Operating smaller argon heads from larger PSU's
Buffo
out why it would be a problem. It's more of a theoretical question,
and it's been bugging me for some time now. Anyway, here goes:
Let's say you have a large-ish argon power supply (say, a spectra
physics 265, or similar...). Now, assuming that you could adjust the
current limit down to, say, 8 amps or so, why couldn't you use that
same power supply to drive a small argon tube like an ALC-60X?
I understand that the current regulation circuits are resistive in
nature (even a pass bank of output transistors is essentially
resistive), and thus if you design the circuit to pass a minimum of 8
amps with a voltage drop of 165 volts, it's going to generate a whole
lot more heat when it's only dropping 95 volts at the same current.
(Because the output transistors are having to drop an extra 70 volts or
so, which means their resistance needs to be greater and thus they get
hotter.) But so long as you provide enough air flow to remove that
extra heat, why should there be a problem?
Note that I'm not suggesting that excessive current be run through the
small tube; I understand that high tube current causes all sorts of bad
things to happen. But if the PSU can regulate the current to a safe
level, why wouldn't it work? What is the danger to the PSU?
Is there some fundamental concept that I'm missing here? Or is the
added heat generation the only problem? Looking at it from the point
of view of ohms law, I just can't see why additional cooling (fans and
heatsinks) wouldn't solve this problem...
So, what am I missing?
Adam
Sam Goldwasser
That about this: Your large power supply is set up for a high tube voltage.
Now, your 60X has a fraction of the tube voltage of the 165 and the power
supply has to be able to handle that difference. With a linear pass-bank
it would likely be impossible since it would all go into heat in the
pass-bank transistors. With a switcher, it might work but you would
need to know the specifications of the power supply. Else, could be
messy. :( :)
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LesioQ
voltage.
I'd hate to see typical starter board being fed with 500V DC waiting
for start pulse.
I have a small SP laser here that shows a tendency to fire the return
hole instead of main discharge path if starter board sees 300V instead
of 250V. Simple add-on board fixed the problem (it's simple with SMPS).
LesioQ
Buffo
> Your large power supply is set up for a high tube voltage.
> Now, your 60X has a fraction of the tube voltage of the 165 and the power
> supply has to be able to handle that difference. With a linear pass-bank
> it would likely be impossible since it would all go into heat in the
> pass-bank transistors.
Hi Sam!
Right. This is basically what I was saying in my first post. The pass
bank would have to dissapate a *lot* of extra heat, because in order to
handle the different forward voltage drop the resistance will need to
be higher, and thus you'll generate more heat. Eventually, the excess
heat will fry the transistors...
But if you can keep the pass bank cool, it should be OK, yes?
As a bit of an extreme example, I've passed several AMPS of current
through a tiny 2N2222 switching transistor without buring it up. But
in order to get rid of the excess heat, I had to submerge the
transistor in icewater! (Ok - I know, it's not a practical solution!)
But with large power transistors, you can mount them on big heat sinks
(or even on TEC's) and then use lots of fans to keep everything cool...
So if you could keep the pass bank from burning up due to excess heat,
it would work?
Adam
Buffo
LesioQ wrote:
> I'd hate to see typical starter board being fed with 500V DC waiting
> for start pulse.
I had forgotten about the start pulse. I suppose an over-voltage on
the ignition pulse could cause some damage to the tube internals - but
even so, that shouldn't harm the power supply, right?
> I have a small SP laser here that shows a tendency to fire the return
> hole instead of main discharge path if starter board sees 300V instead
> of 250V. Simple add-on board fixed the problem (it's simple with SMPS).
Do you mean that it lights on the return path if the *ingnition* pulse
voltage is too high? How did the add-on board fix the problem? (Or
did you just regulate the pulse back to 250V?)
Lighting down the return path is no doubt bad for the tube, but again,
the power supply doesn't really care, right? I mean, you're still
passing the same current through the tube, and the voltage across the
tube should be about the same, right?
Adam
LesioQ
case of overvoltaging the starter.
