resistance of filament warm vs cold

[Gideon Wackers]

I just measured the filament resistance of a IV-11 tube, according to my multimeter that is roughly 5-5,5 ohm (assuming it measures perfectly at such a low resistance). But I calculated something different;

I'm going to wire the filaments of two tubes in series and feed them 5 volt though a resistor. So I used the voltage/resistor divider formula. to calculate the resistor needed for a 2 volt drop (tubes want 2*1,5V) and that is 10 ohms, going back to the voltage divider that gives ma a resistance of 15 ohms for both tubes and 7,5 ohms for each filament. Close you might say but still a 50% difference. 

Options:

-my multimeter sucks

-filament resistance differs because it is cold

-both options

-my calculations are wrong

Well while I wait for an answer I'm going to build a little jig with lego so my tubes will be aligned perfectly straight :)

[Spencer W]

I used this as a reference for mine. If I remember right, it's around 1.15volts for the filament and 100mA. So with 5v power and 2 in series, 27ohm resistor though you want 2 watts.

http://led.linear1.org/led.wiz

Sent from my iPhone

--
You received this message because you are subscribed to the Google Groups "neonixie-l" group.
To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+...@googlegroups.com.
To post to this group, send an email to neoni...@googlegroups.com.
To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/4e980192-1b38-42f2-b092-9e62298f2ea3%40googlegroups.com.
For more options, visit https://groups.google.com/groups/opt_out.

[Gideon Wackers]

http://www.tube-tester.com/sites/nixie/dat_arch/IV-11_2.pdf

IV-11 tubes use 1,5 volt for the filaments according to the datasheets I found

Op vrijdag 6 december 2013 12:16:01 UTC+1 schreef Spencer:

[Gideon Wackers]

I just tried it carefully with a 1,5 volt battery, the filaments do not light up (good thing). but I could not get a segment to light up I only had two 9 volt batteries and that gave me 17,5 volt which is not enough. 

Btw I did notice that when I hooked up the 1,5 volt battery or took it away that the filaments made very soft *ting* sound.

[David Forbes]

I don't know much about that tube. However, I do know:

1. Filaments have lower resistance when cold.

2. VFD filaments should not get bright enough to see clearly. A tiny
glow in a dark room, maybe.

> --
> You received this message because you are subscribed to the Google
> Groups "neonixie-l" group.
> To unsubscribe from this group and stop receiving emails from it, send
> an email to neonixie-l+...@googlegroups.com.
> To post to this group, send an email to neoni...@googlegroups.com.
> To view this discussion on the web, visit
> https://groups.google.com/d/msgid/neonixie-l/4e980192-1b38-42f2-b092-9e62298f2ea3%40googlegroups.com.
> For more options, visit https://groups.google.com/groups/opt_out.

--
David Forbes, Tucson AZ

[Spencer W]

Did you hook the 17.5v to the grid and segment you wanted to light up? It should be a faint glow

Sent from my iPhone

On Dec 6, 2013, at 5:41 AM, Gideon Wackers <gideon....@student.uhasselt.be> wrote:

I just tried it carefully with a 1,5 volt battery, the filaments do not light up (good thing). but I could not get a segment to light up I only had two 9 volt batteries and that gave me 17,5 volt which is not enough. 

Btw I did notice that when I hooked up the 1,5 volt battery or took it away that the filaments made very soft *ting* sound.

--

You received this message because you are subscribed to the Google Groups "neonixie-l" group.
To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+...@googlegroups.com.
To post to this group, send an email to neoni...@googlegroups.com.

To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/79d9649e-48ab-4e65-be11-d36fd1307ba7%40googlegroups.com.

