Re: LED: Light Emitting Diode Question

[goo@cwis.unomaha.edu (Kent Radek)]

wsquires@lonestar.utsa.edu (William H. Squires) writes:

>In article <1994Mar7.005405.16729@presby.edu> sebugg@presby.edu (Stephen Buggie) writes:
>>
>>QUESTION: What resistor should be wired in series with the LED?  I expect that
>>the resistor (ohm) value will differ for the two voltages: +5V and +12V.
>>The higher voltage will require a higher ohms resistance, correct?  Please
>>specify ohms values for each of the two voltages, +5v and +12v.  I presume
>>that
>>I can use even a tiny resistor in series with the LED; perhaps even the 1/8
>>watt size; is this correct?
>>

>  You can figure the resistance value needed with this simple equation:

>  Rled = (Vs - Vd) / Id

>where Vs is the source voltage, Vd is the forward voltage drop of the LED
>(usually around 1.6-1.8 volts or so), and Id is the maximum current rating
>of the LED, usually around 10-20 mA or so. Thus if you have an LED with a
>Vd of 1.7, and an Id of 10 mA, you would use a

>  Rled = (5 - 1.7) / .01 = 330 ohms, a common value. This is for 5 volts;

>For 12 volts, this resistor will be 1030 ohms. Use a 1.1 kilohm resistor, the
>nearest common value. The power rating of the resistor is:

All of the above is correct, except you may have an easier time finding a
1 KOhm resistor than a 1.1 KOhm resistor...

>Prled = (Vd^2)/Rled where Prled is the power dissipation of Rled above. Thus
>you would need at least an 8.76 mW rating on the 330 ohm resistor, and a
>2.63 mW rating on the 1.1 kilohm resistor. Thus both resistors can safely be
>of the little 1/8 W (125 mW) rating without problem.

Here's where you're slightly off.  Prled should be (Vs-Vd)^2/Rled.  In the 5V
circuit, the resistor will dissipate about 33 mW...at 12V, 103 mW...still well
within the 1/8 W rating of the resistors, though.  :-)

This is also why 3 V circuitry is more power efficient than 5 V circuitry.

goo