RF Directional Bridge: Operation versus Source and Detector Impedances

During recent correspondence with Owen Duffy, VK1OD, we discussed RF Directional Bridges and their design.

One question that arose was the effect of the values of the various bridge resistances on a bridge's operation, in this instance the effect of bridge Source and Detector impedances on directional bridge measurements.  

In the various bridge circuits I've seen, Source and Detector impedances (Rs and Rd) seem to ubiquitously be set to Zo (the bridge's reference impedance, typically 50 ohms).  But why?  I could see why this would be the case if one were using an external signal generator with a defined source impedance of 50 ohms as the signal source and measuring the resulting bridge voltage with an external detector having an input impedance of 50 ohms.

But suppose I were designing my own circuit incorporating source and detector with the bridge.  Was it a design requirement that Rs  = Rd = Zo?

As it turns out, no, this is not a design requirement.

Rs and Rd need not equal Zo, the bridge's reference impedance, but their values must satisfy the following equation, assuming all other bridge resistor values equal Zo:

Rs*Rd = Zo^2

This blog post will explain why.

RF Directional Bridge Circuit:

The classic RF Directional Bridge circuit, also known as a Return-Loss-Bridge (or RLB), is shown below:

This circuit is basically a Wheatstone Bridge.  As drawn, above, ZL is the unknown load impedance to be measured, located in the upper right-hand arm of the bridge.  (And although GND is shown at the bottom node of the circuit, it need not be assigned to that node, or to any node at all.)

The bridge is driven by a voltage source Vs with a source impedance Rs (I used Rs in lieu of Zs because it really is a resistive-only impedance).  The source impedance has been set to equal Zo, the bridge's Reference impedance.

The resistors in the bridge arms, R1, R2, and R3, all equal Zo, the bridge's Reference impedance.

Voltage measurement is made across the Detector's impedance, Rd (also equal to Zo).  This measurement is voltage 'Vdet', and its value is proportional to the Reflection Coefficient of ZL, referenced to Zo.

That is,

Vdet = A*(ZL - Zo)/(ZL + Zo)

where 'A' is a constant, and the quantity (ZL - Zo)/(ZL + Zo) represents ZL's Reflection Coefficient, also known as Gamma (Γ).

Directional Bridge Design Requirements:

There are two design requirements for Directional Bridges (that I am aware of):

1.  The "source impedance" looking back into the port to which ZL would be attached (but not currently attached-to) should equal Zo.

2.  The Vdet voltage if ZL = Zo should be 0.  And Vdet should be equal and opposite for ZL values of 0 and Infinity.

The bridge above satisfies these two requirements.  If we remove ZL and calculate (or simulate) the impedance looking into the port to which it had been attached, it equals Zo.  And, given Vs = 1 volt, Vdet is 0v for ZL = Zo, -0.5v for ZL = 0 ohms, and +0.5v for ZL = Open (infinite ohms).

An Equation for Vdet:

To understand how a Directional Bridge's component values affect its operation, we should derive an equation for Vdet that is a function of these components.  We can do this using basic circuit analysis of the circuit, shown below, which results in four equations (also shown below) that we can solve for Vdet:

Pushing pencil on paper to derive an equation for Vdet from these four equations, for me, quickly became prone to error, and so I turned to MATLAB's Symbolic Math Toolbox to ease the pain with the following MATLAB script:

The equation it returned was a complicated one:

vdet = -(Rd*Vs*(R1*R3 - R2*ZL))/(R1*R2*R3 + R1*R3*Rd + R2*R3*Rd + R1*R2*Rs + R1*R3*Rs + R1*Rd*Rs + R2*Rd*Rs + R3*Rd*Rs + R1*R2*ZL + R1*R3*ZL + R2*R3*ZL + R1*Rd*ZL + R2*Rd*ZL + R2*Rs*ZL + R3*Rs*ZL + Rd*Rs*ZL)

Slightly rearranging the equation:

Vdet = (Rd*Vs*(R2*ZL - R1*R3)) /

        (ZL*R1*R2 + ZL*R1*R3 + ZL*R2*R3* + ZL*Rd*Rs +

         ZL*R1*Rd + ZL*R2*Rd + ZL*R2*Rs + ZL*R3*Rs +

         R1*R3*Rd + R2*R3*Rd + R1*R2*Rs + R1*R3*Rs +

         R1*Rd*Rs + R2*Rd*Rs + R3*Rd*Rs + R1*R2*R3)

Our goal is to take the above equation and from it produce a final equation in the form of:

Vdet = A*(ZL - Zo)/(ZL + Zo)

where 'A' is a constant, and the quantity (ZL - Zo)/(ZL + Zo) represents ZL's "Reflection Coefficient", also known as Gamma (Γ).

The numerator Rd*Vs*(R2*ZL - R1*R3) is already in the form of a Constant times (ZL - Zo), if we assume R1*R3 = R2*Zo, in which case the numerator becomes:  Vs*Rd*R2*(ZL - Zo).

Note that if R1 = R2 = R3 = Zo, then the requirement that R1*R3 = R2*Zo is satisfied.

