Hi
I am connected a 6v battery (4x1.5 AA) to a LED an took the voltage reading across it. I expected to read 2v but it read 5.65v. can some one tell me why? isn't suppose to maintain a 2v drop across it?

thx

 

You must put a series resistor into the circuit otherwise you will distroy the poor LED.

I = desired current which is 2-10 mA for normal LED's.

Uled is 1.6-1.8 for red LED's
Yellow and green has somewhat higher values
Blue has 3-4 V

R = (Ubat-Uled)/I

 

And that LED is probably toast. Even if it works, it is damaged.

Have you read this? It explains LEDs and more...

LEDs, 555s, Flashers, and Light Chasers

Bill's Index

 

You read 5.65V because the LED has burnt.

It will drop about 2V (depends on the LED colour and temperature) if it is operated at its rated current.

Thus for a LED which drops 2V and operates at 20mA you will need a (6-2)V/20mA=200 R resistor in series with the LED for a 6V supply.