Transistor switch for VFD tube grid

Question

I am designing a on/off switch for the grid of a VFD tube, and I would like to use a NPN transistor. The requirements are:

• The Grid is on when 12V are applied. At this voltage, 0.85mA flow.
• I want to use a LCU with 3.3V logic.

I already have a solution that uses a high side switch (see my other question here), but now I'd like to see if I can spare a transistor.

I ended up with this design:

^{simulate this circuit – Schematic created using CircuitLab}

And here is my reasoning:

1. If 0.85mA flow in the grid when 12V are applied, I can think it as a 14.1 kOhm resistor.
2. When Q1 is cut-off, R1 and R GRID form a voltage partitor. If I assume that 12V is the voltage applied to R GRID, then 0.85mA flow through it (as per datasheet). This mean 0.85 flow through R1 too. So R1 = (15-12)/0.00085 Ohm, and the next standard 5% resistor value is 3.6k So, when Q1 is cut-off, the grid should be on with 12V on it.
3. When the LCU pin is logic HIGH the Q1 enters its saturation region, current flows through it cutting R GRID out (actually is not completely grounded, the grid will see the collector-emitter saturation voltage, that for the 2N3904 should be 0.2V in this situation, but it's anyway too weak to have any effect.).

Now my question: is there any weak point in this design?

Answer 1

I did some tests, and I found the main weak point of this design: the voltage on R_GRID depends on the current flowing through the grid. And unfortunately, this current is not always 0.85mA, as stated in the documentation. It can vary a lot (yesterday I measured 0.22mA for example) and also depends on the filament voltage and current. So although this design is valid from a theoretical point of view, it isn't very compatible with the real world. I opted for a more flexible solution, which uses a 12V zener at BJT the emitter.

Answer 2

For controlled luminescence and negative incremental resistances involved vs temp, and other variables, it is better to use the CC regulator.