HeNe laser power supply

Posted by Author makecircuits

 HeNe laser power supply

This is the power supply I traced out and measured which is in an Edmond Scientific 0.5 mw. Laser circa probably around 1975. I bought a 1mw. tube (1986) when the old one broke. It is still running just fine. I think it is a rather clever design.

Transformer Secondary |-----||--------------| 710V. rms | .001 | | | ||C---*->|-->|-->|---*-----*--->|-->|-*-->|-->|--*-->|-->|--* ||C | | | | | ||C | +| | .001 | ||C --- /1Meg *---------||----------* ||C | --- | | ||C | *-----* --- | ||C | 4.7uF (4) +| --- .001 ||C | 450 V. --- / | 150K / ||C | --- 1Meg | ||C | | / | / ||C---/|------------*-----* | | +| | | | --- /1Meg | 0 | --- | + | *-----* | Neg. Out Pos. Out | +| | - | --- /1Meg | 0 | --- | | | | / | | *-|<--|<--|<--*-----*---------/|------| | | |------------------------------| Ah! A work of art, eh ?

My notes show that the transformer secondary measures 710 V RMS while in operation. The 710 V transformer output has a voltage doubler (six diodes and 4 caps), but my notes on the schematic say that there is 1750V or 1800 V across all the electrolytics.

The other diodes and the .001`s form a classical voltage multiplier ladder which I think is a tripler (I can`t quite decipher my notes from 10 years ago and I`m not in the mood to try to figure it out). If it is, this gives a starting voltage of 7000V. - 7200V.

Because the .001`s are so small, the multiplier can`t really supply much current. Once the tube lights and current starts to flow, the ladder just becomes some series diodes. Then you`re back to the voltage doubler output, with a few diodes in series. No relay to fire the thing off as in the modern switching supplies. Pretty neat, huh?

All diodes 1N4007. For some reason, they don`t have voltage equalizing resistors as the electrolytic capacitors do. I am guessing on the 1Meg resistors because I don`t want to take the thing apart. 10Meg probably ok too. The .001`s all appear to have 1800 V across them, so they have to be 2kV or better. A rule of thumb is only have half the rated voltage on ANY cap., but you see Edmond didn`t do it and this thing`s still bookin`.

The transformer will probably be a problem. It does not have to be one transformer. You can put tw1o or three secondaries in series, but don`t get carried away since a lower voltage transformer may not be insulated for the higher voltage. If you find a single one (or the lowest current rating of the ones in series) it should have a secondary current rating of probably, oh, I`d say 10ma or better.

The 150K "ballast" resistor is actually constructed from 4 - 33K resistors and one 18K resistor all in series. It doesn`t have to be, but it`s convenient to build it up from several in series. My notes show 600V. across it and a dissipation of 2.4W. and a tube current of 4ma. Of course, building it up this way, they should each be 1W resistors. If they are each 0.5W. they may get a bit too hot. One rule of thumb says to dissipate only half their rating. This resistor should be located close to the tube with AS SHORT a LEAD AS POSSIBLE and as LITTLE capacitance to surroundings as possible. The tube needs to see a high impedance source. This isn`t super critical, but keep the wire down to 1-3 inches and the first few resistors away from any case or ground material.

Now, the ballast resistor, the actual voltage at the doubler output and the tube running voltage determine the tube current. Look at the tube ratings for the running voltage and tube current. My notes say 1150V and 4ma. Since all HeHe`s are pretty similar With a supply of 1750V. and a tube voltage of 1150V. this leaves 600V. across the ballast resistor. 600V. and 4ma better come to 150K or I goofed somewhere. Once you get the transformer picked out, make your best estimate of the output voltage you`ll have. It should be above 1400V for a 0.5mw tube. Any less and you could get into trouble since the ballast resistor should be kept above 50K. Typically, they are 70K, if I remember.

Working the other way... First calculate the "ideal" ballast resistor voltage drop as, E = tube current * 70K. To the tube voltage, add the voltage across the ballast resistor to get the required supply voltage. This would be a minimum supply voltage for that tube. If yours gives a few hundred volts more, you can make the ballast resistor larger to get the required tube current. It`s just that the bigger the ballast resistor, the more power you throw away. If you think you may go to a more powerful tube in the future, then get a higher voltage transformer output for the bigger tube and just use the larger ballast resistor for now.