Calculating multiplexed nixie's RMS current

[Imbanon]

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!

[Adam Jacobs]

Disable the multiplexing, so that only 2 nixies are lit (and those lit
at 100% duty cycle). THEN measure the current draw. You will not be able
to accurately measure the peak current draw of a multiplexed nixie with
a multimeter. . and YES, you definitely need a current limiting resistor
or you will break your nixies.

-Adam

[Imbanon]

I direct drived all 6 of them, delivering 12mA without a problem.
Tested only up to 15mA. Didn't want to go further..

Thanks again

[Cobra007]

If you direct drive them @ 12mA and you want to have them the same
brightness when multiplexed, you will need to increase the current to
36mA (as they are only on for 1/3 of the time). Which means 12mA for 2
tubes at all times, 6mA per tube @ 33% duty = 2mA average current per
tube.

Adam is right, you cannot measure that accurately with a DC
multimeter. I even tried with a Fluke true RMS multimeter and that
gives rubbish readings as well.

Michel

[Nick]

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

[Cobra007]

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

[Nick]

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

[GastonP]

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

[Nick]

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

[Instrument Resources of America]

Actually the signal was applied to a "thermocouple" which was "one
arm" of a bridge circuit. An "identical" thermocouple, which was a
"second arm" in the same bridge circuit, then had D.C. applied to it. A
null was then achieved across the bridge, and the meter actually
measured the D.C. being applied. A very simple and elegant solution to
the measurement issue of both wave shape and frequency.
RMS is the A.C. voltage, regardless of wave shape, that will
produce the ""same heating effect"" in a PURE resistance (think here of
an incandescent lamp filament), as an EQUIVALENT amount of D.C. Applies
to current as well. As an example a 100V RMS wave shape, will produce
the same heating effect, in a pure resistance, as 100V D.C. RMS means
"Root Mean Squared" of which there is a mathematical way of calculating
it. It actually is a mathematical process. If you want more I'd suggest
looking it up on Wikipedia, or other places on the net. Hope that this
helps. Ira.

[Frank Bemelman]

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

Nick

http://en.wikipedia.org/wiki/Root_mean_square

Gaston

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[Imbanon]

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:

> To view this discussion on the web, visithttps://groups.google.com/d/msg/neonixie-l/-/cOKZXWW5GXwJ.

[Cobra007]

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

[Imbanon]

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

[Cobra007]

I can smell a misunderstanding here (from my side, that is).

What are you trying to achieve? I just read your previous posts, it
seems like you're after 2mA average current per tube, so your power
supply should be able to deliver 12mA in total.

I assume the anodes from the 2 tubes go through 1 resistor to the 200V
supply (2 anodes share 1 resistor).

The peak current for 2 tubes should be 6mA * (33.3 / 26.7) = 7.5mA

With 60V across the resistor, you would then need 8k resistors.

However, it is probably a bit less, maybe more like 6.8k as the
voltage is probably less than 60V.

I really liked the spiderweb, that was not a joke, I think it looks
good.

Michel

[Cobra007]

No, it is still not correct because the total power consumption
doesn't add up.

The 12mA DC current for the 200V supply is correct because you want
2mA for each tube.

This means when multiplexing, a pair of tubes must have a maximum peak
current of 12mA, not 6mA. That's where the problem is.

12mA @ 50V would be a 4k resistor.

The easiest thing to do measure the DC current that the 200V power
supply delivers, if that is 12mA you know that each tube has 2mA
average current.

Just use a (200V) parallel capacitor, then a series resistor (100R),
then another (200V) parallel capacitor and then you can measure the
voltage over the series resistor with a normal DMM, which should be
1.2V for 12mA.

You probably need a resistor between 3k3 and 4k.

Now it should be OK :-)

Michel

[Adam Jacobs]

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

>>>> 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

[Cobra007]

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

> >>>> 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

[dr pepper]

Dmms tend to use charge balance techniques for measurements and these
can be way off at ac esp at higher frequencies.