The return path mauintained a discharge of 260V @ 6A while the main
path was 195V @ 16A. These number gave me the idea what to do - I
reduced max PS output voltage by adding additional feedback loop to PWM
IC.
LesioQ
Buffo
> The return path mauintained a discharge of 260V @ 6A while the main
> path was 195V @ 16A. These number gave me the idea what to do - I
> reduced max PS output voltage by adding additional feedback loop to PWM
> IC.
I see. Interesting - I wouldn't have guessed that there would be
*that* much difference in voltage (and current) when lighting down the
bore vs the return path. Cool... Thanks for the input!
Adam
STEVE ROBERTS
meditec switcher designed for alc 68. The pwm circuit would not make
a small enough pulse to run the
tube at 10A, resulting in oscillation and balls of bore material
forming in the lasing path in just a few seconds. Tube voltage was just
too low for the supply compliance. Nasty RF plasmas IN AIR were
floating around inside the psu's case from the oscillation. The inside
of the circuit breakers glowed green from RF arcing when I switched
them off, and I got a RF burn while doing it. Somehow the 2o or so
switching transisters survived.
Fantastic display, but not one I'd like to see again.
Steve Roberts
Lostgallifreyan
news:6wejwor...@blue.seas.upenn.edu:
Why not use power resistors, as in the 'rabbit hutch' SP161 argon I put
together for a friend? Those are cheap, and a bit of DIY with some folded
aluminum sheet and a fan will mount them safely for cooling. Come to that,
if you know the optimal current and tube voltage drop, you can calculate
resistance and build a dummy load. While it won't accurately simulate the
laser tube, surely it will get you closer to a supply modification that
will run it safely? If so, further adjustments can be made once you
exchange the dummy load with the tube. Resistors are cheap, even big ones.
So are fans.
Buffo
Holy crapimoly! RF plasma *in air* inside the PSU?!? Wow...
I see where I made my mistake now... I didn't realise that the current
control was accomplished via pwm; I assumed the output of the pass bank
would be straight DC and whatever wasn't sent to the tube would be
dissapated as heat. But yeah, if the circuit couldn't modulate low
enough, and started oscillating violently with 10's of amps of current
available... That would be icky! (Not to mention disasterous for the
tube.)
Your story is amusing in the abstract, but I agree that it's not
something I'd like to experience in real life. (The concept of "balls
of bore material forming in the laser path" makes for quite a vivid
image!)
Thanks for the insight...
Adam
Buffo
> Why not use power resistors, as in the 'rabbit hutch' SP161 argon I put
> together for a friend? Those are cheap, and a bit of DIY with some folded
> aluminum sheet and a fan will mount them safely for cooling.
Ok - that would remove the heat load from the pass bank and transfer it
to the resistor bank, which makes sense. I guess the only problem
would be finding resistors that could drop 70 volts at 8 amps of
current without melting. That would be, what, roughly half a kilowatt
of heat? I don't think I've ever seen a resistor that can handle more
than about 10 watts... So I would need a crapload of them!
Assuming one could find power resistors that could handle, say, 100
watts or so, then in theory you could put them in series with the tube
and they would give you the extra 70 volts of drop across the PSU,
right?
Thinking about it further, even if the PSU used pwm to control the
current, with the extra resistance in series with the tube the PSU
should function just like it would with the normal sized tube installed
- right?
Adam
Sam Goldwasser
Sure, aside from some of the issues mentioned in other posts, using enough
power resistors to drop the voltage could work.
But it's silly. Suitable power supplies are available dirt cheap. Why
risk a catastrophe in more ways than one? :)
Lostgallifreyan
news:6w3bd3i...@red.seas.upenn.edu:
Oh it's totally silly, I agree. I built it to an someone's plan in a week
on my own for the princely sum of Ł100. With a workload like that I wan't
about to redesign, I just put the supplied stuff together. I had enough to
do getting what I didn't already have.
