Trying to identify the LED in the Coleman Slim-Lantern key-f

[John Popelish]

wylbur37 wrote:

Two questions ...

1. I'm aware that when running an LED on a DC power source, the
polarity must be observed or the LED won't light. But when the rated
voltage of the LED is less than the source voltage and a resistor has
to be used, does it make any difference whether the resistor is
attached to the anode side of the LED or the cathode side?

In series is all that matters. Either side is fine.

2. The AC adapter for my "itty-bitty booklight" is rated "4.8V 500mA",
but it's actually 4.8 volts *AC*, not DC. I hooked up an LED (with a
resistor on the "+" side of the LED) to the socket and found that it
"works". But just because it "lights up" doesn't mean that it's
"right" (in other words, if you overload an LED by running it on a
higher voltage, it'll "light up" too, but you'll be damaging it). So
the question is, is it OK to run an LED on an AC power source? and if
"yes", is it sufficient to have one resistor or do you have to have
one on either side of the LED?

LEDS don't have much reverse voltage capability. Your 4.8 volt AC
supply will produce more than 4.8 volts when it is unloaded. 10 or 20
percent more. It also produces a sine wave that peaks at 1.414 times
as high a voltage in each direction than the effective voltage you
read with your meter. so that LED may have to withstand something
like 7 to 9 volts peak during the half cycle that it is blocking the
current. Some LEDs will handle that and some will not.

You may want to add components to lower this high reverse voltage.
Some possibilities are, a series diode that will prevent any
significant reverse current, an anti parallel diode that will conduct
enough reverse current that the series resistor will waste all the
reverse voltage, both of the above, another LED connected anti
parallel to both do the job of the second diode mentioned, and also
produce extra light, a bridge diode between the AC supply and the LED
resistor combination to convert the AC to rectified DC.

By the way, the transformer in the wall wart doesn't like DC loads
much, either. It will run warmer than it normally would for the same
output power drawn from both half cycles.

--
John Popelish