# A desk study of the Swan 500CX PA

I recall the arrival of the Swan 500CX in Australia, it was regarded highly and talked up quite forcibly on air by the local agent.

At the time, I was still acquiring the knowledge and skills to analyse the PA design in the 500CX, but I recall lots of on air discussions that were disparaging, but were not convincing.

More recently, I have had occasion to perform a desk study of the 500CX PA. The Swan 500CX used a pair of 6LQ6, low cost TV sweep valves. From the GE datasheet, the valves are rated at 30 W plate (anode) dissipation. No safe grid 1 current or dissipation is given, so the safe approach is to regard that they must be operated with zero grid current, Class AB1 in this case. Above is the anode (plate) characteristics from the datasheet with some overlaid load lines which will be discussed below.

## Swan 500CX manual – CW TUNE

Key details gleaned from various parts of the manual of operation in the TUNE mode, key down, are:

• anode supply (loaded): 720 VDC;
• PEP DC input: 360 W.

Assumptions:

• output network efficiency: 90%;
• conduction angle: ideal Class B (ie idle current=0, perfectly linear); and
• interpreted from the anode characteristic graph, minimum anode – cathode voltage: 50 V.

The implied DC current is $$I_{a_{DC}}=\frac{360}{720}=0.5 \text{ A}$$.

Let’s create a model using those parameters. So, we might expect around 240 W output under those conditions. The load line is drawn in red on the anode graph above (current is per valve). Note that the load line grossly crosses the Ec1=0 (grid1=0) line, grid current will flow, this is not Class AB1 and outside the safe conditions specified in the valve datasheet.

Anode dissipation is 100 W (per valve), three times its continuous rating, so it is prudent to limit key down tuning.

Note that driving grid current might severely degrade linearity if the driver is not designed for that purpose.

A back of the envelope calculation to check the calculator: from the chart, Ip=0.790A (one valve), Vp=680V, so $$P_{a_{RF}}=\frac{I_p V_p}{2 \cdot 2}2=264 \text{ W}$$ which reconciles with the calculated $$P_{a_{RF}}=263.3 \text{ W}$$. Power output is 90% of 264 W, or 238 W.

## Swan 500CX – AB1 CW TUNE

Let us look at a similar model, but with the drive and load adjusted to operate in Class AB1 (safe grid 1 conditions, better linearity). Power output is quite a deal (27%) lower. So, if you were to adjust the PA for maximum output under those conditions, and anode load would be 1174 Ω.

In fact, good practice is to overcouple the output to reduce the risk of valve cutoff on peaks. In practice, the load might be adjusted to more like 1050 Ω (which of course would reduce output power a little), but for this study, let’s stay with the power optimal value of 1174 Ω.

The load line is drawn in blue on the anode graph above (current is per valve).

## Swan 500CX – AB1 SSB telephony with PA tuning

If one changes to SSB telephony, no other changes, the DC supply voltage is higher (less load, less power supply sag), an let’s assume it is 800 V for this discussion. If the PA tune and load controls are left as adjusted in the last configuration (AB1 CW TUNE) then the anode load remains 1174 Ω.

Adjusting the model for those conditions… The load line is drawn in magenta on the anode graph above (current is per valve), it is parallel to the blue line (though it might look a little different due to the accuracy of the scanned chart), same anode load resistance.

In this case, the PEP input is $$P_{input}=800 \cdot 0.4074=326 \text{ W}$$ (somewhat shy of the 550W specification) and $$P_{output}=216 \text{ W PEP}$$.

Though anode dissipation of 86 W per valve, these are PEP conditions and the average dissipation under SSB telephony will be quite a deal lower.

Note that the load line crosses into +ve grid voltage and grid current. The ALC circuit is activated by rectified grid current and should reduce drive so that the grid is not driven +ve for long. The ALC circuit is AC coupled, so is probably not effective in key down CW. The above model shows that ALC limiting the drive in that situation to -ve grid voltage reduces output power to 193 W PEP.

## What is the maximum power available in AB1 SSB telephony?

If you examine the anode characteristic graph, saturation is about 0.58A @ 50V. If you have a 950V supply under lightly pulsed audio drive, Ip=0.580A (one valve), Vp=900V, so $$P_{a_{RF}}=\frac{I_p V_p}{2 \cdot 2}2=261 \text{ W}$$, Power output is 90% of 261 W, or 235 W. Claims of higher power raise the question of whether the signal is distorted.