More Notes on Directional Couplers for HF -- the Bruene Coupler, Part 2

This is Part 2 of my notes on the Bruene Directional Coupler.  Part 1 is >>here<<.  I strongly suggest reading Part 1 -- it explains a more easily understood variation of Bruene's circuit.

So, assuming you're familiar with the basic principles of the Bruene coupler, let's move on to Bruene's design!

Let's start with the schematic of the Bruene Coupler:

(click on image to enlarge)

This coupler was used in the Collins 302C-3 Directional Wattmeter and it was also described in an article written by Warren Bruene in QST magazine ("An Inside Picture of Directional Wattmeters," QST, April, 1959).

This architecture uses the same principles as the ones described in Part 1:  to generate a voltage representative of a Forward Wave on a transmission line, a sample of the line current is added to a sample of the line voltage.

And to generate a voltage representative of a Reflected (or Reverse) wave on a transmission line, a sample of the line current is subtracted from a sample of the line voltage.

What differentiates Bruene's design from the other variations is that those variations keep the circuitry generating the voltage sample independent of the circuitry generating the current sample.  Which is to say, with those other designs I could take a voltmeter and independently measure the voltage sample and the current sample.  I cannot do that with Bruene's circuit -- voltage-sample and current-sample generation are bound together!

You're probably looking at the above schematic right now and wondering what the heck I'm talking about.  You see two voltage dividers, so they must be generating the voltage sample of the Transmission Line voltage, while the transformer (with the two 10 ohms resistors) is generating a positive and negative voltage related to the current on the Transmission Line.  Everything's copacetic, right?

But take a closer look.

Vfwd and Vref. which are the sum and differences of the voltage and current samples, are measured at the divider-taps of the two capacitive voltage dividers.  So the voltage dividers can't just be generating samples of the Transmission Line's voltage.  They must somehow be involved with summing in the Line current samples, too!

The key to the summing are the two diodes. But how are the diodes doing this?

Each diode is actually behaving as a "shunt rectifier" (see Chapter 2 in this >reference<).  And this shunt-rectification process is differencing (rather than summing) the voltage-sample and the current sample that are at the diode's Cathode and Anode, respectively, with the result appearing at the diode's Cathode (the junction of the two voltage-dividing capacitors).

To illustrate this concept better, I've done some basic simulations using LTSpice.  My simulation-circuit is a much simplified version of the part of Bruene's coupler that generates Vref.  Please note that:

1. I use the same capacitor values for the voltage divider as Bruene.  Vsource is the voltage across the transmission line.  It's a large amplitude because the voltage divider has a huge division ratio.
2. The voltage created by the current-sampler, originally generated by a current source (the transformer) driving its current through the resistors, has been replaced by its Thevenin equivalent, but I've made the series-R very small to keep its effect from confusing this discussion.  I've called this voltage source Vinduced.
3. I've added a "switch" between the Diode's anode and Vinduced so that I can look at the voltage sample (at the voltage-divider tap) for a full 360 degree cycle without it being influenced by the current sample.

Below is my first simulation.  To illustrate the shunt-diode's operation, we will first simulate Vinduced as a DC voltage source of 0 volts.  For the shunt-diode I've chosen a 1N916 (a silicon diode).  The actual circuit uses Germanium diodes (I believe).

I've annotated the schematic below to describe the basic functions of each part of the simulation circuit.

(click on image to enlarge)

So what's going on in the plots?  Let's look...

(click to enlarge)

I've defined an "Offset" voltage across C2.  It can also be thought to be across C1, too, and it's effectively the amount that Vsample has been shifted.  I'll explain this further a bit further down the post.

Looking at the plots, at 100nS we connect Vinduced to the diode and so at the 150 ns mark, when the voltage at Vsample is driven negative by Vsource, the diode's cathode connected to Vsample will start to go negative during the negative half-cycle of Vsource.  Because the diode's anode is tied to a 0V source, the diode will turn ON when its cathode voltage drops about 0.7V below its anode voltage.  I.e. when Vsample is approximately -0.7V.

When the diode turns ON, it effectively "clamps" the voltage at Vsample to -0.7 volts.

However, Vsource is still going negative.  So we have the "+" terminal of C2 (see simulation schematic) clamped at -0.7V and C2's "-" terminal (attached to Vsource) continuing down, which causes charge to flow onto C2's plates.   When Vsource reaches its negative peak, this charge can be calculated from the equation Q = C*V:

Q(max) = C2*Vsource(pk)

Actually, in the real world the voltage is slightly less than this by Vf, the forward voltage  drop across the diode.

Now, as Vsource starts to rise again from its negative most transition, the diode turns off, but not immediately, which is the "real-world" diode turn-off effect you see on the Red line in the plot above.

With real-world diode effects, the plots can get a bit confusing, so let me introduce a simpler example that uses an "ideal" diode.  I'll define the diode to have a Vf of 0V:  if its Cathode is at all negative with respect to its Anode, it is ON, and if its Cathode is at all positive with respect to its Anode, it is OFF.

Shunt Rectifier with Vinduced as a DC source and an Ideal Diode:

Here's the circuit and the simulation:

 (click on image to enlarge)

 

Looks a bit cleaner, doesn't it?

So what is happening?

From 0-100ns the diode's Anode is floating because the switch is open, and it is essentially out of the circuit.  Vsample is just Vsource divided down by the capacitive voltage divider (C1 and C2), and therefore Vsample goes between -3.4V and +3.4V.  Its average value is 0V.

At 100ns the switch turns ON, and the diode's Anode is connected to Vinduced, which here is a 0VDC source.

From 100ns to 150ns, Vsource is positive and thus Vsample is still positive.  The diode is OFF (it is back-biased:  Cathode positive, Anode at 0VDC), and Vsample is still just Vsource divided down by the capacitive voltage divider (C1 and C2).

But starting at 150ns, Vsource goes negative and it tries to take Vsample below 0, too.  But Vsample cannot go negative, because as soon as it begins to go below 0V, the diode turns ON and clamps it to the voltage at its Anode, 0VDC.

So as Vsource goes negative, Vsample sticks at 0V, and the voltage across C2 (1.7pf) gets larger and larger, until, at 175ns when Vsource is at its negative peak, the voltage across C2 (Vc2) is at its max:

Vc2(max) = 1000 V.

At this point C2 will have its maximum charge on its plates.  From the formula Q = C*V, we know that:

Qc2 @ max V = C2 * Vc2(max) = C2*Vsource(peak)

(...given the voltage at the diode's Anode is 0VDC and that the diode is ideal with a 0V Forward-voltage drop.)

Also, note that the voltage across C1 (500pf) is 0V (Vsample is clamped to 0V).  Therefore the charge on C1 is 0 coulombs.

So at 175ns, when Vsource has reached its most negative excursion, the maximum charge is on the two caps:

Qmax = Qc1 + Qc2 = 0 + C2*Vsource(peak)    (equation 1)

Now, as time moves on from 175ns, Vsource just starts to rise up from its negative peak. Vsample, which was at 0 V, starts to go positive and the diode immediately cuts off (because Vsample = Vsource + Vc2(max)).  The diode is now effectively out of the circuit.

The diode is back-biased and out of the circuit.  There's no path for the charge of Qmax on C2 to go but to C1 (there's no other path but to C1).  So, as time continues, no charge will by lost by the C1/C2 combo, and no charge will be added -- the diode never turns on again.  And so the total charge on C2 and C1 (=Qmax) will flow back and forth between C2 and C1 as Vsource goes up and down.

No charge is being added, no charge subtracted.  Therefore, at all times (after we've first charged C2 via the diode clamp), the charge across C1 and C2 must satisfy this equation:

Qc1 + Qc2 = Qmax

And therefore (using Q = C*V):

C2*(Vsample - Vsource) + C1*Vsample = Qmax

Rearranging:

Vsample = Vsource*C2/(C1+C2) + Qmax/(C1+C2)   (equation 2)

So, if we know Qmax (and we do), we can calculate what Vsample will be for any value of Vsource.

