Good advice ^

This isn't making sense on the power supply front...

The article specifies a 125-0-125 volt transformer. To set that up in LT we calculate the peak voltage which is 176. That's an absolute.

Where does this 120ma come from ? The circuit shows 12 individual stages, all with 11k cathode resistors. Assuming each stage has the cathode at 0 volts then we have a current draw of (100/11000)*12 which is 110ma, OK lets say 120ma all in.

Next up and he says that the circuit is running at -/+100 volts (or 200vdc in total)

So lets now set the simulation up to emulate that. The load has to be 200/0.12 which is 1.666k in total. For our split supply that means 833 ohms per rail.

Check the maths. V supply is 100 volts. I = 0.12 amps R = approx. 833 ohms

Plugging these numbers into the simulation and we need around 91 ohms for the four resistors. They would need to be at least 2 watt rated.

Here are the voltages and the load current (load current scale shown at the right).

If you now edit the simulation (right click the schematic and select

**Edit Simulation CMD**) and change the time it starts saving data to 29.8s then you can look at the ripple.

This is showing the ripple voltage of B+ and the current in R2. You can see that even after 30 seconds that there is still a definite slant to the voltage... its still not stabilised. Try setting the sim for 300 seconds in total and to start saving data at say 299.6. The sim will take a few seconds to run, the progress is shown at the bottom left.