance (i.e. imaginary) values, and then use Excel to calculate Γ:

(click on image to enlarge)

Here's how these eight Reflection Coefficients look when plotted on the Complex plane:

(click on image to enlarge)

If we remove the annotation from the image above, our points look like this:

(click on image to enlarge)

Looks like they form a crude circle, doesn't it?

If we calculate Γ for all possible Zloads of 50 ohms resistive and anything reactive, that is, Zload = 50 + jX, where X ranges from -∞ to + ∞, and we add these values to our plot of Γ above, it would become:

It's a circle!  This is a circle of "constant resistance".  That is, it is the locus of all possible Γ values calculated from the equation Zload = 50 + jX, where R, the resistance of Zload, is fixed at 50 ohms.

Now, if we were to calculate Γ for all possible values of Zload = R + jX (i.e. both R and X range independently from -∞ to + ∞) and then plotted the results on the complex plane, they would plot into a region bounded by (and including) a circle of radius 1.0, centered at 0,0:

(click on image to enlarge)

Let's add a few more "constant resistance" plots of  Γ to our plot of Γ for Zload = 50 + jX.  I'll chose Zload = 150 + jX and Zload = 16.67 + jX, where again, X ranges from -∞ to + ∞:

(click on image to enlarge)

We have three circles, each circle representing Γ for a Zload with "constant" Resistance and all possibilities of Reactance.

Lets look at what happens to our plot of  Γ if we add additional series Reactance (either inductive or capacitive) to an existing value of Zload.  I'll again use the example of Zload = 50 + j200.

First, let's put  Γ for Zload = 50 + j200 on our "Resistance" circle:

(click on image to enlarge)

What happens if we add series reactance?  We are creating a new Zload:

Zload = 50 + j200 + jX, where X is the reactance we're adding.

When we add jX to our Zload, we are going to move Γ along our "resistance" circle.  The movement will be clockwise if we add inductance and counter-clockwise if we add capacitance.  If we add just enough series capacitance to our "j200" term, we will cancel it out and be left with Zload = 50 + j0, a perfect match.  But if we add too much capacitance, we will continue to move counter-clockwise away from that point of perfect match.  To illustrate these points:

(click on image to enlarge)

How does the value of the reactance we're adding correlate to the amount Γ moves along the circle?   We are adding a series reactance to Zload.  What happens if I add a series-element that is a short circuit (zero ohms)?  Nothing happens -- Γ doesn't move (that is, the impedance at that point doesn't change).  So if I add a series element that has a very small reactance, that is, it looks almost a short circuit (such as a large capacitor or a small inductor), there should be little movement along the circle.

In contrast, the more the series element looks like an open-circuit (that is, the higher its reactance, e.g. small capacitor or large inductor), the greater will be Γ's movement around the circle.

(Also note that in the plot above, Γ at its limit (when jX is either -j∞ or +j∞) is at coordinate 1,0, on the circumference of the unit-circle.)

But suppose we add reactance in parallel with Zload, rather than in series as was done above?  How do we visually represent this?

First, recognize that any representation of Zload impedance in series form (i.e. R + jX, where R is resistance and X is reactance) can be transformed into an equivalent parallel form in which:

Where Y is Admittance, G is Conductance, and B is Susceptance.

Yload is simply 1/Zload.  Therefore, G and B can be easily calculated by inverting the equation for Zload (Zload = R + jX) which results in these two equations:

G = R/(R2 + X2), and

B = -X/(R2 + X2)

Let's calculate the equivalent Yload admittances for our example of nine impedances (all with R = 50 ohms) and add them to the table that's earlier in this post:

(click on image to enlarge)

The first thing you should note:  although R = 50 for all nine values, G is not constant.  But take a look at G when Zload = 50 + j0 (our ideal match).  It's equal to 0.02 Siemens.

So...what happens to Γ if, say,  we hold Yload's Conductance (G) constant but vary Yload's Susceptance (B)?

As an example, what happens if we fix G at 0.02 Siemens?  Let's calculate Γ for nine different Yloads, each with its G term set to 0.02 Siemens:

(click on image to enlarge)

Plotted on the complex plane, the coordinates for these 9 Γ are:

(click on image to enlarge)

Now, again with G fixed at 0.02 Siemens, let's plot Γ for Yload = 0.02 + jB, where B ranges from -∞ to + ∞:

(click on image to enlarge)

This is a circle of "constant admittance" (in this case, G = 0.02).  Note that it looks exactly like the circle of "constant resistance" that we drew above for Zload = 50 + jX, except this new circle is shifted to the left-hand side of the complex-plane.

