Return Loss Bridge – some important details

Articles describing how to make a Return Loss Bridge are pretty common, but they don’t often spell out component values that are critical to accuracy.

Above is a schematic for discussion. It is somewhat simplified, but it is complete and will work.

To analyse the circuit, we can use the mesh currents method. Mesh currents i1, i2 and i3 are annotated on the schematic.

The mesh equations are easy to write:

\(
vs=(zs+zref+zref) \cdot i1-zref \cdot i2-zref \cdot i3\\
0=-zref \cdot i1+(zref+zref+zd) \cdot i2-zd \cdot i3\\0=-zref \cdot i1-zd \cdot i2+(zref+zd+zu) \cdot i3\\
\)

This is a system of 3 linear simultaneous equations in three unknowns. In matrix notation:

\(
\begin{vmatrix}i1\\i2\\i3 \end{vmatrix}=
\begin{vmatrix}
zs+zref+zref & -zref & -zref\\
-zref & zref+zref+zd & -zd\\
-zref & -zd & zref+zd+zu
\end{vmatrix}^{-1} \times
\begin{vmatrix}1\\0\\0\end{vmatrix}\\
\)

which we can write in Octave as

\(
\begin{vmatrix}i1\\i2\\i3 \end{vmatrix}=
\begin{vmatrix}
zs+zref+zref & -zref & -zref\\
-zref & zref+zref+zd & -zd\\
-zref & -zd & zref+zd+zu
\end{vmatrix} \setminus
\begin{vmatrix}1\\0\\0\end{vmatrix}\\
\)

Lets solve it in GNU Octave (since we are going to solve for some different input values) for vs=1V.

First pass: zs, zref, zd are 50Ω, and zu is short circuit (1e-300 to avoid division by zero) for calibration and 25Ω for measurement.

zs=50
zref=50
zd=50
#s/c cal
zu=1e-300
A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu]
x=A\[1;0;0]
vcal=(x(2)-x(3))*zd
#zu=25
zu=25
A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu]
x=A\[1;0;0]
vm=(x(2)-x(3))*zd
rl=-20*log10(abs(vm)/abs(vcal))

This gives calculated rl=9.5424dB which is correct.

Now lets change zs to 40Ω.

zs=40
zref=50
zd=50
#s/c cal
zu=1e-300
A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu]
x=A\[1;0;0]
vcal=(x(2)-x(3))*zd
#zu=25
zu=25
A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu]
x=A\[1;0;0]
vm=(x(2)-x(3))*zd
rl=-20*log10(abs(vm)/abs(vcal))

This gives calculated rl=9.0852dB which is wrong.

Now lets change zs back to 50Ω and zd to 40Ω.

zs=40
zref=50
zd=40
#s/c cal
zu=1e-300
A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu]
x=A\[1;0;0]
vcal=(x(2)-x(3))*zd
#zu=25
zu=25
A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu]
x=A\[1;0;0]
vm=(x(2)-x(3))*zd
rl=-20*log10(abs(vm)/abs(vcal))

This gives calculated rl=9.0852dB which is wrong.

So these examples show that accuracy depends not only on the three bridge components being zref, but the Thevenin equivalent source impedance must be zref, and so must the detector zd.

Common failings are:

  • the source is not well represented as a Thevenin equivalent circuit;
  • the Thevenin equivalent source impedance is not zref;
  • the detector does not present a load of zref at the bridge (eg possibly due to imperfections in a common mode choke); and
  • an unbalanced common mode choke.

The Return Loss Bridge is a deceptively simple thing… it does take careful attention to all details to obtain accurate results.