Date: Mon, 18 Mar 2002 21:09:40 -0500
Reply-To: "Mr. Pants" <mrpants@MINDSPRING.COM>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: "Mr. Pants" <mrpants@MINDSPRING.COM>
Subject: Re: High beam blue LED
Content-Type: text/plain; charset="iso-8859-1"
The first blue LED's weren't available to the public until about 1991-1992
and even then they were quite dim by even that days standards. It wasn't
until late 1995/1996 that ones were developed that would have been bright
enough to use in an automotive environment.
While I'm still coming up to speed on the vanagon, I can (hopefully) give
anyone the knowledge they'll need to replace a incandescent lamp with an
LED.
First and foremost, unlike a lamp, which is pretty much a resistor/inductor
that gives off light, the D in LED is diode, a semiconductor switch which
conducts in only one direction. It has a anode (positive) and a cathode
(negative). Place a battery correctly across it and once the voltage reaches
a certain point (which is different for different materials used to make
them) it turns for all practical purposes into a short (which just so
happens to give off light) This is why you need a dropping resistor, to make
sure that current doesn't surge through the led once it's biased on. Normal
red, green and yellow LED's usually take about 1.7 volts to turn them on.
New material LED's often take more then 3 volts to turn them on (Blue,
White)
Ok here comes the math. For a car, I'm assuming a normal 12v system, and
I've picked a random blue LED from Radio Shack, Cat.#: 900-7170. The VF
(forward voltage drop across the diode) is listed as 3.7 Volts @ 20ma (.02
amps)
Mr. Ohm taught us that Voltage = Current times Resistance or V = I * R
In our circuit, we'll drop 3.7 volts across the LED, leaving 12 v - 3.7 v =
8.3 volts left that we have to handle via the resistor (so the LED doesn't
blow it's top)
Also, since this is a series circuit, we know that current is equal at all
points (trust me on this one) so that the resistor will also have 20ma
running across it. Since we have the voltage and the current, we can figure
out the resistance we need.
Once again V = I * R, so we also know that resistance equals voltage divided
by current R = V/I so we get
8.3 volts /.02 amps = 415 Ohms.
Now you're not going to find a 415 Ohm resistor at Radio Shack, but that's
ok because long ago people took such things into account and created the
standard value of resistors we have now. The most common type, the E12
series, is set up so that at worst the worst you'll ever be off from a given
value is 10%. I happen to know that 470 Ohms is in that series, and will do
just fine (in fact since the charging system in a car can get over 14 volts,
it gives us some added safety as well)
The last thing we need to figure out is the power rating of the resistor we
need. This is measured in watts, and in a DC circuit the formula is Watts
equals Volts times Amps or W = V * A
The voltage across the resistor is 8.3 Volts and the current is .02 amps or
8.3 *.02 = Watts
From this we get 0.166 Watts or 166mW. The most common resistors usually
come in 1/4 (250mW) or 1/2 (500mW) ratings. While the 1/4 would be ok, I
usually like to factor in an extra level of safety (in this case I seem to
remember Radio Shack charges the same price for both styles) so I'd go with
the 1/2 watt.
I hope the above makes sense. If it didn't or was way too long winded,
please forgive me in advance. If anyone needs help with a conversion, email
me direct and I'll help you with anything I didn't explain well enough.
Cheers,
Tom Corbitt
84 GL owner (yet unnamed)
----- Original Message -----
From: "John McLean" <jaymac@INTERNETCDS.COM>
To: <vanagon@GERRY.VANAGON.COM>
Sent: Sunday, March 17, 2002 5:48 PM
Subject: Re: High beam blue LED
> > It isn't actually an LED. They didn't make blue LEDs in those days.
> >
>
> I'll have to admit I would'nt know the difference between incandescent and
LED. However
> everyone (including my Bentley - copyright 1991) refers to these indicater
bulbs as LED's.
>
> After reading the responses that corrected my error I went back to the
Bentley to have a
> look again. In the picture displayed they show what I assume is a real
LED (page 90.19) and
> a point is made of the polarity of the bulb. I did not notice this on
the bulb that I used
> in my own repair. The bulb I used only had two thin wires that were
exactly alike. It
> would appear that what I installed was in fact an incandescent.
>
> John McLean 87 Vanagon GL
