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HELP PLS. Automatic Led Emergency Light

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The BD140 switches on the light when the power fails. As long as power is present, the voltage at the base of that PNP transistor is higher than its emitter. A PNP requires a voltage lower than its emitter. When the power fails, the base is brought low by the 560Ω resistor connected to ground. Then the BD140 conducts, and the lights come on.

The 12 LEDs are connected in parallel, with series resistors on each to limit current draw. The LM317 and the associated transistor network is a current-limited voltage regulator, and is used to charge the +4.5V battery safely from the +9V source. This +9V comes from the transformer and bridge circuit on the left, which gets its AC power from the wall outlet.
 
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When there is no 230VAC and no 9VAC then both 560 ohm resistors turn on the BD140. The LEDs can light.
----------Maybe---------
Then the 230VAC is present and the 9VAC is present then one resistor pulls down to ground and the other pulls up from 12 volts. The base of the BD140 us at 6 volts while the emitter is at 6 volts. The BD140 is off so the LEDs will not light.
----------Maybe---------Why the maybe? When there is AC power I think the base of the transistor should be pulled up harder.
 
Probably wouldn't hurt to drop R15 to 470Ω if you are having trouble with that - but I think the circuit will be off because a rectified and filtered 9V transformer is usually more like 10V-12V, so half that is around 5V-6V, and the PNP will be off around 3.8V (4.5V - .7v) so it looks like there's a margin for error.

Additionally, that "4.5V" is probably three nicads or nimh, so it's probably barely 4V on that 4.5V rail even under charge (nicads and nimh are only about 1.2V per cell, so 3.8V).

If you have a dead cell that voltage could rise, and the light would come on. Is that what's happening?
 
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