> After spending several days playing with the negative voltages and testing > it, it doesn't work as I was hoping. It seems that a lot of people use AC > instead of DC and I can't find an example that uses DC for negative voltages. > This was the closest I could find, > http://threeneurons.wordpress.com/vfd-stuff/ , but it looks like its AC.
Yeah, that's AC. > I was working with 2 tubes in series with a 30 ohm 2W resistor that has +5V > applied to the first filament and then the other end is grounded, and the > grid/anode was fed +30V with a HV5812 shift register. I'm not quite sure what that lashup is, my best guess is one end of one filament is connected to +5V, the other end to a resistor, the other end of the resistor to the other filament, and the other end of the other filament to ground. That's a kind of silly lashup. I'd connect +5V - filament - filament - resistor - ground. This would put the filaments at a slight positive bias, so grounding other elements would give solid cutoff. You don't really need a resistor when running two IV-17 filaments in series, as they're rated for 2.4 volts, so two of them in series would need 4.8 volts - close enough to 5 volts. If you wanted to use a series resistor to give them more precisely the right voltage, you'd only need 4 ohms. > I put the -27V on the first end and GND on the other end of the filament, it > lit up pretty bright on the filament and I tried with different resistor > values but even the 2W resistors were heating up and the brightness of the > segments wasn't there. You're dropping a lot of voltage there:
Note that you want the resistor connected to ground, and the filament connected to -27V, not the other way around. The IV-17 wants 47mA of current*, which is the same as 0.047 amps. You want to drop the difference between your negative supply (-27 volts) and the filament voltage (2.4 volts), so you'll want the resistor to drop 24.6 volts. Dividing 24.6 volts by 0.047 amps yields 523 ohms. 510 ohms is a common value, and would be close enough. That resistor would be dissipating 0.046 amps times 24.6 volts, or nearly 1.2 watts - enough to make a 2 watt resistor fairly warm! Since your resistor is dissipating more than ten times the power of the filament, and you need to make the filament hot, you're going to burn a lot of power this way (this is one of the reasons some of us like to use AC filament drive - then you can use capacitors to limit the filament power instead of resistors, and capacitors don't just waste power as heat). Ideally, you'd have a supply that's 2.4 volts different from your -27V supply. So if you had a -27V supply and a -24.6V supply, that would work with no resistor. Similarly, a -27V supply and a 29.4V supply. Normally, you'd split the difference and have a -25.8V supply and a -28.2V supply, so the average filament voltage would be -27V, but it's not critical. > I then tried -27V on the first filament and different combinations of -27V > and the +5V with resistors on the other end of the filament and the anodes > were grounded, but nothing was working or it was very dim segments. Going from -27V to +5V only makes the problem worse - now you're trying to get 2.4 volts out of a 32 volt difference. -27V to ground is better. You want to use the voltage that's closest to -27V available (as long as it's >= 2.4 volts different). Since you only have the single negative supply, you're pretty much stuck with either the big watt-burning resistor or going to an AC filament drive. > While I would love to get this working, It looks like I should use a second > MCU to handle the display/boost and another to handle all the other > operations. I don't think another MCU is going to help with this problem. You could use the MCU clock output to derive your AC filament drive, or some I/O pin that you have generating a pulsing signal. * the current is key, that's what creates the heat that makes the filament thermionically emit electrons. - John
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