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Jarrod (http://www.jarrodkinsey.org/) was recently wondering about the relation between current, voltage, tube size etc. in a CO2 laser tube, so I thought I'd try and measure the tube in my laser cutter.
The tube was originally 40W, but has quite a few hours on it so the output power is probably much lower - around 25W, according to my previous measurements. Discharge length is 64cm and the bore is 8mm.
Measuring the current was easy enough - a 10Ω resistor in series with the panel meter and measure the voltage across it. I didn't want to put my meter directly in the current path.
Voltage was a little more tricky. I made up a voltage divider with a string of ten 47MΩ high-voltage resistors and a 1kΩ bottom resistor, giving a divison ratio of about 460,000:1.
Since the laser tube is being driven directly from the rectified output of the high-frequency transformer in the power supply, there is rather a large AC component in both the current and voltage. However, I'm only measuring the DC average signal, to keep things simple. I also measured readings both with increasing and decreasing "power" (knob setting) in case there were any differences - there weren't.
Here's a rather waffly video showing the process and results:
Graph of current vs. voltage:
Note how the resistance characteristic changes from positive to negative at higher currents. There is also a small kink at around 11mA - I'm not sure what this is. Possibly a transition to a different discharge mode.
Graph of resistance vs. current:
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