It's a bit of a mind game this problem. I was wondering why use rms values instead of average values. In the end, the result should be the same I guess.
I had another go at this problem through another approach. The rms power is defined as the heat dissipated by a resistor due to a voltage V(t) that is the same as if the resistor was powered by a DC voltage. So I thought why not take that as a beginning. In direct drive you want 2mA through each tube. The tube voltage (as measured) is about 145V and power supply voltage is 200V. This leaves 55V across the anode resistor. For direct drive, the dissipated heat in the anode resistor would be 55V * 2mA = 110mW = Pdc. When you go to multiplexed mode, all powers will stay the same because you want to achieve the same tube brightness. Therefore, the loss in the anode resistor will also stay the same. The rms voltage across the resistor is 55V * sqrt(T1/T) with T1/T = 0.267, Vrms will be 28.42V Then we can calculate the resistor as R = (Vrms ^ 2)/Pdc which is (28.42 ^ 2)/0.11 = (surprisingly) 7.3k So it looks like your resistor is correct. The only thing is that the voltage across the tube will slightly increase due to the higher current, so it's not 100% correct but pretty much. This calculation seems ok to me unless I missed something. Michel On Mar 14, 8:10 pm, Imbanon <imba.a...@gmail.com> 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 <mic...@xiac.com> 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 <imba.a...@gmail.com> 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" <bemel...@franktechniek.nl> > > > 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: neonixie-l@googlegroups.com > > > > 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 <n...@desmith.net> 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 neonixie-l@googlegroups.com. > > > > To unsubscribe from this group, send email to > > > > neonixie-l+unsubscr...@googlegroups.com. > > > > 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 neonixie-l@googlegroups.com. To unsubscribe from this group, send email to neonixie-l+unsubscr...@googlegroups.com. 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