This morning my study room light popped and thought of throwing some oil into this fire by mentioning this equation:
[image: \operatorname{R}(T) = \operatorname{R}(T_0)(1 + \alpha\Delta T)] copied straight from the all-knowing wikipedia. The full page is here<http://en.wikipedia.org/wiki/Temperature_coefficient> I think we have the same nixie clock Gideon :-) Alex. On Saturday, December 7, 2013 9:43:28 PM UTC, Adam Jacobs wrote: > > Hi Alex, > I think that you are describing the same thing that I tried to describe. > Yes, V/I is substituted in place of R into the equation as you recommend > and your number is the same as what the math produces. It's easy to say > 'why not just' and then proceed with the common sense understanding of an > electronics topic, but that understanding breaks down rapidly when things > get more complicated. Maybe I spelled it out too slowly (and I'm sure my > broken equations in the first email didn't help). :) > Also, normally we talk about Impedance when it is alternating current. > Filaments are normally AC. I think filaments may have a small reactive > component at certain frequencies, but I doubt they do at 60hz. Usually I > don't worry about significant reactive components unless I'm building a new > antenna for transmission. > > -Adam > > On 12/7/2013 2:04 AM, Alex wrote: > > 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+...@googlegroups.com <javascript:>. > To post to this group, send an email to neoni...@googlegroups.com<javascript:> > . > 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. > > > -- 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. 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