[Gideon Wackers]

http://www.tubeclockdb.com/vfd-tubes/100-simple-vfd-tester.html

this is what I did

But I'll try it again tomorrow, maybe one of the connections was bad because I had to hold the wires against the batteries at 4 different points :P

Op vrijdag 6 december 2013 16:11:42 UTC+1 schreef Spencer:

[Gideon Wackers]

Yeah I could not see them glow in a dim lit room so :)

Op vrijdag 6 december 2013 16:08:03 UTC+1 schreef nixiebunny:

[Adam Jacobs]

Hi Gideon,
  You're doing it wrong. :)
We do not normally refer to filaments by their impedance, but rather by their power draw. What is the equivalent resistor value of a 100watt incandescent lightbulb? There isn't one, because filaments and resistors behave differently. For starters, a filament has a much lower impedance when it is cold than after it has warmed up. Instead, we identify a filament by the current draw. I don't know offhand what the current draw of an IV-11 filament is, but the datasheet will have it. I think it is roughly 100ma. Same thing with the filament voltage, the datasheet will have the exact number for the real math. I'm going to call it ~1.5v.
  Going back to ohm's law, we know that I=V/R. The formula for a resistor divider without a load is:
R_1=(V_1*R_2)/(R_1+R_2)

Now, normally there would be load in parallel with R_2. However, in our case, R_2 *is* the load.
So, our equation becomes:
R_1=(V_1*R_L)/(R_1+R_L)

We don't know the equivalent load resistor, only the current draw, so we use substitution algebra:
R_1=(V_1*(V_L/I_L))/(R_1+(V_L/I_L))

V_1 is the supply voltage, in my case 5v. V_L (normally V_out) is the filament voltage, in my case 1.5v.
R_1=(5*(1.5/0.1))/(R_1+(1.5/0.1))

Also, the reason you weren't seeing success using those batteries probably is because you didn't tie them both to a common potential. There is no assumed potential difference between two separate dry cells, I think.

-Adam

[Adam Jacobs]

Oops. Somewhere in my wrestling with the MS Office Equation editor, I messed the equation up. :)

 

 

On 12/6/2013 11:59 AM, Gideon Wackers wrote:

1, 5v 100ma is correct. It could be because i am viewing this on my phone but i do not think your formula is going to work. R_1 is before and after the = sign.

I will look at it tomorrow morning

Op 6 dec. 2013 20:41 schreef "Adam Jacobs" <jacob...@gmail.com>:

Hi Gideon,
  You're doing it wrong. :)

We do not normally refer to filaments by their impedance, but rather by their power draw. What is the equivalent resistor value of a 100watt incandescent lightbulb? There isn't one, because filaments and resistors behave differently. For starters, a filament has a much lower impedance when it is cold than after it has warmed up. Instead, we identify a filament by the current draw. I don't know offhand what the current draw of an IV-11 filament is, but the datasheet will have it. I think it is roughly 100ma. Same thing with the filament voltage, the datasheet will have the exact number for the real math. I'm going to call it ~1.5v.

  Going back to ohm's law, we know that I=V/R. The formula for a resistor divider without a load is:

R_1=(V_1*R_2)/(R_1+R_2)

Now, normally there would be load in parallel with R_2. However, in our case, R_2 *is* the load.
So, our equation becomes:
R_1=(V_1*R_L)/(R_1+R_L)

We don't know the equivalent load resistor, only the current draw, so we use substitution algebra:
R_1=(V_1*(V_L/I_L))/(R_1+(V_L/I_L))

V_1 is the supply voltage, in my case 5v. V_L (normally V_out) is the filament voltage, in my case 1.5v.
R_1=(5*(1.5/0.1))/(R_1+(1.5/0.1))

Also, the reason you weren't seeing success using those batteries probably is because you didn't tie them both to a common potential. There is no assumed potential difference between two separate dry cells, I think.

-Adam

 

On 12/6/2013 3:06 AM, Gideon Wackers wrote:

I just measured the filament resistance of a IV-11 tube, according to my multimeter that is roughly 5-5,5 ohm (assuming it measures perfectly at such a low resistance). But I calculated something different;

I'm going to wire the filaments of two tubes in series and feed them 5 volt though a resistor. So I used the voltage/resistor divider formula. to calculate the resistor needed for a 2 volt drop (tubes want 2*1,5V) and that is 10 ohms, going back to the voltage divider that gives ma a resistance of 15 ohms for both tubes and 7,5 ohms for each filament. Close you might say but still a 50% difference. 

