RE: LED rules of thumb

["Stu Worrall" <stuart@xxxxxxx>]

Complete electronics thicko here im afraid but what if I wanted to put
20 in a line to go down the garden?  Im assuming I will need a
resistor
for each LED and a 12V supply.  So the big question for me is do they
go
in parallel or series and can I get away with this many on one line?

Thanks

Stu Worrall

> -----Original Message-----
> From: Bricknell, Tony [mailto:tony.bricknell@xxxxxxx] 
> Sent: 17 September 2002 14:41
> To: ukha_d@xxxxxxx
> Subject: RE: [ukha_d] LED rules of thumb
> 
> 
> 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

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