The equations above do this for an ideal diode with 0V attached to its anode.  We can derive a more general equation for Qmax (rather than use equation 1 above) that takes into account the voltage connected to the diode's Anode and also its Forward-voltage drop (i.e. a non-ideal diode):

Qmax = C2*(Vsource(pk) + V(anode) - Vf(diode)) + C1*(V(anode) - Vf(diode))   (equation 3)

Using equations 2 and 3, we can now calculate Vsample for different voltages connected to the diode's Anode, per the table below (I'll keep the diode ideal, though, with Vf = 0):

(click on image to enlarge)

You can see that "Vsample without diode" ranges from -3.4 to 3.4V.  If we look at, say, the diode with 0V attached to its anode, the range of Vsample now becomes: 0V to 6.8V -- a shift of 3.4V, the peak voltage of the non-diode Vsample.  So Vsample's average voltage is now 3.4 volts instead of 0, and if we were to filter out the AC waveform on Vsample we would measure 3.4 VDC.

The previous example was with a DC source connected to the diode's Anode and demonstrates how the shunt-rectifier can create a DC voltage from an AC signal.  What happens if the Anode is instead connected to an AC source representing the voltage from the current-sample?


Vinduced as an AC source:

So all is fine and good if Vinduced is a DC source.  Let's change Vinduced from a DC voltage source to something instead the simulates what is actually happening in the Bruene Coupler -- an AC source of the same frequency as Vsource, but with a phase offset.

We should still see a DC offset voltage generated on the capacitors in an analogous way to when Vinduced was a DC source. But the level of this offset voltage should now depend on the phase difference between the Vinduced and Vsource.

To keep the analysis simple, I won't be changing the amplitudes of Vsource or Vinduced, just their phase relationship.  And note that the amplitude of Vsource has been chosen so that, if the diode were removed, Vsample = Vinduced in amplitude.  

Let's examine how the voltage offset changes with phase (see the RED line in the plots below).  With the two waveforms 180 degrees out of phase, LTSpice simulates this value to be 6.8V:

(click on image to enlarge)

With the two waveforms 180 degrees in phase (0 degrees delta), LTSpice simulates this value to be 0V:

(click on image to enlarge)

 

With the two waveforms 45 degrees out of phase, LTSpice simulates this value to be 2.61V:

(click on image to enlarge)

With the two waveforms 180 degrees out of phase, LTSpice simulates this value to be 4.8V:

(click on image to enlarge)

 

And finally, with  the two waveforms 135 degrees out of phase, LTSpice simulates this value to be  6.28V:

(click on image to enlarge)

Summarizing the DC offset calculated by the simulations:

But is this DC value that LTSpice calculates actually equal to the difference (or sum) of the two sine waves representing our voltage sample and current sample?  This is a requirement that must be met if we are to calculate Vfwd and Vref by adding or subtracting the voltage and current samples.

Let's check this by doing some separate math, using the same peak voltages and phases that I used in the LTSpice examples above...

If summing sine waves, I can use the following equations to calculate the the resulting amplitude of the summed waveform (amplitude can be expressed as Vrms, Vpp, etc.  I'll use Vpeak, which, from our simulations, is 3.4V).

Knowing the amplitudes of the sine waves and the different phases, I can plug these numbers into Excel and solve for c, the summed-amplitude (called Vc in the table, below):

(click on image to enlarge)

The "LTSpice" column is 180 degrees out of phase with with "formula" column because the "shunt-diode" circuit is actually a differencer, not a summer, which is equivalent to adding 180 degrees to the phase shift "α" in the formula for c.

If we compare the table results from the math with our simulation results (see the earlier table), we find that they are exactly the same values (after first taking into account the shunt-diode's additional 180 degree phase shift).  Which means that the shunt diode circuit is indeed summing (or differencing) the voltage waveforms of the voltage sample and the current sample!

And therefore, it is functionally equivalent to the earlier Bruene coupler variants that we looked at in Part 1.

Now that I've shown that the shunt-rectifier sums or differences the voltage and current samples, I'll refer you back to Part 1 for a recap of circuit operation in Transmission Line and non-Transmission Line environments.

The only thing to keep in mind is that, because the shunt-rectifier is actually a differencing circuit, not an adder, the side of the coupler that, in part 1, generated Vfwd (that is, the left side of the circuit in the diagrams), now generates Vref.

And the opposite side (the right side of the circuit in the diagrams) now generates Vfwd.

Other than that, the same analysis applies!


Bruene Variant, Daiwa SWR Meters:

While looking around the shack I came across a Daiwa CN-620B wattmeter that I pickup up several years ago.  Opening it up, I discovered it's a variation of the Bruene Coupler that uses Bruene's "shunt rectifier" method of creating a DC voltage from the voltage and current samples.

(click on image to enlarge)

Inside the Daiwa CN-620B (click on image to enlarge))

 The main differences that I can see between it and Bruene's design are:

• Daiwa uses two transformers, one for Vref for Vfwd, compared to Bruene's one.
• The Daiwa transformers have fewer windings than Bruene's (I believe the latter has 60).

A number of other Daiwa wattmeters (e.g. CN710, CN720) use the same or similar design.  Curiously, when I look at the values of the components on the CN-620B's PCB, they are significantly different than the values listed in the schematic! Don't know why.


Links to my Directional Coupler blog posts:

Notes on the Bruene Coupler, Part 2

Notes on the Bruene Coupler, Part 1

Notes on HF Directional Couplers (Tandem Match)

Building an HF Directional Coupler

Notes on the Bird Wattmeter

Notes on the Monimatch

Notes on the Twin-lead "Twin-Lamp" SWR Indicator

Calculating Flux Density in Tandem-Match Transformers


And some related links from my Auto-Tuner and my HF PA posts:

Auto Tuner, Part 5:  Directional Coupler Design

Auto Tuner, Part 6:  Notes on Match Detection

Auto Tuner, Part 8:  The Build, Phase 2 (Integration of Match Detection)

HF PA, Part 5: T/R Switching and Output Directional Coupler


Bruene Coupler References:

Bruene, Warren, "An Inside Picture of Directional Wattmeters," QST,  Apr., 1959.  Includes both a good explanation of the Monimatch operation and a design for a directional wattmeter whose directional coupler topology would later be known as the "Bruene Coupler."

Collins 302C-3 Directional Wattmeter, PDF Manual containing schematic.

Rush, James, Jr., "The Mini-Mono-Monimatch," QST, Mar., 1965.  Although called a Monimatch in the title, the design is actually more similar to Bruene's directional coupler.

Bold, Gary, ZL1AN, "The Bruene Directional Coupler and Transmission Lines," PDF. This PDF gives an excellent explanation of the Bruene Coupler.

Kiciak, Paul, N2PK, "An HF In-Line Return Loss and Power Meter," PDF.  Constructions details of a power meter using a Bruene Coupler.  Contains an explanation by the other of why he prefers the Bruene coupler of the Tandem-Match Coupler.  Also interesting because the author separates the voltage-sampler from the current sampler and uses the differential inputs of an AD8307 to do the required addition (or subtraction) to get FWD and REF voltages.

(This web page could be useful for understanding the sampling method used in the N2PK meter:  http://www.g3ynh.info/zdocs/bridges/magdiff/part1.html )

Lewallen, Roy, W7EL, "A Simple and Accurate QRP Directional Wattmeter," QST, Feb, 1990, PDF.  Interesting variant of the Bruene coupler.  Roy uses two transformers for the voltage sample in lieu of capacitor voltage dividers.

White, Ian, G3SEK, "Inside a Directional Wattmeter," RadCom, Sept., 2002, PDF.  Discussion and a bit of analysis of Bruene coupler.  Includes Bruene's phase-relationship diagrams.

http://www.g3ynh.info/zdocs/bridges/reflectom/part1.html  interesting analysis

http://www.g3ynh.info/circuits/Diode_det.pdf  Diode detectors -- includes some info on shunt detectors, which is what Bruene's design uses.


