power-off alarm

316

07:44:14

This is a very simple circuit I designed.It gives out a sharp alarm(LED lights up for a second) when the switch breaks open(to indicate the power is down).
The problems be:
1.It consumes 3mA continuously when the power is up.This energy is wasted,though not a lot,can it be eliminated?
2.There is a voltage drop on the transistor between the capacitor and the led,so this circuit won't work with a 3V LED or a 3V buzzer.
3.How can I make the LED light up longer without using a battery or a super capacitor?

published 9 years ago

CatTV

9 years ago

Ooops. While reading the description I mixed up mA and MA. Lol. 😜

BillyT

9 years ago

Because the biasing of a transistor requires a bit of current, the only answer in a simple circuit appears to be to use a FET; http://everycircuit.com/circuit/5719910546407424

vlad_fine

9 years ago

Хуесос

Yanghelium

9 years ago

ugh12 year olds

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