A cheap old analogue meter would probably give you a better approx
reading.

Another way I have used is just supply the thing from a bench power
supply and look at the current draw difference between no tubes on and
one or more on, either multiplexed or not, the power supplys (your
clocks supply that is) evens out the pulses into a more constant draw
and provides a good measurement point, you just need to to subtract
the current pulled by all the support electronics.

> > >>>> For more options, visit this group athttp://groups.google.com/group/neonixie-l?hl=en-GB.- Hide quoted text -
>
> - Show quoted text -

[Adam Jacobs]

Is there a reason why the nixie can not be placed in a non-multiplexed
state for long enough to take the reading? I'm a software guy, so I
would just bodge a version of the code that turned on a single nixie and
left it on. Then I'd measure the current with a DMM. Ta-da. :) Done. The
only gotcha is if you're running a ton of current (like 8ma) then you'll
pretty promptly cook the nixie. One more reason for going with lower
current.

-Adam

[Cobra007]

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

[Cobra007]

Interestingly, I just realize, if you work this further out you come
to the following formula:

Rmux = Rdc * (T1/T)

Rdc is the anode resistor in direct drive (55V / 2mA = 27.5k)
Rmux is the anode resistor in a multiplexed system = 27.5k * 0.267 =
7.3k :-)

Michel

[Charles MacDonald]

On 12-03-15 05:46 AM, Cobra007 wrote:

> 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.

Since we are talking a Neon device, the voltage across the tube will try
to stay the same, with the current adjusting if needed. That is why
Neon bulbs were used as Voltage reference devices in days of Old.

--
Charles MacDonald Stittsville Ontario
cm...@zeusprune.ca Just Beyond the Fringe
http://users.trytel.com/~cmacd/tubes.html
No Microsoft Products were used in sending this e-mail.

[Cobra007]

Yes, I think I mentioned "slightly" increase rather than a mayor
increase.

I have measure this on another nixie tube and came to the following
voltages:
0.5mA : 120V
1.0mA : 125V
1.5mA: 130V
2.0mA: 133V
3.0mA: 140V
4.5mA: 150V

His tube current will increase from 2mA to about 7.5mA, so according
to the above measurements, the increase in tube voltage will be
playing a role.

Michel

On Mar 16, 10:06 am, Charles MacDonald <cm...@zeusprune.ca> wrote:
> On 12-03-15 05:46 AM, Cobra007 wrote:
>
> > 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.
>
> Since we are talking a Neon device, the voltage across the tube will try
> to stay the same, with the current adjusting if needed.  That is why
> Neon bulbs were used as Voltage reference devices in days of Old.
>
> --
> Charles MacDonald                 Stittsville Ontario

> cm...@zeusprune.ca              Just Beyond the Fringehttp://users.trytel.com/~cmacd/tubes.html

[Dekatron42]

Many manufacturers write that you will have to contact them for the
special curves you need when you are going to multiplex their Nixies
since they do not usually print that information in the databooks.

These sheets show you that the Nixie will have an increased turn-on
voltage corresponding to the increased current when run in switched
mode. This is the same as when a neon voltage stabilizer tube is used,
the voltage increases somewhat when the current increases, you can
check the OB2 voltage regulator tube for instance.

The curves for most Nixies when used in multiplexed mode are not
linear so if you can't find those curves you'll have to make the
measurements yourself and take into account the spread between
different Nixies to draw the curve. Some of these special curves have
a voltage span of approximately 10-30V for a certain current through
the Nixie, so there is an upper and a lower limit for the turn-on
voltage corresponding to the current used.

This book: http://www.oldtimeradio.de/BU7908.php "Electronica 171 -
Elektronische Anzeigebauelemente" by Winfired Müller contains a few of
these curves for the ZM-series of Nixies.

/Martin

[Cobra007]

That makes sense indeed.

What would the reason be to choose for a higher voltage rather than a
lower anode resistor? You can achieve a higher tube current by either
raising the voltage or lowering the resistor, so what is the advantage
of raising the voltage? Is it because ionization will be quicker or
doesn't have that anything to do with it?

