# how to wire 12 blue leds? with 9 volt ?not enuf info on packaging?

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rickharris5 years ago
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.
jconner3 (author)  rickharris5 years ago
so it gives me the resistor but how do i wire 12 leds in series just one resistor? resistor to each?
5 years ago
5 years ago
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.
5 years ago
Using ICENG's figures it produces this - exactly what he said!

Needs correct figures I was just illustrating the process.
5 years ago
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
5 years ago
Reasonable point
5 years ago
NOTE in the reply below I have inserted values for the LEDS YOUR values may be different so use your correct values.
jconner3 (author)  rickharris5 years ago
it sais a 3300 ohm resistor..
iceng5 years ago
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

jconner3 (author)  iceng5 years ago
soo 9 v battery 90 ohm resistor and just series every two?
5 years ago
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
jconner3 (author)  iceng5 years ago
so resistor is on negitive of the series
5 years ago
You pick Negative, in-between or positive side,
Builders choice.

-il -----|<|------|<|-----/\/\/\/\-----+9V

If you don't find a 90 try a 100 ohm

A

jconner3 (author)  iceng5 years ago
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
lemonie5 years ago

Not enough info in the question.

L