Harvey is correct on just about everything, but one (he is almost correct 
on that one, too).

At 12:37 PM 7/29/96 -0500, you wrote:
>
>2)  BUT !!!!!!!  I vaguely recall from a class taken almost 30 years ago - 
>that the current draw of the diode at applied voltages above the device design
>point (apparently 2.5 volts in this case) is only limited by the value of the
>series resistor, so you want to define the conditions you want at the diode

Yes, for ideal diodes.

>(volts [characteristic of the LED] and ma. draw to stay within the design
>power dissipation of LED) and select a resistor that limits to current as
>desired.  Otherwise, if my memory is correct - if you provided the diode with
>an unlimited current source at 2.5 +  volts, it would draw unlimited current
>and burn itself out unless some kind of current limit was imposed.  That's
>primarily  what the series resistor is for-current limiting, and secondarily,
>to drop the excess source voltage from our car battery/alternator.   Memory
>says that an LED is a reverse biased diode, and when you apply a voltage in
>excess of the breakdown voltage, it begins to conduct (and thus radiate) but

No.  an LED is a forward-biased diode.  You're thinking of the Zener Diodes.
(The world of science/engineering lost Mr. Zener a few years back -- his
discovery
has made a great many things possible, among which is the solid-state
voltage regulator).  Hence the rest of this discussion needs to be modified.
I'll work on it some....

>without current limit up to the limitations of the p/n junction to handle the
>current and associated power dissipation.  Individual designs for LEDs have
>different breakdown voltage (2.2 v. min is typical as I recall) and design
>power dissipation that is related to brightness.  The power dissipation is
>basically the current through the LED times the voltage across it. Check the
>data sheet for the device in quesiton. In fact now that I think about it more
>- the LED will hold a constant voltage drop across itself independent of
>applied voltage applied higher than the breakdown voltage.  Of course, the

This is quite true.  It goes for forward-biased diodes, as well as Zener
diodes (even LEDs, which are not-so-ideal forward diodes).  Forward biased
diodes just display this property under normal operation.  Silicon-based
ones characteristically display .7 Volts of drop, Gallium based ones
display .2 Volts, and whatever LEDs are made of, they get about 2.6V
drop.

>higher the applied voltage beyound the drop across the diode, the higher the
>current drawn, the brighter the device until it exceeds power dissipation and
>burns out!   
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