Michel

[Dekatron42]

It all has to do with how you design the circuit around the Nixie, in
the old days the transistors did not have perhaps more than 120V Vce
as a maximum, usually lower than that, around 80-100V. With a lower
Vce you have to use either a cathode bias voltage to meet the
transistor requirements or youll have to use both a positive and
negative voltage to get the bias which the transistor would work with.
Then if you lower the resistor or raise the voltage is up to what you
want to achieve and what transistors you have at hand. It also has to
do with what turn-on time (ionization and de-ionization) that you want
to acheive. Some of the Nixies datasheets for multiplexing specify a
100uS pulse as the shortest ionization pulse width in these datasheets
and roughly 200-400uS for de-ionization pulse width this was also due
to how the transistors were able to switch these voltages and not only
due to the Nixies. Modern transistors are able to switch these
currents and voltages faster than the older types so you could
possibly have shorter pulses which acheives the same effect. The most
common way of looking at the pulses and currents/voltages involved, by
these books and datasheets, is by using the average voltage/current
and not the RMS values since the pulses are square pulses and not
sinusoidal waves or triangular waves. Here's a nice sheet showing RMS,
Mean and their relationships: http://www.yokogawa.com/gmi/pdf/Bulletin/Bull-DMMglossary-E.pdf
.

There are quite a few examples in the book I mentioned above on how
the Nixies were designed to be driven in those days. This is not the
only book showing how this is suppsed to be done, but all that I have
found have been written in German, there are possibly some english
books to which show these details, probably from Philips or Valvo as
they published a lot of design examples. I have been quite happy with
the German books as they are easier to find and I guess were also
printed in larger quantities and not so collectable as the English
ones, which I guess is the reason why they are harder to find.

/Martin

[John Rehwinkel]

> What would the reason be to choose for a higher voltage rather than a
> lower anode resistor? You can achieve a higher tube current by either
> raising the voltage or lowering the resistor, so what is the advantage
> of raising the voltage? Is it because ionization will be quicker or
> doesn't have that anything to do with it?

Ionization will be quicker with a higher voltage. Mike M. (threeneurons) actually went to the
trouble to measure the turn-on behaviour of a nixie:

http://threeneurons.files.wordpress.com/2011/08/timing01.jpg

I'm thinking of duplicating his setup and trying it with different anode voltages and a few
different nixies (I'm guessing little nixies ionize faster). I may also try using a decimal point
as a "primer" and even comparing nixies with and without ambient light.

Another advantage of higher voltage is better current regulation - since more of the drop
is across the anode resistor and less across the (dynamic, negative, varies with the digit lit)
resistance of the nixie, the current for different digits will be more similar.

A disadvantage of higher voltage is lower efficiency - more of your voltage is used up in
the anode resistors as heat. This can be combated by using a current-limited supply, which
can give wonderful efficiency but adds a good deal of complexity (current limiting a negative
resistance load which is being continually switched ain't easy). Alternatively, a scheme like
David uses in his watches (raising the voltage to strike, then lowering it) strikes a different
balance between ionization time, efficiency, and complexity.

Another disadvantage is the voltage impressed on the "off" cathodes if the anode voltage is
turned on but the digit is blanked. This can lead to ghosting (especially if the transistors are
zener clamped or leak at high voltage) or failure of the cathode switches if they can't withstand
the voltage. An obvious way to address this is don't enable the anode switch for blanked
digits, which requires a few wiring and/or software considerations, depending on the design.

