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FrozenGate by Avery

Playing with my new 40W laser

Wow, that's awesome!
What does the power supply draw? I have read they can be up to about 20% efficient, that should give you an absolute maximum power rating i guess.
 





bobhaha said:
...Since the CO2 lasers power is dependent on its length...
Click to expand...

Meh, sorta. Most of the chinese 40W tubes were 1m a few years ago.

A closer approximation of output power would be tube current. If it were indeed 70-80W, it would likely be operating somewhere around 28mA while a 40W would be somewhere around 18mA. It might be they've given you a 60W tube with a 40W supply.

It is safe to measure the current with any standard multimeter, just be sure it's unplugged when you insert it or remove it from the circuit. Also, be sure to measure current on the low side - between tube negative and PSU negative.
 
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Ok, well I did some measurements...

I'm getting kinda strange results...

Idle (Power supply) --------- Active

Current ------ 17mA -------- 2.5A
Power --------36.2W ------- 426W
Voltage ------ 237V -------- 237V
Tube Current - 0mA -------- 21mA

Considering 20% max theoretical efficiency that would be 77.96W. Is it safe to say it is greater than 40W?

What would you conclude from these results?
 
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1.3m and your readings are pretty consistent with a 80W tube. Lucky bastard :p
 
The tube current suggests somewhere between 40 and 50W.

The 20% max is the efficiency of the resonator, not the wall-plug efficiency. Your PSU probably isn't 100% efficient. :p

Try measuring it on a make-shift calorimeter. (ask teh google)
 
Cyparagon said:
The tube current suggests somewhere between 40 and 50W.

The 20% max is the efficiency of the resonator, not the wall-plug efficiency. Your PSU probably isn't 100% efficient. :p

Try measuring it on a make-shift calorimeter. (ask teh google)
Click to expand...

I was going to try that... BlueFusion and Things told me to shoot the laser into a can filled with water for a minute and observe the difference in heat. Would this be the best way to do it?
 
Things said:
Only way to do it properly for a laser cutter is with a ZnSe combiner. I was considering one for mine, but they are mega expensive at around $90. Would save a lot of time and effort (and many burns to yourself) during alignment though.
Click to expand...

Yep, this kind of lens is very expensive :) if you are intrested I can sell mine for 75$
 
bobhaha said:
BlueFusion and Things told me to shoot the laser into a can filled with water for a minute and observe the difference in heat. Would this be the best way to do it?
Click to expand...

No, nothing you can do with spare parts is the best way to do it :p. Ideally, you want a highly insulated container, constant stirring, and very precise ways of measuring the volume of water and the temperature of water. It would help to "calibrate" it with a power resistor, too, since the thermal capacity of the system changes slightly with the hardness of water, and the container it is in.

Maybe the easiest way to do it is to just shove a known power into it (via power resistor), then compare with the laser.
 
Cyparagon said:
The tube current suggests somewhere between 40 and 50W.

The 20% max is the efficiency of the resonator, not the wall-plug efficiency. Your PSU probably isn't 100% efficient. :p

Try measuring it on a make-shift calorimeter. (ask teh google)
Click to expand...

Also I /kinda/ took the PSU into account. I took out the idle power consumption from the PSU, when I calculated the power based on efficiency..

No, nothing you can do with spare parts is the best way to do it . Ideally, you want a highly insulated container, constant stirring, and very precise ways of measuring the volume of water and the temperature of water. It would help to "calibrate" it with a power resistor, too, since the thermal capacity of the system changes slightly with the hardness of water, and the container it is in.

Maybe the easiest way to do it is to just shove a known power into it (via power resistor), then compare with the laser.
Click to expand...

I can try that... would styrofoam be a good enough insulator?
 
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Not how this would work with a laser, but this is how we used to do it in the the RF world:

Back when GHz range power meters were expensive I ran RF through 100m or so of very lossy coax (RG-174) in an esky filled with a measured amount of ice water at zero degrees C. Most of the RF was converted to heat, and you can calculate the calories (and power in watts) from the temperature rise of the water.

Laser wise, I just read that 10600nm is almost totally absorbed by water so the same principle may well work there?
 
getting a bigger tube than orderd would be nice :D , My 60 W Tube is 1.2m Long .
 
That is an excellent video demo. :beer:
I would love to have a CO2 laser as well.. But I fear the space I have in S.K. is far too small to have one.
 
Nice vid I want a co2 laser but I would rather have a wavelength in the 1000nm range than 10.6um because I cant even see that with a camera :( I understand that its better for burning though because platic and glass absorbs it wheras 1000nm it much more like visible light
 
wow that is something
Amazing laser indeed but why cant we see the beam ?
 
hoo7h said:
wow that is something
Amazing laser indeed but why cant we see the beam ?
Click to expand...

I believe in the video he says its a wavelength that is much longer than what the human eye normally perceives(between 380-750nm). The wavelength is also blocked by normal plastics and glass.
 
AUS said:
Not how this would work with a laser, but this is how we used to do it in the the RF world:

Back when GHz range power meters were expensive I ran RF through 100m or so of very lossy coax (RG-174) in an esky filled with a measured amount of ice water at zero degrees C. Most of the RF was converted to heat, and you can calculate the calories (and power in watts) from the temperature rise of the water.

Laser wise, I just read that 10600nm is almost totally absorbed by water so the same principle may well work there?
Click to expand...

I've tried this before with my laser, I find it is absorbed too well. All 40~watts ends up being absorbed by such a small volume of water most of it just boils away and takes the heat with it.
I can imagine it working well if a beam expander was used to distribute the light across the surface, but then you get some losses from the optics.
 
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