# Optimal Zo for TLT sections challenge – a solution

Optimal Zo for Guanella balun sections left the reader with a though exercise, a transmission line transformer used by PA0V in a 144MHz power amplifier output network. The pair of tabs to the left are driven by FET drains, the upper pink centre conductor is grounded, the lower end connecting to C1 is the output to a nominal 50R load. The network shown near OUT is for fine load adjustment. There are two coax sections making this TLT, shields bonded all the way around and the centre conductors connected as shown. What is the optimal value of Zo for each the coax sections? Above is a pic of the PA, and we are looking at the network to the right of the dual FET.

## Analysis

The U shaped sections of line are of solder soaked braided coax, they are bonded to each other and to the underlying PCB track. Think of it as a s/c TL stub, it resonates with the output capacitance of the FETs, so it presents a high impedance to common mode current from the drains, ie there is very low common mode current on the stub formed by the shields.

So, the shields are connected to the FET drains, and driven in differential mode and a substantial differential mode current flows on the stub formed by the shields.

Now lets talk about the inner of those coax sections (in differential mode).

They are short electrically, about 30°, so for a first pass lets assume there is insignificant attenuation and phase shift along the lines, and therefore that Vout=Vin and Iout=Iin for each section of line.

Lets call the RF voltage on the lower drain V, and that on the upper drain then is  -V (they are in push-pull).

The differential voltage impressed by the upper drain on the inner conductor of the coax nearest to the board (the end where the inner is bonded to ground) is +V. At the lower end of that coax, assuming Vout=Vin, the shield is at +V potential, and due to the differential voltage in the coax, its centre conductor is at +V++V=2V.

The cross connection of the lower and upper coax sections means that in our simple model, the same current I flows in both sections.

Now lets look at the top coax layer. The cross connection for the lower coax section connects +2V to the inner conductor, and as mentioned already, the shield potential is -V, so the differential voltage on the top coax section is +2V–V=3V. At the other end of that coax section, the shield is at +V as mentioned, and with differential voltage +3V, the voltage on the inner conductor wrt ground is +V++3V=+4V.

This +4V point is intended to be loaded with approximately 50+j0Ω (give or take the small adjustment in the following network), and so we can calculate I=+4V/50.

We have calculated that the differential voltage and current in the lower coax section is V and 4V/50, so Z=V/(4V/50)=50/4=12.5Ω. To minimise transformation due to standing waves, Zo of the lower coax section should be 12.5Ω.

We have calculated that the differential voltage and current in the upper coax section is 3V and 4V/50, so Z=3V/(4V/50)=3*50/4=37.5Ω. To minimise transformation due to standing waves, Zo of the upper coax section should be 37.5Ω.

In fact, there is a small amplitude and phase change on each line section. It depends on length  (~30° in this case) and standing wave ratio, so there is a small departure from the simple model, least if the optimal coax is used. The approximate solution is adequate for selection of appropriate transmission line in this example.

Were I designing this, I would use two parallel 25Ω sections for the bottom layer (12.5Ω together) and a single 35Ω section for the upper layer. This minimises standing waves on those line sections which broadens the matching system. Of course, changing the dimensions of the overall shields may require some change to the other dimensions of the s/c stub so formed.