Subject:
|
Re: Homemade LED Bricks
|
Newsgroups:
|
lugnet.robotics
|
Date:
|
Fri, 12 Nov 1999 03:53:12 GMT
|
Viewed:
|
933 times
|
| |
|
|
Keith Miller wrote:
> basicly.. it's i*r=v where i = current in amps and r = resistance in ohms
> and v = voltage in volts..
> so..
> If we want to find out the resistor's value.. we re arrange it like so..
> r = v/i
> so 10 millamps (typical led current) = .01 Amps..
> And at 9 volts it becomes
>
> r = .01/9
> r = 900 ohms
Thanks for the enlightenment. I remembered it went something like that.
But don't you have to allow for the forward voltage drop across the LED,
typically 1.6 volt (still from memory)? That is, the voltage should be 9 - 1.6
= 7.4 volts.
So the resistance is 7.4/0.01 = 740 ohm.
But resistors only come in 680 or 820 ohm, as I understand.
So my guess of 680 ohm which is closer to 740 ohm, would have worked.
Well it's kinda good to know that the RAM which I have is still ok :-)
--
C S Soh
http://web.singnet.com.sg/~cssoh
... where air is power
|
|
Message is in Reply To:
 | | Re: Homemade LED Bricks
|
| basicly.. it's i*r=v where i = current in amps and r = resistance in ohms and v = voltage in volts.. so.. If we want to find out the resistor's value.. we re arrange it like so.. r = v/i so 10 millamps (typical led current) = .01 Amps.. And at 9 (...) (25 years ago, 12-Nov-99, to lugnet.robotics)
|
23 Messages in This Thread:
 
  
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
