Walt, Thanks for catching that. After I send the e-mail I was thinking about it and realized that I probably made that mistake. I'm glad you corrected it for the record.
If you find that driver, let us know. - Peter. On 04/01/2011 12:13 am, [email protected] wrote: > Peter, > > Thanks for mentioning the power dissipation calculations when using a > dropping resister with a LED. Unfortunately, there is an error in your > calculation. You multiplied the current (0.02 amps) by the voltage drop > across the LED (2 volts) to come up with 0.04 watts. This is power > dissipated by the LED. For a > 9 volt supply, the voltage drop across the resistor is 7 volts and the power > dissipated by the resistor is 7 x 0.02 = 0.14 watts. A 1/8 watt resistor > will get very hot, indeed, and will probably melt plastic. For safety's sake > I would probably use a 1/2 watt resistor. I have been looking for a good, > inexpensive LED driver which will not waste all the supply power in the > resistor. > > Walt Jopke -- Peter Vanvliet ([email protected], or [email protected]) Houston, Texas "It is easy to give up; anyone can do that..." http://pmrr.org/ (my model railroad - RSS feed <http://pmrr.org/rss.xml>) http://fourthray.com/ (my company) http://houstonsgaugers.org/ (model railroad club) -- [Non-text portions of this message have been removed] ------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/S-Scale/ <*> Your email settings: Individual Email | Traditional <*> To change settings online go to: http://groups.yahoo.com/group/S-Scale/join (Yahoo! ID required) <*> To change settings via email: [email protected] [email protected] <*> To unsubscribe from this group, send an email to: [email protected] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
