I have the following LEDs that I need to wire in my NBv3 saber:
Tri-Cree B/B/W:
Blue = 3.47V (1A)
Blue = 3.47V (1A)
White = 3.15V (1A) (FoC)
Green Accent LED = 2.2V (25mA) (on when saber is not asleep)
Blue Ring Switch LED = 3.3V (20mA)
I know the formula for Ohm's Law is simple math to calculate and gives me these resistor values (before any rounding to what's available):
Tri-Cree B/B/W:
Blue = (3.7V - 3.47V) / 1A = .23 Ohms ---------- .23 Ohms * 1A^2 = .23W
Blue = (3.7V - 3.47V) / 1A = .23 Ohms ---------- .23 Ohms * 1A^2 = .23W
White = (3.7V - 3.15V) / 1A = .55 Ohms ---------- .55 Ohms * 1A^2 = .55W
Green Accent LED = (3.7V - 2.2V)/ .025A = 60 Ohms ---------- 60 Ohms * .025A^2 = .0375W
Blue Ring Switch LED = (3.7V - 3.3V)/.02A = 20 Ohms ---------- 20 Ohms * .02A^2 = .8W
First question --> Is my math correct?
Second question --> Does it present a problem that the amperage when everything is on at once is 3045mA when the battery is rated at 2400mAh? Or does that just mean that if everything were on at the same time for long enough, the battery would last less than an hour?
Thanks!
Bookmarks