Date: Mon, 29 Jul 1996 02:10:43 -0400
Sender: Vanagon Mailing List <vanagon@vanagon.com>
From: Cetin Seren <cseren@fore.com>
Subject: Re: Re- basic electronics q
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!
Cetin Seren Direct: 412-635-3479
Software Development Engineer Main: 412-772-6600
Outbound Technology Group fax: 412-635-3350
FORE Systems, Inc. email: cseren@fore.com
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