Other references of generally interest:

http://www.g3ynh.info/zdocs/bridges/Xformers/part_1.html  great discussion on current-transformers for directional coupler applications

http://www.g3ynh.info/zdocs/bridges/Xformers/part_2.html Part 2 of current-transformers

http://www.g3ynh.info/zdocs/bridges/Xformers/part_3.html  And part 3, the last part, of current-transformers

http://www.g3ynh.info/zdocs/bridges/index.html  Indexes numerous topics.  Lots of great info to be found here!

http://www.richtek.com/assets/AppNote/AN008_EN/AN008_EN.jsp  Common-Mode choke model


Final Caveats:

As always, I might have made a mistake in my equations, assumptions, or interpretations.  If you see anything you believe to be in error, or if anything is confusing, please feel free to contact me.

More Notes on Directional Couplers for HF -- the Bruene Coupler, Part 1

In a previous post I looked at how the popular "Tandem-Match" Directional Coupler worked.  Recall from that analysis:

1. I analyzed the Tandem-Match coupler in terms of "lumped" circuit elements, not distributed elements.
2. The coupler works by taking a sample of the voltage across the transmission line at a single point and a sample of the current through that point.  
3. The Tandem-Match Coupler creates both of these samples with transformers.
4. The voltage at the "Forward Port" (that we measure on our meter) is calculated by adding the current and voltage samples.
5. The voltage at the "Reflected Port" is calculated by subtracting the current and voltage samples.
6. The voltage at the "Reflected Port" is 0 when the load is a real resistance equal to the value of the resistors terminating the measurement ports (Forward and Reflected Ports).  Thus, the resistor values should be selected to be the same as the characteristic impedance, Zo, of the transmission-line system into which the the coupler is inserted.
7. I explained operation of the coupler both in terms of Forward and Reflected waves and also without using waves, instead in terms of only the voltage driving the coupler and the load at its output port.

Another popular directional coupler topology for HF is known as the Bruene Coupler.  This coupler was used in the Collins 302C-3 Directional Wattmeter and it was also described in an article written by Warren Bruene in QST magazine ("An Inside Picture of Directional Wattmeters," QST, Apr., 1959).

I'll analyze the original design as well as some of the variants it inspired.  Again, I will use lumped-element analysis, and I'll present explanations in terms of forward and reflected waves and also in terms of the voltage being applied to the coupler and the load connected to its output port.

My goal is to present you with an explanation that is understandable -- this will not be a rigorous analysis.  So, when possible, I will take advantage of models and assumptions that will simplify the math and hopefully help you grasp the underlying principles.

So here we go!  Let's start with the schematic of the Bruene Coupler:

(click on image to enlarge)

There are two capacitive voltage dividers; one is used for generating the Vref voltage, the other is used for the Vfwd voltage.  Let's call these the voltage samples, but in fact, they are more than that because of the two diodes connected to them.

Line current passing through the 1-turn primary of a 1:60 turn transformer and induces current in the 60-turn secondary.  This current, in turn, is transformed into voltage as it passes through the two 10 ohm resistors.   One voltage is positive with respect to ground, the other is negative.  These two voltages represent the current sample.

Now the explanation gets a bit tricky.  Take a look at the schematic -- we're measuring Vref and Vfwd at the voltage-divider taps.  This means that the two caps comprising each voltage-divider aren't simply generating a voltage divided down from the line voltage, they are also involved in adding or subtracting the current-sample voltages, too, so that, when we measure the voltage across either 500 pf cap, it's actually the sum or difference of voltage and current samples.

It wasn't obvious to me how the summing/differencing of voltage and current samples at the voltage-divider nodes was being accomplished.  It turns out the diodes linking the voltage-dividers and the current samples play a crucial role -- they are creating a DC voltage at the voltage-divider "tap" to which their Cathodes are connected.  But not as series rectifiers -- the diodes instead are serving as "shunt" rectifiers.

The DC voltage they create is a function of the phase and amplitude differences between the voltage and current samples.  I'll explain the role these caps and diodes play in more detail as it's interesting, but it's also a  a bit complex.  So instead let me kick off this post with an analysis of a Bruene variant that isn't quite so daunting.

(I'll get back to Bruene's original circuit, but it will be in another blog post -- look for Part 2!).

And before I get any further into this discussion, let me also make this important point:

I will first look at the Bruene Coupler in terms of Forward and Reflected waves. The Bruene Coupler, being made of "lumped elements" (in the first example below: 2 capacitors, 1 transformer, and 1 resistor), is only looking at the voltage and current present at its output port.  It has no idea what the load is, or even how the load is connected to the port.  The load might be a resistor or other component simply clipped onto the output connector with test leads, or it might be a length of transmission line with a load (either known or unknown) at its other end.

Irrespective of what the load actually is (transmission line, clipped-on component, or whatever), the Bruene Coupler gives us voltage readings that can be interpreted in terms of Forward and Reflected waves.  It is important to remember:  these readings should only be interpreted as representing actual Forward and Reflected waves when the Bruene Coupler is connected in a transmission line with the same characteristic impedance, Zo, as the Coupler's designed-for target impedance!

I'll start this analysis assuming the Bruene Coupler is inserted into a transmission line of the designed-for (target) characteristic impedance, Zo.  I will follow that with a look at its operation in a non-transmission line environment.


Bruene Coupler Variant, ZL1AN

ZL1AN has written an excellent article explaining a variation on Bruene's original coupler design, and I strongly recommend you take a look at it.  Tthe Heathkit HM-102 (and I believe also the Drake W4) used this version of the circuit.  I'll simplify the analysis a bit by assuming that the current-sampling is done with an ideal transformer.

Here's ZL1AN's circuit:

(click on image to enlarge)

Let's take this circuit and add a bit more information...

(click on image to enlarge)

In the image above:

1. "V" is the voltage across the transmission line at our "point" of measurement.  We will assume that the voltages are the same at the input and output ports of the coupler -- there is no drop through the coupler.
2. "I" is the current on the transmission line through that point.
3. I've replaced the resistor load (of resistance "R" ohms) across the transformer secondary with two series resistors of value R/2.
4. The voltage at the junction of these two resistors is Vc, because the voltage at the center-tap of the transformer is Vc, and these two resistor form a divide-by-2 voltage divider across the secondary that essentially places their junction at the same voltage as the center-tap.
5. Current "I" through the transformer primary induces a current "Is" in the secondary.
6. Is = I/(2*N), and it flows in the opposite direction of I (per the transformer "dots" that I've shown).
7. We assume that no current flows out either the Vfwd or Vref measurement ports.
8. Therefore there is no load on our capacitive divider's voltage "Vc" (that is, there isn't another path from Vc to ground that parallels the path through C2).
9. Therefore the voltage at Vc is:

Vc =  V*(1/(jw*C2)) / ((1/(jw*C1))+(1/(jw*C2)))

If  1/(jw*C1) >> 1/(jw*C2), this simplifies to: Vc = V*C1/C2

 (click on image to enlarge)

Vfwd and Vref are easily calculate from the series addition of Vc and the appropriate voltage generated by the current sample:

Vfwd = Vc + Is*R/2

Vref = Vc - Is*R/2

Substituting in our equations for V and Is, we get:

Vfwd = V*(C1/C2) + I*R/(4*N)   (equation 1)

Vref =  V*(C1/C2) - I*R/(4*N)   (equation 2)

Analysis Using Forward and Reflected Waves

Let's first analyze this circuit in terms of Forward and Reflected waves passing through our coupler.

The Forward and Reflected waves each have a voltage and a current: Vf and If are the voltage and current of the forward wave, and Vr and Ir are the voltage and current of the reflected (or reverse) wave.

 (click on image to enlarge)

The total voltage on the line, V, at any point is the sum of Vf and Vr at that point.

V = Vf + Vr

And the total current on the line, I, at any point is the difference of If and Ir at that point (they subtract because Ir is flowing in the opposite direction of If).

I = If - Ir

We also know:

If = Vf/Zo

Ir = Vr/Zo

Where Zo is the characteristic impedance of the transmission line.

