Yes, you're right Adam, I have to agree that my previous posts were indeed very strange :-)
It would have been helpful if he had posted a schematic as I mixed up the current per tube and the total current a few times. The thing that is commonly correct is that the power supply's average current (12mA) should be the same in direct drive as well as multiplexed to achieve equal tube brightness. To achieve that, he would need approximately 7.5mA per tube (based on the 33.3 / 26.7 dutycycle). If he uses one resistor per tube, the resistor would be around 6.8k. The common node current through 2 tubes would then be 15mA (which averages to 12mA over the 33.3 / 26.7 duty cycle). Since you're running with a 7.5k resistor already, going down to 6.8k is not going to make a significant difference in tube brightness. If 7.5mA currents per tube will kill the tube, I do not know. This is still a good trick to measure your average power supply current with a DMM: Use a relatively large (and at least 200V) parallel capacitor, then a series resistor (100R), then another relatively large parallel capacitor and then you can measure the voltage across the series resistor with a normal DMM, which should be 1.2V for 12mA. Michel On Mar 15, 2:57 am, Adam Jacobs <[email protected]> wrote: > 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 > >>>> -- > >>>> You received this message because you are subscribed to the Google Groups > >>>> "neonixie-l" group. > >>>> To view this discussion on the web, > >>>> visithttps://groups.google.com/d/msg/neonixie-l/-/cOKZXWW5GXwJ. > >>>> To post to this group, send an email to [email protected]. > >>>> To unsubscribe from this group, send email to > >>>> [email protected]. > >>>> For more options, visit this group > >>>> athttp://groups.google.com/group/neonixie-l?hl=en-GB. -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To post to this group, send an email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/neonixie-l?hl=en-GB.
