No, it is still not correct because the total power consumption doesn't add up.
The 12mA DC current for the 200V supply is correct because you want 2mA for each tube. This means when multiplexing, a pair of tubes must have a maximum peak current of 12mA, not 6mA. That's where the problem is. 12mA @ 50V would be a 4k resistor. The easiest thing to do measure the DC current that the 200V power supply delivers, if that is 12mA you know that each tube has 2mA average current. Just use a (200V) parallel capacitor, then a series resistor (100R), then another (200V) parallel capacitor and then you can measure the voltage over the series resistor with a normal DMM, which should be 1.2V for 12mA. You probably need a resistor between 3k3 and 4k. Now it should be OK :-) Michel On Mar 14, 11:05 pm, Cobra007 <[email protected]> wrote: > I can smell a misunderstanding here (from my side, that is). > > What are you trying to achieve? I just read your previous posts, it > seems like you're after 2mA average current per tube, so your power > supply should be able to deliver 12mA in total. > > I assume the anodes from the 2 tubes go through 1 resistor to the 200V > supply (2 anodes share 1 resistor). > > The peak current for 2 tubes should be 6mA * (33.3 / 26.7) = 7.5mA > > With 60V across the resistor, you would then need 8k resistors. > > However, it is probably a bit less, maybe more like 6.8k as the > voltage is probably less than 60V. > > I really liked the spiderweb, that was not a joke, I think it looks > good. > > Michel -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To post to this group, send an email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/neonixie-l?hl=en-GB.
