Building a simple VFD clock with 4x IV-22 tubes. The heaters run at 1 to 1.32v @ 100mA each.
Whats the best way to supply power to the heaters?
All the heaters need to be in parallel so that their potential is the same relative to the grids and segments.
Thats ~1.3v @ 400mA. for 5V supply a resistor would need to dissapate 1/2W. That seems wasteful.
Is it possible to use the MCU to generate a PWM signal to drive the heaters directly? Then how does the current get limited (is P now 2W = 5v * 400mA?) The heater in VFD tubes is both a heater and a cathode, it has to be hot and at 1v potential to emit electrons, right, but not too hot to limit its life?
Maybe there is a way to generate AC waveform to eliminate luminance slant of segments on one side?
Also, as a side question, while I have attracted the attention of VFD knowledgeable people: For internally non-multiplexed tubes such as IV-22 what is the purpose of the grid? When would you ever want to keep it low on a single digit tube? Am I OK just wiring all the grids together into VBB (the 30V supply)?
Update: Thanks for the comments. To clarify further I will be powering this from a low voltage PSU, 12 or 9V DC so I won't have a transformer to tap from. Any voltages higher than supply will be made using Boost DC converters.
Best Answer
A 400mA buck converter would probably be easiest. As below, alter the resistor values to get your desired nominal filament voltage. Simple, cheap and tiny.
You could also whack the filaments with the full input voltage (anodes off) at very low duty cycle but that's kind of ugly and if anything sticks on (such as your micro freezing) your VFD filaments will quickly die. At 9V in to get 1.2V RMS at the filaments your duty cycle would be
\$\frac{t_{ON}}{t_{OFF}+t_{ON}} = \frac{1.2^2}{9^2} \approx 1.5 \% \$