I'd use a switched supply as a power converter. >85% efficiency and an
output set to whatever was wanted. If it had direct current regulation, so
much the better. Are these cheap though? The few I've seen were at places
like RS and had nasty price tags on them. Several hundred pounds sterling.
Any idea what kind of place in the UK might be found selling one cheap?
What kind of equipment might result in these things becoming orphaned?
Buffo, if a switched supply is monitoring DC volts or amps output it would
use a sensing resistance anyway, probably. More esistance is just more
resistance. It's just wasteful. :) But it IS cheap. You can even get 250W
power resistors easily from suppliers like RS Components in the UK. (I only
cite them as an example..)
The ultimate cheap resistor would be made of nichrome wire coils as
supplied for a few pounds or dollars, for repairing elements in bar fires.
Those can be cut to lengths of required resistance. While it's different
when cold, you can base calculation on knowing that one complete length
will dissipate 1 kilowatt at its intended supply voltage (AC). You can get
accurate calculations from that. Given a good former to hold the wire, you
might not even need a fan to make it safe, just a means of reflecting the
heat away. Easier to use forced air cooling though. The hotter it runs, the
more demands are made of any mounting you use.
What interests me most about the use of resistance is the dummy load,
rather than a permanent part of a PSU. I don't know about tubes much, but
if they are in any way responsive to changes in current as resistance wire
is, you might save a tube during PSU setup and testing. I guess they aren't
though. I saw Pat and others here talk of negative resistance, something
you'll never see in a resistor.
Buffo
> > Sure, aside from some of the issues mentioned in other posts, using
> > enough power resistors to drop the voltage could work.
> >
> > But it's silly. Suitable power supplies are available dirt cheap.
> > Why risk a catastrophe in more ways than one? :)
Sam - you're right, of course. I wasn't seriously thinking about
performing the experiment, but I was trying to understand the
fundamentals behind why it was a bad idea. It relates to my initial
confusion over several comments in the FAQ about PSU's failing when
they are forced to run a tube at lower than normal voltage, either
because of low pressure or because the tube was re-gassed to krypton or
mixed gas and now the tube voltage is different. I had a hard time
understanding why the PSU would fail just because it was operating at a
lower voltage. But it's starting to make more sense now.
Lostgallifreyan wrote:
> Buffo, if a switched supply is monitoring DC volts or amps output it would
> use a sensing resistance anyway, probably. More esistance is just more
> resistance. It's just wasteful. :) But it IS cheap. You can even get 250W
> power resistors easily from suppliers like RS Components in the UK.
Hmmmm.... I didn't realise that such resistors were so readily
available. (I need to get out more!) Your nichrome wire example also
makes sense, though I agree that it would be cumbersome, slightly
dangerous, and grossly inefficient.
> I saw Pat and others here talk of negative resistance, something
> you'll never see in a resistor.
Groan... You *had* to bring that up. (grin) I've *never* been able
to wrap my head around the concept of "negative resistance"... I can
understand positive resistance. I can understand very, very small
positive resistance. I can accept the idea that once a tube lights,
the plasma conducts as if it were a dead short.
But even with a dead short, you've still got the resistance of the wire
leading to and from the tube. (Not to mention the minor internal
resistance of the PSU itself.) Even if we're only talking about a
tenth of an ohm or something, it's still a positive value. I just
don't understand how you could ever get a negative value for
resistance. (And if you plug negative values into your standard
electrical formulas, you get answers that seem to violate the law of
conservation of energy!)
Adam
Lostgallifreyan
>> you'll never see in a resistor.
>
> Groan... You *had* to bring that up. (grin) I've *never* been able
> to wrap my head around the concept of "negative resistance"... I can
> understand positive resistance. I can understand very, very small
> positive resistance. I can accept the idea that once a tube lights,
> the plasma conducts as if it were a dead short.
