Sorry Adam, I sent a reply to you and not the group... I do get your approach now (shouldn't read such things before finishing the morning coffee), but it still seems a slight unnecessary over complication? If the current is specified as 100mA and voltage is 1.5v then surely that gives the manufacturers specified resistance, when running hot. (I still argue its resistance and not impedance due to the lack of significant inductance and capacitance?)
This would give a mfg resistance of 1.5/0.1 = 15ohm? If you treat two of them as one 30 ohm resistor with 3v over it then you can get your 2v drop at 100ma do you not need a 2/0.1 = 20 ohm resistor? A soft start arrangement would be kind, and I am sure I have seen it used on VFD filaments somewhere, but I doubt their resistance changes that much to as require one as they don't get too hot (compared to a incandescent filament...) I would still always run it though a bench PSU with a current limit set, then just play with some resistors (or a wire wound pot / rheostat) until the figures work out... - Alex On Friday, 6 December 2013 19:41:13 UTC, Adam Jacobs wrote: > > Hi Gideon, > You're doing it wrong. :) > We do not normally refer to filaments by their impedance, but rather by > their power draw. What is the equivalent resistor value of a 100watt > incandescent lightbulb? There isn't one, because filaments and resistors > behave differently. For starters, a filament has a much lower impedance > when it is cold than after it has warmed up. Instead, we identify a > filament by the current draw. I don't know offhand what the current draw of > an IV-11 filament is, but the datasheet will have it. I think it is roughly > 100ma. Same thing with the filament voltage, the datasheet will have the > exact number for the real math. I'm going to call it ~1.5v. > Going back to ohm's law, we know that I=V/R. The formula for a resistor > divider without a load is: > R_1=(V_1*R_2)/(R_1+R_2) > > Now, normally there would be load in parallel with R_2. However, in our > case, R_2 *is* the load. > So, our equation becomes: > R_1=(V_1*R_L)/(R_1+R_L) > > We don't know the equivalent load resistor, only the current draw, so we > use substitution algebra: > R_1=(V_1*(V_L/I_L))/(R_1+(V_L/I_L)) > > [image: R_1=(V_1*(V_L/I_L))/(R_1+(V_L/I_L))] > > V_1 is the supply voltage, in my case 5v. V_L (normally V_out) is the > filament voltage, in my case 1.5v. > R_1=(5*(1.5/0.1))/(R_1+(1.5/0.1)) > > Also, the reason you weren't seeing success using those batteries probably > is because you didn't tie them both to a common potential. There is no > assumed potential difference between two separate dry cells, I think. > > -Adam > > -- 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 post to this group, send an email to neonixie-l@googlegroups.com. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/48e8ec1f-7894-4569-a89e-07a0021c4544%40googlegroups.com. For more options, visit https://groups.google.com/groups/opt_out.
