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I'll keep it brief. I've dropped a flashlight on the floor, after which the casing broke a bit. The flashlight normally works with a single 1.5V 'D' battery. I was looking into it and tried these steps in a chronological order (number 1 is the one I tried first):

1) Connected a 1.5V 'D' battery (with juice left in it) directly to the circuit with no luck. 2) Connected my power supply set at 3V (that's the MIN voltage I can set it at) directly to the circuit and the LED worked. 3) Connected x2 1.5V 'D' batteries, as a test and worked again.

So after the drop the LED lights up only with x2 batteries at a total of 3V instead of its original 1.5V.

There is nothing wrong with the PCB (it's a very basic design); to test this I repeated 1-3 with wires directly touching the LED. Same results again. So I guess something must have 'changed' in the LED. Is this possible? It's all a bit strange and interesting at the same time :). Some photos follow.

This is my first post here, so I hope all is clear!

the LED

PCB 1

PCB 2

power supply

DIY test kit :)

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    \$\begingroup\$ How much did you pay for the flashlight? \$\endgroup\$ – Andy aka May 19 '20 at 8:29
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If the flashligt worked on a 1.5 V D-cell it has to have a DCDC boost converter otherwise it cannot make a LED light up as most LEDs need at least 3 V for that.

The drop could have caused a fault in the DCDC converter. Depending on what and how a component breaks in such a DCDC converter, the input voltage might still reach the output. In that case, applying 3 V at the input might indeed light up the LED.

The circuit you see might be simple as it is just a switch. My guess is that there is a more complex circuit close to the LED.

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