The wonderful Series - Parallel transformation

By Miguel R. Ghezzi (LU 6ETJ - Argentina)
SOLVEGJ Comunicaciones

One of the most complicated areas for the Novice amateur is related with impedance matching; area that embraces question such as the one " Pi " of our old tube transmitter, transmatch, adaptation between the stages of those amplifiers and it is linked with the mysteries of the Hairpin of our directional one.

As well as the laws of Ohm and Kirchoff, together with the Thevenin and Norton's theorem  allow to solve a net or mesh easily, if we can to understand the simple Series - Parallel transformation, we will be able to solve many usual situations on the radio amateur activities, especially  understanding how it is that this works?.

Look the two circuits of the fig. 1. could you measure impedance on terminals of each one?. Surely that if. Now, then it would be possible that do both have the same value of Impedance?. The answer is YES...

This means that circuit of the fig. has an homologous that is the one of the fig. B. We can see that although one is constituted by two series elements, the other one has two elements in parallel, since both it has same impedance these circuits are equivalent.

We will make an exception:
We know that the Impedance of a circuit (unless it is purely resisitivo) it will depend on the frequency, for this, when we made the mensuration we carry out to a certain frequency. If we changed the frequency of mensuration we would find that both no longer have the same impedance, so that the circuits A and B are equivalent on a unique frequency.

The interesting thing is that with a couple of formulas we will always be able to convert a circuit like that of the fig. A in one as that of the fig. B and vice versa, and this it will be very useful to understand the operation of numerous circuits for impedances matching.

If we know the values of a series circuit and we want to discover those equivalent parallel values  we will do with the following one:

Si conocemos los valores de un  circuito serie y queremos averiguar los valores equivalentes en paralelo empleamos la siguiente:

To get the parallel equivalent of two series elements.

If we know the values of a parallel circuit and we want to discover those value equivalent in series we use this other one:

To get the equivalent serie circuit of two elements in parallel.

A practical case:

The circuit of fig. 2A contain one resistance that "accidentally" it is 50 Ohms in series with one 100 Ohms inductive reactance. If we apply the formula it will transform into the parallel circuit of the fig 2B. Now, our equivalent circuit has a 250 Ohms resistance in parallel with a 125 Ohms inductive  reactance. What it would happen if we connect in parallel a -125 Ohms reactance capacitor to this circuit?. Both reactancias, like we know, will cancel out, and it will be visible only the 250 Ohms resistance. Notice that the the inductor in the fig 2D remain physically in series, since those of the figures B and C are only their equivalent ones.
Do you realize that we have achieved? We are able to convert the 50 Ohms resistance in one of 250 Ohms...! Is it or not a matching of impedances?".
This matching net, one of the simplest, is the "L" net. It has a great matching capacity, for example, if we change the value of series inductance from 100 to 500 Ohms, the resistance equivalent parallel that we would get it would be in order of 5000 Ohms, with it we would very easily adapt an end fed half wave dipole to a 50 Ohms line.

Let us see now what it happens if we connect a series capacitor whose reactancia is -100 Ohms (that is to say the same as the reactancia of the previous example, but capacitiva this time ) to the 50 Ohms resistance. Nothing especial, except that now will have a parallel equivalent circuit formed by a capacitor and a resistance. Canceling the equivalent capacitive reactancia with an inductive of same value, again we get a matching. It is also a "L" net, but, in this case, the element in series is a capacitor and the element in derivation an inductor.

Which is the difference among them?

An important difference is that the "L" net with series inductance and capacity in derivation have lowpass filter characteristics, this favors us if we want that matching network helps to cancel harmonic radiations. On the contrary the one that has as series element the capacitor and as element in derivation the inductor, has higpass filter characteristic, fact that we can also take advantage, for example, to avoid broadcasting band signals in the front end amplifier receiver when it is operated in 160 or 80m.

Up to now we have connected to the 50 Ohms resistance a capacitor or an  inductor in series and in both cases the resistance equivalent obtained it was bigger than 50 Ohms  achieving a " raise " impedance transformation. We could think that connecting an element in parallel to the 50 Ohms resistance and finding, this time, their series equivalent, the result would be the inverse one. It is right...! If we apply the formula, we will see that the series equivalent circuit for this circuit offers us a smaller equivalent resistance.


But if we look fig. 3D we see that on 250 Ohms side, (high impedance side),  is the inductor in derivation and on low side impedance the series capacitor; the same as in this example. Is enough obvious, then, we are in front of a symmetrical behavior.

Common applications

The first application that we are thinking is just a transmatch, also the matching inter stage circuit of an amplifier, but an employment of this, not so evident, is antennas matching by means of "Beta Match" also well-known "Hairpin" match...

A several elements directional antenna, usually have an impedance smaller than 50 Ohms on the excited element, and it is convenient to adapt it to transmission line, that generally is 50 Ohms.

Hairpin or Beta Match

As we have seen in the example of the figure 3A, if we connect a capacitor in series with one given resistance, the parallel equivalent resistance increases. The parallel capacitive reactance equivalent that is useful for the transformation can be canceled by an inductor achieving this way the wanted matching. Applying this notion to our problem, we see that:

An antenna whose longitude is smaller than necessary one for its autoresonance it will present a capacitive component on its feeding point, then, intentionally shortening the excited element  we produce an complex impedance similar to that on figure 3A, so that their equivalent one parallel have a 50 Ohms resistance. Canceling the capacitive reactance with the corresponding inductor we get the match. The one "Hairpin" is not another thing that the inductor connected in derivation to achieve the wanted effect, with that which we have found another application not so evident of the "L" net.

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