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May 2001, Week 2

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Subject:	Re: OT: Electronic Circuitry Questions
From:	Dennis Heidner <[log in to unmask]>
Reply To:	Dennis Heidner <[log in to unmask]>
Date:	Fri, 11 May 2001 04:01:11 -0500
Content-Type:	text/plain
Parts/Attachments:	text/plain (22 lines)

Rich,  I haven't seen the reply that Steve has made.

The LED should have a published spec for the  voltage drop (Vfwd) and "MAX"
 voltage drop.  The term forward is general added to indicate that the
device is wired and biased so that it should operate (light up).  It will
also have a spec for the current that will flow when the device has the
correct voltage across it.  We can call that Ifwd.

Your resistor value should then be      Rdrop = (17.5  - Vfwd)/Ifwd

Or using the value that Steve provided (820 ohms).
   Vfwd is typically 2 volts -- for most LED's,  (higher for some laser
devices and the new super bright LED's)

820 = (17.5 - 2.0)/Ifwd   or Ifwd = 15.5/820 = (rough) 19 milliampere's
current.

The resistor will see about a 1/3 watt heat generation.

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