So, substituting and rearranging, equations 1 and 2 become:

Vfwd = (Vf+Vr)*(C1/C2) + (Vf/Zo - Vr/Zo)*R/(4*N)

Vref =  (Vf+Vr)*(C1/C2) - (Vf/Zo - Vr/Zo)*R/(4*N)

Regrouping terms:

Vfwd = Vf*((C1/C2) + R/(4*N*Zo)) + Vr*((C1/C2) - R/(4*N*Zo))

Vref = Vf*((C1/C2) - R/(4*N*Zo)) + Vr*((C1/C2) + R/(4*N*Zo))

Notice what happens if we select our components such that they meet the following requirement:

C1/C2 = R/(4*N*Zo) = K    (equation 3)

The two equations reduce down to:

Vfwd = Vf*2*K

Vref = Vr*2*K

So, if we select C1, C2, R, and N such that the satisfy the relationship above, then the voltage we measure at Vfwd is solely related Vf, the voltage of the Forward wave, and the voltage we measure at Vref is solely related to Vr, the voltage of the Reflected (or Reverse) wave!


Analysis in a non-Transmission Line Environment:

It's instructional to analyze the operation of the directional coupler just in terms of the components themselves, the voltage applied to the coupler, and the load at its output port without using the concepts of waves.  After all, the coupler consists of lumped-elements, so there's no real reason to think of its operation in terms of waves.

So let's draw our circuit like this, where V is the voltage source driving the input of the Bruene Coupler and Zload is connected to its output.

We'll use the same assumptions that we used above.  Therefore equations 1 and 2 still hold:

Vfwd = V*(C1/C2) + I*R/(4*N)   (equation 1)

Vref =  V*(C1/C2) - I*R/(4*N)   (equation 2)

This time, rather than expressing V and I in terms of waves, we will note that "I" is simply "V" divided by Zload:

I = V/Zload

If we substitute this equation into equations 1 and 2, we get:

Vfwd = V*((C1/C2) + R/(4*N*Zload))   (equation 4)

Vref = V*((C1/C2) - R/(4*N*Zload))     (equation 5)

Now let's take equation 3:

C1/C2 = R/(4*N*Zo) = K    (equation 3)

Let's take that second half:

R/(4*N*Zo) = K

And rearrange it:

R/(4*N) = K*Zo  (equation 6)

Where Zo can be considered to be our "Target" impedance.  Usually this is selected to be the impedance of the transmission line, but it needn't be.

Now let's plug equations 3 and 6 into 4 and 5 and reduce.  We get:

Vfwd = V*K*(Zload + Zo)/Zload

Vref = V*K*(Zload - Zo)/Zload

Well, these are interesting equations.  If Zload equals our target Zo that we selected our components for (e.g. 50 ohms), then Vfwd = V*2*K and Vref = 0.

We could use these equations as they are, but note what happens if we take these two voltages and divide one into the other.  We get something that ought to look familiar:

Vref/Vfwd = (Zload - Zo) / (Zload + Zo)

Which is the definition of Reflection Coefficient!

So we can use the voltages measured at Vref and Vfwd to determine an impedance relationship (i.e. imbalance) between the actual Zload and coupler's "target" impedance of Zo.  And this impedance relationship is exactly the same as the Reflection Coefficient.

SWR is easily calculated from the Reflection Coefficient:

SWR = (1 + |Reflection Coefficient|)  / (1 - |Reflection Coefficient|)

Note that, although SWR implies the presence of Forward and Reflected waves, we have no guarantee that what we measure (or calculate) to be SWR or Forward/Reflected power is actually what is happening in our system.   I'll quote G3YNH:

"[The bridge] can only infer the existence of reflected power from the difference between the actual load impedance and the target load impedance. To understand this point, consider an SWR bridge designed to balance when the load is 50+j0Ω. If we connect this bridge directly to a 100Ω load resistor, it will declare an SWR of 2:1. The resistor is not reactive however, and so will absorb all of the power delivered to it and reflect none. The 2:1 SWR reading is only true when the bridge sees an impedance magnitude of 100Ω (or 25Ω) at the input to a 50Ω transmission line. The bridge is just an impedance bridge, it has no special psychic powers, and its readings are only true when it is inserted into a line having the same characteristic resistance." 

So, summing up:

It should be evident from the above analysis that we don't need to rely on the concepts and Forward and Reflected waves to understand how the Bruene coupler operates.

Our typical "Bruene" SWR meter is really just calculating a relationship between the load at its output port (Zload) and its own design parameters (i.e. C1/C2 = R/(4*N*Zo) = K).  And this relationship is equivalent to the Reflection Coefficient if "Zo" in the relationship: "C1/C2 = R/(4*N*Zo) = K" is the same as the characteristic impedance of the transmission line, should the coupler be connected to a transmission line.

It's  important to note that Zload could be a load connected directly to the coupler's "OUT" port with a couple of wires, or it could be the impedance "presented" to the port by a long length of transmission line (an impedance determined, at that point, by the interaction of the Forward and Reflected waves).

The coupler doesn't care "how" the load is connected to its OUT port.  It's just looking at the voltage across the OUT port and the current through the OUT port.  It doesn't know anything else about the load except for this voltage and current relationship at its OUT port.  For example, if the OUT port happens to be connected to a transmission line, the coupler has no knowledge of the line's Zo. It doesn't even know if there's a transmission line attached, nor that the impedance it sees at its OUT port might be due to the interaction of Forward and Reflected waves.

For this reason, never assume that the meter reading is the actual Reflection Coefficient, Γ, or that the SWR reading is the actual SWR reading of the line.  It might not be.  We are really just measuring the relationship between Zload (as it appears at the OUT port) and the design parameters of the coupler.  Only if the "Zo" in the design relationship "C1/C2 = R/(4*N*Zo) = K" equals the actual characteristic impedance, Zo, of the transmission line would we truly be measuring the Reflection Coefficient.


And on that note, I'll end this non-Transmission Line analysis!


Frequency Insensitivity:

If you refer back to the equations for Vfwd and Vref, you should notice something interesting:  there are no j*w terms (where omega (w) = 2*pi*frequency).  This means that the voltage-divider voltage, Vc, and the voltage generated by the current-sample via the transformer are both constant over frequency.

Of course, in the real world nothing is perfect and there will be effects due to strays and parasitics.  Never the less, if designed correctly (to deal with strays), the frequency response should be flat over a broad range of frequencies.

Which leads to an interesting observation:  If the voltage divider is independent of frequency, why use capacitors?

Well, one doesn't need to use caps, we could just as easily use inductors (whose "jw" terms will cancel), or even resistors!

Which leads me to another variant of the Bruene coupler, which can be found in G3SEK's "In Practice" column in the September, 2002 issue of Radcom...

Bruene Coupler Variants, G3SEK:

One of the coupler's described in G3SEK's column looks very similar to the coupler described by ZL1AN, but there are a few differences:

The first is that a resistive voltage divider replaces the capacitive voltage divider.

The second is that the voltage sample from the resistor divider now feeds the junction of two resistors instead of the transformer center-tap.

I'll skip analysis -- the process is no different that what we've done earlier in this post.

Frankly, I don't know if it's better to feed the voltage-sample to the common-point between two resistors, as done above, or to the center-tap of the transformer secondary, as done by ZL1AN.  Our concerns are:  what is the effect on Directivity, and what is the effect on Frequency Response?


Other Riffs on the Same Theme:

Vc can feed both the resistors and the transformer center-tap.  (Any negative effects?  I don't know.)

The common point between the two resistors could be tied to ground.  But I'm not sure I'd recommend this -- it puts the two resistors in parallel with C2, which means that the frequency response of Vc will no longer be flat.  (The secondary of the transformer acts as an auto transformer, and thus, if Vc feeds its center-tap, it will look like a low impedance (i.e. short) to its two ends.  Which is to say -- it doesn't act as a common-mode choke when feeding the center-).

Here's another interesting variation, found on G3YNH's website (worth a visit!).  A single core is used, but the secondary consists of two windings that are not interconnected.  Thus, two voltage dividers are necessary in order to create voltage samples for the two independent windings.

Analysis is similar to the other variants.  Voltage and current samples add on the left-hand side.  And they subtract on the right-hand side.  I don't know what the advantage is to doing it this way, though.  But one advantage might be that the transformer's secondary, no longer center-tapped, doesn't act like a "shorting" auto-transformer to Vc's common-mode path to ground.  Now, there is some impedance in the path (due to the inductance of each of the secondary coils), but I'm not sure how effective this would be, as it will depend upon the resistance "R" of the two resistors which parallel these coils.

