Matching

Gerard Borg
ENGN4545
2007

Matching

In networks we learned that maximum power transfer from source to load occurs when the output impedance of the source is equal to the complex conjugate of the impedance of the load. The output impedance is that appearing in the Thevenin (or Norton equivalent circuits). Here we apply this rule to the design of LC matching networks used for the matching of a source impedance to a load impedance. We only cover a fraction of the matching network possibilities here (for further reading see Bowick).

Matching networks guarantee maximum transfer of power from a source to a load. As a result they appear everywhere even for matching one amplifier stage to another. First however we learn the elementary facts about reactances and resonant circuits.

RESONANT CIRCUITS AND Q

As we saw in the chapter on impedance, L's and C's have internal losses. If a perfect L and a perfect C are connected in series then at resonance their impedance is zero. If they are connected in parallel their impedance is infinite. Clearly neither is the case if either the L or the C has losses. Reactances that have internal losses are often described by the term lossy. Moreover, even perfect L's and C's used in matching networks involving resonant circuits between resistive source and load impedances are loaded by these resistors.

A useful quantity to describe the loading effect on a component is the Q. The Q of a perfect reactance, X, connected in series with a loss resistance R is defined as Q = X/R. If the loss resistance R' is connected across the reactance X' then Q = R'/X'. The physical meaning of Q is exposed through the following formula,

Q = &omega (Energy Stored in X)/(Power lost in R)

By the energy stored we mean the mean oscillating electromagnetic signal energy present in the reactance under excitation while the power lost is dissipated in the resistance.

For a perfect lossless reactance Q is infinite. One way to measure Q at a particular frequency, is to connect the reactance under test in series with a lossless reactance of opposite sign at that frequency and resonate the two by scanning the signal frequency around the centre frequency as shown in the following figure.



The measured curve would look as follows,



The Q is obtained by the formula Q = fo/(f2-f1) where f1 and f2 are the 3dB frequencies or the frequencies where the signal power in the lossy reactance drops to a half on either side of the resonant frequency. The resonant frequency is that at which the reactance looking into any of the above circuits is zero. Thus the bandwidth of the circuit is determined by Q.

Notice that Q is quite a general quantity capable of characterising lumped reaactances such as inductor and capacitor components as well as resonant cavities. It is also a useful quantity. As we shall see Q directly determines the stopband rolloff of certain filters. We now use it to design matching networks.

Ex 9. Derive an expression for the Q of a reactance if R' appears in parallel with X'. I.E. Prove that Q = R'/X'.

Ex 10. Show that R' = (1 + Q2)R and X' = (1 + Q2)X/Q2.


THE L-NETWORK

A common and simple to design matching network is the L network. It employs the minimum number of reactive elements in order to match a source to a load with no other bells and whistles. All the possible varieties are shown in the following figure.



Usually Zs and ZL, the source and load impedances to be matched, are not resistances (but they must contain a finite resistive part). Just the same, the matching reactances Xp and Xs (p for parallel and s for series) are usually opposite in sign (hence the different colours in the figure). i.e. if one is inductive the other is capacitive. The choice between these configurations depends on which of Zs or ZL is the greater. The parallel or shunt element always appears across the impedance with the higher (series) resistance.

To get an idea of how to employ Q to do such a design consider the following example.

EXAMPLE

Suppose that Zs = 100 + j126 Ohms and ZL = 1000 Ohms in parallel with 2 pF. Design a matching network for 100 MHz.

Solution

A sensible choice in this case would be a circuit in the form of the first above with Xs inductive (in this way it can absorb the reactive part of Zs which is inductive) and Xp capaciive so it can absorb the 2 pF. As a result we can do the design for the 100 Ohm and the 1000 Ohm and forget about the reactive parts of the source and load impedance which can be absorbed into Xs and Xp later on.

It is easy to see what we are aiming to do. We will choose Xp to divide down the 1000 Ohms to 100 Ohms (since shunting a resistor with a reactance always reduces its value). Of course this will also produce a reactive part which in this case will be capacitive. The 2 pF will make it even more capacitive. So we then choose Xs to cancel out both this combined capacitive reactance and the inductive (+j126 Ohm) of the source impedance Zs.

Exercise. What if the inductive part of Zs were too large to cancel?

The first trick is the following. If the resistance looking into the shunt pair Xp and ZL has to be equal to the real part of Zs (100 Ohm) and the reactances are equal and opposite (viz equal complex conjugates have equal real parts and equal and opposite imaginary parts), then the Q's of the elements made of Zs, Xs and Zp, Xp must be equal. Using the result derived in Q2, we may therefore write

Q = (1000/100 - 1)1/2 = 3.

A bit of a low value for some applications (but never mind we'll deal with this in a minute. Here we have no choice). Given Q we can immediately determine the magnitude of Xp from Xp = 1000/Q = 333 Ohm hence Zp = -j333 Ohms. At 100 MHz this amounts to Cp = 4.8 pF. Since there is already 2 pF shunting (-j796 W) the load we only need Cp = 2.8 pF. Similarly Xs = Q*100 = 300 Ohms hence Zs = j300 Ohms. At 100 Mhz this is Ls = 477 nH. Since j126 Ohm correeponds to 200 nH at 100 MHz, we only need Ls = 277 nH. We finally obtain the following network...





PI and T NETWORKS


In order to build higher Q networks (higher than the L-network Q) , PI or T-section networks provide a solution. The following figure shows the possible PI configurations. The colours are meant to indicate possible opposite values in the sign of reactance.



Below are the T-networks.



An obvious way to design a PI matching network is shown in the following figure. In this trick, the PI network is decomposed into two back to back L-networks whose Q can be chosen. A virtual resistance R terminates each sub-network in the centre. The idea is that the left circuit is coupled to the resistor R via an L network. The circuit at the right is effectively fed from a circuit of source resistance R via its own L network.



The virtual resistor is used to program the Q as follows,

Q = (Rmax/R - 1)1/2

where Rmax is the maximum of the load or source resistors. R must be smaller than both the load or source resistors because it is connected in a series arm and must have a low value to produce a high Q. Notice that this is the result of Q2 again because both L-networks have the property that R with Re(Zs) on the left and R with Re(Zp) on the right are in series shunt pairs.

EXAMPLE

Consider again the case of the above example. Now design a PI matching network with a Q of 20 that (unlike in the above example) can block the flow of DC.

Solution

In principle, any one of the following circuits could do the job keeping in mind that XS1 or XS2 or both must be a capacitor in order to block the flow of DC.



The reactance values can all be determined in one fell swoop and we can think about their signs later. Firstly, R = 1000/(1 + Q2) = 2.5 W. Once again, forgetting about the 2 pF and the +j126 W, we can say that for the right circuit, Xp2 = 1000/20 = 50 and Xs2 = 20*2.5 = 50. The Q for the left circuit must be given by Q = (Rs/R - 1)1/2 = (100/2.5 - 1)1/2 = 6.25. Therefore Xp1 = 100/6.25 = 16 and Xs1 = 6.25*2.5 = 15.6.

For the first circuit, Xp2 || 2 pF = j50 W . So Xp2 becomes j47. Xs = -j65.6 W and Xp1 = 100/6.25 = j16W. To cancel the +j126 W, insert -j126 W (12.6 pF).

We obtain the following circuit.



Exercises