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I just have a question about this simple circuit or example: http://www.555-timer-circuits.com/flashing-led.html. I just don't get 1) why do they use 9V battery to power a single red LED light. 2) why there need to be a 1k resistor before the led?

Is there anyway to power a 12V LED (current: 0.08A) by 12V battery efficiently (optimal brightness) or i have to use a slightly higher voltage?

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I suspect that they use a 9 volt battery since it is a handy, readily-available power source.

The electrical characteristics of LEDs (and diodes in general) are a bit odd - they don't follow Ohm's Law. The voltage across an ordinary ("bare") LED depends primarily on its colour, and will vary only slightly with current (for a red LED, the voltage will be about 1.8 volts). Therefore it is necessary to include something in an LED circuit to limit the current to a safe value - the 1K resistor in the circuit you linked to serves as the currrent limiter.

An LED advertised as "12 volts" or "5 volts" will include a series resistor or other current-limiting device, and can be connected directly to the advertised voltage with no other components.

A "bare" red LED will have a voltage drop of about 1.8 volts, yellow is about 2 volts, and green is about 2.2 volts. Blue and white are about 3.3 volts (white LEDs are really blue LEDs with a yellow phosphor).

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  • \$\begingroup\$ that means the LED will have the same brightness if it is powered by 5V battery instead right? One more thing, does that circuit consume battery significantly (since i make it blink to prolong battery life)? \$\endgroup\$ – user49148 Sep 2 '14 at 3:51
  • \$\begingroup\$ No. If you power the circuit with 5 volts instead of 9, the current through the LED will be reduced, which will result in less light. With the 9 volt battery, the LED will pass about 7 mA, but will only pass 3 mA with a 5 volt supply. If you change the 1K resistor to 420 ohms, the LED will then pass 7 mA with a 5 volt supply, so should have the same brightness as the original circuit with a 9 volt supply. \$\endgroup\$ – Peter Bennett Sep 2 '14 at 4:04

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