V_out = (V_in*(V_out/I_L))/(R_1+(V_out/I_L)) R_1=((V_in*(V_out/I_L))/V_out)-(V_out/I_L) On 12/6/2013 11:59 AM, Gideon Wackers wrote:
1, 5v 100ma is correct. It could be because i am viewing this on my phone but i do not think your formula is going to work. R_1 is before and after the = sign.I will look at it tomorrow morningOp 6 dec. 2013 20:41 schreef "Adam Jacobs" <[email protected] <mailto:[email protected]>>:Hi Gideon, Ā 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. Ā 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)) 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 Ā On 12/6/2013 3:06 AM, Gideon Wackers wrote:I just measured the filament resistance of a IV-11 tube, according to my multimeter that is roughly 5-5,5 ohm (assuming it measures perfectly at such a low resistance). But I calculated something different; I'm going to wire the filaments of two tubes in series and feed them 5 volt though a resistor. So I used the voltage/resistor divider formula. to calculate the resistor needed for a 2 volt drop (tubes want 2*1,5V) and that is 10 ohms, going back to the voltage divider that gives ma a resistance of 15 ohms for both tubes and 7,5 ohms for each filament. Close you might say but still a 50% difference.Ā Options: -my multimeter sucks -filament resistance differs because it is cold -both options -my calculations are wrong Well while I wait for an answer I'm going to build a little jig with lego so my tubes will be aligned perfectly straight :)-- You received this message because you are subscribed to theGoogle Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected] <mailto:[email protected]>. To post to this group, send an email to [email protected] <mailto:[email protected]>. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/4e980192-1b38-42f2-b092-9e62298f2ea3%40googlegroups.com. For more options, visit https://groups.google.com/groups/opt_out.
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