Hello,

This might be a stupid question, but more I think about it more it confuses me.

Consider an LED of 2 V with current rating 20 mA

To connect it to a 12 V supply circuit is as follows


R = (12 - 2) / 0.02 = 500 Ohm

And to connect it to a 24 V supply



R = (24 - 2) / 0.02 = 1.1 K ohm

But what if U use same circuit used for 12 V in 24 V supply. LED will burn out, but I don't understand how



My math here is

current = ( 24 - 2) / 500 = 44 mA

Circuit is capable of passing 44mA.

If I connect a 2V LED to 2V, 20 mA supply, if will glow. Same happens if I connect it to 2V 2A supply. It only draw rated current.
Can't I consider same in the case of above circuit??

If my above calculation is correct 0.044 * 500 = 22 V will be dropped across resistor and remaining 2 V will be across LED, that's safe for LED. Since 2V is not more than the forward voltage of LED, it shouldn't draw much current https://www.sparkfun.com/tutorials/219 and shouldn't burn out

I know what I am saying above is wrong, but I can't figure out where??

Thanks

 

An LED is different than many other common devices such as light bulbs or motors. Those devices will draw a current that is roughly proportional to the voltage supplied, at least within a range of voltage near the nominal voltage. A 10W lightbulb will draw a little less current at 100V than at 120V, and a little more current at 140V (maybe even enough to burn it out).

By contrast, as you raise the voltage applied to an LED, it won't conduct at all until you reach a minimum voltage. As you raise voltage above that minimum voltage, the current increases quickly and will reach a destructive level in only about another 0.5V. So if your LED starts to make light at 1.9V, it'll likely be destroyed at 2.4V as the current goes beyond the 20mA rating of the LED. It can't control itself and will conduct as much as it can until it blows like a fuse.

 

You did not include a datasheet for the LED. There is usually a suggested current, a max continuous current and a peak current.
You can get higher "effective" intensity by pulseing the LED so the eye doesn't see it.

 

You stated that the current rating is 20ma. Yet you are giving it 44ma. What do you expect?

 

Let's use your voltages with the forward voltage of the LED already discounted. 12 volts becomes 10 volts and 24 volts becomes 22 volts. That said - - - :

10V ÷ 500Ω = 0.02A (20 mA) LED survives
22V ÷ 1100Ω = 0.02A (20 mA) LED survives.
22V ÷ 500Ω = 0.044A (44 mA) LED dies. We shed a tear for the lost smoke.

The LED (assuming standard type of LED) can't handle that much current. They typically operate between 5mA and 20mA. If you push it much beyond 20mA gets hotter than it can handle. The hotter it runs the shorter its life. If you push it with 22mA it will probably survive, but it's lifespan will be reduced. 25mA and it's getting much shorter much faster.

IF the LED is connected to a supply that is constant current regulated at 20mA then no matter what voltage you're applying, as long as the current doesn't exceed 20mA then you're safe. Just understand that a constant current supply will regulate the voltage to match the load requirements at 20mA. Meaning if you put your LED in series with a 500Ω resistor then your constant current supply will supply 12 volts. It does that to maintain the 20mA setting. It won't apply 24 volts because on the 500Ω circuit, 24 volts will push 44mA through the circuit.