- John

[Imbanon]

First of all, thank you all for your support. Feels great to have some
people with knowledge behind my back.
So many replies since I had time to check the group last time that I
don't know where to start :)

I do not have a schematic for my design, as it is my own design that I
pretty much pull out of my head as I go. That often shown like a bad
idea, making me to change a lot of things afterwards. Just like today,
I had to rewire the whole 'spider web'. And that was the second time I
had to do it!
And I have to clear out that I have a common anode resistor for 2
tubes, making a total of 3 anode resistors for all 6 tubes. That means
that the current has to be double (resistance cut by half). I hope
that it now explains the rounded 8mA (7.74 to be more precise) on the
anode resistor. It cuts to two nixies, giving appox 3.87mA peak
current to each tube. So then 3.87*sqrt(0.267)=2mA RMS
Anyhow, I do believe that I have to use the RMS values when working
with multiplexed designs, rather than average. Just like someone
already explained why, because of the power dissipation. I really hope
it is that way :)

Well that's all for now. I hope that my setup with anode resistors is
finally done. If not - I still have another week to use all the fancy
expensive oscilloscopes! So if anyone doesn't agree with this, please
say the word :)

Cheers

[Imbanon]

Oh and I accidentaly reported your post as spam. Sorry about that!
Dunno how that effects anything though..

On Mar 16, 9:13 am, Dekatron42 <martin.forsb...@gmail.com> wrote:

[Adam Jacobs]

Why are you sharing 1 anode resistor across two tubes? :) Is board space
at that much of a premium?

>> Elektronische Anzeigebauelemente" by Winfired M�ller contains a few of

[Imbanon]

I wasn't home at the time, and all I had were three 10k trimpots :)
Why is that bad? I would leave it that way, if that's ok.

> >> Elektronische Anzeigebauelemente" by Winfired M�ller contains a few of

[John Rehwinkel]

> And I have to clear out that I have a common anode resistor for 2
> tubes, making a total of 3 anode resistors for all 6 tubes.

It seems to me that would only work if you only selected a cathode for one tube at a time. Otherwise (if you tried to light both tubes at once), only one tube would light, pulling the
anode end of the resistor down to the maintaining voltage, which would be insufficient
to light the second tube (because it is now below the striking voltage).

- John

[Adam Jacobs]

I think John's right. There's no reason you couldn't theoretically do
it, provided you jumped through the extra hoops. Design decisions like
this one have a tendency to add complication.

-Adam

[Imbanon]

Actually that is exactly what I am doing. I have 3 anode control
circuits, each controlling 2 nixies. I am doing this because I lack
digital I/O pins. Should I then change my design to one anode resistor
per tube? I would still have only 3 anode drivers..

[Adam Jacobs]

You could do this.. It's not ideal in my opinion, but as long as you are
careful to never turn on more than one of the paired nixies at a time,
you could get away with multiplexing this way. I would set the anode
resistor at something like 10-15k to start with.

I think that this looks like a great reason to learn how shift registers
work, though. :)

-Adam

[John Rehwinkel]

> Actually that is exactly what I am doing.

Which? Lighting only one tube (per anode resistor) at a time? In that case, it should work just fine.

- John

[Imbanon]

I think my post was a bit unclear. I AM turning on two at the time x)
I have 3 pins controlling 6 nixies. That means one pin turns on 2
nixies at the same time. Each lit nixie is controlled by separate
K155ID1 (two muxes in total).
Or am I having problems understanding you..
Cheers

[Adam Jacobs]

Right, each anode driver powers 2 nixies. You need to be careful to have
one of the pair's K155ID1's set to blank at any given time, though.
Never sink more than one of the K155ID1's at any given time and this
arrangement will work fine.

-Adam

[Imbanon]

Blank as in set to 0?

My sequence is pretty much this;
set bits to first mux
set bits to second mux
turn on one anode driver
wait for 3ms
turn off anode driver
blanking period of 200us
repeat this n-times

I don't really touch muxes' bits when I am not displaying anything
(nixies turned off). I only set them again when I want to display time
again.

[Adam Jacobs]

Instead, change to:
- turn off first pair's anode driver
- set the first 74141 to BLANK (set the input to A)
- set the second 74141 to BLANK (A)

- turn on first pair's anode driver
- set the first 74141 to the desired value
- set the first 74141 to BLANK
- set the second 74141 to the desired value
- set the second 74141 to BLANK
- set the first 74141 to the desired value
(etc)

get it working with one pair before trying to make the others work.