Options:

-my multimeter sucks

-filament resistance differs because it is cold

-both options

-my calculations are wrong

Well while I wait for an answer I'm going to build a little jig with lego so my tubes will be aligned perfectly straight :)

[Charles MacDonald]

On 13-12-06 06:06 AM, Gideon Wackers wrote:

> -filament resistance differs because it is cold

Filament resistance is different when cold, which is why Incandescent
Light bulbs generally fail at turn on if they are going to fail.

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

[Gideon Wackers]

Owwww, so 20 ohms, well I have 10, 1 watt, 10 ohm resistors for three clocks so I'll make two extra pairs with some funky arrangement of 10 ohm, 22 ohm, and 47 ohm 1/4watt resistors. I only need 0,2W but, well I don't trust things being pushed to their limits. 

thanks for catching my silly mistake before I blow tubes. 

And about the tube not lighting up... Let me guess, the pins are counted from the bottom view and not top view. I think I hooked up two segments. When I was doing it I was already like; hmm why doesn't it look like those wires are going to the grid?

sleep deprivation, it's a thing I believe :)

[Gideon Wackers]

Yeah 9/10 times when a light bulb breaks it goes when you turn it on.

 I do have a theory about them blowing in the early morning or late at night with an extra loud *POP* to just scare people. 

Op zaterdag 7 december 2013 02:56:24 UTC+1 schreef charles:

[Alex]

Sorry Adam, I sent a reply to you and not the group...

I do get your approach now (shouldn't read such things before finishing the morning coffee), but it still seems a slight unnecessary over complication? If the current is specified as 100mA and voltage is 1.5v then surely that gives the manufacturers specified resistance, when running hot. (I still argue its resistance and not impedance due to the lack of significant inductance and capacitance?)

This would give a mfg resistance of 1.5/0.1 = 15ohm? 

If you treat two of them as one 30 ohm resistor with 3v over it then you can get your 2v drop at 100ma do you not need a 2/0.1 = 20 ohm resistor?

A soft start arrangement would be kind, and I am sure I have seen it used on VFD filaments somewhere, but I doubt their resistance changes that much to as require one as they don't get too hot (compared to a incandescent filament...)

I would still always run it though a bench PSU with a current limit set, then just play with some resistors (or a wire wound pot / rheostat) until the figures work out...

- Alex

[Gideon Wackers]

Well I'll keep an eye on the discussion about the filament resistor(s) to see what the final opinion is. 

In the mean time I discovered that tube pins are labeled from the bottom... and got my first fluorescent light (bright as hell even at 17,5 volt) 

http://i.imgur.com/RZjMkUf.jpg 

sorry for the messy picture but holding wires to three batteries and operating the camera at the same time is hard to do. 

[Adam Jacobs]

Hi Alex,
  I think that you are describing the same thing that I tried to describe. Yes, V/I is substituted in place of R into the equation as you recommend and your number is the same as what the math produces. It's easy to say 'why not just' and then proceed with the common sense understanding of an electronics topic, but that understanding breaks down rapidly when things get more complicated. Maybe I spelled it out too slowly (and I'm sure my broken equations in the first email didn't help). :)
  Also, normally we talk about Impedance when it is alternating current. Filaments are normally AC. I think filaments may have a small reactive component at certain frequencies, but I doubt they do at 60hz. Usually I don't worry about significant reactive components unless I'm building a new antenna for transmission.

-Adam

--
You received this message because you are subscribed to the Google Groups "neonixie-l" group.
To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+...@googlegroups.com.
To post to this group, send an email to neoni...@googlegroups.com.

To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/48e8ec1f-7894-4569-a89e-07a0021c4544%40googlegroups.com.

[AlexTsekenis]

This morning my study room light popped and thought of throwing some oil into this fire by mentioning this equation:

 copied straight from the all-knowing wikipedia. The full page is here

I think we have the same nixie clock Gideon :-)

Alex.

