rollerce:
The thing I'm having trouble coping with at the moment is why everyone is freaking out at me for experimenting... It's a spare LED out of a bag of 10 that I got for 5$.
One of my classmates misread lab instructions and, instead of swapping the input signal connections to a 741 op amp, he swapped the power leads.
The current surge through the now-forward-biased ESD protection diodes caused the IC to literally explode with a loud pop that everyone in the room heard, and a large chunk of the plastic casing broke off and flew upward, almost hitting the ceiling.
A 741 op amp is also less than $0.50. If he had been less lucky and not wearing glasses, his eyes could have been a lot more expensive.
I'm still confused...
If i'd measured the battery to be 4.2v, I would have put a series resistor in. I measured the battery at 3.37v, which i knew to already be less than the forward voltage of the LED, so I saw no harm in powering the led from it for a moment or two.
You keep harping on about the battery at full charge, BUT IT WASN'T! The battery I used was at 3.4v. Please tell me what resistor makes sense to use when powering an LED with a forward voltage of 3.8V in that circumstance. Someone said earlier that they were surprised the LED fired at all, so why would I have put a series resistor in, AT THAT VOLTAGE?
The fact that the LED lit that brightly means that your assumption about its forward voltage value was wrong. Accept that, and try to find out what chain of events or reasoning led to that mistake so that you do not repeat it. What made you think the forward voltage was 3.8V? Did you get it from a website? A datasheet? Did you measure it in some way?
- While the battery "discharge" (what is the english word ?) , the led will be less bright
Discharge is correct, though the proper verb form to use in this case is discharges (While the battery discharges...).