Will this work??????
#16
Voltage = Current X Resistance). With the 330ohm resiter there will be a voltage drop of 6.6V, Idealy a 270 ohm resistor will give you the 5.4 volt drop needed but with the % of tolerance 270 ohm resistor may allow too mutch current to pass so the next higest value is 330 ohm.
By the way I know ohms law, may be you should study it a bit more.
http://www.the12volt.com/ohm/ohmslaw.asp
#17
erm......
wouldn't it be 135ohms and the best resister to use is a 150 ohm 1/4 watt?
The leds are in parellel so you have to work on the Rtotal and then presume, as per kirchoffs laws, that the current =0
You would also only need to use one resistor
.
wouldn't it be 135ohms and the best resister to use is a 150 ohm 1/4 watt?
The leds are in parellel so you have to work on the Rtotal and then presume, as per kirchoffs laws, that the current =0
You would also only need to use one resistor

.
#18
No, because in that aplication you are running the current of both LED's through 1 resistor and in my aplication the resistors are only seeing the current of 1 LED each.
#19
Not to give all my secrets away but there are lots of sites out there with LED calculators. First example I came too.
http://ledcalc.com/
http://ledcalc.com/
#21
Not to give all my secrets away but there are lots of sites out there with LED calculators. First example I came too.
http://ledcalc.com/
http://ledcalc.com/
#23
#26
#27
Sorry, I was refuring to your first post, not the one with thw schematic. You are correct in it is a bit simpler but for the reason I all ready stated, the addition of 1 resistor I think is a better solution.
Again, I apologize for my misunderstanding of your post.
I try my best helping others out when I can but I will make mistakes once in a while.
Again, I apologize for my misunderstanding of your post.
I try my best helping others out when I can but I will make mistakes once in a while.



