Hi all,
I have a bit of electronics background but not much with circuitry. I'm trying to make a board of LEDs and came accross a snag in this experimental LED automotive project.
I'm working with a voltage of 14v and LEDS with a forward voltage drop of 1.9 and the LEDs take 18mA. I have a line of 7 and a line of 6 elsewhere.
Line of 7 gets a 39ohm resistor and line of 6 gets a 150ohm resistor to limit the current. Now all is good, but I have two inputs, one for bright and one for low (the above being for the bright mode). I experimented, then put a 560ohm resistor on the set of 7 for when they should be dark. Now the problem is understanding how to calculate a resistor that will work on the set of 6 that will match the intensity of the 7.
I figure....
7(led) * 1.9 (forward v) = 13.3 (.7 left over)
.7/.018 = 39ohm
So I tried calculating what the 560 is actually doing -- is this right?
.7/A = 560
A = .00125
I know that my target amperage is .00125 on the set of 6 for low mode.
So I calculate for 6:
6x1.9=11.4 (2.6 leftover)
2.6/.00125 = 2080Ω
Why when I use the 2080 on the row of 6 it seems much less bright than the row of 7 using a 560Ω resistor? Where am I off in my calculations?
I have a bit of electronics background but not much with circuitry. I'm trying to make a board of LEDs and came accross a snag in this experimental LED automotive project.
I'm working with a voltage of 14v and LEDS with a forward voltage drop of 1.9 and the LEDs take 18mA. I have a line of 7 and a line of 6 elsewhere.
Line of 7 gets a 39ohm resistor and line of 6 gets a 150ohm resistor to limit the current. Now all is good, but I have two inputs, one for bright and one for low (the above being for the bright mode). I experimented, then put a 560ohm resistor on the set of 7 for when they should be dark. Now the problem is understanding how to calculate a resistor that will work on the set of 6 that will match the intensity of the 7.
I figure....
7(led) * 1.9 (forward v) = 13.3 (.7 left over)
.7/.018 = 39ohm
So I tried calculating what the 560 is actually doing -- is this right?
.7/A = 560
A = .00125
I know that my target amperage is .00125 on the set of 6 for low mode.
So I calculate for 6:
6x1.9=11.4 (2.6 leftover)
2.6/.00125 = 2080Ω
Why when I use the 2080 on the row of 6 it seems much less bright than the row of 7 using a 560Ω resistor? Where am I off in my calculations?
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