[Gideon Wackers]

PV-electronics?

Yeah I recommended his kits to many people. He now has a very cheap IN-12 kit by the way.

But 20 ohms should be correct right? Because then I'll just stick two strings of -22ohm resistors in series- in parallel. I hope it is clear what I mean

[AlexTsekenis]

Yes and yes. You need the arrangement you described if using standard 1/4 W resistors. You do not need a soft start circuit.

Alex

[Gideon Wackers]

I think I'm going to warm up the soldering iron and stick some tubes in that protoboard (well do that three times) 

Op zondag 8 december 2013 11:05:08 UTC+1 schreef AlexTsekenis:

[Mark Moulding]

I've been following this discussion for a couple of days now, and thought I knew what was going on, but I just went to my bench a physically put my hands on an IV-6 VFD tube.  The filaments for these tubes are specified at 0.85-1.15 Vrms (nominal 1V), with a current draw of 50 mA.  Note that the spec sheet says Vrms, implying an AC filament drive.  I know this matters for the long, skinny tubes like the IV-18 8-digit tubes, but for single digits DC is just fine.  Fortunately, ACV rms is *exactly* equal to DCV when going into a pure resistive heater; in fact, that's one commonly used simplified definition of RMS, and in the olden days the physical method used to make a true-RMS meter.

I agree with other posters that the reactive components of this tube should be nearly nil, but I went ahead and measured it anyway.  It turned out to be 1.7 uH - higher than I expected, but still an insignificantly low value.  So I measured the resistance of the filament cold with a four-wire ohmmeter, so that lead resistance would not be significant.  This turned out to be 5.1 ohms, which at 1V filament supply would cause a current of 196 mA - about four times the specified draw.

I then set my bench supply to 1.00 VDC, and connected it directly to the filament.  The power supply's current meter read 0.05 A, or 50 mA - right at spec, but not with very good resolution, so I then put a milliamp meter in series.  I've previously measured the internal resistance of this meter, and know it to be 0.1 ohm, so it wouldn't significantly affect the reading.  Sure enough, it read 51.2 mA, which is *plenty* close enough for any practical use.  Therefore, the hot filament resistance is just about 200 ohms.

This means that there's a surge when power is first applied, which might (does, actually) shorten the life of the tube.  As one poster mentioned, light bulbs and other filament-based devices fail most frequently when power is first applied; this is mostly due to the filament rapidly changing tension and shape as it heats up.  (If you fire up a Numitron tube, you can actually see the filament sag as it begins to glow.)

I've found that a very easy way to give a soft start, and simultaneously solving the problem of getting a 1V supply for the filament, is to simply place an 80 ohm (I use the standard 82 ohm value) resistor in series with the filament, and run it off the 5V supply.  The 82 ohm resistor is by far the largest part of the current path, so even if the tube filament started out with zero resistance, the maximum current is limited to 5/82 = 60mA - only 20% over spec, instead of 400%.  (Older stepper-motor drives used to use this technique, as well.)  Of course, there's a penalty, as always: this is quite inefficient, since 4/5 of the filament power is being dissipated - wasted - in the dropper resistor, but if you've got plenty of 5V power around, who cares.  (For this tube, there's still only 200 mW dissipated in the resistor, so a standard quarter-watt resistor is fine.  Those old stepper drives had big honking power resistors, though...)

Sorry for the dissertation.  I accidentally bought a large batch of these a couple of years ago, and to make back my money (except for time, of course...) broke it into small batches for eBay.  I sold them with a little data sheet I printed up that included the original Russian data sheet, plus an English-language text sheet and a quick-test jig using a 1.5V cell and a couple of 9-volt batteries.  Since I had so many, I also spent a bit of time playing with them, and got to know them pretty well. They really are considerably easier to use than Nixies, just because of the lower voltage requirements, and I think they're longer-lasting as well.  I don't find them nearly as attractive as a nixie's warm glow, though.  Numitrons, on the other hand, I like a lot...
~~
Mark

[threeneurons]

Put a new battery in your calculator: 1V/0.05A = 20 ohms

Don't feel too embarrassed. If you look at the old Yahoo nixie forum, I have a long trail of on-line oops's.