So all we need to do is to whip the denominator into the shape of a second Constant times (ZL + Zo).  You can see that the 16 denominator terms have been grouped into 8 that contain ZL and 8 that don't contain ZL -- the latter will become, in some way, the 'Zo' part of the equation (ZL + Zo).

Arriving at Vdet's Relationship with Rs and Rd, the Source and Detector Impedances:

From LTSpice simulations it appeared that the values of Rs and Rd resulted in a circuit that met the two "Directional Bridge Design Requirements" mentioned, above, if the values of Rs and Rd satisfied the following equation:

Rs*Rd = Zo^2

and with R1 = R2 = R3 = Zo.

But a demonstration with LTSpice is not a mathematical proof.  This goal is easily achieved, though, by substituting Rs*Rd = Zo^2 and R1 = R2 = R3 = Zo into the above equation for Vdet.  Doing so, the Vdet equation reduces to:

Vdet = (Vs*Rd/(2*(2*Zo + Rs + Rd)))*((ZL - Zo)/(ZL + Zo))

We can see that this equation is now in the form of 

Vdet = A*(ZL - Zo)/(ZL + Zo) = A*Γ

where A = Vs*Rd/(2*(2*Zo + Rs + Rd)).

An LTSpice Simulation Example:

Let's let Zo = 50 ohms, Rs = 1 ohm and Rd = 2500 ohms (thus satisfying Rs*Rd = Zo^2), and Vs = 1 volt.  Here are the results of an LTSpice simulation of this circuit:

You can see that the port impedance measured looking into the port to which ZL would attach is 50 ohms, and that Vdet = 0 when ZL = Zo, and +/- 0.481 volts when ZL = Infinite or 0 ohms, respectively.

Taking it to the Limit:

So now we know that Rs and Rd need not always equal Zo.  What happens if we take these two values to their limits by reducing Rs to 0 and increasing Rd to infinity (i.e. we remove Rd)?

The equation for Vdet reduces to:

Vdet = (Vs/2)*(ZL - Zo)/(ZL + Zo)

We will see something interesting if I add a signal-port (and ground) in lieu of ZL and rearrange this circuit by flipping all components (except for the voltage source) upside down to become:

I've replaced ZL with a test port to which a load (ZL) would attach.  If a transmission line were attached to this port, the voltage at the R3/Port node would represent the sum of the forward voltage wave's amplitude (Vf) and the reflected voltage wave's amplitude (Vr) at this point.

That is, the voltage at the test port equals Vf + Vr.

Note that Vf travels out the ZL port onto the transmission line, and Vr, representing reflection of Vf from the far-end of the transmission line, comes into the port from the outside world.

Vr, arriving at the port from the other end of the transmission line, sees R3 as the transmission line's near-end termination.  And so its power absorbed, with no re-reflections, by R3, because the latter's value equals the transmission line's Zo.

The voltage at the R1/R2 node is simply a simulcra of Vf traveling out the ZL port.  Assuming the ZL port is attached to a transmission line of Zo and that there are no re-reflections of Vr from R3 (because R3 = Zo), Vf will always be Vs*R3/(Zo+R3), or Vs/2, given R3 = Zo, irrespective of the value of Vr.

Given that R1 and R2 both equal Zo, the voltage at their common node is also Vs/2, the same as Vf going out the port.  

Vdet is the difference between these two nodes, or 'Vf+Vr' - Vf.  The result is:  Vdet = Vr.

One of the definitions of Reflection Coefficient is  Γ = Vr/Vf.  So if we normalize our measured Vr by Vf (the voltage at the R1/R2 node), we will have measured Gamma, the Reflection Coefficient of the impedance ZL, measured at the test port.

We arrive at the same result if we used our previous equation, derived above:

Vdet = (Vs/2)*(ZL - Zo)/(ZL + Zo)

Substitute Vf for Vs/2 and Γ for (ZL - Zo)/(ZL + Zo).  The resulting equation becomes:

Vdet = Vf*Γ

And therefore:

Γ = Vdet/Vf

One more note regarding the circuit, above:  because R1 and R2 play no role in the impedance seen looking into the test port from the outside world, they can be any value, not just Zo.  The only rule is that their resistance values be the same so that the "simulated" Vf  at the R1/R2 node equals Vs/2.

And why, you might wonder, do R1 and R2 play no role in the port's impedance?  This is simply the result of the Superposition Principle, which states that, when analyzing how a circuit reacts to a voltage or current source, the effects of all other voltage and current sources connected in the circuit should first be removed.

So, to remove the effect of these other source, the other voltage sources are replaced with shorts, and the other current sources are replaced with opens (i.e. the current sources are simply removed).

So, to analyze the impedance that Vr sees as it enters the test port, first short Vs.  Clearly, the two series resistors R1 and R2 are now shorted to ground, and the impedance looking into the bridge's test port from the outside world consists of a single resistor, R3, shunting the port to ground and thus terminating it with an impedance Zo.

Standard Caveat:

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

And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.