That covers the variants that I've seen that are obviously similar to the ZL1AN topology (and thus will analyze in a similar fashion).  I'll introduce below a few more variants that stray a bit further afield (but not by much).

The models presented thus far are in some cases simplified models of the actual circuits.  I've left off components that might be related to frequency compensation, or detection, or other functions deemed secondary, because I felt they would detract from understanding the underlying theory of operation.  If you're interested in more details, please click on the links I've provided!

As to the positives and negatives for each topology, I wish I had some answers, but I don't. If you have any experience or thoughts on the subject, please feel free to let me know, either via comments to the post or email.

Continuing on...


Bruene Variant, W7EL:

Here's an interesting take on the Bruene Coupler, published by W7EL in the Feb, 1990 issue of QST magazine.

Per the article, this version has +/- 7% accuracy over the range of 1 to 432 MHz.  Quite impressive!

At first glance the design looks similar to the Tandem-Match coupler, but it really is a variant on the Bruene topology; the current sample is either added-to, or subtracted-from, the voltage sample.  (With the Tandem-Match coupler, it's the voltage sample that is either added-to, or subtracted-from, the current sample).

The two transformers to ground create two voltage samples of the same polarity and whose value is V/N, where V is the voltage on the line.

The "series" transformer samples the line current, I, and its secondary generates a current that is I/N in amplitude.

If "I" is flowing from left-to-right in the diagram below (into the 1-turn primary's "dot"), the secondary current runs from left-to-right (out of the secondary's "dot").  This current creates a positive voltage of amplitude I*51/N across the left-hand resistor and a negative voltage (w.r.t. ground) across the right-hand resistor of amplitude -(I*51/N).

The voltage samples generated by the two voltage-sampling transformers are in series with their respective resistors, and thus Vfwd is the sum of the positive voltage across the left-hand resistor and the positive voltage across left-hand secondary, while Vref is the sum of the negative voltage across the right-hand resistor plus the positive voltage across the right-hand secondary.

The only negative that I can see with respect to the design is that you need to wind 3 transformers!


Bruene Variant, N2PK Power Meter:

N2PK cleverly used the differential inputs of the AD8307 Log Amplifier to do the summing and differencing of the voltage and current samples in his homebrew Power Meter.

That's it for the analysis of Bruene variants!  Analysis of the actual Bruene design will follow in Part 2...


Links to my Directional Coupler blog posts:

Notes on the Bruene Coupler, Part 2

Notes on the Bruene Coupler, Part 1

Notes on HF Directional Couplers (Tandem Match)

Building an HF Directional Coupler

Notes on the Bird Wattmeter

Notes on the Monimatch

Notes on the Twin-lead "Twin-Lamp" SWR Indicator

Calculating Flux Density in Tandem-Match Transformers


And some related links from my Auto-Tuner and my HF PA posts:

Auto Tuner, Part 5:  Directional Coupler Design

Auto Tuner, Part 6:  Notes on Match Detection

Auto Tuner, Part 8:  The Build, Phase 2 (Integration of Match Detection)

HF PA, Part 5: T/R Switching and Output Directional Coupler


Bruene Coupler References:

Bruene, Warren, "An Inside Picture of Directional Wattmeters," QST,  Apr., 1959.  Includes both a good explanation of the Monimatch operation and a design for a directional wattmeter whose directional coupler topology would later be known as the "Bruene Coupler."

Collins 302C-3 Directional Wattmeter, PDF Manual containing schematic.

Rush, James, Jr., "The Mini-Mono-Monimatch," QST, Mar., 1965.  Although called a Monimatch in the title, the design actually is more similar to Bruene's directional coupler.

Bold, Gary, ZL1AN, "The Bruene Directional Coupler and Transmission Lines," PDF. This PDF gives an excellent explanation of the Bruene Coupler.

Kiciak, Paul, N2PK, "An HF In-Line Return Loss and Power Meter," PDF.  Constructions details of a power meter using a Bruene Coupler.  Contains an explanation by the other of why he prefers the Bruene coupler of the Tandem-Match Coupler.  Also interesting because the author separates the voltage-sampler from the current sampler and uses the differential inputs of an AD8307 to do the required addition (or subtraction) to get FWD and REF voltages.

(This web page could be useful for understanding the sampling method used in the N2PK meter:  http://www.g3ynh.info/zdocs/bridges/magdiff/part1.html )

Lewallen, Roy, W7EL, "A Simple and Accurate QRP Directional Wattmeter," QST, Feb, 1990, PDF.  Interesting variant of the Bruene coupler.  Roy uses two transformers for the voltage sample in lieu of capacitor voltage dividers.

White, Ian, G3SEK, "Inside a Directional Wattmeter," RadCom, Sept., 2002, PDF.  Discussion and a bit of analysis of Bruene coupler.  Includes Bruene's phase-relationship diagrams.

http://www.g3ynh.info/zdocs/bridges/reflectom/part1.html  interesting analysis

http://www.g3ynh.info/circuits/Diode_det.pdf  Diode detectors -- includes some info on shunt detectors, which is what Bruene's design uses.


Other references of generally interest:

http://www.g3ynh.info/zdocs/bridges/Xformers/part_1.html  great discussion on current-transformers for directional coupler applications

http://www.g3ynh.info/zdocs/bridges/Xformers/part_2.html Part 2 of current-transformers

http://www.g3ynh.info/zdocs/bridges/Xformers/part_3.html  And part 3, the last part, of current-transformers

http://www.g3ynh.info/zdocs/bridges/index.html  Indexes numerous topics.  Lots of great info to be found here!

http://www.richtek.com/assets/AppNote/AN008_EN/AN008_EN.jsp  Common-Mode choke model


Final Caveats:

As always, I might have made a mistake in my equations, assumptions, or interpretations.  If you see anything you believe to be in error, or if anything is confusing, please feel free to contact me.

Notes on Directional Couplers for HF

Many years ago, when I was a student studying Electrical Engineering, one of my professors mentioned that the Standing-Wave Ratio (SWR) on a transmission line could be calculated by first finding (and measuring) a voltage maximum on a transmission line and then moving 1/4 wavelength away and measuring the voltage there.  If there were a standing wave, there would be a voltage minimum at this point.  And these two values (voltage maxima and minima) could then be used to calculate the standing wave ratio:

SWR = |Vmax| / |Vmin|

And I remember thinking...the inexpensive Radio Shack SWR meter that, as a teenager, I'd been using with my ham radio was only about 5 inches long.  Nowhere near the 20 meters (65 feet) of length required if SWR were to be measured on 80 meters the traditional maxima/minima way.  How did it work?  Something else must be going on inside those meters, but what?

Wait a sec...to measure SWR on 80 meters,

shouldn't these connectors be 65 feet apart?

Many years later, I finally decided to dig a little deeper...

On HF, SWR is measured using "lumped-element" directional couplers.  By "lumped-element" I mean that the components of the coupler and their interconnections are all much much shorter than the wavelengths of the frequencies being measured.

There seem to be two main topologies for these lumped-element couplers.  I'll call these two topologies the 1) Tandem  coupler (or bridge), and the 2) Bruene coupler, or bridge (after Warren Bruene, of Collins Radio).

The Tandem coupler topology was popularized for ham radio use by John Grebenkemper in his article, "The Tandem Match -- an Accurate Directional Wattmeter", which appeared in the January, 1987 issue of QST magazine.

(QRPers might recognize this as the "Stockton" bridge, named after David Stockton, G4ZNQ, for his article "A Bi-directional Inline Wattmeter," in the Winter 1989/90 issue of Sprat.)

The Bruene directional coupler is an older design, and it made its first appearance (to my knowledge) in Bruene's article "An Inside Picture of Directional Wattmeters," in the April, 1959 issue of QST.

Here's an example of the Bruene directional coupler:

Being lumped-element couplers, both of these designs operate by sampling the line voltage and line current at a single point (not two points 1/4-lambda apart!).  Both sample the line current with a transformer.  Their difference is how they sample line voltage -- line voltage is sampled with a second transformer in the Tandem coupler, while Bruene uses a capacitive divider to sample this voltage.

