What is the current during BaTiO3 voltage breakdown?

In summary, the conversation discusses an experiment on dielectric breakdown using 0.3mm of BaTiO3 between two electrodes. The breakdown occurred at 97kV/mm, or approximately 29kV at 0.3mm thickness. The current used was from a dc high voltage generator with a current limit of 4 mA and a maximum voltage of 35 kV. The relationship between μJ/m and current is not clear, but it was the only information on joules used in the experiment. The conversation also mentions the possibility of negative resistance and the use of similar compounds in voltage protection devices. High voltage sources can cause damage to the dielectric before actual breakdown occurs.
  • #1
HelloCthulhu
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I've been researching dielectric breakdown for a while and came across this interesting experiment:

https://tore.tuhh.de/bitstream/1142..._nano_to_millimeter_scale_TUB_Doc_version.pdf

0.3mm of BaTiO3 was placed between two electrodes and reached its dielectric breakdown at 97kV/mm. At 0.3mm thickness this was probably around 29kV. I'm trying to understand what the current was, but so far I only know that a dc high voltage generator HCN 140-35000 with a current limit of 4 mA and a maximum voltage of 35 kV was used during the experiment. I'm not familiar with the relationship between μJ/m and current if there is one, but it's the only information on joules used I could see. Any assistance is greatly appreciated!

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  • #2
https://ieeexplore.ieee.org/document/38354
Expect negative resistance after breakdown. Apparently the current limit prevented avalanch breakdown. 35K at 4 mil was 150 Watts which is a pretty hefty high voltage source. Similar compounds used in voltage protection devices have an avalanche breackdown that drop line voltage to a few volts. Usually used for electronics capacitors, dielectric has problems at high voltage. (Dielectric is damaged before voltage breakdown.)
 
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