-Adam

[Adam Jacobs]

You really ought to take the time to at least do a back of the napkin
schematic. :D

I'm trying to decide how this arrangement works out to being more pin
efficient than the standard way. So, you must have 8 GPIO pins delegated
to driving 74141 inputs, and then 3 GPIO pins delegated to the anode
drivers. That is 11 pins.

With a standard 1x6 mux (6 anode drivers, one 74141), we use 10 GPIO pins.

Until you show us exactly what you've got going on, we're going to be of
limited help to you. Also, you'll find that going through the exercise
of drawing out the schematic may help you solve your own design problems. :)

-Adam

On 3/16/2012 1:17 PM, Imbanon wrote:

[David Forbes]

My nixie watch has zero anode resistors. I added a current sense circuit to the HV supply feedback circuit. I only light one tube at a time, using a total of 16 cathode drivers in TD62083 parts. It works fine. See my website-google nixie watch theory to find the page.

David Forbes
http://www.cathodecorner.com/

> --
> You received this message because you are subscribed to the Google Groups "neonixie-l" group.

[Adam Jacobs]

That's a clever idea. I'll have to give it a try sometime.

[Cobra007]

I don't use anode resistors either for the obvious reason that they
consume power and therefore reduce the overall efficiency of the
circuit. If the circuit is not powered by batteries, this is not
really a problem, it will only increase the electricity bill :-).

Michel

On Mar 17, 7:39 am, David Forbes <dfor...@dakotacom.net> wrote:
> My nixie watch has zero anode resistors. I added a current sense circuit to the HV supply feedback circuit. I only light one tube at a time, using a total of 16 cathode drivers in TD62083 parts. It works fine. See my website-google nixie watch theory to find the page.
>

> David Forbeshttp://www.cathodecorner.com/

[Cobra007]

Hi Imbanon,

I followed your calculations for quite a bit and was also wondering
why your measurements are so much different than your calculations. My
calculation suggest that you need a 7.3k resistor for each tube. This
means if you have 2 tubes with a common anode connected to 1 resistor,
and switch both tubes on at the same time (which you say you do), then
you need to half the resistor. In that case, 3.65k.

Now, I was also puzzled why my resistor calculation results in half of
your resistor calculation. The answer is in fact quite simple, and I
will try to explain to you.

First of all, we need to agree that the tube power needs to be the
same in both direct drive as well as multiplexed mode to guarantee
equal tube brightness. Therefore Pdc = Prms

Pdc = Udc * Idc
and also
Prms = Urms * Irms

Do you notice the problem already?

Pdc = 145V * 2mA = 0.29W

Prms = 145*sqrt(T1/T) * Irms

So, Irms should be 0.29 / Urms = 0.29 / (145*sqrt(T1/T)) = 3.9mA

Your calculation was based on the fact that Irms had to be 2mA, but
this is not correct because it needs to be 3.9mA.
The peak current through the resistor is then 3.9mA / sqrt(T1/T) =
7.5mA per tube (15mA if 2 tubes are switched on at the same time).
This is exactly what you get if you take the average currents. You
want 12mA average current (6 tubes * 2mA). which is 12mA / (26.7 /
33.3) = 15mA.

Unknowingly, to calculate Prms you were multiplying Irms * Udc, not
Irms * Urms.

Does this make sense?

I do agree with the rest of the guys that 6 resistors (7.3k) would be
better that 3 resistors (3.65k), it will split the current more evenly
through the 2 tubes.

Michel

[Imbanon]

Hey Cobra! Thanks you for your explanation. I see what you did there,
and you are quite correct. But the thing is that I did not calculate
anything using power. You are completely right with the brightness,
but I do not want the same brightness as with direct drived nixies. I
lean towards tube life, and I think that the only way to achieve this
is by getting the correct current, and that would be around 2mA.

