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Jareeb, A resistor in series with a LED is not there to change the voltage; it is there to limit the current that flows through the LED. To tell what value of resistor you need, you will have to tell us the rated current it takes to illuminate your LEDs.
I am assuming that the LED is 3.4V?
In that case you need to consider the current through the LED to calculate the right resistance..
Example: Input Voltage = 12V; LED Voltage = 3.4V, In that case: Resistor Voltage = Input Voltage - LED Voltage = 12V-3.4V = 8.6V
The next step would depend on the LED Current...
Example: Assume LED current = 10mA, than Resistor Value = Voltage Resistor/LED Current = 8.6V/10mA = 860 Ohm
Hope this Helps...
Note: Keep in mind that LED's are basically a short circuit if connected directly to the power supply. LED's require some form of current limiting (resistor in series) in order to avoid short circuit,
the standard calculating method for required resistor is: (Input Voltage - LED Voltage) / (LED Current).
okay so i use 8.6v/20ma = 430 ohm and how should i do the circuit i might be having 20 + leds in my computer case. and will they be able to tho a switch in there to?
heres the leds **broken link removed**
and is this the resistor i need? **broken link removed**
is the a simple circuit i could make to make them flash? just an idea
I would put 10 groups of two LEDs in series, for a total forward drop of 2*3.6V = 7.2V. The resistor would be dropping 12.0-7.2 = 4.8V. This means that the resistor is E/I = 4.8/0.02 = 240Ω. The power dissipation in the resistor P = IE = 0.02*4.8 = 96mW, so 10ea 1/4W 240Ω resistors would work fine.
A 555 timer chip could be used to flash the ten stings...
I dont know if the Yellow or Red wire is +12V, but whichever has the +12V on it would go to one end of the resisor. The other end of the resistor goes to the anode of the first LED. The cathode of the first LED to the anode of the second LED, the cathode of the second LED goes to the Black lead (0V).
Technical Data:
Material : InGaN
Size(mm) : 5mm
Shape : Round
Lens Color : Water Clear
Emitted Color : Blue
Intensity Typ : 9,000mcd
Viewing Angle : 15 - 20 Degree
Forward Voltage : 3.3v-3.6v
Forward Current : 20mA
The forward voltage is between 3.3 and 3.6 V so 3.4 is a good number to work with give or take. They don't say I max is 20 mA but I would assume so. I would look at resistors between 270 and 330 Ohms but as Mike suggest 240 Ohms would work fine. Just keep in mind you really don't need to run the LEDs at their maximum current.
Each little circuit will look like the attached image. Also as to the LED polarity, generally one leg will be longer than the other and exit the bezel where there is a small flat in the round. That is the cathode in most cases (not all but most) and will go to your common (The Negative).
Hope that helps.
<EDIT> I seriously doubt you will exceed your PSU 12 Volt current limits. You are looking at less than an amp with 20 of them. In fact at 20 mA each LED 1.0 Amp would support 50 LEDs. </EDIT)
Those cheap LEDs have a very narrow viewing angle (15 to 20 degrees). They might not be seen if they are not pointing directly at you but then they might blind you. Use LEDs that have a viewing angle of at least 40 degrees instead.
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