# EB104 alternative output circuit

One of the very important designs of HF broadband MOSFET power amplifiers was that of Helge Granberg in Motorola application note EB104.

This article offers an explanation of how the the alternative output circuit at Fig 5 of EB104 works.

Let’s look at the schematic diagram of the PA. Let’s run a simple initial load line model of the amplifier (albeit in tube terminology). Above, we find that if we assume a saturation voltage for the MOSFETs of 0V, a 40.823V supply delivers 600w into exactly 5.555Ω load for ideal Class B. As noted, 5.555Ω is 50/9, so an ideal transformer with 1:3 turns ratio should provide that.

The drain current in an ideal Class B stage is a half wave of sine wave. If we were to perform a Fourier analysis of such a wave of say 1A peak amplitude, we would find that the fundamental RF component is 0.5A. (There are higher order harmonics, but they are not of interest to us in our frequency domain analysis at the fundamental frequency.)

In this case, the calculator results tell us that the drain current amplitude is 29.40A and the amplitude of the fundamental RF current is 14.70A or 10.39Arms. Both FETs are driving such a current but of opposite phase. Above is a diagram of the ideal Class B current wave shape, fundamental RF component and DC component (albeit in tube terms).

Let’s deal with T2, the magnetising impedance is sufficient that we will ignore the current flowing in its windings for an approximate solution.

Now to the alternative arrangment shown in Fig 5.

Granberg offers a simplification of the alternative to T3 as a cascade of a transformer and 1:1 Guanella balun.

Performing a frequency domain analysis of that simplified circuit, we have two current sources supplying a sinusoidal current to the coupling network.

Let’s annotate the diagram with currents relative to the current into the 50Ω load.

We can simplify the analysis by making some assumptions and depending on coax with well developed skin effect in TEM mode:

1. the purpose of the ferrite sleeves is to increase the common mode impedance of the coax section and make its common mode current (the current that flows on the outer surface of the shield) insignificant, we will assume zero;
2. in the interior of the coax, at any point along the coax, the current on the inner surface of the outer conductor is exactly equal to that on the outer surface of the inner conductor but of the opposite phase (ie flowing in the opposite direction); and
3. the coax sections are so short the the current phase is uniform along the section. Above is from Fig 5. Starting at the right we can annotate current into the upper load terminal (the open arrow) and then observing the rules above, we can annotate currents through the whole network, arriving at the conclusion that the current flowing from the current sources representing the MOSFETs is 3I or three times the 50Ω load current, and therefore the impedance transformation is 1:3^2 from source to load, and the drain to drain load impedance is $$Z_{dd}=\frac{50}{3^2}=5.555 Ω$$.