Electrical schoolwork...

[Ameer Antar]

All you have to do is figure out how much voltage the resistor should take from the full batt. voltage (the remaining goes to the LED), and you also need to know how much current is needed for the LED (the current through the resistor and LED will be the same since they're in series). 

An example is:

V = I*R
(13.6V - 2.0V) = (.015A)*(R ohms)
(batt. voltage - LED voltage) = (current)*(resistor)

so R (ohms) = 11.6V / .015A = 773 ohms

You may need a couple resistors to make the right value. Make sure you calculate the proper current for the LED (15mA is usually good). About the LED voltage, 2.5-3V seems a little high for red; usually more like 1.7-2V. Unless this is some special kind of LED. Good luck.

-Ameer

---Original Message---
Date: Tue, 10 Feb 2004 00:16:18 -0500
From: "Louis-Alain RICHARD" <laraa at sympatico.ca>
Subject: Electrical schoolwork...
To: <quattro at audifans.com>
Message-ID: <000601c3ef95$0409e9f0$680bfea9 at LARAA>
Content-Type: text/plain; charset="us-ascii"

Hi there,

If I were to install some red 3V LEDs to backlight the numerous switches
around the instrument panel of my 1983 urQuattro, what is the resistance
value I must introduce to this circuit to lower the 12V to something more
acceptable?

Resistance value of the LEDs: 150 000 ohms 
Ideal voltage at the LEDs: between 2.5V and 3V.
Electrical voltage at the switches: 13.6V, permanent, switched by the
ignition, not the IP rheostat.

My electrical knowledge is so far away (20+ years), I don't remember anymore
how to compute series resistance in a simple circuit like this!

Please light me... and my car!

Louis-Alain
1983 Quattro
85-D-900463

------------------------------