What is the Wavelength of Corona Discharge in High Voltage Testing?

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SUMMARY

The discussion focuses on capturing and analyzing the wavelength of corona discharge during high voltage testing (20-35 kV) using a Canon EOS T7 (2000D) DSLR. Participants suggest using a telescope and a diffraction grating to generate the optical spectrum, with recommendations for gratings such as the G1200TU26.7GB and G1200TU36.9GB, which have 1200 grooves/mm. The importance of groove angle and the choice between transmission and reflection gratings for UV analysis are also emphasized, with specific references to Thorlabs products for optimal efficiency in the 235-400 nm range.

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  • Understanding of high voltage testing principles
  • Familiarity with optical spectroscopy techniques
  • Knowledge of diffraction gratings and their specifications
  • Experience with camera-based spectrometry
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  • Research the use of the Theremino open-source spectrometer software for webcam integration
  • Investigate the effects of groove angle on diffraction efficiency
  • Explore the specifications and applications of Thorlabs diffraction gratings
  • Learn about the calibration of spectrographs using known wavelength sources
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Electrical engineers, optical scientists, and researchers involved in high voltage testing and spectral analysis of corona discharge phenomena.

js2020
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TL;DR
I would like to measure the wavelength of light source
Hi,

I am doing high voltage testing (20-35 kV or more) where I am trying to capture corona discharge. My DUT is in a dark test room with no light. I am using a Canon EOS T7 (2000D) DSLR to capture the corona. What I would like to know is the wavelength of the corona so I can do further investigations. The light it not too bright as you can tell form the picture attached.

Please note: In the picture attached, the green light is coming from a small LED that was on a different measurement device in the room. The LED has since been covered, so the green light is no longer present. I included this image because the green light may makes it a little clearer the size scale of things. This image is of a 10 AWG wire connected to a 16 AWG wire using a ring terminal. The connection is in a beaker filled with transformer oil to eliminate discharges at that point. The other end of the 16 AWG wire has a small plastic connector attached and is in air. The corona is occurring where the wires are inserted in the connector. The purplish color on the beaker is not corona from the connector inside but a reflection from air. All of this is sitting on a shiny white insulator which is also the reason for the small reflection at the bottom of the wire.

Any suggestions on how to measure the wavelength of this light while not actually in the room is much appreciated. I can run cables to any equipment inside the room but objects in the room must be fixed in place.
 

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I expect it will be broadband UV excitation, but if you analyse the spectrum you will see the spectrum of ionised atmospheric gasses, and maybe the metal of the wire.
https://en.wikipedia.org/wiki/Aurora#Colors_and_wavelengths_of_auroral_light
You will need a telescope and a grating to generate the optical spectrum.
The wide UV spectrum will be more difficult to analyse due to transmission of UV.
 
Good idea. I'll see if I can find a grading for a camera lenses specifically. If not, I'm sure I can find a way to attach one. I think the tricky part would be calculating the wavelength though. I would need the distance to the sensor I'm guessing? With such a short distance, I think too rough of an estimate might really skew the calculations.
 
You can calibrate a spectrograph with different chemical composition LEDs that produce known colours.
 
Baluncore said:
You will need a telescope and a grating to generate the optical spectrum.
Ah ok I see what you mean now! I found an open source spectrometer software that can run on windows and work with many webcams. Now it's just a matter of building the telescope and selecting a webcam.
Here is the link in case anyone else is ever interested. Just search for spectrometer
https://www.theremino.com/

Since I'm interested in mainly UV, do you have any suggestions on on key things to keep in mind? So far I've found that the higher number of groves/mm, the smaller the wavelength it can resolve. In my case I'm considering 1200 grooves/mm which they claim is good down to 235 nm. I know probably won't be able to see this low anyway with the camera, but I'd still like to pay a little more and get the better item so I can always upgrade or reuse for more purposes later. How does the groove angle effect things and why should I consider this? I noticed that one is 26.74 degrees and another is 36.87 degrees with the same 1200 gooves/mm.
I am considering the G1200TU26.7GB and G1200TU36.9GB due to the size and grooves/mm (1200/mm).
https://www.dynasil.com/product-category/diffraction-gratings/transmission-gratings/
 
For wide-band you must use reflection from a metallic surface, not transmission.

The groove angle is the approximate tool dimensions.
26.74° = arctan( 1 / 2 );
36.87° = arctan( 3 / 4 );
 
Baluncore said:
For wide-band you must use reflection from a metallic surface, not transmission.

The groove angle is the approximate tool dimensions.
26.74° = arctan( 1 / 2 );
36.87° = arctan( 3 / 4 );

What are the transmission gratings for? I also found some through Thorlabs shown below. From looking at their efficiency plots, it looks like I would actually want either 300 groves/mm (GTU13-03) or 600 groves/mm (GTU13-06) as these offer the highest efficiency from 235-400 nm.
https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=1122

Ok I think I understand about the groove angle. Is Case a or Case b the correct angle it's measuring?
 
Ok actually I think both of those are wrong. I reviewed the grating tutorial on Thorlab's website earlier. They have blaze and transmission listed separate so I thought this angle was not the angle discussed for braze. I just downloaded the PDF drawing of this and it specifically lists it as braze angle. So I understand that now and see both of my previous drawings were incorrect.

I believe this image gamma is the angle it's referring to, right?
 

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