That’s right, I don’t know and have never known how much watt output my lasers have in reality.
But the test clearly indicates that my tube is losing some of its original power, and more important that I know where the boundaries of the machine/tube are.
I just found the test from when the machine was new, the curve geometry is completely similar just in a higher area.
You can quicker and easier find the difference of your output effect with a proper power meter, can you recognize the “power curve” I talk about?, When does your curve fall or when does it not rise anymore?
I see the curve along with the peak… I think mine was maximum wattage out at about 85% power or almost 18mA. I have had a lot of 2nd thoughts about what the working maximum and do not exceed maximums are for the supplied tube.
I have not done this with the new tube… I have a Cloud Ray 40W to replace the original OMTech tube, supposedly 50W but more like 40W.
There are more variables to tube life… I know mine was out in the garage at 115 deg heat for a big part of previous summer and a lot of 100 plus days this summer…
I also know that it lases at a lower threshold when it’s at a lower temperature than it does when warmer.
Not only is the mA draw nonlinear, but neither is the output power. Maybe some of this is from it being a negative resistance device.
I did not have a voltmeter on the machine when I did the power/mA testing. For the new tube, I will note the hv values to see if it really relates to anything useful.
I did expect with both at 50% power, a continuous line drawn would show a higher average voltage than a scanned graphic where most of the engraving wasn’t being lased… It was the opposite…
@ednisley has done some of these he plotted… don’t know if he used the acrylic or not… seems to me he did…
I have read your tests, very interesting I think, although I am not particularly deep into the electronic background.
If I have understood your test and summary correctly, my own “observations” and “statement” are correct.
I cannot measure the tube voltage along with the current, but I’m reasonably sure it does not increase enough to compensate for the flattening of the current, which means the tube output power must either flatten or decrease. Your beam penetration measurements suggest the latter, with the effective power decreasing as the PWM increases beyond 55%.
Your recent measurements show the penetration at 15% is about ⅔ of the maximum at 55%, so the effective power variation isn’t nearly as much as the PWM percentage variation would suggest.
If your instruments will resolve in the millivolt range, the HV power supply already includes a 100 Ω SMD resistor between the cathode lead and common: a 10 mA tube current puts the cathode wire 1 V above ground.
Yep, a mA meter is basically measuring voltage across a shunt. A shunt can go from the high side, supply to load, or low side, load to ground. A shunt often has good heat dissipation for a consistent measurement, select ohm value to suit current, 5amp, 50amp, 100amp, etc.
That’s similar to the one that came with the Cloudray power supply I bought as a backup and plugs into the RJ45 jack on the supply.
AFAICT it’s the same circuitry as the built-in display on the original (failed) & replacement OMTech / JYE supplies in my laser. The RJ45 jack has a pin going to the 100 Ω current-sense resistor in the supply, so I assume it’s displaying that voltage scaled to represent tube current.
The current adjustment knob only works with the 100 and 150 W power supplies. I’m willing to believe it scales the current downward from whatever the internal trimpot sets, because I doubt it would increase the setpoint. Seems like a bad idea in either case, as all your Material Library power settings would suddenly stop producing the right results.
Nope, just the reverse-engineered sketches from K40 power supplies. They’re similar to the HV supplies in larger machines, but definitely not the same.
I’d expect the pot in the display to feed a scaled version of the trimpot setting back into the supply, where the usual comparator would balance it against the sampled current. But it could be a complete replacement for the internal trimpot, using a bypass available only in the 100 and 150 W supplies.
