RE: LED rules of thumb

["Bricknell, Tony" <tony.bricknell@xxxxxxx>]

This is good, but remember to take into account the voltage drop of the
LED(s).

Example:  Using a 12V power source, your chosen LED needs 20mA.
Manufacturers data sheet shows the LED's to have a forward voltage drop Vf
=
2.1V (this is the voltage the LED will want to maintain across it, and
will
vary from manufacturer to manufacturer and colour to colour).

So, your resistor needs to drop the excess of 12 - 2.1 = 9.9V

R = V/I = 9.9 / 0.02 = 495 ohms.

You can't (easily) get a 495 ohm resistor, but you can get 470 ohms and
560
ohms.
Transposing to see what current we get with 470 and 560 ohms...

I = V/R = 9.9 / 470 = 0.021 A (21mA) for the 470 ohm
          9.9 / 560 = 0.018 A
(18mA) for the 560 ohm

If you're still with me, for multiple LED's just add up their forward
voltage:

12V supply, three series LED's of Vf = 2.1V, forward current 20mA

Voltage for resistor to drop = 12 - 2.1 - 2.1 - 2.1 = 5.7V

R = V/I = 5.7 / 0.02 = 285 ohm

-----Original Message-----
From: Jon Payne [mailto:jgpayne@xxxxxxx]
Sent: 17 September, 2002 14:21
To: ukha_d@xxxxxxx
Subject: Re: [ukha_d] LED rules of thumb


Use Ohms law:

V=IR, so R = V/I

e.g. for 20mA at 12v

R = 12v / 0.02A = 600 ohms

However, the white/blue LEDs may have a large voltage drop (haven't
checked)
which I guess may effect things - anyone?

jon

Yahoo! Groups Sponsor

ADVERTISEMENT

http://www.automatedhome.co.uk
Post message: ukha_d@xxxxxxx 
Subscribe:  ukha_d-subscribe@xxxxxxx 
Unsubscribe:  ukha_d-unsubscribe@xxxxxxx 
List owner:  ukha_d-owner@xxxxxxx
List of UKHA Groups here - http://groups.yahoo.com/group/UKHA_Grouplists/

Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.