Date: 7/28/96 3:41 PM
Sender: Vanagon Mailing List <vanagon@vanagon.com>
From: daniel.houg@sunny.health.state
Subject:
> Been watching this one from the sidelines,
coward! blunder in here with the of rest of us fools!!! :)
=============
Dan and the rest of you - GAWD NO! What if one of my former professors were
on the list and should see?! :-) but yeah, I probably am the "prince of
fools" hisself!
15 volts no load - my experience is that a 12 VAC output, rectified and
filtered yields an open circuit output of about 16 vdc (12 x 1.414) which
rapidly sags down upon load, depending upon load and
diode/transformer/capacitor ratings. The only sure way is to go back and
measure the circuit under load.
Throwing caution to the winds -
1) for accuracy (but not practicality's sake) there are two operating
conditions to consider, engine on and engine off. The supply voltage will
vary between 12.6 v engine off and anywhere between 13.6 =>14.4 volts engine
on. Therefore to be accurate you must subtract the voltage drop across the
LED at the desired current draw from either of the supply voltage conditions
to determine the required voltage drop across the resistor.
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
(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
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
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! Kind of like a Zener diod with a sloppy "knee" in the I/V curve.
You could probly even stack LEDs to for a voltage regulator for low current
draws in lieu of using Zener diodes with the added benefit of built in "pilot
lights".
go back to my "critical things" of the previous post:
the critical things here are:
1) Does the LED light to an acceptable brightness?
2) Is the LED gettting excess current that could cause it to fail
prematurely, and - do we care if it fails prematurely, and if so, HOW
prematurely?
3) Is the 1/2 watt resistor running too hot - to the point where it could
fail and be a fire hazard???
