Hello!

hahaha well, the main reason I want to see this working is generating high
voltage, but, It would be so funny watching it catching fire!

Yes, now I understand that the MOSFET must be on the reverse side, drain
and source.

Gregebert, Am I okay with this quote?


*"To see the Booster working, The PIC needs to send and OUTPUT via RC2 (in
this example) so, this little current gets to the Gate pin of the MOSFET so
it start's going on with the Switching mode (Helped with Q1 BC856), sending
current to the inductor L1 which generates picks of high voltage that are
then stored on the capacitor and later given to the nixie tube?" *

THANKS!

El sáb., 11 jul. 2020 a las 23:45, gregebert (<gregeb...@hotmail.com>)
escribió:

> Well, if the schematic symbol for U1 is correct, this will be a
> "catch-on-fire" inverter, because the internal diode of the MOSFET will
> create a complete circuit thru the battery and inductor, leading to high
> enough current to overheat the inductor or battery.
>
>
> Sarcasm aside, I think the intent is for U1 to be a NMOS with the drain
> and source/substrate connections reversed. Then it becomes a flyback
> converter. It looks like Q1 is in the circuit to provide faster shutoff of
> U1. You would need to run some SPICE simulations to make sure it's working
> as expected, and you would need a good model for the PIC output, such as an
> IBIS model.
>
> This circuit has no feedback loop, so the output voltage will vary wildly
> with load. Theoretically, the output voltage will be infinite with no load,
> and a minimum of around 3V out at maximum load (battery voltage minus
> diode-drop).
>
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