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Need help understanding VFD Filament
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Ian.M:
It sounds like you were thinking of grid bias in a small signal amplifying stage, which in most cases were run with the grid negative with respect to the cathode so they would draw negligible grid current.   However, even there one could choose to get the negative grid bias by a DC grounded grid and a resistor in series with the cathode so that the cathode current through it would develop a positive bias on the cathode.  One also usually had the advantage of an indirectly heated cathode with an insulating oxide layer between it and the filament, and as long as the cathode circuit impedance was low compared to the filament-cathode leakage current, that gave you a lot more flexibility in biassing. 

Also it should be noted that the active digit of a VFD runs at a significant positive grid voltage, usually equal to the anode voltage, which provides much of the acceleration needed to get enough K.E. into the electrons to effectively excite the phosphor, but as a result the active grid draws a significant current.
janoc:

--- Quote from: Ian.M on September 14, 2019, 02:09:08 pm ---Also it should be noted that the active digit of a VFD runs at a significant positive grid voltage, usually equal to the anode voltage, which provides much of the acceleration needed to get enough K.E. into the electrons to effectively excite the phosphor, but as a result the active grid draws a significant current.

--- End quote ---

Yup! Going to toot my own horn a bit:
https://janoc.rd-h.com/archives/433

At the bottom are pictures of one of the VFDs I have, that thing takes 600mA easily and that's not even at full intensity, running multiplexed (i.e. only 2 characters are on at the time - each grid is common for a column - two characters stacked on top of each other). If that thing is fully on, it is over 1A of current!

VFDs are pretty power hungry beasts.
dom0:
You can just drive the filament with an H bridge (3ish V filaments are common and go well with the voltage losses of a bipolar transistor H bridge connected to a 5 V supply) and simply switch the polarity after each multiplexing iteration. Doesn't need a transformer, doesn't flicker and has no brightness drop.
Zero999:
0.85V is a very low voltage.

I think winding a custom transformer is the only option. I don't see how a centre tap is that difficult. In the basic circuit I posted previously, it would need to have 2 turns on the secondary and 56 on the primary. The primary centre tapped and each end is driven alternately, so the effective ratio is 28:2.

Another winding and voltage doubler could be added for the HV DC.
janoc:

--- Quote from: dom0 on September 15, 2019, 12:12:53 pm ---You can just drive the filament with an H bridge (3ish V filaments are common and go well with the voltage losses of a bipolar transistor H bridge connected to a 5 V supply) and simply switch the polarity after each multiplexing iteration. Doesn't need a transformer, doesn't flicker and has no brightness drop.

--- End quote ---

And you are totally missing the point of why the center tapped transformer is there. It is not because the engineers don't know how to generate AC voltage across the filament by other means. The transformer is there to have a floating neutral point where to apply a positive bias voltage to the filament so that you don't get segment ghosting from segments that are not completely off.

VFD is a directly heated triode, if the grid is at ground potential you still have some current flowing between anode and cathode, so even the nominally off segments will be faintly glowing. You need to make the grid negative with regards to the cathode to prevent the current from flowing (think depletion mode FET). One way to do that is simply to raise the cathode (filament) voltage above the ground, so when the grid is grounded by the microcontroller, it is negative relative to the cathode, thus repelling the electrons being emitted and blocking the current from flowing.

How do you do that with an H-bridge that isn't floating with regards to the rest of the drive circuit?

And another thing is that a VFD needs a high anode voltage, typically around 40-60V. If you have a transformer there already, that is a matter of adding an extra winding, problem solved. No need to build a separate inverter for that.
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