Why are you trying to achieve 8ma of current on IN-14's? Nominal supply
current for that tube is 2.5ma, everything else is providing excessive
current. Now, when multiplexing, lots of times we like to use excessive
current to make the display brighter, but I wouldn't kill myself trying
to achieve precisely 8ma - especially if it is a completely arbitrary
number. In my multiplexed IN-14 designs, I use 180vdc supply and a 1x6
mux. I never overdrive the tubes, I'm pretty happy with the brightness
as is... and the tubes will last a great deal longer. Do as Mike says,
run the experiments and see for yourself.
Also, be careful about taking Michel's advice... Lots of the things he
says seem very strange to me. Do you have a schematic for your design?
If this is a 1x6 multiplex, then you are only lighting one tube at a
time. (or 2x3 is two tubes at at time) I don't follow how he is arriving
at 48ma of supply current. :S
-Adam W7QI
On 3/14/2012 2:10 AM, Imbanon wrote:
IN-14 strike at 170V, but when multiplexed this should be a bit
higher. That's why it's set to 200 volts. It then drops to 140V
according to the datasheet, but in reality, I measured 144. So if I
take 200-140 it's 60 volts across the anode resistor, giving the peak
of 8mA.
But to be honest, I am really confused with this. By my calculations,
with 26.7% duty cycle per tube, for current of 2mA, I should have a
peak of 3.864mA ( 2/sqrt(0.267) = 3.864).
So with my supply stable at 200V and anode resistors of 7.5K, I should
get the 8mA peak on one tube, or 16mA on two tubes, but I really
measure current of 6.4mA alltogether that goes from my supply. How is
this possible? Why should my supply give me 48mA when I need only
6.5mA for two tubes at a time? By the way, I am using blanking period
of 200us, so maybe the current really settles by this time, so the
supply needs to give enough current for only 2 tubes.
Can someone clear this out to me?
And about that spider web.. it isn't really as messy as it looks in
the video. It's just a matter of viewing angle. And everything is
organised by cable color.
Thanks
On Mar 14, 1:24 am, Cobra007<[email protected]> wrote:
Wow, I like that spider web you created there!
How exactly did you estimate that a 7.5k resistor would result in a
8mA tube current? Honestly, I do not know the nominal voltage of the
tube but I don't think it will be less than 150V. In that case, you
have a maximum of 50V across your resistor which would only be 6.7mA.
If you measure 5.5mA, the voltage across the resistor would be 41.25V
so in that case, your resistor should have been between 4.7k and 5.2k
to come to 8mA. My best guess is 4.7k. Try one tube and see if the
value is then closer to 8mA for that tube. Also check that your 200V
stays stable and can supply the required 48mA.
Michel
On Mar 14, 10:56 am, Imbanon<[email protected]> wrote:
Got my hands on some older Tektronix oscilloscope and a Fluke 199c. I
did quite a lot of measurements, even with the current probe. I
learned a lot about the tubes and their behaviour, but didn't really
solve my problem.
I ended up calculating my anode resistors (around 7.5k), that should
give a peak of 8mA, but gives 5.5mA measured with a scope. You can see
the result in the video below. The quality isn't at it's finest, but
it's better than nothing!
Check it out and tell me what you think.
Also, the supply is set to 200V. It that too much?http://youtu.be/p7QNEL8s4l4
Thanks everyone
On Mar 6, 10:10 pm, "Frank Bemelman"<[email protected]>
wrote:
AC DMM’s always excluded the DC component, if I am not mistaken. For a
mainly
troubleshooting tool (citation needed), that is not a bad choice. After all,
many AC signals
found in circuits have a DC offset. Assuming sinewaves makes the design of
the meter
easier (cheaper).
I would not expect a different behaviour from a DMM that is TRUE RMS. Nice
to have
that AC/DC switch though, on the Tek meters. But I’m still a Fluke only guy
;-)
Frank
From: Nick
Sent: Tuesday, March 06, 2012 4:03 PM
To: [email protected]
Subject: [neonixie-l] Re: Calculating multiplexed nixie's RMS current
Yes, RMS has only one physical definition, but in the case of DMMs the
actual implementation is obfuscated.
"true" RMS in a DMM context is an RMS calculation that does not assume a
sine wave - most cheaper DMMs do indeed assume a sine wave input.
Then there are "true RMS" (and indeed "ordinary" RMS) DMMs that may or may
not include any DC component, or at least in the Tek case, give you the
choice.
Old meters indeed did use to measure the heat produced in a resistor - the
definition of the "RMS value" used was that of the DC voltage that would
give the equivalent heating effect to the signal under inspection.
Nick
On Tuesday, March 6, 2012 2:16:45 PM UTC, GastonP wrote: Actually there is
only a definition of RMS, not subject to
"trueness" :)
http://en.wikipedia.org/wiki/Root_mean_square
AFAIK, the old instruments that gave a true-"true RMS" output measured
the heat generated by the signal when applied to a resistor. That way
the waveform shape did not affect the measurement, and they were able
to measure with the DC component included, something fake-"True RMS"
instruments can't do.
Many of the existing instruments assume sinusoidal signals and thus
are subject to gross errors.
Gaston
On Mar 5, 6:15 am, Nick<[email protected]> wrote:> On Monday, March 5, 2012
8:46:42 AM UTC, Cobra007 wrote:
Yes, you're right Nick, the Fluke is indeed AC coupled. I didn't
expect that to be honest as it undermines the definition of "true RMS"
but a simple battery test shows 0V RMS :-).
Its not a commonly known problem, even among professional EEs. One of my
DMMs, a Tektronix DMM916, has the option to include/exclude any DC
component as required. I've had "forthright" discussions with some over
what theoretically constitutes true-RMS vs. what they expect/want in
actuality.
Nick
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