When light falls on the
phototransistor, it begins to conduct up to about 1.5 mA, which pulls down the
voltage at the lower side of the resistor by 1.5 V, turning off the transistor,
which turns off the LED. When it's dark, the transistor is able to conduct
about 15 mA through the LED. So, the circuit uses only about 1/10 as much
current while the LED is off. One thing to note about this circuit: We're using
a red LED. That's because the voltage drop across the transistor allows less
than the full 3 V across the LED. The full three volts is really only marginal
for driving blue LEDs anyway, so two-point-something really doesn't cut it.
SOURCE:http://www.evilmadscientist.com/article.php/nightlight#comments
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