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LED forward voltage

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Berserk87

New Member
ive seen/am seeing a lot of guides to random electronics projects that shows leds in line with a 5V source, then a resistor.

which usually ends up being 5V @ ~20mA.

but looking at led spec sheets, the max forward voltage (for ultra brights) is 3.8V.

am i missing something?

do you count the leds ~1.5V voltage drop when your looking at it. so it ends up being 3.5V forward if your connecting it to 5V?
 

Diver300

Well-Known Member
Most Helpful Member
The normal thing to do is to take the typical forward voltage drop and subtract that from the supply voltage.

Single red LEDs are never more than about 2 V forward voltage drop. What colour are the ultrabrights that you are quoting?

If you have a red LED with a drop of 2 V and you want to light if at 20 mA from a 5 V supply, there is (5 - 2) = 3 V across the resistor, so you want 150 Ω.

On a 12 V supply, it is usual to put a few LEDs in series. So to light four of the same LEDs from 12 V the total drop is 8 V, so there is 4 V left on the resistor, which should be 200 Ω.

You could go to 5 LEDs but not on a car battery because the voltage regulation is poor, so the brightness would vary too much.
 

Berserk87

New Member
ok, i was just double checking.

ive been up for a long time, and obvious things are going over my head... :p

edit:

o ya, im using ultra bright blue/green/white.
 
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audioguru

Well-Known Member
Most Helpful Member
If your LEDs have a forward voltage of 3.8V (seems like a max rating, not a typical rating) then with a 5V supply the resistor will have 5V - 3.8V= 1.2V across it.
Ohm's Law is used to calculate the value of the current-limiting resistor. For 20mA, R= 1.2V/20mA= 60 ohms. Use 62 ohms as the nearest standard value.
 
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