For measuring the phase of the reflection coefficient in the 360 degree range using the Analog Devices AD8302, the company application note (https://ez.analog.com/cfs-file/__key/communityserver-wikis-components-files/00-00-00-01-03/UsetwoAD8302tomeasure360degrees_2D00_RevA.pdf) proposes using two such devices and applying the signal to one and its quadrature to the other. Since I am using the device in a directional coupler, a reliable way of producing the phase shifted signal is with a second set of windings in the directional coupler transformers. But this change upsets the balance of the two transformers and it requires a more refined approach. So, let's begin.
This is a basic directional coupler with a second set of windings with exactly the same turns ratio connected in the opposite direction to the first winding. We start by writing the voltage and current wave equations for a lossless transmission line:
\begin{equation}V(z)=V ^{+} _{0}(e ^{j\beta z}+\Gamma e ^{j\beta z})\end{equation}
\begin{equation}I(z)=\frac {V ^{+} _{0}}{Z _{0}}(e ^{j\beta z}-\Gamma e ^{j\beta z})\end{equation}
Where:
\begin{equation}\Gamma=\frac {Z _{L}-Z _{0}}{Z _{L}+Z _{0}}\end{equation}
\begin{equation}\beta=\omega \sqrt {LC}\end{equation}
L and C are the line inductance and capacitance per unit length and z is the distance of the source from the antenna and usually a negative number.
Transformer T1 (current sense transformer) primary carries the current to the load (antenna) but has a very small voltage drop across it. So, the secondary can be treated as a high impedance current source where the voltage across it is a function of the resistance in the secondary and the induced current from the primary. Transformer T2 (voltage sense transformer) primary has the full output voltage across it and it induces a voltage on its secondary. So, the secondary can be treated as a low impedance voltage source where the current through it is a function of the resistance in the secondary and the induced voltage from the primary.
Since the circuit is linear, by the principal of superposition, we can first replace T1 with an open (since it is a current source) and write down the voltage equations and then replace T2 with a short (since it is a voltage source) and do the same. The current through the primary of T1 induces a current through each of the secondaries. It is effectively the same as if the secondary had twice as many turns. So, the current through each secondary is one half of the current if there was no second secondary.
Replacing T2 with a short:
\begin{equation}V _{11}=-\frac {R _{m1} R _{n1}}{R _{m1}+R _{n1}} \times \frac {I(z)}{2n}\end{equation}
\begin{equation}V _{12}=\frac {R _{m2} R _{n2}}{R _{m2}+R _{n2}} \times \frac {I(z)}{2n}\end{equation}
\begin{equation}V _{21}=-\frac {R _{m1} R _{n1}}{R _{m1}+R _{n1}} \times \frac {I(z)}{2n}\end{equation}
\begin{equation}V _{22}=\frac {R _{m2} R _{n2}}{R _{m2}+R _{n2}} \times \frac {I(z)}{2n}\end{equation}
Replacing T1 with an ope:
\begin{equation}V _{11}=\frac {R _{n1}}{R _{m1}+R _{n1}} \times \frac {V(z)}{m}\end{equation}
\begin{equation}V _{12}=-\frac {R _{n2}}{R _{m2}+R _{n2}} \times \frac {V(z)}{m}\end{equation}
\begin{equation}V _{21}=-\frac {R _{m1}}{R _{m1}+R _{n1}} \times \frac {V(z)}{m}\end{equation}
\begin{equation}V _{22}=\frac {R _{m2}}{R _{m2}+R _{n2}} \times \frac {V(z)}{m}\end{equation}
Note the distinction that T1 is a current sensor and the total current available to it is the current through the main line as long as it represents a small drain from the line. But T2 is a voltage source and as long as it does not run out of compliance, its current capability is "unlimited".
By the principal of superposition:
\begin{equation}V _{11}=-\frac {R _{n1}}{R _{m1}+R _{n1}} \left [\frac {R _{m1}I(z)}{2n}- \frac {V(z)}{m} \right ]\end{equation}
\begin{equation}V _{12}=\frac {R _{n2}}{R _{m2}+R _{n2}} \left [\frac {R _{m2}I(z)}{2n}- \frac {V(z)}{m} \right ]\end{equation}
\begin{equation}V _{21}=-\frac {R _{m1}}{R _{m1}+R _{n1}} \left [\frac {R _{n1}I(z)}{2n}+ \frac {V(z)}{n}\right ]\end{equation}
\begin{equation}V _{22}=\frac {R _{m2}}{R _{m2}+R _{n2}} \left [\frac {R _{n1}I(z)}{2n}+ \frac {V(z)}{m}\right ]\end{equation}
By substituting the wave equations, we have:
