Higher power PS on lower power tubes - high voltage discussion

[ARuk (Ante)]

Hello,

I have been lurking this forum (along with several others) from time to time when I was trying to solve issues on my own, and most often than not I would find a solution.

I have finally reached a point where no search would help answer my question and seeing that this community has a lot of users who spare no keyboard, I decided to turn here.

So after short introduction, here goes my first post:

I have been using 80W tube and 80W power supply.
New tube I got is 90W and manufacturers don’t have 80W tubes anymore.

Looks the same, dimensions are the same, and I can’t tell what changed exactly that it now provides 10W more.

For whatever reason, manufacturers hide detailed technical data and do not keep the old ones, so I cannot compare current and voltage numbers of the old and new tube.

What bugs me is this - to get most out of the tube, while not sacrificing longevity (longevity is imperative) would it make more sense now to use 80W power supply or 100W?
Also, what would happen if I were to use lets say 130W power supply?

When using PS with higher power than tube rated power, I’ve seen posts where people don’t get any beam output when using less than 20% or over 80% power in settings. *1
Everybody keeps talking about current and how stronger PS will shorten tube life if used with more current than the tube is designed for to handle. I get that.

Power is voltage times current.

What happens when a tube is rated for lets say 24mA (for long life), 19kV ignition voltage and 17kV working voltage, and it gets connected to a PS that provides lets say 40kV and 36kV working voltage, while at the same time using 24mA or less?

It should generate more power, but I am inclined to think that it comes under a cost. How would it affect the tube long term?
Has anyone actually tested and played with this?

*1 Seems to me that it is voltage related and I’ve never seen anyone discuss this.

[bernd.dk (Bernd)]

The question is interesting but also a bit academic
First, when we talk about standard Chinese laser tubes, we know nothing about their real values ​​or quality. Without having tested it, however, I think that when they are the manufacturers who sell quality tubes with the associated technical specifications and a reasonable guarantee, then we can count on their specifications.
Energy output, ie the optical / mechanical effect of a laser tube is proportional to its physical dimensions. Beyond that, there is an area of ​​an unknown tolerance in terms of effect. The gas mixture in a CO2 laser tube reacts with the current with which it is exposed. There is a certain ratio of current and the effect that the gas “can give” in the tube, above this level you get no additional effect but breaks down / destroys the gas instead.
If you want to know how much power a given laser tube has, do a test. Increase the output power by an e.g. factor 10 and record when you get the most penetration.
The picture shows a test with a nominal 50 Watt laser tube, from left to right the energy is increased by 10%. At 80%, the maximum penetration is achieved, then it actually goes back a little in effect.

[ARuk (Ante)]

Interesting, I knew of such penetration testing methods but never did them myself because I didn’t see the need to. (And I just took for granted that power = penetration.)

What’s the take on significantly increasing voltage, while keeping the current in safety zone check?
Would you say that gases would break down at accelerated pace despite the presence of catalyst?
I am under impression that current would help speed up gas decomposition, but I am unsure of what I think of voltage…

In either case, I am happy to learn!

[bernd.dk (Bernd)]

@ARuk

I didn’t see the need to.

I think it is extremely important to know where the boundaries of one’s tool are. Take a drill and different drills for example, revolutions (and the angle of its cutting surface) are decisive for the result. If you use the same parameter that applies to a drill, intended for steel and use it for a concrete drill, then it will destroy it very quickly.
Why should we use 24mA when 18mA is the maximum effective output?
As far as I know, the volts are relatively “statically” stable, you have an ignition voltage of approx. 20 KV and then the actual operating voltage of approx. 16 KV (at a 40Watt tube). The actual regulation takes place with screw up and down for the current (mA), … but then it also becomes too electric for me
I bet Russ Sadler has an explanation for your questions, search him on YouTube if you do not know him in advance. But be prepared, you must set aside some time for this magnificent man.

[jkwilborn (Jack Wilborn)]

When a gas tube lases, it exhibits a trait known as ‘negative resistance’. Meaning that for an increase in voltage across the tube the current will drop. No passive device exhibits this behavior. I don’t think your normal ohms law and power laws really work as expected with this type of device.

If you are thinking at that level then you must remember there is no ‘power’ control of the laser. It is on or off depending on the state of the pwm signal. There is no 50% power, when it lases, it is full power. What you are reading is the rms value of it, whether mA or kV, unless you can keep it on long enough to get an accurate reading. The illusion of power is that of the pwm control over time.

My chemistry of what’s happening here kind of poops out. It’s a very erratic signal, the hv.

I have a mA and kV meter on my machine, both rms.

Russ Sadler video on HV power supplies.

RDWorks Learning Lab 97 Understand Your HV Power Supply via Florida

Good luck…