Your hot resistance measurement procedure is correct, and that's what's really important.

On Sunday, December 8, 2013 10:14:55 AM UTC-8, Mark Moulding wrote:

I then set my bench supply to 1.00 VDC, and connected it directly to the filament.  The power supply's current meter read 0.05 A, or 50 mA - right at spec, but not with very good resolution, so I then put a milliamp meter in series.  I've previously measured the internal resistance of this meter, and know it to be 0.1 ohm, so it wouldn't significantly affect the reading.  Sure enough, it read 51.2 mA, which is *plenty* close enough for any practical use.  Therefore, the hot filament resistance is just about 200 ohms.

~~
Mark

[Gideon Wackers]

Well the first board is filled with tubes, and my head is filled with headache from peering at the board.. The enameled wire was very hard to solder even after burning off the enamel layer. Although I do not dare to show the abomination that I call "soldering" the tubes are all connected. Don't worry I know how to solder, but the enamel wire was giving me a very hard time. The nice thing about the enamel wire was that it was easy to go through the forest of component legs. I'm off to bed. 

[AlexTsekenis]

For any diameter of enamel-coated wire larger than a strand of hair, the burnt enamel should be removed to expose the copper. Otherwise solder wetting will be poor and the soldering process frustrating.
You can do this using fine sandpaper, a file, or a sharp knife. Burning the enamel first makes removal easier. For thin wires the soldering iron is adequate. Larger diameters require a lighter. Even large diameters a small blowtorch. Tin the wire prior to soldering using a generous amount of flux. This will also show you if you did a good job scrapping off the burnt enamel.

Alex

[Mark Moulding]

On Sunday, December 8, 2013 10:50:20 AM UTC-8, threeneurons wrote:

Put a new battery in your calculator: 1V/0.05A = 20 ohms

Oh - duh! I think the rest of the discourse was correct.  (Sadly, no calculator was used - I guess it's my brain that needs the new batteries...)
~~
Mark

[Alex]

I have a vague feeling that enamel wire fumes are fairly toxic, hence I used to sandpaper them first to remove most of it prior to a roasting with a lighter. My workplace used to have an enamel wire stripper, which was a funky bit of kit - 3 blades that spun round :-)

- Alex

[Mark Moulding]

There's a type of enamled wire that's intended to be soldered through.  It used to be used for the Vector Wiring Pencil, and is still available intended as transformer wire.  I bought the smallest spool I could (1 pound - at 30 gauge that's nearly a mile!), and still use the Vector Pencil for some projects.  My wife and I sometimes put together creative projects that include lots of LEDs mounted without circuit boards, and this type of wiring works really well by just wrapping a couple of turns right around the LED leads, and soldering it in place.  A couple of the brand names of this wire (Google "solder strippable magnet wire") are Solidon and Soderex.

In theory, the insulation is even intended to act as flux, so you can just use a solder pot to do the soldering.  In practice, I found that it is a little more tedious than soldering regular (stripped) wire.  In fact, I abandoned it back in the 70's/80's for wiring up tons of IC's, because there were occasional cold or even non-conductive joints, which were a pain to track down.  Mostly, this is because I didn't heat the joints long enough to cook off all the insulation, but it was still a pain.  Nowadays, I only use it for specific applications, and I make sure that I cook the joints longer than my natural inclination, but my success rate is pretty nearly 100%  On something like tube leads, where overheating the component isn't too big a concern, it should work fine.

According to WikiPedia, most magnet wire now has this type of insulation, bu not older wire.  This may be, but the brands I mentioned above are specifically intended to be soldered this way.

There also used to be a solvent specifically intended to strip insulation, but I've never used it.  It may not be available any longer, because it probably wasn't the healthiest stuff around.
~~
Mark

[Adam Jacobs]

Yes, but why use enamel coated wire in the first place? I've only ever used it for winding inductors and transformers. I hate the stuff. Dipping in molten solder usually works to remove the enamel, with a clean-up pass using sandpaper. 