\begin{equation}V _{11}=-\frac {R _{n1}}{R _{m1}+R _{n1}} \left [\frac {R _{m1}}{2n}\frac{V _{0} ^{+}}{Z _{0}}(e ^{j\beta z}-\Gamma e ^{j\beta z})- \frac {1}{m} V ^{+} _{0}(e ^{j\beta z}+\Gamma e ^{j\beta z})\right ]\end{equation}
\begin{equation}V _{12}=\frac {R _{n2}}{R _{m2}+R _{n2}} \left [\frac {R _{m2}}{2n}\frac{V _{0} ^{+}}{Z _{0}}(e ^{j\beta z}-\Gamma e ^{j\beta z})- \frac {1}{m} V ^{+} _{0}(e ^{j\beta z}+\Gamma e ^{j\beta z})\right ]\end{equation}
\begin{equation}V _{21}=-\frac {R _{m1}}{R _{m1}+R _{n1}} \left [\frac {R _{n1}}{2n}\frac{V _{0} ^{+}}{Z _{0}}(e ^{j\beta z}-\Gamma e ^{j\beta z})+ \frac {1}{m}V ^{+} _{0}(e ^{j\beta z}+\Gamma e ^{j\beta z})\right ]\end{equation}
\begin{equation}V _{22}=\frac {R _{m2}}{R _{m2}+R _{n2}} \left [\frac {R _{n2}}{2n}\frac{V _{0} ^{+}}{Z _{0}}(e ^{j\beta z}-\Gamma e ^{j\beta z})+ \frac {1}{m}V ^{+} _{0}(e ^{j\beta z}+\Gamma e ^{j\beta z})\right ]\end{equation}
By setting:
\begin{equation}k _{1}=\frac{R _{m1}}{2nZ _{0}}=\frac{R _{m2}}{2nZ _{0}}=\frac{1}{m}\end{equation}
We get:
\begin{equation}V _{11}=-\frac {k _{1}R _{n1}V _{0} ^{+}}{R _{m1}+R _{n1}} \left [e ^{j\beta z}-\Gamma e ^{j\beta z}- e ^{j\beta z}-\Gamma e ^{j\beta z}\right ]\end{equation}
\begin{equation}V _{12}=\frac {k _{1}R _{n2}V _{0} ^{+}}{R _{m2}+R _{n2}} \left [e ^{j\beta z}-\Gamma e ^{j\beta z}- e ^{j\beta z}-\Gamma e ^{j\beta z}\right ]\end{equation}
\begin{equation}V _{21}=-\frac {k _{2}R _{m1}V _{0} ^{+}}{R _{m1}+R _{n1}} \left [e ^{j\beta z}- \Gamma e ^{j\beta z}+ e ^{j\beta z}+\Gamma e ^{j\beta z}\right ]\end{equation}
\begin{equation}V _{22}=\frac {k _{2}R _{m2}V _{0} ^{+}}{R _{m2}+R _{n2}} \left [e ^{j\beta z}- \Gamma e ^{j\beta z}+ e ^{j\beta z}+\Gamma e ^{j\beta z}\right ]\end{equation}
And after simplification we have:
\begin{equation}V _{11}=\frac {2k _{1}R _{n1}V _{0} ^{+}}{R _{m1}+R _{n1}}\Gamma e ^{j\beta z} \end{equation}
\begin{equation}V _{12}=-\frac {2k _{1}R _{n2}V _{0} ^{+}}{R _{m2}+R _{n2}} \Gamma e ^{j\beta z} \end{equation}
\begin{equation}V _{21}=-\frac {2k _{2}R _{m1}V _{0} ^{+}}{R _{m1}+R _{n1}} e ^{j\beta z} \end{equation}
\begin{equation}V _{22}=\frac {2k _{2}R _{m2}V _{0} ^{+}}{R _{m2}+R _{n2}} e ^{j\beta z}\end{equation}
And dividing the respective equations we get:
\begin{equation}\frac {V _{11}}{V _{21}}=-\frac {k _{2}R _{m1}}{k _{1}R _{n1}}\Gamma \end{equation}
\begin{equation}\frac {V _{12}}{V _{22}}=-\frac {k _{2}R _{m2}}{k _{1}R _{n2}}\Gamma \end{equation}
By substituting k1 and k2 in these equations, with minimal math we conclude that:
\begin{equation}\frac {V _{11}}{V _{21}}=-\Gamma \end{equation}
\begin{equation}\frac {V _{12}}{V _{22}}=-\Gamma \end{equation}
But, there are second condition that need to be met which are:
\begin{equation}\frac{R _{m1}}{2nZ _{0}}=\frac{R _{m2}}{2nZ _{0}}=\frac{1}{m}\end{equation}
We have two obvious choices. Either:
\begin{equation}R _{m1} = R _{m2} =R _{n1} = R _{n2} = Z _{0} \ \ and\ \ n = \frac {m}{2}\end{equation}
Or:
\begin{equation}R _{m1} = R _{m2} = R _{n1} = R _{n2} = 2Z _{0}\ \ and\ \ n = m\end{equation}
The choice is terminating the transformers in 100 ohms and keeping the turns ratio of the two transformers the same or cutting the turns ratio of T1 in half and terminating the transformers in 50 ohms. The impact on the primaries is the same, so the decision rests on the impact on the secondary circuits. Clearly, building a transformer with half as many turns is simpler. Also, since most systems are 50 ohm systems, any downstream attenuator need to have an input resistance of 100 ohms and output resistance of 50 ohms. These attenuators are not realizable in general for different values of attenuation with real passive devices (require negative resistance).
The choice is made, 50 ohm termination (a Pi network attenuator with a 50 ohm input and output impedance) and T1 secondary with half the ratio of T2 primary.
An interesting byproduct of the two windings in opposite direction is that in the case of T1, the voltage reflections and in the case of T2, the current reflections into the primary are cancelled. Hence, there is no impedance reflection into either primary.
One last point, the negative sign in the equation for Gamma is important but since the measurement is made in software, in case of error, it is easily correctable, not that I am planning on a mistake.
No comments:
Post a Comment