>
> But even with a dead short, you've still got the resistance of the
> wire leading to and from the tube. (Not to mention the minor internal
> resistance of the PSU itself.) Even if we're only talking about a
> tenth of an ohm or something, it's still a positive value. I just
> don't understand how you could ever get a negative value for
> resistance. (And if you plug negative values into your standard
> electrical formulas, you get answers that seem to violate the law of
> conservation of energy!)
>
> Adam
>
>
My guess is it's equivalent in some way to an endothermic reaction,
regarding the laws of thermodynamics, but I agree, it's a noodlebaker. I'm
hoping that Pat or Sam or Steve or someone will pitch in here with a nice
simple explanation. I'll be Googling, but I might not find anything better
than anyone here can point me to.
spam@frontiernet.net C what I mean
>> I saw Pat and others here talk of negative resistance, something
>> you'll never see in a resistor.
>
> Groan... You *had* to bring that up. (grin) I've *never* been able
> to wrap my head around the concept of "negative resistance"... I can
> understand positive resistance. I can understand very, very small
> positive resistance. I can accept the idea that once a tube lights,
> the plasma conducts as if it were a dead short.
>
> But even with a dead short, you've still got the resistance of the wire
> leading to and from the tube. (Not to mention the minor internal
> resistance of the PSU itself.) Even if we're only talking about a
> tenth of an ohm or something, it's still a positive value. I just
> don't understand how you could ever get a negative value for
> resistance. (And if you plug negative values into your standard
> electrical formulas, you get answers that seem to violate the law of
> conservation of energy!)
>
In a standard DC circuit if you increase the voltage across a resistance the
current draw thru that resistor will increase proportionately. E/R=I, good
ole' Ohm's law. With an ion tube the same applies while generating
plasma.... kinda most of the time, (some times it is non linear). In a
plasma tube you can have areas of operation where the current actually
decreases as the voltage applied goes up. It displays characteristics of
having negative resistance.... Therefore the term negative resistance. If
"E" go up and "I" goes down, then R must have gone negative relative to it's
last measured point.. It is not a real negative resistance that you can
measure with an Ohm meter. It is an effect that you can quantify
dynamically....
That work ya?
Craig
Lostgallifreyan
news:nI9ug.5382$oa1...@news02.roc.ny:
It only falls compared to what you might expect by Ohm's law, hence
'relative' negative resistance. I've been reading. I had to, the idea was
too shocking to me not to. If it were absolute, people would be exploiting
plasma tubes as an energy source.
R.X.
> Ok - that would remove the heat load from the pass bank and transfer it
> to the resistor bank, which makes sense. I guess the only problem
> would be finding resistors that could drop 70 volts at 8 amps of
> current without melting. That would be, what, roughly half a kilowatt
> of heat? I don't think I've ever seen a resistor that can handle more
> than about 10 watts... So I would need a crapload of them!
>
> Assuming one could find power resistors that could handle, say, 100
> watts or so, then in theory you could put them in series with the tube
> and they would give you the extra 70 volts of drop across the PSU,
> right?
I was doing exactly this at some point, before using a switcher for
good reasons. The problem with a linear regulator is that the
transistors typically can handle only part of the voltage, and with
a tube with less voltage, the voltage across the bank is higher.
For example, in a typical linear PS, you get like 160V after
rectifier, so that say 60V need to be killed for a 100V tube. Assume
that the transistors can just take that (eg 2n3055 are pretty bad in
this respect). Then, the moment you attach a say 80V tube instead, the
transistors see 80V, and this may be too much even though their
power dissipation rating is not exceeded.
Having a series resistor makes things more complicated. While less
power is dissipated and there is less voltage across the bank at
full current, the voltage across the passbank will strongly vary
with the current, and may be too high at small currents. I once ran
into this sort of problem by running a 160V NEC laser tube directly
off rectified 220V, which is like 310V. I had a series resistance
of some 10Ohms, which dissipated 1kW into the air at full 10Ams,
and the voltage over the passbank was like 310-160-100=50V. But at
2amps it was like 310-160-20=130V, and this can make quite a
difference for the transistors.
