I, and others, keep posting this LED calculator site - It has all the information you need All you need to do is read and apply.
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so it gives me the resistor but how do i wire 12 leds in series just one resistor? resistor to each?
Same site different page
That's a great example where the wizard is out and out wrong. There is no headroom to control the current, and the brightness will be a strong function of battery volts and device temperature.
Using ICENG's figures it produces this - exactly what he said!Needs correct figures I was just illustrating the process.
Yes, but what it shows is a lack of intelligence in the "wizard" which should surely at least post an advisory on the lack of headroom.Steve
NOTE in the reply below I have inserted values for the LEDS YOUR values may be different so use your correct values.
it sais a 3300 ohm resistor..
Blue LED = 3.6 V @ 20 ma Put two LEDs in series with a 90 ohm resistor to 9 VDC Do this six times. There and here you are......A
soo 9 v battery 90 ohm resistor and just series every two?
Only in Hollister :-) Two blue use 7.2 V Resistor gets to see the left over voltage 9 - 7.2 = 1.8 V Resistance = V / I = 1.8 V / o.02 A = 90 ohms Power of resistor = V * I = 1.8 V x o.02 A = o.036 W less them ¼ Watt Even less then 1/8 Watt.............. A
so resistor is on negitive of the series
You pick Negative, in-between or positive side, Builders choice. -il -----|<|------|<|-----/\/\/\/\-----+9V If you don't find a 90 try a 100 ohm A
please tell me the easyest way to wire 12 blue leds to a 9 volt battery i dont need the math just the help on the way to do it
Not enough info in the question. L