Date: Tue, 18 Apr 2000 23:20:13 -0700
Reply-To: "Keith & Lynn C." <clayard1@NETCOM.CA>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: "Keith & Lynn C." <clayard1@NETCOM.CA>
Subject: Re: Third brake lights (tack-free)
Content-Type: text/plain; charset=us-ascii
Parallel them. Around 15 ma draw thru each one, calculate 1.4V across each one,
use a 1kohm resistor, 1/4W as a series current limiter for each. 1 per LED. This
way, if one goes out, you don't have to play the "Christmas light game" and guess
which one is bad. High intensity LEDS are available but they draw a little more
current. Would need a lower value resistance, but probably 1/2 W would be needed
Cheers,
Keith
81 Westy 230k km
David Beierl wrote:
> At 10:37 4/18/2000, Chris Smith wrote:
> >Wiring your own is fairly easy. Each LED should get around 1.5 to 2.2v
> >(check on package) if you wire them in series (positive to negative) in a
> >daisy chain, all you need to do is to divide the voltage available by the
> >voltage of each LED to get the number of LEDs in each string. For most
> >LEDs, 7 would be the correct number (2v times 7= 14v)
>
> Ok, yes, you can wire them in series. Shoulda thought of that. But you
> *absolutely* need to have a current-limiting series resistor, because these
> things basically act like switches, on or off. If you don't use the
> resistor you'll either see darkness or learn what burning epoxy smells like...
>
> I'd use a maximum of 5 reds in series, myself, on a 12v automotive
> circuit. Assume the red LEDs are 1.8v each, totalling 9. The formula is
> Rlimit [ohms] = (Vsrc - Vled) [volts] / I [amps]. If you use milliamps,
> the answer is in kilohms.
>
> So for 20 ma at 14.3 volts, you need 260 ohm, quarter watt. (Worst case)
> At 13.5v you get 17 ma.
> At 12.5 v you're down to 13 ma.
>
> If you use six @ 1.8v, total 10.8v. For the same conditions, resistor
> would be 175 ohms, quarter watt. That would give you:
> 14.3v >> 20 ma
> 13.5v >> 15 ma
> 12.5v >> 10 ma
>
> For seven @ 1.8v, total 12.6v. Resistor 85 ohms, quarter watt.
> 14.3v >> 20 ma.
> 13.5v >> 11 ma.
> 12.5v >> no light.
>
> So you see that the bigger your series string is, the more light variation
> you will get from different system voltages (but smaller power dissipation
> by the resistor). For contrast:
>
> One LED @ 1.8v. Resistor 625 ohms, half watt.
> 14.3v >> 20 ma.
> 13.5v >> 19 ma.
> 12.5v >> 17 ma.
>
> david
>
> David Beierl - Providence, RI
> http://pws.prserv.net/synergy/Vanagon/
> '84 Westy "Dutiful Passage"
> '85 GL "Poor Relation"
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