Some time ago I'd determined that the filament of the VFD was being driven with a 5V (pk-pk) square wave. Since square-wave AC is pretty much the same as DC from a power perspective, I connected the VFD filament to my bench supply and slowly brought up the voltage while monitoring the current.
|4.5 Vdc||150 mA|
|5.0 Vdc||170 mA|
|5.5 Vdc||190 mA|
Thus it appears the hot filament has a resistance of about 30 ohms. As I expected, this is quite a bit more than the cold resistance, which measures about 7 ohms.
How is the 5 Vac filament voltage generated? The round, gray device just to the right of the top center of the board is a transformer with three isolated windings. One lead of the primary connects to the unregulated positive output of the power supply (~7.5 VDC under normal load), while the other is driven through a transistor to ground in some manner. A center-tapped secondary winding connects to the two ends of the VFD filament, and the center tap is connected to a Zener diode and a capacitor.
That looks a LOT like the configuration found in the excellent description of VFD operation provided by Noritake-Itron:
That explains two of the transformer's windings; what about the third? This board drives the grids and anodes to +5V to illuminate a segment of a digit, and drives them to -25V to keep the segment or digit dark. I expect that the third winding also provides these voltages, with the filament transformer center tap connected to the -25V through the Zener. I'm guessing the logic also runs from the +5V supply, but that's a guess. Why are there four taps on the transformer? I won't know that until I trace the rest of the circuitry.
Another area to examine is the circuitry driving the transformer. This appears to be a flyback configuration, but I haven't ruled out a forward converter circuit. I also want to determine how the output voltages are regulated, as the unregulated output from the transformer will vary with the line voltage and load.
There's always more to investigate!