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Old 08-29-2015, 05:31 AM #1
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Default Quick Hene Question

Hi guys, I recently got my first HeNe and have just finished building a power supply based on the following schematic (from The Professor's Homebuilt Laser Page)


I know its not the most efficient power supply, and it's definitely not going to be permanent, but I happened to have all the necessary parts on hand, so I thought I'd give it a go.
I connected the cathode of my tube (the big metal can electrode, according to the FAQ), to the ground/centre tap and the ballast resistor (2 megaohms!), to the anode, and all was working nicely, although the beam seemed a little dimmer than expected.
After letting it run for a while though, I noticed a drop in power output, down to the level of nothing but a very faint dot, barely visible on the opposite wall. Being paranoid that I'd managed to get the polarity wrong, I swapped the anode and cathode, grounding the electrode without the can, and connecting the output of the rectifier to the metal can electrode. I switched it on (for no more than about 5 seconds, just in case), and I noticed a BIG increase in power output, from a faint dot, to something several times a commercial cat-teaser (and of a much nicer colour!).
I also noticed that with the metal can electrode (cathode?), connected to ground/the centre tap, the other electrode got quite hot, which of course, the cathode is supposed to do! The metal can electrode stayed totally cool, while the other electrode got quite warm to the touch.
So, I guess my question is this, is there any way that either the metal can electrode is actually the anode, or that I've been stupid and ground is effectively positive in this circuit?
Also, I'm sorry if this is a stupid question, I just really want to make sure I keep my new HeNe nice and healthy!

EDIT: I realise I left out the details of my setup. I'm running from a rectified 15kV neon sign transformer to give 10600VDC, assuming around 1600V operating voltage for the tube, that leaves around 9000V being dropped across the 2 megaohm ballast for a tube current of around 4.5mA, this is on a 30mm by 240mm tube. My filter capacitors are a touch over 0.5uF.

EDIT 2: I did another quick test run in both polarities, just long enough to take a photo of the beam. I really don't want to run it for more than a few seconds at a time unil I've got this polarity issue figured out, but I thought a few pictures might help.



The first picture is the low, barely visible output when the metal can (cathode?), is connected to the centre tap/ground which. The second much brighter dot is that produced when I reverse the polarity, attaching the metal can electrode to the ballast resistor side.
I also noticed a slight ghost beam in this polarity, and a rather significant amount of light output from the HR.
Thanks in advance for any and all help, this laser is absolutely hypnotic!



Last edited by alf638; 08-29-2015 at 12:40 PM. Reason: Forgot some details, and adding pictures
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Old 08-30-2015, 12:30 AM #2
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Default Re: Quick Hene Question

Yes, this is perfectly possible.

On (high power) vacuum tubes in general, the anode gets hot and needs to be heatsinked. This is because electrons move from negative to positive and deposit their kinetic energy onto the anode. The cathode is heated in your typical old fashioned vacuum tube to stimulate it to emit electrons, but this is not the case in a HeNe laser.

Like x-ray tubes, RF amplifier tubes etc the anode is the part that gets hot and needs to be heatsinked. The assumption that the larger part needs to be grounded (as with most solid state devices) does not hold true there.

You can, however, solve this quite easily: just reverse the diodes and capacitors in your supply circuit. You will be driving the cathode of the tube with a negative voltage, but have the 'large' part at ground potential in case you would accidently touch it.
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Old 08-30-2015, 09:10 AM #3
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Default Re: Quick Hene Question

Alright, thank you! I tried reversing the diodes (the capacitors aren't polarised), and as you predicted, I got the bright output with the metal can electrode connected to the centre tap.
So, would I be correct in saying that with the diodes in their original polarity I was effectively passing the negative component of the AC signal, causing the small electrode to act as a cathode and the large electrode connected to the centre tap to act as the anode, thus driving the tube in reverse, and having the centre tap/ground act as an "electron sink"?
As is, with the diodes reversed (arrow pointing towards the transformer), I'm getting the bright output with the large electrode connected to the centre tap and the small electrode connected to the ballast resistor, is there any way this is unhealthy for the tube? I'm just a little worried, as it seems to me that in this configuration the negative part of the AC wave would be passed to the small electrode, which I believe to be the anode, which would be damaging, right?
I bet I'm just over-complicating this, if the old configuration gave a very weak beam, and this one gives a much brighter beam, this has to be the right way around, right?
I'm sorry if this seems really silly, I'm still studying electricity and whatnot at university, and we haven't really covered much on AC, so some of these concepts are still pretty new to me, thanks so much for your help!
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