If you have a low voltage source, say 5 V, it is common to use parallel strings of LED and resistor combinations, such as the right circuit in the second image. This works, it is just a lot of wasted power. Using a single LED will mean a much higher voltage drop across the resistor. If you have a high voltage source, say 12 V, it is common to use parallel strings of series LEDs, such as the right circuit in the first image. Each LED and resistor combination will have a total voltage drop equal to each other LED and resistor combination, but each LED can have a different current, depending upon the size of the resistor. Had you placed a resistor greater than 200 ohms in series with the LED it would have been fine: \$\dfrac = V_R / R\$. It will shine very brightly for a second, then overheat from the high current draw and burn out. An LED has almost no internal resistance, so connecting it directly to a battery will put the full battery voltage through the LED. However, each LED has a maximum voltage and current rating. In general, the more current through an LED, the higher its voltage drop will be, and the brighter the LED will shine. Unless you are just messing around and learning, you should always use a series resistor with an LED to control the current through it. Common 3 mm or 5 mm LEDs are more in the 40 mA range. Also, different kinds have vastly different current ratings, from 20 mA to 2 A. White LEDs are typically around 3 V while red LEDs are typically around 2 V. LEDs have some pretty nasty chemicals inside of them that aren't meant for inhalation.Įvery type of LED is different. Also, if you burn out an LED, don't breathe in the fumes from it. Similarily, you could probably get away with connecting 3 LEDs in series across a 9V battery, but I wouldn't recommend it. Any lower and they probably won't light up. Any higher voltage will burn the LEDs out. To use the LEDs without a resistor, you connect all 3 in parallel across 3 V (a 3.3 V coin cell battery, or 2xAA batteries in series). The current to voltage ratio is usually directly related: as one goes up or down, the other goes up or down.įor a 9 V battery with virtually no resistance, the LED current would have been very high. So if 20 mA are passed through the LED, it will "drop" 3 V. If an LED is listed for a specific voltage, that means it will typically "drop" that much voltage at some listed current - typically 20 mA. To start, yes, that LED is definitely burned out.
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