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 view this discussion on the web, visit https://groups.google.com/d/msg/neonixie-l/-/TaNgBU7JeL8J. 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.