In other words, although the two topologies appear different, their fundamental principles of operation are essentially the same.

Let's get a better understanding of how these couplers (or bridges) operate.  I'll focus on the Tandem Coupler:

The "Tandem" Directional Coupler, in a Transmission Line Environment

Before getting into the discussion of the Tandem Coupler in an environment with Forward and Reflected waves, let me make an important point:

The Tandem Coupler, being made of "lumped elements" (in this case, two toroidal transformers and two resistors), is only looking at the voltage and current present at its output port.  It has no idea what the load is, or even how the load is connected to the port.  The load might be a resistor or other component simply clipped onto the output connector with test leads, or it might be a length of transmission line with a load (either known or unknown) at its other end.

Irrespective of what the load is (transmission line, clipped-on component, or whatever), the Tandem Coupler gives us voltage readings that can be interpreted in terms of Forward and Reflected waves.  It is important to remember:  these readings should only be interpreted as representing actual Forward and Reflected waves when the Tandem Coupler is connected in a transmission line with the same characteristic impedance, Zo, as the Tandem Coupler's designed-for target impedance!

Therefore, for this discussion I will assume that the Directional Coupler is inserted into a transmission line of the designed-for (target) characteristic impedance.

 Tandem Directional Coupler (W1QG)

Let's consider a transmission line with two independent signals on it.  One signal (let's call it the Forward signal) is traveling left-to-right on the line, while other signal (let's call it the Reflected signal) is traveling right-to-left.

These signals have voltage amplitudes of Vfwd and Vref, and, because they are independent of one another, their currents Ifwd and Iref are simply the signal voltages divided by the characteristic impedance of the transmission line.  That is:

Ifwd = Vfwd/Zo

and

Iref = Vref/Zo

From the Principle of Superposition, we know that V at any point on the transmission line is simply Vfwd+Vref.  And the total current at any point on the line, I, is equal to Ifwd - Iref (they subtract because the currents flow in opposite directions).

So now let's insert the tandem coupler into the transmission line at that point, as shown in the drawing below.  Note that I'm simplifying the drawing and not including source or load impedances (at either end of the transmission line) for this analysis.   

So what's going on inside the directional coupler?

First, note that the total current, I, flows through L1.  Therefore the current through L2 is I/N, or (Ifwd-Iref)/N, where N is  the turns ratio.

Similarly, the total voltage across L3 is V.  Therefore, the voltage across L4 is V/N, or (Vfwd+Vref)/N.

To see this, first let's define the orientations of the primary and secondary voltages and currents in each transformer, given the "polarity dots" of their windings:

Next, let's assign values to these voltages and currents based upon the transmission line's forward and reflected voltages and currents:

The measurement ports for this device are across R3 and R4.  So let's calculate the voltages across these two resistors.

For simplification of the calculations, I am going to assume the following:

1. The voltage drop across L1 is negligible.
2. The current through L3 is negligible.  

To demonstrate how this coupler works, I am going to assume that the two transformers are ideal transformers.  Also, I would like to use the Principle of Superposition to calculate the voltages across R3 and R4.  But to use this technique, I need to somehow model the transformers as independent voltage or current sources.

Typically an ideal transformer can be modeled as two dependent sources:

I will use the circuit on the left to model the voltage-sensing transformer (with the voltage source's value changed to V1/n, because the 1-turn winding is the secondary of the voltage-sensing transformer).  And I will use the circuit on the right to model the current-sensing transformer.  But I need to convert each two-dependent source model to one-independent source model.

Let's start with this model showing the Tandem Match's two transformers each replaced with two dependent sources:

The first step -- if I assume that the dependent current source representing the primary of the voltage-sensing transformer introduces negligible current into the transmission line (compared with the current on the line if the Tandem Match were not inserted into the line), then I can replace this source with an open (i.e. representing 0 current).

(Note:  Given that the transmission line's current at the point of measurement due to the load's impedance at that point equals V/Zload', where V is the voltage across the transmission line and Zload' is the load impedance as measured at that point, and given that the current added to the transmission line by the voltage-sense transformer model's dependent current source is V/(2*Zo*(N^2)), then the dependent current source has negligible effect on the transmission line's current if 2*Zo*(N^2) >> Zload'.  (I'll derive this equation later in the post).)

And if I assume that the dependent voltage source representing the primary of the current-sensing transformer introduces a negligible voltage drop into the transmission line (compared to the voltage across the transmission line at the point of measurement if the Tandem Match were not inserted into the line), then I can replace this source with a short (i.e. representing 0 volts).

(Note:  And given that the transmission line's voltage at the point of measurement due to the load's impedance at that point equals I*Zload', where I is the transmission line's current and Zload' is the load impedance as measured at that point, and given that the voltage added into the transmission line  (in series with the load) by the current-sense transformer model's dependent voltage source is I*Zo/(2*N^2), then the dependent voltage source has negligible effect on the transmission line's voltage if Zo/(2N^2) << Zload'.  (I'll derive this equation later in the post).)

The figure below shows the two new models:

(Note that the voltage-sensing transformer's model now represents the correct n:1 turns ratio).

The transformers have now been converted to models each containing a single dependent source.  To convert these two dependent sources to independent sources, recognize that in this application, if these dependent sources have negligible impact on  the current or voltage upon which either depends, then neither affects the other, and thus, for all intents and purposes, these two sources are essentially independent sources, and I can proceed with my analysis using the Principle of Superposition.

I will replace the L1/L2 transformer with a current-source and the L3/L4 transformer with a voltage-source, like this:

To calculate the voltages across R3 and R4, I'm going to use the Principle of Superposition and first calculate the contribution of the voltage-source to the voltages across these two resistors (with the current-source replaced with an open circuit, per the Principle of Superposition.

I will then calculate the contribution of the current-source to the voltages across these two resistors, with the voltage-source replaced with a short.

Let's start with the contribution of the voltage-source to the voltages across R3 and R4.  In the figure below I've replaced the current source with an open circuit and I've defined the voltage polarities across R3 and R4.

Before starting this calculation, let's first assume that R3 = R4 (which, in fact, they do).

Thus, the voltage of the voltage-source is equally divided between the two resistors.  So, the "V" contribution to Vr4 is:

Vr4(V) = (Vfwd+Vref)/(2*N)

Similarly, the "V" contribution to Vr3 is:

Vr3(V) = -(Vfwd+Vref)/(2*N)

Note:

1. Vr3's minus sign is because the actual polarity is opposite of the polarity I arbitrarily defined in the picture above.
2. The factor of 2 is because the voltage across each resistor is 1/2 of the voltage-source's value.

Now, let's calculate the contribution of the current-source, "I", to the voltages across R3 and R4.  Per the Principle of Superposition, I first must replace the voltage source with a short, as shown below:

Because R3 = R4, the current divides equally between  the two resistors, and the voltage across each resistor, due to "I", is:

Vr3(I) = -R3*(Ifwd-Iref)/(2*N)

Vr4(I) = -R4*(Ifwd-Iref)/(2*N)

I can now calculate the total voltage across each resistor which, because of the Principle of Superposition, is the sum of the voltages I've just calculated.  Starting with Vr3:

Vr3 = Vr3(V) + Vr3(I)

Substituting, we get:

Vr3 = -(Vfwd+Vref)/(2*N) - R3*(Ifwd-Iref)/(2*N)   (equation 1)

We can do the same for Vr4.  The result is:

Vr4 = (Vfwd+Vref)/(2*N) - R4*(Ifwd-Iref)/(2*N)    (equation 2)

[Sidebar:  Note that equations 1 and 2 allow us to express Vr3 and Vr4 in terms of the line V and I (where V = Va+Vb and I = Ia-Ib, as we defined earlier in this post).  

Vr3 = - (V + I*R3)/(2*N)    (equation 1a)

Vr4 =   (V - I*R4)/(2*N)    (equation 2a)

In other words, Vr3 is an expression in which the voltage-sample and current-sample voltages are summed, while Vr4 is an expression in which the voltage-sample and the current-sample voltages are differenced.]

OK, let's get back to simplifying equations 1 and 2...

