There's a large difference - 3-fold or often much more - between cold resistance and hot in, for example, a nichrome filament. (It was this effect, in fact, that Mssrs Hewlett and Packard used to stabilize the output level of the Wein bridge in the their first product.) In my projects that involve vacuum tubes applied to modern products (Numitron clocks and digital dashes, and guitar effects units), I take great care to limit the inrush current.
The simplest solution is just a big resistor in series with the filament supply, which is at least somewhat effective. A better choice is a current-limited supply. For example, for a vacuum tube filament rated at 300 mA, I used a supply limited to that value. The tube takes somewhat longer to heat up, but its lifespan is greatly increased - the life-limiting factor becomes cathode emissions, rather than filament failure. In a different application driving Numitron display tubes, my firmware keeps the filaments heated to just below the point of visibility when off, bringing them up to the desired current when they're to be visible. This is all done with PWM, because that's easy to do in both software and hardware. This also follows RCA's recommendations for lighting up Numitron tubes (although, considering the era, they recommend using a pull-down resistor on TTL outputs, rather than PWM). I should note that an old clock I built using very rare DTF-104B tubes, with no special handling at all for reduced thermal shock, is still going strong after 10 years as my bedroom clock; for tube filaments, The Numitron products from RCA are very robust, and it's not clear that limiting the inrush current is of great importance. For your project, a LM317-based current limiter would be an easy thing to implement, and that's certainly a possibility for the filament current supply. If you need to come up with a separate supply anyway for the 2.4-volt filament, one option is to make it current-limited as well. The easiest way though, assuming that you're using multiple tubes, is to just hang a resistor in series with each, and drive them off of the 5-volt supply. 5 volts - 2.4 volts = 2.6 volts, at 47 mA. R=E/I, so the necessary resistor would be 2.6/.047, or 55 ohms. The next standard size is 56 ohms. Power is V^2 / R, which is 0.12W - a quarter-watt resistor would work easily. So, short answer: hang a 56-ohm resistor in series with each filament, and run them off the 5-volt supply. (You're supposed to use AC for the filaments for uniform brightness, but in my experience with single-digit tubes like the IV-17, it didn't make a whit of difference.) ~~ Mark Moulding -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+unsubscr...@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/718ed677-70f9-4a07-a1f9-c9de90493b35o%40googlegroups.com.
