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))
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
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