I have a 16,000 mcd LED but it takes 80 mA.

I do have LEDs that are 1,520 mcd and they take 20 mA.

If I connect 2. of them in Series I know they will take 20 mA.

If I bunch then together will they not give me 21,040 mcd Brightness?

 

No. Even if they could all be run directly from the same voltage supply (which they can't, because of their forward voltage difference), 16,000 + 2 x 1,520 = 19,040 mcd.
Because of the non-linear response of the human eye I douibt you'd notice any brightness difference between 16,000 mcd and your hoped-for 21,040 mcd anyway.

 

I am so so sorry I did the math wrong.

I have an LED that is 16,000 mcd and takes 80 mA.

I do have LEDs that are 1,520 mcd and they take 20 mA.

If I take 11 of the 1,520 mcd LEDs and connect them in Series all 11 of them will only take 20 mA.

So if I bunch them together will they not give a Total of 16,720 mcd?

 

1520 x 11 = 16,720 Yes light adds up.
When you series up 11 LEDs the voltage needed is 11x higher. Voltage adds up.
Your eye and eras are logarithmic. So, you will not see it at 11x brighter. It will seem about 2x.

 

I have a 6. Volt Battery Pack and my LED takes 3.2 Volts at 80 mA.

Am I Right the Math to find out the Resistor I need goes like this.
6. Over 80 = 7.5 Ohms?

 

Hi bif.
Rser = (Vsupply - Vled drop)/ Led current

(6V-3.2V)/0.08A = 35R

E

 

No. First, the units for calculations are volts amps and ohms.
Next your calculation is wrong. 6/80 = 0.075 ohms
There will only be 2.8 volts across the resistor as there is 3.2 volts across the LED. So the calculation will be 2.8 volts/ 0.08 amps (80 mA in amps)
This gives a resistance value of 35 ohms.
Les.

 

I think I see how my Math is Wrong?

5 mA would be 0.005 and 10 mA would be 0.010 am I Right on this?

So 70 mA would be 0.070 and 80 mA would be 0.080 am I Right on this?

 

I think I see how my Math is Wrong?

5 mA would be 0.005 and 10 mA would be 0.010 am I Right on this?

So 70 mA would be 0.070 and 80 mA would be 0.080 am I Right on this?

Correct.

 

I am so so sorry I think I have it Right this time.

The Math goes E Over I = Ohms.

So it would be E Power Source 6. Volts Over I LED Amps 0.08 = 75 Ohms.

Do I have it Right?

 

So it would be E Power Source 6. Volts Over I LED Amps 0.08 = 75 Ohms.

Hi bif
NO.
Did you read my post #6??????

Rser = (Vsupply - Vled drop)/ Led current

E

 

Your resistor will be dissipating about 1/4 Watt, so a 1/2 Watt (or greater) rated one would be needed.

 

Would I be Right my Math is to find out the Resistance for the Whole Circuit when it is up an running not for the Resistor I need?

 

So my Power Source is 6. Volts Minus 3.2 Volts for the LED = 2.8 Volts Drop.

Then I do 2.8 Over 0.08 = 35 Ohms.

 

35 ohms is correct but will need to be rated at 5 watts minimum.
Are you sure that LED requires 80ma?
I've seen 16000 mcd LEDs that only require 20ma.

EDIT: 0.5 watt resistor needed.

 

35 ohms is correct but will need to rated at 5 watts minimum.

Huh?

P = V^2 / R = 2.8 x 2.8 / 35 = 0.224W.

or

P = I^2 R = 0.08 x 0.08 x 35 = 0.224W

A half Watt resistor is fine.

 

Oops missed the decimal point.

 

If I take 11 of the 1,520 mcd LEDs and connect them in Series all 11 of them will only take 20 mA.

I have a 6. Volt Battery Pack and my LED takes 3.2 Volts at 80 mA.

LED's are "Current Driven".
LED's have "Forward Voltage Drop" (Vf)
IF you run 1 LED from a 6V source with a 3.2Vf on 20mA you need a resistor of 140Ω at 120mW. (1/8th watt)
Don't forget about wattage.
If you TRY to run 2 LED's from a 6V source IN SERIES - you don't have enough voltage. LED's may light but they will be dimmer than desired.
If you run 11 LED's who's Vf is 3.2 EACH you have a total Vf of 35.2Volts Dropped. Your power source would need to be higher than 35V.
ASSUME you have a 48VDC source and are running 11 LED's in series then:
(48V-35.2Vf) ÷ 20mA = 640Ω
48V X 20mA = 960mW. A 1W resistor will run hot, but will work. You'd for sure want a 2 watt resistor.
IF YOU BUILD a 36VDC power source then you'll need:
(36V-35.2Vf) ÷ 20mA = 40Ω. . . . . . 36V X 20mA = 720mW. Still, nearly 1 watt. In this case a 1 watt resistor may be sufficient. WITH adequate ventilation. Scorching of the PCB may happen over time. If the resistor is mounted elevated off the board by a half inch it will cool better and less likely to damage the PCB. The extra lead length will also aid in cooling. The trade-off of having a component mounted so high is the likelihood you may have the resistor eventually fail from even a minor vibration. There is greater likelihood of damage to the component.

Bottom line - don't forget wattage.

 

If you have a 6V source and want to run 11 LED's with 3.2Vf then the best way is to run them is to run all 11 in parallel with EACH LED having its own resistor.
If you have a 12V source then you can safely run 3 LED's in series on a single resistor and you can have multiple parallel sets of 3 (+ one resistor each). In other words:
3 LED's & one resistor of 120Ω @ 240mW (I'd opt for a half watt resistor for margin of safety)
3 LED's & one resistor of 120Ω @ 240mW
3 LED's & one resistor of 120Ω @ 240mW
For a total of 9 LED's. If you want to run specifically 11 LED's you can add a fourth string of just two LED's with appropriate resistors. OR you can run one string of 3 and two strings of 4.

 

It is a Super Bright LED from SparkFun.com the Part Number is 11118.

Maybe I am Wrong maybe it does Take 20 mA can somebody help?