Date:         Tue, 18 Sep 2007 15:32:00 -0400
Reply-To:     Mike S <mikes@FLATSURFACE.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Mike S <mikes@FLATSURFACE.COM>
Subject:      Re: LVC adding LED interior lighting
Comments: To: David Kao <dtkao0205@YAHOO.COM>
In-Reply-To:  <837611.94088.qm@web82704.mail.mud.yahoo.com>
Content-Type: text/plain; charset="us-ascii"; format=flowed

At 12:43 PM 9/18/2007, David Kao wrote...
>Unless the two LED devices are really exactly identical and remain
>identical
>throughout the entire service life it is really not a good idea to
>connect
>them in serial. It can be done but if they are not exactly identical
>the
>voltage drop on each one of them may not be 6V equally.
>
>If one has lower internal resistance than the other, it will receive
>less than
>6V and the other will get more than 6V. Even each has its own current
>limiting
>resistor one will bear more voltage than the other and will be
>brighter too.
>One will die earlier than the other as well.

LEDs run on current, not voltage. When connected in series, they will
always operate at their own individual voltage drop. That is necessary
and desired.

The voltage drop across an LED will vary from about 1.7 volts in a red
LED to about 3.5 in a white one, and will also vary with individual
LEDs. As long as sufficient resistance is provided to avoid exceeding
the LEDs maximum current rating (usually 25-30 mA, most have their
light output rated at 20 mA) , it will be fine. Note that current will
be exactly identical through every device in a series connection. So,
the voltage across each of two LEDs in series doesn't really matter,
current does, but the current will always be identical.

So, worst case, figuring 14V from the alternator and 2 LEDs with 1.6
volt drops and max 20 mA current, a total resistance of (14V - (2 *
1.6V)) / 0.02A = 540 ohms or more is safe. The resistor will dissipate
10.8V * .020A = 0.216W, so with common values, a 560 ohm, 1/4 W
resistor should be used.

For a high brightness white LED, like the Kingbright WP7104RWC/Z, the
worst case calculations are (14V - (2 * 2.7V) / 0.020A = 430 ohms. 8.6V
* 0.02A = 0.172W, so 430 ohm, 1/4 watt.

If someone wants to connect two LEDs which have internal dropping
resistors, as was originally suggested, one only needs verify that the
current through the series doesn't exceed about 20 mA. There are a few
LEDs top out at a max of around 10 mA, but they're rare.