I fried my LED, why ?


I fried my LED when I connected it like this ?
Can anyone explain me why ?

I’m 2 days into this,and i dont know why this happened :’(
Thanks :slight_smile:

Your resistor is shorted out by the tracks on that bread board, so electrically it is not in the circuit. Tracks on the top and bottom are joined horizontal, the others are joined vertically.

Move your circuit into the vertical rows.

Did it get real bright? It looks like you connected the LED directly to the 9 volt battery should have made a noticeable flash!

try hooking it up without the breadboard. Go positive terminal (the smaller clip) battery to the long lead of the LED, the short lead to the resistor, and the other end of the resistor to the negative battery terminal.

Thanks for the answers, but i still dont get it.
No, it blinks once, and it dies.
With the same resistor, when i connect it normaly in the middle of the breadboard, it works perfectly.

Why does the LED dies ? How the hell I shorted it out ? :smiley:

I’m my brain the situation is like this:
The current enters the + rail of the breadbord, then it hits the resistor, then the LED should glow, and trought the catode of the LED the current goes into the - rail , and vuolaaaa … it shoud work, but it doesnt :slight_smile:

Ok, i will not connect the + and - rails of the breadbord with a component ever again, but i need some deeper understanding. Thanks :slight_smile:

Here is one more drawing :slight_smile:

Thanks guys for the fast replays

----------------------- minus
 --         +-R-+
 |          |   |
----------------------- plus

The plus and minus rails each are one long strip. Your resistor is shorted by the plus strip (as explained by grumpy_mike); so the current does not flow through the resitor (route of least resistance)

The current enters the + rail of the breadbord, then it hits the resistor, then the LED should glow, and trought the catode of the LED the current goes into the - rail , and vuolaaaa ... it shoud work, but it doesn

No. The current enters the - rail then hits a junction between the resistor and the rest of the strip. So the current splits between going through the resistor and on through the rest of the strip. The amount of current that goes through the resistor before meeting up with the strip again, is technically known as fuck all. So the current never gets cut down by the resistor. The current sneaks past the resistor because both ends of the resistor are connected to electrically the same point.

Basically that second drawing is not a drawing of what you have actually made. If you made that drawing then it would work.

Thank you so much. The answer is very simple, I learned something, and paid the knowledge with 4 LED . I totally neglected the fact that under the resistor lies a wire, so the resistor had no function. 9V killed the LED's , lesson learned.

Thank you once again. BTW I'm totally new to electronics, I have a Bch degree in political sciences, and it doesn't help me very much with this stuff.

: )

Thanks for the feedback, it will help us answer similar questions in the future. It is good to know exactly how you express how you see the situation. Have a Karma point.

A typical LED gives normal luminescence at 1.5V and 15mA. The power it can absorb is about 22.5 mW (1.5V * 15mA).

In your 9V-300R-LED circuit, the series current is (9-1.5)/300 = 0.025A = 25 mA. The continuous power loss in the LED is 1.5V*25mA = 37.5mW which is 1.67 times (37.5/22.5) of mormal power dissipation (22.5mW).

The LED will survive for a while, and after that it is going to be fried due to an effect known as thermal runaway?

Giving a typical LED 25mA instead of 20mA isn't going to damage it I don't think - unless everything is already very hot and close to failing already. You definitely don't get thermal runaway with a series resistor that's taking most of the supply voltage - the current change is severely limited by the resistor.

If you had a 1 ohm resistor and a 1.52 volt supply with a 1.50 volt LED, then as the LED heats up and its forward voltage drops the current can increase a very great deal - that could then go into runaway.

Basically if you have a series current limiting resistor that's got a reasonable proportion of the supply voltage across it (10% or more), runaway is unlikely.

"is technically known as fuck all."

lol !