Recall that:

Ifwd = Vfwd / Zo

and

Iref = Vref / Zo

where Zo is the characteristic impedance of the transmission line.

If I substitute these into equations 1 and 2, the equations become:

Vr3 = [Vref*(R3 - Zo) - Vfwd*(R3 + Zo)] / (2*Zo*N)

and

 Vr4 = [Vfwd*(Zo - R4) + Vref*(Zo + R4)] / (2*Zo*N)

Still pretty complex.  But we can simplify even further...

We've already defined R3 to be the same value as R4.  Let's also define these values to be the same resistance as the characteristic impedance, Zo, of the transmission line (for example, they could be 50 ohms, if the transmission line were 50 ohm coax).  That is:

R3 = R4 = Zo

This is an important equality.  If R3 and R4 don't equal Zo, the Tandem Match circuit will not correctly calculate forward and reflected voltages.

Substituting this equality into our two equations, the equations reduce to:

Vr3 = -Vfwd/N            (equation 3)

and

Vr4 = Vref/N            (equation 4)

In other words, selecting R3 and R4 to equal Zo (and assuming the voltage drop across L1 and the current through L3 are both negligible), then:

1. The voltage across Vr3 is solely a function of Vfwd (the signal moving from left-to-right on the transmission line).
2. The voltage across Vr4 is solely a function of Vref (the signal moving from right-to-left on the transmission line).

Note:  By measuring Vr3 and Vr4, we now know the voltages Vfwd and Vref.  Thus, we also know Forward and Reverse Power:

P(forward) = Pa = (Vfwd^2))/Zo = (Vr3^2)*(N^2)/Zo

P(reverse) = Pb = (Vref^2)/Zo = (Vr4^2)*(N^2)/Zo

Calculating the effect of the dependent voltage and current sources on Transmission Line voltage and current:

The transformers in the Tandem Match each be modeled as two dependent sources:

Earlier in this post I mentioned that if we assume that the two-dependent source models for our transformers have negligible impact upon the transmission line current and voltage, then we can model each transformer as a circuit containing one independent source.

Let's calculate when this assumption is true.  First, let's define the voltage polarities and the current directions based upon the voltage polarities and current directions defined for the transformers:

Therefore, the current-sense transformer's voltage polarities are:

Recognize that the voltage across the secondary of the current sense transformer (with the voltage-sense transformer removed from the circuit) is simply the current of the model's independent current source times R3 and R4 in parallel, which, because R3 = R4 = Zo, means that the secondary voltage is (I/N)*(Zo/2), which can be expressed as I*Zo/(2N).

If the secondary voltage is I*Zo/(2N), then the primary voltage is simply this value divided by N (because it is a 1:N transformer).

Therefore the voltage inserted by the current-sense transformer's primary into the transmission line is:

Vpri = Vsec/N = I*Zo/(2*N^2)

The voltage across the Zload' (the load impedance seen by the transmission line at the point of measurement, assuming Vpri is negligible) is I*Zload'

Therefore, I can state that Vpri is negligible if I*Zo/(2*N^2) is significantly smaller than I*Zload'.  In other words, for  the assumption to be true, the following condition must be met: Zo/(2*N^2) << Zload'.

Next, the voltage-sense transformer (current directions as defined in the transformer circuit, above):

For the voltage-sense transformer, the current added to the transmission line's current by the dependent current source representing the voltage-sense transformer's n-turn primary winding (i.e. Ipri) is the secondary's current divided by N:

Ipri = Isec/N

Isec can be shown to be (if we ignore the effect of the current-sense transformer) (V/N)/(2*Zo), given that R3 in series with R4 are the load of the voltage-source representing the transformer's secondary (ignoring the current-sense transformer). And, because R3 = R4 = Zo and the voltage source representing the secondary has a value of V/N, where V is the transmission line's voltage at the point of measurement.

Therefore:

Ipri = Isec/N = V/(2*Zo*(N^2))

The transmission line's current, if we assume the effect of Ipri is negligible, equals V/Zload'.

Therefore, Ipri is negligible compared to the transmission line's I if V/Zload' is significantly greater than V/(2*Zo*(N^2)), which is met if (2*Zo*(N^2)) >> Zload'.

Analyzing the "Tandem" Directional Coupler in a "Non-Transmission Line" Environment

As stated earlier, this coupler is actually a "lumped-element" device.  That is it's really looking at the voltage and current relationship at, essentially, a single point on the transmission line.  And we can therefore analyze it as a "lumped-element" circuit.  That is, we can analyze its operation as a circuit consisting of a combination of transformers, L's, C's and R's, and no transmission lines or traveling waves.

The drawing below illustrates this concept.  Zload is the load at the coupler's output port.  Let's assume the connection between the coupler and Zload is very short with respect to the wavelength of the signal being measured (it isn't a transmission line):

Note that I haven't shown a driving voltage-source (nor the source-impedance of this source).  It's actually there, connected to the left-hand IN port.  But all I care about is the voltage level, V, at the IN port.


Let's again simplify the analysis by first assuming that the voltage drop across L1 and the current through L3 are both negligible.

Clearly, if there is no voltage drop across L1, then the voltage across Zload is the same as the voltage at the IN port and across L3's N turns.  Let's call this voltage V.
And therefore current I through L1 is simply:

I = V / Zload.

Because L1/L2 and L3/L4 are both transformers, the current through L2 is I/N, and the voltage across L4 is V/N.

We can substitute ideal current and voltage sources for the two transformers (as we did before), and the circuit now looks like:

Using the Principle of Superposition, we can separately analyze the contributions of the voltage source and the current source to Vr3 and Vr4 and then sum these contributions together.

Doing this, we get:

Vr3 = -V/(2*N) - (I*R3)/(2*N)       (equation A)

and

Vr4 = V/(2*N) - (I*R4)/(2*N)        (equation B)

Let's express these in terms of V.  To do this, first recognize that:

I  = V / Zload

Using this relationship, we get:

Vr3 = -V*(R3 + Zload) / (2*N*Zload)      (equation C)

and

Vr4 = V*(Zload - R4) / (2*N*Zload)       (equation D)

Note that V is an unknown, so let's eliminate it from our equations.  To do this, let's rearrange equation C:

V / (2*N*Zload) = -Vr3 / (R3 + Zload)

Substituting this into equation D, we get:

Vr4 = -Vr3*(Zload - R4) / (Zload + R3)   (equation E)

Well, this is interesting.  Note the term:

(Zload - R4) / (Zload + R3) 

Let's say we've designed the coupler for use in a system with a Zo of 50 ohms. So R3 and R4 should be selected to equal the Zo of 50 ohms, and thus:

Vr4 = -Vr3 * (Zload - 50) / (Zload + 50)   (equation F)

(If Vr4 is negative it just means that it is 180 degrees out of phase from Vr3.)

Note the term:

(Zload - 50) / (Zload + 50)

This is the definition of the Reflection Coefficient, Γ (in which 50 is Zo, the characteristic impedance of a transmission line system that this directional coupler might be used in):

Γ = (Zload - Zo) / (Zload + Zo)  

And therefore:

 Γ = -Vr4 / Vr3    (equation G)

So, in other words, if we normalize Vr4 with respect to Vr3, the resulting voltage equals the Reflection Coefficient!

If you are familiar with the inexpensive SWR meters used for Amateur Radio, this is essentially what they do.  You are normalizing the magnitude of Vr4 with respect to the magnitude of Vr3 when you adjust Vr3 to deflect the SWR meter to full-scale before flipping the switch to measure Vr4.  When you do this:

• Full-scale meter deflection equals a |Γ| of 1.
• Zero meter deflection equals a |Γ| of 0.
• And a mid-scale meter deflection equals a  |Γ| of 0.5

(Note:  the SWR Meter is only using the magnitude of Gamma (the magnitude of Gamma is also known as rho).  There is no angle information.)

How does |Γ| relate to SWR?

SWR = (1+|Γ|) / (1-|Γ|)

So if we know the magnitude of Γ (and we do), we therefore know SWR from the above formula.

