The UK Home Automation Archive

Archive Home
Group Home
Search Archive


Advanced Search

The UKHA-ARCHIVE IS CEASING OPERATIONS 31 DEC 2024


[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

RE: LED rules of thumb


  • To: <ukha_d@xxxxxxx>
  • Subject: RE: LED rules of thumb
  • From: "Stu Worrall" <stuart@xxxxxxx>
  • Date: Tue, 17 Sep 2002 14:55:21 +0100
  • Mailing-list: list ukha_d@xxxxxxx; contact ukha_d-owner@xxxxxxx
  • Reply-to: ukha_d@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


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.

Home | Main Index | Thread Index

Comments to the Webmaster are always welcomed, please use this contact form . Note that as this site is a mailing list archive, the Webmaster has no control over the contents of the messages. Comments about message content should be directed to the relevant mailing list.