So before all I would check the transistor voltage specification
to start with. More precisely, what is relevant is the so-called
safe operating area (SOA). This is the region in the voltage/current
diagram that is safe (ie no second breakdown occurs). The point is
that for many high voltage transistors, the current they can take
at elevated voltages is much less than what corresponds to the max
power dissipation. Eg I was using 400V transistors that can dissipate
250W each, and can take 30A or so at 10V. But at 100V they can take
only a few 100mA! I don't remember any more the details, but
assuming for simplicity that they can take 100mA at 100V, this gives
just 10W max dissipation at 100V in contrast to 250W at low voltage.
So in total I ended up having 20 of them in parallel.
To make a long story short, when using a sizeable series resistor
in addition to the passbank, make sure that the SOA of the transistors
is not exceeded for _all_ intermediate currents; and this is not a
simple matter of power dissipation only.
spam@frontiernet.net C what I mean
"Lostgallifreyan" <no-...@nowhere.net> wrote in message
news:Xns9801BF5BAAD...@140.99.99.130...
Yep.. now you know!
Buffo
> So before all I would check the transistor voltage specification
> to start with. More precisely, what is relevant is the so-called
> safe operating area (SOA). This is the region in the voltage/current
> diagram that is safe (ie no second breakdown occurs). The point is
> that for many high voltage transistors, the current they can take
> at elevated voltages is much less than what corresponds to the max
> power dissipation.
Great information! Thanks! I didn't realise the current capacity
would drop off so much when the voltage increased. And RE: SOA... I
see now that there's a lot more to be looked at beyond just the waste
heat! (I never even considered that. Sigh....)
Adam
Buffo
C what I mean wrote:
> Negative resistance relating to Ion tubes made easy :
>
> In a standard DC circuit if you increase the voltage across a resistance the
> current draw thru that resistor will increase proportionately. E/R=I, good
> ole' Ohm's law. With an ion tube the same applies while generating
> plasma.... kinda most of the time, (some times it is non linear). In a
> plasma tube you can have areas of operation where the current actually
> decreases as the voltage applied goes up. It displays characteristics of
> having negative resistance.... Therefore the term negative resistance. If
> "E" go up and "I" goes down, then R must have gone negative relative to it's
> last measured point.. It is not a real negative resistance that you can
> measure with an Ohm meter. It is an effect that you can quantify
> dynamically....
Hey Craig;
Thanks for the explaination. Ok - that makes more sense. Basically at
some point the extra voltage applied does something funny to the plasma
and it doesn't carry as much current, so even with higher voltage the
current flow drops off. (Though I think I would have rather called it
"variable resistance", or "increased resistance with voltage" rather
than negative resistance, but that's just me....)
But here's the rub: I've read that the main reason for having a large
resistor on a HeNe head is to overcome the negative resistance of the
tube. Now, how does that series resistor change anything? I mean, if
you reach a certain point where the plasma tube tends to start
exhibiting negative resistance, it's still going to have an effect on
the total current flow, right?
Or is the balast resistor sized in such a way that it's effect on the
tube current is several orders of magnitude smaller that that of the
tube's negative resistance?
Adam
spam@frontiernet.net C what I mean
"Buffo" <buffo...@gmail.com> wrote in message
news:1153071259.9...@s13g2000cwa.googlegroups.com...
OK.. that is a question I can only guess the answer to since I have not
played with HeNe tubes much....
My guest is this: Before the tube on the HeNe starts the supply puts out a
lot of open circuit voltage. At some point the HeNe ionizes and becomes a
much lower resistance then when it was off (negative resistance term
inserted here). That ballast resistor will drop a pretty fair amount of
voltage when the current starts to flow. I guess that takes the hit
partially away from the supply. Makes HeNe supplies less costly and the
drop across the resistor is most likely faster then the supply could
regulate down to.
Again, that is just a guess. I remember years.. ok ok.. decades ago...
that I was told the ballast resistor needed to be right next to the laser.
I don't know why, but that is what I remember..
Anyway.. if I am off base here someone please chime in!
Craig