Remember that the meter, when set to measure Vr4, measures |Γ| over the range from 0 to 1 when the meter is first adjusted to have Vr3 drive it to Full Scale.  We can create an alternate scale, using the above equation, to give us SWR.

|Γ|    SWR

    0     1.00 

 0.1     1.22

 0.2     1.50

 0.3     1.86

 0.4     2.33

 0.5     3.00

 0.6     4.00

 0.7     5.67

 0.8     9.00

 0.9   10.00

 1.0   infinite

Note, for example, that 50% of Full-scale (|Γ| = 0.5) represents an SWR of 3.

Here's a meter and scale I found on the internet (from PY2OHH).  You can see that its SWR markings follow the table above.  E.g. SWR of 3:1 is 50% of the meter's FS.

Summing up:

From the above discussion, it should be evident that we don't need to use concepts of Forward and Reflected waves to understand how the Tandem Match coupler (and SWR meters based on it) operate.

Our typical SWR meter is really just calculating a relationship between the load at its output port (Zload) and its own terminating resistors, R3 and R4.  And this relationship is equivalent to the Reflection Coefficient if R3 and R4 have the same value as Zo of a Transmission Line.

Zload could be a load connected directly to the coupler's "OUT" port with a couple of wires, or it could be the impedance "presented" to the port by a long length of transmission line (an impedance determined, at that point, by the interaction of the Forward and Reflected waves).

The coupler doesn't care "how" the load is connected to its OUT port.  It's just looking at the relation between the voltage across the OUT port and the current through the OUT port.  It doesn't know anything else about the load except for this voltage and current relationship at its OUT port.  For example, if the OUT port happens to be connected to a transmission line, the coupler has no knowledge of the line's Zo. It doesn't even know if there's a transmission line attached, nor that the impedance it sees at its OUT port might be due to the interaction of Forward and Reflected waves.

For this reason, never assume that the meter reading is the actual Reflection Coefficient, Γ, or that the SWR reading is the actual SWR reading of the line.  It might not be.  We are really just measuring the relationship between Zload (as it appears at the OUT port) and the values of R3 and R4.  Only if R3 and R4 equal the actual characteristic impedance, Zo, of the transmission line would we truly be measuring the Reflection Coefficient.


Continuing on...

Here's something else we could do with Vr3 and Vr4:  if we know Vr3 and Vr4, we could also solve for Zload, our unknown load.  Here's how...

Let R3 = R4 = Rt, where Rt is the value we select for our termination resistors at the FWD and REF ports (they could be equal to Zo, but they could be anything else, too.  It only matters that R3 = R4).

Substituting and solving equation E for Zload:

Zload = Rt *(Vr3-Vr4) / (Vr3+Vr4)    (equation H)

Note that these variables and voltages represent complex numbers -- they have both magnitude and phase.  

The above analysis assumes Vsource and Rsource are unknown (often the case if discussing radio transmitters).  But suppose they are known (e.g. if you're using a signal generator).  We can do a similar analysis:

Lumped-element analysis for a known Vsource and Rsource:

Suppose Vsource and Rsource are known (for example, suppose we are using a signal generator with a known 50 ohm source impedance and a known signal level).  Let's rewrite the equations for Vr3 and Vr4 in terms of these new values.  First, we know that

I = Vsource / (Rsource + Zload).

V =  Vsource * (Zload) / (Rsource + Zload).

Substituting these two equations into Equation A and B from above and solving, we get:

Vr3 = -Vsource*(Zload + R3) / (2*N*(Zload + Rsource))

and

Vr4 = Vsource*(Zload - R4) / (2*N*(Zload + Rsource))

If Rsource = R3 = R4, these two equations simplify to:

Vr3 = -Vsource / (2*N)   (equation 3)

and

Vr4 = Vsource*(Zload - Rsource) / (2*N*(Zload + Rsource))   (equation 4)

What do these two equations tell us?

Vr3 will be constant, independent of load.  That is, Vr3 is solely a function of Vsource.

And, because Γ = (Zload - Zsource) / (Zload+Zsource)  [Note -- this is per Wikipedia.  I prefer to use Zo in lieu of Zsource, but if we assume Zo = Zsource, then they are equivalent.].

Vr4 = Γ*(Vsource / (2*N))

 
So Vr4 is directly proportional to a transmission line system's Reflection Coefficient.  Not surprising.  And if we substitute Vsource = -Vr3 * (2*N) (from equation 3) into the equation for Vr4, then

Vr4 = Γ * (-Vr3)

or

Γ = -Vr4 / Vr3

Which is the same as equation G from our first lumped-element analysis. 

Links to my Directional Coupler blog posts:

Notes on the Bruene Coupler, Part 2

Notes on the Bruene Coupler, Part 1

Notes on HF Directional Couplers (Tandem Match)

Building an HF Directional Coupler

Notes on the Bird Wattmeter

Notes on the Monimatch

Notes on the Twin-lead "Twin-Lamp" SWR Indicator

Calculating Flux Density in Tandem-Match Transformers


And some related links from my Auto-Tuner posts:

Auto Tuner, Part 5:  Directional Coupler Design

Auto Tuner, Part 6:  Notes on Match Detection

Auto Tuner, Part 8:  The Build, Phase 2 (Integration of Match Detection)

HF PA, Part 5: T/R Switching and Output Directional Coupler


Links and Articles for Further Reading:

• Explanation of the Bruene Coupler

• N2PK Power and Return Loss Meter (PDF) -- good article and discussion of Bruene vs. Tandem Couplers

• N2PK Power and Return Loss Meter (web page)

• Ellis -- RF Directional Coupler

• The Tandem Match -- A Closer Look 

• LP-100 Wattmeter -- QEX article by N8LP that contains a good discussion on directional coupler design.

• If you can access the ARRL's site for QST back issues, you can also look up Bruene's April, 1959 article and Grebenkemper's January, 1987 article (and further Tandem match discussion in January, 1988, and July 1993).

• US Patent 2285211, Korman, Radio Frequency Wattmeter, Granted June 2, 1942.  Similar to the Tandem Match coupler

Other notes...

1.  A note from Dick Benson, W1QG: 

One sad, but true, reality is to make one of these work well takes more than the usual amount of black magic.

Strays are everything.   The VNA is indispensable in maximizing performance.  

(VNA = Vector Network Analyzer.  E.g. HP 8505, 8753, 3577, etc.)


2.  3/28/2019.  The comment below was made to a different post in this blog.  I thought it worthwhile to add it here.)

Good day Jeffrey, 

I recently visited your site on this coupler. I was in the process of presenting a class on reflection coefficient measure and decided to build a coupler for lab work for the students. Accomplished at low frequencies, 1 MHz-30 MHz, it is straight forward to obtain a sufficient Vinc and Vrefl voltage to operate and trigger a low cost scope. 


As you present in your documentation, the value of GAMMA is the NEGATION of the usual GAMMA definition. Surprise comes about when you attempt to explain the NORMAL action of the reflected voltage across a short, It should be 180 degrees out of phase with the incident voltage so NO voltage is present across the short. An open circuit is exactly opposite and Vinc should be in phase with Vref. However, for both cases it is the exact opposite for this coupler. 

I think this would be a worthwhile item to mention in the blog, particularly when one decides to build the coupler and actually go in there and measure the PHASE of the voltages; Vrefl and Vinc. A bit of head scratching at first! 

Thanks, Alan Victor, W4AMV

This is a good point: the circuit topology as described above results in Vfwd being proportional to -Vr3 (per the mathematical derivation), rather than +Vr3.  I will note that if it is important to the user that Vfwd be the same polarity as Vr3, this could probably be accomplished by changing the phase of one of the transformer windings (I will leave the analysis to the reader).  

Caveats:

These notes are meant to aid the reader in understanding how directional couplers work.  But this is not a rigorous analysis.  I've made a number of assumptions to simplify the math.  For example, the two transformers within the Tandem Match coupler are assumed to be ideal.  In the real world, they are not.  A complete analysis would take into account, for example, their inductances and the mutual coupling between primaries and secondaries.

Finally, I could easily have made a mistake in my interpretations and/or analysis.  So take everything with a grain of salt.  If there's something here that doesn't look right to you, please feel free to email me.