Adding either inductance or capacitance in parallel with Zload moves us along this circle.  Adding inductance will move us counter-clockwise and adding capacitance moves us clockwise.

(click on image to enlarge)

How does the value of the reactance we're adding correlate to the amount Γ moves along the circle?   We are adding a parallel reactance across Zload.  What happens if I add a parallel-element across Zload that is an open circuit (infinite ohms)?  Nothing happens -- Γ doesn't move (that is, the impedance at that point doesn't change).  So if I add a parallel element that has a very large reactance, that is, it looks almost an open circuit (such as a small capacitor or a large inductor), there should be little movement along the circle.

In contrast, the more the parallel element looks like a short-circuit (that is, the lower its reactance, e.g. large capacitor or small inductor), the greater will be Γ's movement around the circle.

(Also note that in the plot above, Γ at its limit (when jB is either -j∞ or +j∞) is at coordinate -1,0, on the circumference of the unit-circle.)

For the sake of symmetry (because we did it above for circles of "constant-resistance"), let's add two more "constant-conductance" circles to the one above:
(click on image to enlarge)

And let's add our previous Γ plots of circles of constant-resistance:

(click on image to enlarge)

In addition to Γ plots of circles of constant-resistance or circles of constant-conductance, I could also plot Γwith Zload's reactance term (X) held constant and varying R from -∞ to + ∞, or I could plot Γwith Yload's susceptance term (B) held constant and varying G from -∞ to + ∞).  The resulting curves would not be circles, though.  Let's look at them:

Here are plots of Γin the complex-plane with  R held constant (circles) and with X held constant (not circles):

(click on image to enlarge)

This is a Smith Chart. (I hope the circles look familiar!).  With the addition of the "constant-reactance" curves, I can easily plot Γ for a Zload = R + jX.  (This is made even easier with Smith-charting programs such as Smith V3.10, which I will  use throughout this blog post).

For example, let's plot Γ for an impedance of Zload = 25 + j25:

(click on image to enlarge)

I can plot the same point in terms of admittance:
(click on image to enlarge)

And here are the admittance and impedance charts, combined:

(click on image to enlarge)

The point represents a unique Γ, but the load associated with it can be represented either as an impedance (Z=R+jX) or as an admittance (Y=G+jB).  The two forms are equivalent.

One more quick topic:  circles of constant SWR.

SWR can be calculated from Γ:

SWR = (1 + |Γ|) / (1 - |Γ|)

That is, SWR depends on the magnitude of Γ.  But the magnitude of Γ is also rho, ρ (see discussion at the beginning of this post).  And so we could also write the equation in terms of ρ:

SWR = (1 + ρ) / (1 - ρ)

(Note:  some authors use ρ in lieu of Γ to represent the Reflection Coefficient.  I believe this is a mistake.  I prefer to represent the Reflection Coefficient with Γ and its magnitude with  ρ).

The SWR for any  ρ (that is, |Γ|) can be plotted as an overlay on the Smith Chart.  For any given ρ, the set of impedance (or admittances) that result in this ρ will lie on a circle centered at (0,0) and with radius ρ.  Here are some examples:

(click on image to enlarge)

That's it for the tutorial.  To recap, the important points are these:
First, a load's Reflection Coefficient (Γ) can be plotted on the complex-plane.  This plotted point will have a magnitude (also known as ρ), when referenced to the center of the plane (0,0), that is between (and inclusive of) 0 and 1.
Second, the Smith Chart itself is simply an overlay over this plot of the Reflection Coefficient, and this overlay identifies the impedance (or admittance) of the load associated with Γ at any point.
Third, if we add capacitance or inductance in series with our load we can visualize this as moving Γ along Smith Chart circles of constant-resistance (the direction will depend on the sign of the reactance: inductive or capacitive).  And in an analogous fashion, if we add capacitance or inductance in parallel with our load, we will move Γ along circles of constant-conductance.
Fourth, when we add capacitance or inductance to our circuit, this movement of  Γ (and thus the change in impedance or admittance) is only along these circles.  It is not along the other lines representing constant-reactance or constant susceptance.
Fifth, circles of constant SWR can be plotted on the Smith Chart.  These circles are centered at (0,0) -- the chart's Zo, and their radius is ρ, the magnitude of Γ.

Resources: