Depending on what the pot is connected to, can't you use just part of the rotary range and compensate electronically for the fact that the resistance range will be from something more than zero to something less than 25k?
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Messing with the ABS is dangerous (particularly so given your inexperience) and is likely to invalidate your insurance. Why do you want to it anyway?
You'd probably have to bypass the processor (that multi-legged component half way across the pcb) and re-configure connections to the 5 wires feeding the LED array. Without knowing the specs of either the processor or array the chances of success are slim :(.
In the present context it's an oscillator which drives the piezo and has positive feedback (to maintain oscillation) obtained by sampling the current through, or oscillation amplitude of, the piezo.
See, for example, Fig 9 of this article.
If you did have sensors placed as shown, how far apart would they be? Metres? Kilometres? Bearing in mind that radio signals travel at around 30cm per nS, any method involving time delays would need extremely accurate timing of very short delays.
To get maximum output from a piezo it needs to be driven at its resonant frequency. That frequency may not be constant if the loading on the piezo varies. Therefore, a standard way to drive a piezo is to make it a frequency-controlling part of a self-resonant circuit. Google should find you...
Absolutely not.
Pond pumps for domestic use are designed for low voltage in the interests of safety. Mains voltage and water are not good companions. Surely Belgium has somewhere (a Garden Centre?) selling 12V pumps?
A 3 inch (8 Ohm) speaker (cheap transistor radio type) will typically handle about 1W of power, so a 1W amplifier should suffice for that.
We don't know the current draw of the toy, but I doubt it would be as much as 500mA. The fuse would go between toy and converter (not battery). The converter...
The black blob hides a miroprocessor. It's possible (I'm only guessing) that it monitors current produced by the solar panel and only turns on the LED when it detects a reduction in current, which would normally occur at twilight. There is no such reduction if you just insert a fully-charged...