If I remember correctly a lot of the older/cheaper photo flash unit (like that in a disposable camera) work as follows:
A push button (shutter button) short circuits a capacitor through a pulse transformer that produces a high voltage impulse incident on the xenon flash tube. This ionizes the gas, allowing it to rapidly discharge a high energy capacitor, all the while producing a bright flash.
If you simply replace the the shutter button with your automotive shock sensor then this might be the only change in existing circuitry you need to make.
So to make this more clear, assuming your shock sensor has two terminals, one that would normally be tied to Vcc and the other ground, just attach these leads in parallel with the shutter switch. Don't worry about having one end tied to ground, as your shock sensor is just acting as a mechanical switch and it doesn't care what potential you have on what end.
A push button (shutter button) short circuits a capacitor through a pulse transformer that produces a high voltage impulse incident on the xenon flash tube. This ionizes the gas, allowing it to rapidly discharge a high energy capacitor, all the while producing a bright flash.
If you simply replace the the shutter button with your automotive shock sensor then this might be the only change in existing circuitry you need to make.
So to make this more clear, assuming your shock sensor has two terminals, one that would normally be tied to Vcc and the other ground, just attach these leads in parallel with the shutter switch. Don't worry about having one end tied to ground, as your shock sensor is just acting as a mechanical switch and it doesn't care what potential you have on what end.
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