My favorite protoboard wire is the old 80-wire PATA ribbon cables. In Seattle, I can buy the ribbon cable for $1, giving me quite a bit of good wire for the price. The wires easily separate with fingernails and the insulation easily strips with fingernails too. Avoid the 40-wire version, those use stranded wire instead of solid-core.

-Adam

--

You received this message because you are subscribed to the Google Groups "neonixie-l" group.
To unsubscribe from this group and stop receiving emails from it, send an email to neonixie-l+...@googlegroups.com.
To post to this group, send an email to neoni...@googlegroups.com.

To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/3bd64c59-b0f7-4be1-a4ec-4ced96bbff34%40googlegroups.com.

[Tomislav Kordaso]

Enameled wire is pain to strip mechanically. Back in my school days we used a pill of Aspirin and push the end of enameled wire shortly onto it with hot solder iron. Aspirin would melt and eat away the enamel, easing the tining process. Fumes are not nice to inhale, though.

@Gideon if you prefer protoboards, you can try kynar wire. A 30-gauge spool of it is not expensive, strips off easily and solders even easier.

Tomislav

To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/CA%2BAY7RydOr5RoK2EA%3D-p%2B_1cDOg8V8bpJWwud5BMQreKt63d7w%40mail.gmail.com.

[Adam Jacobs]

Wow, great tip about the aspirin! I'll try that on my next transceiver build. :)

-73 Adam W7QI

To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/CAChpj6zH1nRMH1GCTdcjpWqfHy39bj7FoTBC-vyELaZW7doZQA%40mail.gmail.com.

[chuck richards]

Yes, the aspirin trick is a great one on the enameled
wire! I think of this place every time I use it.

I've built many circuit cards on Vector #3677 plugboards.
They are about 4.5 inch x 9.5 inch, and they have
a 22/44 edge connector, and a hole pattern for 3 columns
of ics.

I use #30 ga wirewrap wire as soldered jumpers.
I run the jumpers on the top side, leaving the
solder side being just the solder joints.

One really nice advantage of the Kynar insulation is
that it has a very small amount of "shrink-back" when
soldered. I have found that many other wires such as
stranded wires from ribbon cable tend to have lots
of shrink-back, which can be a real pain.

I also tend to juice up most of my solder joints with
a little extra flux to make the solder flow easily.
I use some special non-conductive flux for these boards
with the tiny close-spaced traces.

Yeah, they are through-hole parts, and that stuff is
plenty small enough for me! I have not worked on SMD
yet, and do not especially relish the thought of it.

Chuck

$4.95/mo. National Dialup, Anti-Spam, Anti-Virus, 5mb personal web space. 5x faster dialup for only $9.95/mo. No contracts, No fees, No Kidding! See http://www.All2Easy.net for more details!

[Quixotic Nixotic]

On 9 Dec 2013, at 00:01, Mark Moulding wrote:

> There's a type of enamled wire that's intended to be soldered through. It used to be used for the Vector Wiring Pencil, and is still available intended as transformer wire. I bought the smallest spool I could (1 pound - at 30 gauge that's nearly a mile!), and still use the Vector Pencil for some projects. My wife and I sometimes put together creative projects that include lots of LEDs mounted without circuit boards, and this type of wiring works really well by just wrapping a couple of turns right around the LED leads, and soldering it in place. A couple of the brand names of this wire (Google "solder strippable magnet wire") are Solidon and Soderex.

This kind of wire on the vero-style spools is also called Roadrunner and is available from Fartnell, eg:

http://uk.farnell.com/roadrunner/rrw-a-105/wire-0-19mm-assorted-pk-4/dp/5017245 - is an example of the wire
http://uk.farnell.com/roadrunner/rrp-103/wiring-pencil/dp/145223 - is the wiring pencil

Wire comes in two different diameters.

John S