I will make a schematic and share it, hopefully tomorrow, if I find
some time.

My first design was 1x6, but I wasn't happy with the brightness. It
also required some higher current peaks, which can't be good for the
tubes. 2x3 mux design gives better brightness, but uses one more pin.

And can someone please explain how does one blank a 74141? Adam, what
do you mean by setting the input to A? Setting a logic 1 to the A
input pin by the datasheet? That only sets the output to "1". My
design so far didn't use blanking with the 74141's. I just turned all
anodes off for 200us, and I never had any ghosting problems or
anything like that.

Cheers

[Cobra007]

If you take 7.3k resistors rather than 3.65k resistors, the tubes will
only be 6dB less bright which is not significant. If that increases
tube life, I would say, that is the better choice as you pointed out
already.

For blanking, I think Adam means to send a hex number (0x0A) to the
74141 (D=1; C=0; B=1; A=0). According to the datasheet, this would
lead to a correct blanking of the tubes.

For my clock I am designing a module as I am not really a fan to use
these types of old TTL logic. It's a 24 pin module that fits into a
DIP24 IC socket. It basically mimics the 74141 but has high voltage
output mosfets (240V) and the 4 inputs can be latched, so you don't
need extra latches as required by the standard 74141. It also offers a
blanking input, either by writing 0x0a or using a dedicated pin (which
is convenient for PWM dimming). It can be interfaced with MCU or
arduino.

Michel

[threeneurons]

I'd start with 6.8K anode resistors (assuming 3,6 to 7.3 is your target range) and see how it works out. Keep a close eye on it for a couple of months, if its running 24/7. If the current is too low, you'll start to see partial illumination on the the digits over time. Then you need to increase the current. Its easier to do this by tacking resistors in parallel, to existing resistors, than by complete replacing resistors.

On 74141s any code over 9 will keep all cathodes OFF. So, you can blank it that way. There is a possibility of leakage as the parts age. This will show as "ghosting". Again, this may not show up right away. Only time will tell. Since you're already switching the anodes, why not use the anode circuit to blank the display. Its only a software modification. Also, since you're  muxing, the supply voltage will be higher. Nominally 200V, so any leakage problems, will show up earlier.

[Dylan Distasio]

I'm very interested in hearing more about this module...Are you saying you are having this custom manufactured?  If so, how have you found a way to do this economically?  

[Cobra007]

Hi Dylan,

It is actually a small circuit board that I have manufactured together
with the nixie watch circuit boards to save costs. The circuit board
is the size of a DIP24 chip and has an SMD 4514 multiplexer that
drives 10 high voltage SMD mosfets. There is 1 spare mosfet on the
board that is not driven by the multiplexer, it can be used to drive
the colon lamps for example, not really a dedicated purpose but I had
board space left over :-). It's pretty basic really, but very
convenient to use.

Michel

[Adam Jacobs]

I would be interested in this as well. Are you planning to market and
sell these modules?

[Cobra007]

That should be fine Adam, I should have 50 modules in a couple of
weeks so I do have a few spares :-).

Michel

[Imbanon]

Hey all

I found some time to make a schematic. It only shows the two 74141,
nixie tubes and anode drivers.
I also tried blanking nixies via 74141. I would have never guessed
that you meant on hex code, as I am doing this on an arduino :)
Blanking nixies with 74141's give me a lot of leaking (or at least I
think thats leaking), so I'm not so sure about using this method. But
I do think that it's maybe possible to divide the leaking to the rest
of the nixies by adding a resistor for each anode, instead of sharing
them. See for yourself in the links.

http://www.mediafire.com/download.php?bbx4z4k5vjul56b
http://www.mediafire.com/download.php?ym4s96yeusrm9sy

So I guess my next move should be to remove the trimpots, and replace
them with actual resistors for each anode.
But what should I do with the blanking? To be honest, I would leave
the setup as is, because it seems to work fine. But if you guys think
I can do something better to get a longer tube life, I will make
changes to the schematic immediately.

