Date:         Wed, 13 Aug 2003 09:01:08 -0500
Reply-To:     Larry Alofs <lalofs@RCN.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Larry Alofs <lalofs@RCN.COM>
Subject:      Re: LEDs in 12V systems
Comments: To: "Mike D." <md03@XOCHI.COM>
Content-Type: text/plain; charset=us-ascii; format=flowed

Mike D. wrote:
> For those of you experimenting with replacing 12V bulbs with LEDs,
> here is a quick lesson.
>
> LEDs come with two ratings, a "Vf" (voltage drop forward) and a
> current/amps rating ("If")
>
> Most LEDs have Vf of around 2.5 volts (ultrabrites tend towards 3.5v)
> and If of 0.02 A (20 milliamps), although there are some weird ones
> out there so read the label.
>
> LEDs are not like normal lightbulbs: if you over-drive (pass too much
> current through) them a little bit they will get brighter, but if you
> really overdrive them they will actually get dimmer.  It's one of
> those weird quantum mechanical effects that boggles the mind, but is
> true.
>
> So basically, in a circuit you treat an LED as if it has a fixed
> voltage drop and wire a resistor in-series to limit current.
>
> The resistor value is calculated as
>   R = (V - Vf)/If
>
> Where V is the driving voltage (in a car, 12v), Vf is the LED's
> voltage drop, and If is the LED's current (expressed in Amps, not
> milliamps!).
>
> So for the typical bright LED (3.5V, 20mA) in a 12V system you get
>
> R = (12 - 3.5)/.02  <-- remember use Amps, not milliamps!
> R = 8.5/.02
> R = 425 ohms.
>
> Feel free to adjust R up or down by 10% or 20% to tweak the
> brightness as you see fit.
>
> If you want more brightness, you can also wire a set of 4 3V LEDS in
> series without a resistor at all.
>
This last idea is a little questionable since the voltage supply could
be around 14 V in a car.  This may produce very high currents if no
limiting resistor is used.
   The current would not be infinite as the equations imply, because the
LEDs have some internal resistance and the forward voltage drops vary
somewhat with current.

Larry A.