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 :-).
Michel On Mar 5, 6:08 pm, Nick <[email protected]> wrote: > Digital MMs sample and require a periodic waveform to give accurate results > - even so-called "true RMS" DMMs do this. The whole business of what > constitutes a "true RMS" reading is beyond the scope of this note - e.g. > how is any super-imposed DC level incorporated in the calculation - most > so-called true-RMS DMMs are AC coupled for the purposes of RMS > calculations, i.e. they ignore any underlying DC offset. > > Anyway, as the pulses are essentially square waves, the average current > (M:S ratio x peak) is a good first approximation to the RMS value. > > As the tubes are multiplexed, the current waveform will confuse any DMM - I > suspect even good Flukes and Tektronix ones will have the same problem. > > What you need is an analogue MM, like an old AVO or something with a needle > - the analogue movement smooths and averages the pulses - set it on a volts > range and measure the drop over the anode resistor - then measure the anode > resistor accurately using your DMM, and calculate the average current that > way (if the analogue meter you are using isn't a valve or FET-input type, > don't forget to compensate for the input resistance in parallel with the > anode resistor - it'll be something like 20,000 ohms/volt). Otherwise you > will need a 'scope to measure the voltage pulses and do the calculations > from those. > > Lot to be said for analogue MMs... > > Nick > > > > > > > > On Sunday, March 4, 2012 10:26:23 PM UTC, Imbanon wrote: > > > Hi all! > > > I have a question about multiplexed nixies (2x3 - 2 turned on at a > > time). > > First of all.. I cannot get a 2mA (or at least I think so) on my > > IN-14s. It lead me to completely remove the anode resistor! Can a > > nixie tube be harmed if it does not have an anode resistor? Without > > any resistors, I can get up to 1.8mA measured with my multimeter in DC > > mode. > > > So I figured to try to calculate it. I think that the multimeter in DC > > mode shows average readings (that's true, right?). So with the formula > > that the average current equals Vpp*T1/T, in which T1=4ma, T=13.6 and > > Iavg=1.8mA, Vpp equals 6.12mA. > > Is that really possible? I would say that the current would be much > > higher. My 555 supply is capable to deliver at least 15mA at 200V > > (tested). > > So with Vpp I calculated that by the RMS formula Irms=Vpp*sqrt(T1/T), > > RMS current is 3.32mA, which is impossible by my judgement of > > brightness. > > > I will hopefully get my hands on a scope this week to check out the > > real peak current. But is there anything I can do before, or even if I > > get a chance to use a scope? > > > Many thanks! -- 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.