Thanks!

[Imbanon]

Could you share a few pics? Also, how much will they cost? :)

[Adam Jacobs]

Would you mind converting that schematic to some kind of image file? Is
that an eaglecad file?

blanking on the 74141 will cause leakage if the supply voltage is too
high. Are you using real 74141's or the russian kind?
From that picture, I'm not sure if that is leakage.

-Adam

[Imbanon]

Sorry about that. I thought it was an universal shematic file type,
'cause they are all .sch

Here's the pic http://www.mediafire.com/download.php?5903q1ur3inc729
Cheers

[Adam Jacobs]

_MUCH_ better.

Okay, firstly this design can work - there's nothing fundamentally wrong
with it. Like I said, you'll need to be able to make cathode-side
blanking (74141) work if you're going to stick with only 3 anode drivers.

1) Your anode drivers aren't quite right. The resistor on the base of
the NPN should be more like 10k (R1). You're missing the pulldown on the
base as well. Add a 10k between the base of the NPN and GND. The
resistor between the collector of the NPN and the base of the PNP should
be more like 1M. The resistor between the base of the PNP and the
collector of the PNP should be more like 10k.

2) The reason that you are getting current leakage with the 74141 is
because your HV supply is too high. Instead of 215v, use 180v.

-Adam

[Imbanon]

Please take note that I am not using PNP's at all, only NPN's :)

[Adam Jacobs]

I'm starting to get the impression that this is a parts-box project.
Have you ever heard the expression that when the only tool you have is a
hammer, everything starts looking like a nail?

[Imbanon]

Why would you say something like that? What's wrong with using
trimpots instead of resistors when you're not home with your stuff? Or
using NPN based anode drivers.. Why would I NEED to make a so popular
NPN-PNP anode driver, when I first thought of something like this,
with parts I already had. And it all seems to work.
And sorry, no I haven't heard that one. Nothing like that in my
language.

[Adam Jacobs]

Ok, Good luck.

[Cobra007]

It's an interesting combination those 2 npn transistors. The signal on
the base is actually inverted compared to "normal" npn/pnp
transistors. When T2, T3, T5 are switched off, the tubes are actually
on :-). That's also where your leakage problem comes from if you ask
me. You choose 220k resistors, which means the lowest possible anode
voltage is half the supply voltage (215V / (220k + 220k)) * 220k =
107.5V. You better choose a lower resistor that goes to the collector
of T2, T3, T5 (maybe 100k). This will bring the anode voltage down to
about 68V.

I don't have a price yet for my modules, they won't be expensive of
course but I need to wait until I have a finished product.

Michel

[threeneurons]

On Thursday, March 22, 2012 11:12:21 AM UTC-7, Adam Jacobs wrote:

The resistor between the base of the PNP and the collector of the PNP should be more like 10k.

More like between base and emitter, to speed up turn-off time.  

[Nick]

On Friday, 16 March 2012 19:22:05 UTC, Imbanon wrote:

Oh and I accidentaly reported your post as spam. Sorry about that!
Dunno how that effects anything though..

It doesn't - the mods clean up behind you ;-) 

[Imbanon]

Indeed, I see you got it right :)
Anyhow, I got it all fixed. I just lowered the power supply to 180V.
Of course, I had to recalculate all the anode resistors (finally have
individal anode resistors), but the blanking now works just fine.
But still, I don't understand why I should blank one nixie in an anode
pair, like Adam suggested. Also, sequence like that will cut the duty
cycle by half, meaning I would need to increase the current peak, and
lower the tube life. It does seem to work fine as is.


> Ok, Good luck.
Thanks a lot!

[Adam Jacobs]

I just took another look at the schematic you provided, you're right.
You don't need to do cathode side blanking. For some reason, I thought
that you had both nixies in the pair connected to the same 74141 and
would need to blank cathode side in order to address a single nixie.
Since you have each nixie in the pair connected to a separate 74141, you
should be fine.

-Adam