Subject:
|
Re: Need Electrical Help
|
Newsgroups:
|
lugnet.trains
|
Date:
|
Mon, 20 Aug 2007 16:53:58 GMT
|
Viewed:
|
16539 times
|
| |
|
|
"Jason J Railton" <j.j.railton@btinternet.com> wrote in message
news:Jn2wvH.BoL@lugnet.com...
*snip*
> Let me check I understand the theory of this - you want 1.5V across the
> LED,
> which means the resistor has to take the other 7.5V. You then pick a
> resistor
> that gives you a 20mA current at that figure of 7.5V. Then you
> double-check
> the power so you don't cook the resistor?
>
> So two LEDs in series would drop 3V, so you'd work out the resistor values
> based
> on the remaining 6V but still with 20mA going through the circuit?
>
> Jason R
You are correct,
In your case (assuming those LEDs are 1.5v and not your typical 2.1v
variety),
Vl would be 1.5v + 1.5v, or 3v
So in our formula:
R = (Vs - Vl)/I
(9v - (1.5v + 1.5v)) / .020A = 300 ohm
and...
P = (Vs - Vl) * I
(9v - (1.5v + 1.5v)) * .020A = .12w (so again, use a .25w [1/4
watt])
-Rob
www.brickmodder.net
|
|
Message is in Reply To:
 | | Re: Need Electrical Help
|
| (...) Let me check I understand the theory of this - you want 1.5V across the LED, which means the resistor has to take the other 7.5V. You then pick a resistor that gives you a 20mA current at that figure of 7.5V. Then you double-check the power so (...) (19 years ago, 20-Aug-07, to lugnet.trains)
|
17 Messages in This Thread:
 
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
