On Sun, 26 Dec 2010 22:20:45 -0600, G. Morgan
wrote:

wrote:

On Sun, 26 Dec 2010 21:17:20 -0600, G. Morgan
wrote:

(Don Klipstein) wrote:

Brightness of LEDs is controlled by how much current flows through them
- same as with dimmable CFLs. All that is necessary is to make the
current control circuitry in the LED "bulbs" compatible with and making
use from the usual dimmers, similar story as with CFLs.

I was taught (in college) that once break-through voltage is
equaled, that's it, on or off. Now I have to learn WTF went wrong
in my imaginary circuit if that is true.

KRW is right that there is some difference in the voltage drop across
the junction as the current rises., It is not a big difference but it
is a difference.(1.79 - 1.96 in the ones I tested over a range of 3ma
to 15ma) It could just be heat doing it at the high end. I did notice
in my flashlight tests that have no internal esistor that a very small
series resistance starts dropping the current very fast so they may be
using this "wall" of a heated junction to regulate the current,
similar to a nichrome wire.

The only way to get an LED to work is put a resistor in series.
Are you sure the (bulb, lamp, LED) was not a 'package'? They sell
them both ways, people who want only the LED (specialized), and
people who just want it to work. The later is what I think is
causing this confusion.
No, an LED will light very nicely with no resistor if the voltage is
close enough to the forward drop of the led. Excede that voltage by
very much and the current goes up real fast, very quickly exceding the
maximum power dissipation of the LED, and popping it.

If I set my lab supply to 15MA and connect a standard red LED I get
1.9 volts drop across it. I can stiffen that power supply so it can
supply 20 amps of power, and as long as I do not increase the voltage
above 1.9 volts, the current does not excede 15ma
Below 1.45 volts, the LED does not light at all, and it varies in
brightness quite linearly from 1.5 to 1.9 volts

With an 82 ohm resistor in series,changing the input voltage from 1.5
to 14, the voltage drop across the LED goes from 1.5 to 1.9 volts,
with the current reaching 15ma. (roughly.02 watts) while the poor
resistor is shedding 1.875 watts.

Without the resistor,25 ma of current produces a 1.91 volt drop,30ma
produces 1.98 volt drop,35ma is 2 volts, and going to 50 ma goes to
2.06

So doubling the current from 25ma to 50ma only raises the voltage drop
by .08 volts, which is roughly 4% on this 20ma rated device (I believe
that is the spec)