Let me start by saying I get confused screwing in light bulbs. I am in the final stages of a project and I am stymied.
I want to use a LED out of a novelty flashlight that runs off two 3V 2016 watch type batteries. I want to use a 120VAC/6VDC adaptor to eliminate the (replacement) batteries. I am assuming that the LED is a 6VDC. (2 X 3VDC =6VDC) The output of the 6VDC adaptor is actually 9.3VDC. (Likewise, a 4.5VDC adaptor output is 7.3VDC.)
I am soldering wires from the 120VAC/6VDC adaptor directly to the legs of the LED. I have burned out 3 LED so far and have stopped.
How can I make the LED work? I was informed locally to install a 120 ohm resistor to one of the LED legs, did so and there is no change in the adaptor output. Is the 6VDC 400ma adaptor too hot and is that burning out the LED?
I want to use a LED out of a novelty flashlight that runs off two 3V 2016 watch type batteries. I want to use a 120VAC/6VDC adaptor to eliminate the (replacement) batteries. I am assuming that the LED is a 6VDC. (2 X 3VDC =6VDC) The output of the 6VDC adaptor is actually 9.3VDC. (Likewise, a 4.5VDC adaptor output is 7.3VDC.)
I am soldering wires from the 120VAC/6VDC adaptor directly to the legs of the LED. I have burned out 3 LED so far and have stopped.
How can I make the LED work? I was informed locally to install a 120 ohm resistor to one of the LED legs, did so and there is no change in the adaptor output. Is the 6VDC 400ma adaptor too hot and is that burning out the LED?