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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 Research Park, 5800 Corporate Drive URL: http://www.fore.com Pittsburgh, PA 15237-5829


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