Cathodyne PI Gain <1 On Load Line Chart?

[robrob]

I know cathodyne phase inverters always have a gain of slightly less than unity of around .96 but I can't figure out how that can be shown on a load line chart.

Here's the load line chart for a 5E3 cathodyne PI with 250V B+3 plate suppy, 56k tail and plate load resistors and 1.5k cathode resistor:



It shows a 1 grid volt swing of 179v - 123v = 56v which is shared between the cathode and plate for 28v each, which implies a voltage gain of 28. How would we show the true gain of .96 on this chart?

[Helmholtz]

Not knowing the answer to your question yet, but have you considered that Eg is not the stage's true input voltage but the difference between grid and cathode potential? The true input is cathode AC voltage + Eg AC voltage and this must be greater than cathode AC voltage, causing a gain <1.

[Malcolm Irving]

The Eg shown on the characteristics is the voltage difference between grid voltage and cathode voltage, Eg = Vg – Vk

Using the AC load-line, if we move to the point where Eg = -1.5 V using the AC load-line we find plate current (=cathode current) = 0.7 mA

So Vk = 0.7 mA x 56 kohms = 39.2 V and Vg = Eg + Vk = -1.5 + 39.2 = 37.7 V

If we swing the input the other way to obtain Eg = -1 V using the AC load-line we find plate current (=cathode current) = 1.0 mA

So Vk = 1.0 mA x 56 kohms = 56 V and Vg = Eg + Vk = -1 + 56 = 55 V

Gain = (change in Vk) / (change in Vg) = (56 – 39.2)/(55 – 37.7) = 16.8/17.3 = 0.97

The only way I can think to show this on the load-line is to put some coloured dots on the line and label them with the values of Vg and Vk at that point.

[R.G.]

You beat me to it, Malcolm.

You're absolutely correct. It's the difference between grid and cathode that makes the current change, and the cathode voltage is flying around at about 0.96 of the grid signal. The plate is merely converting cathode current into voltage by Ohm's law on the plate resistor.

@ others: a split-load phase inverter is best thought of as a cathode follower first, then as a phase splitter. The inverted signal at the plate is a symptom, not a cause.

[robrob]

Thanks Helmholtz, Dr. Irving and R.G., I really appreciate your complete and understandable answers. I finally understand how we can alter the cathodyne's load resistors and get more voltage swing output and yet not increase gain.