Lightning compared to piezo electric element

AI Thread Summary
The discussion focuses on building a lightning detector that measures electromagnetic charges from lightning bolts. A piezoelectric element from an electric lighter is used to generate a small spark for testing, producing 20 to 25 kV. Participants discuss the need for proper terminology, emphasizing that lightning emits an electromagnetic pulse rather than a charge. They suggest that the relationship between the spark and lightning can be estimated proportionally, considering the air's conductivity at approximately 3 million V/m. Calibration of the detector is crucial, with recommendations for using a wideband detector and peak sensing circuits for accurate measurements.
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I'm building (or rather, debugging) a lightning detector that detects the electromagnetic charge emitted by a lightning bolt.

I need a good way to generate a little lightning as a test signal and I'm currently using a the piezo electric part of an electric lighter which generates a small spark.

The piezo electric element generates about 20 to 25 kV and produces a spark that is about 3 mm long. If I hold it two meters from my antenna, at what distance does an average lightning bolt need to be to produce the same amount of energy in my antenna?

Of course a lot of estimations and averages will need to be used, but roughly?
 
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You need to sort out your terms of reference, I think. A lightning strike 'emits' an EM pulse and not a "charge". You can save yourself a lot of time and effort by learning more basics before launching out on haphazard measurements, imo.
 
Pulse is probably the word yes.

I know it is a difficult question to answer but someone might know how to make such an estimation, I hope.
 
The relationship will be proportional, assuming the air becomes conductive at the same electric field strength. This is about 3 million V/m, or 30,000 V/mm. Your 20-25kV was probably able to generate a current over a larger 3 mm due to geometric considerations (the field is concentrated near sharp edges of a conductor).

To find the voltage for 2m, just multiply 25[kV] by 2[m]/3[mm]
 
Calibration can be difficult but what you need is a wideband detector with a peak sensing circuit. Something along the lines of a peak programme audio circuit. Having calibrated it with low level signals, you just put a large attenuator at its input to measure high levels.
[Edit: a log amp would be useful for showing a wide range of signal values (dB scale).]
 
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