You're putting AC across the tube? In addition to what Paul said, think 
about the scenario in the negative half-cycles. There you've got ten 
'anodes' (the display digits, normally cathodes) each with their own 
resistor pouring current through one 'cathode' (the normal anode box/grid), 
which is definitely not sized for that current flow nor designed to have 
the glow on it bombarding its surface with lots of energetic particles. So 
all bets are off on tube behaviour in my view.

In either polarity of the cycle,  the common electrode might be seeing 
22.5mA through it if your initial calculation holds. But very likely it 
doesn't, because that calculation assumes the normal tube maintaining 
voltage which I would have no confidence in being the case under these 
conditions. If the maintaining voltage drops significantly when the tube is 
run like this, then your current flow will be even more than you calculate. 
Maybe that's how you get to 9W.

Ouch. Wouldn't bother putting that tube in a clock!

Jon.


On Wednesday, March 31, 2021 at 10:34:06 PM UTC+1 Paul Andrews wrote:

> You should calculate the anode resistor you need for one segment. Use 
> that, then connect all the cathodes to ground. I have done this many times 
> accidentally.  Now the hand waving part: Imagine the connection between the 
> anode and cathode is a resistor and you connect all of the cathodes 
> together - you are putting all of those resistors in parallel. You are 
> limiting the current on each one to 2.25mA, so you are pumping 22.5mA 
> through the one tube. I get that to be about 2W - (230-140)*0.0225.
>
> On Wednesday, March 31, 2021 at 4:11:49 PM UTC-4 gregebert wrote:
>
>> Very interesting; thanks for posting.
>>
>> You might want to try successive numbers of lit cathodes, say 01, then 
>> 012, then 0123, etc and see how the current increases, and also see if it 
>> changes over time due to heating.
>>
>> I dont recall seeing this behavior with segmented tubes, like the 7971. 
>> In fact, on my clock I have a current regulator on each cathode, and 
>> another one for the anode, for every tube.....that works out to 128 current 
>> regulators on that clock.
>>
>> On Wednesday, March 31, 2021 at 12:18:08 PM UTC-7 Bill van Dijk wrote:
>>
>>> Just a guess, but I think by lighting them all up you get some kind of 
>>> “super ionization” in the tube, which could then increase the current and 
>>> heat dissipation.
>>>
>>>  
>>>
>>> Bill
>>>
>>>  
>>>
>>> *From:* neoni...@googlegroups.com [mailto:neoni...@googlegroups.com] *On 
>>> Behalf Of *Yohan Park
>>> *Sent:* Wednesday, March 31, 2021 3:15 PM
>>> *To:* neonixie-l <neoni...@googlegroups.com>
>>> *Subject:* [neonixie-l] Lighting all digits at the same time: Why does 
>>> this happen?
>>>
>>>  
>>>
>>> I have a few Chinese QS30-1 tubes lying around which don't have much 
>>> value for me and was wondering how it would look to have all digits lit at 
>>> the same time.
>>> So I looked up the specs which say 170V and 2.25mA
>>> So I calculated the needed resistor to have it hooked up to 230V which 
>>> is a little below 27K
>>> I then connected a 27K resistor to one cathode and it lit perfectly fine 
>>> (230V AC so the anode also glows).
>>> So I then connected 10x 27K resistors to all the cathodes and plugged it 
>>> in.
>>> Holy Moly! The thing lit up like crazy and was drawing over 9 Watts and 
>>> was getting VERY hot. So I turned it off again after a few seconds.
>>> Can anyone tell why it's behaving like that?
>>>
>>> -- 
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>>> https://groups.google.com/d/msgid/neonixie-l/bffb114c-9810-4d65-8e98-ee13f6c860fcn%40googlegroups.com
>>>  
>>> <https://groups.google.com/d/msgid/neonixie-l/bffb114c-9810-4d65-8e98-ee13f6c860fcn%40googlegroups.com?utm_medium=email&utm_source=footer>
>>> .
>>>
>>

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