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Simple question regarding voltage drop on LED's

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danrogers

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Hi all, a real simple question here...

I have 3 LEDS in series powered from 12v, that I want to drive @ 50ma each. two different strings,

A - forward voltage of 2.3v x 3 = 6.9v (using a 110ohm resistor)
B - forward voltage of 3.7v x 3 = 11.1v (using a 18ohm resistor)

What is the power that will be dissipated over these resistors? I need to know what power rating to buy. I have tried a couple of online calculators but the results seem to vary somewhat!

Thanks :D
 
12v-6.9v=5.1V on resistor
12v-11.1v=0.9v on resistor.
Power is voltage across resistor x current in resistor.
or
Power is (Voltage * Voltage)/ resistance

0.9 volts is not much! the 3.7V is temperature dependent. It is also +/- 30% or more. Depending on the LED the 3.7 volts might be 3.0 or 4 volts.
 
Power = I² R

I² = 0.05² = 0.0025

So I² R = 0.275 W for the 110 Ω resistor, and a lot less for the 18 Ω resistor.

On string B, having 11.1 V of LED voltage drop might be a little dodgy with a 12 V supply, because small percentage variations in the voltage drop or the supply will result in large percentage changes in current. If the supply is a car's battery voltage, it will get to 14.4 V or so when the engine is running. Even if the forward voltage is accurate, that will give 183 mA and blown LEDs. You either need shorter strings or a constant current circuit.
 
In the first case (A)
The voltage across the resistor is 5.1volts.
To find the power dissipated in the resistor W = V²/R = (5.5 x 5.1)/110 = 0.236 watts
A 1/4 watt resistor will do, but will get a bit hot.
A 1/3 or 1/2 watt would be better.

In the second case (B)
The voltage across the resistor is 0.9 volts.
To find the power dissipated in the resistor W = V²/R = (0.9 x 0.9)/18 = 0.045 watts
A 1/4 watt resistor will do, no problems
However, the current through the LEDs will be very sensitive to changes in the supply voltage.

JimB
 
Power = Voltage X Current.

So take a look at the datasheet for the LED's. Find their forward voltage drop. This is critical in the calculation. Depending on what their forward voltage drop is you can then know what type of voltage you will drop over the resistor.(add them up because they are in series) Using V = IR you should be able to figure out the rest.
 
So take a look at the datasheet for the LED's. Find their forward voltage drop.
No.
Find their range (minimum and maximum) of forward voltages. If you calculate using the "typical" voltage drop then some LEDs will burn out and other LEDs will be dim or not light.
 
hey all, thanks for the replies, very informative!

I'm just trying to work stuff out without the on-line calculator that I used. Getting some strange results I think?

I'm going to bring the forward voltage down a bit, so.....

string a - 3.2v x 3= 9.6v 12-9.6v= 2.4v to drop. each LED is rated to 50ma, so 2.4/0.150= 16Ω (sounds a bit low?) power, 2.4v x 0.150a = 0.360W
string b - 2v x 3 = 6v 12-6v= 6v to drop. 50ma current in each so 6/0.150= 40 (also sounds low?) power, 6v x 0.150a = 0.9W <---surely not?


EDIT! what an idiot I am lol, the current wont add up will it!

So, I get:

A: 48 Ω, 0.12w across resistor
B: 120 Ω, 0.3w across resistor

does that seem more sensible ;)
 
Last edited:
Almost.
Try it this way:
Each string is in series and you want 50ma to flow in the string.
Using the 2.3 volt forward voltage drop X 3 is 6.9 volts.
So you have 5.1 volts you want to drop across the resistor at 50 ma.
So using R=E/I you get 5.1/.050 = 102 ohms - use 100.
Power = IxE so .05 X 5.1 = .255 watts. Or I^2 R = .0025 X 102 = .255 watts. So use a 1/2 watt resistor.
Try the other one now.
 
Hi Ron,

The datasheet for the rgb packages that I'm using state average & max ratings, average is 2v for red, and 3.2v for g&b. Max ratings are 2.6v and 4v respectively.
I found that at average they are pretty bright but I wanted to drive them a bit harder, or should I stick to the averages?

Right, so string b at 3.2v... 12-9.6= 2.4v, 2.4/.05= 48ohms
0.05x 9.6= 0.48watts. or 0.0025 x 48 = 0.12watts

that look ok?

thanks for the help
 
Last edited:
The datasheet for the rgb packages that I'm using state average & max ratings, average is 2v for red, and 3.2v for g&b. Max ratings are 2.6v and 4v respectively.
Now we are getting down to what AG was talking about. The datasheet should have a maximum, typical and minimum Vf specification. This is the voltage "dropped" across the diode when it is turned on. So for the green and blue string not everyone made will turn on with your 12 volts supply. (you would have to be very unlucky, but it could happen) Or they might be dimmer than you would like because there would be little current flowing through them. What I'm trying to say is that for the blue green ones you should only use two per string.
Now on power. To avoid burning them out to soon you should use the minimum Vf and calculate the resistor to keep the current below the maximum forward current spec. from the datasheet. Then check again at maximum Vf to make sure you still have a reasonable amount of current.
Right, so string b at 3.2v... 12-9.6= 2.4v, 2.4/.05= 48ohms
0.05x 9.6= 0.48watts. or 0.0025 x 48 = 0.12watts
If you use P=IE it is the voltage across the resistor not the LED that determines the power. So it is .05 X 2.4 or .12 watts.
 
What is the maximum allowed current in the LEDs? Do they need some form of cooling to operate at the maximum allowed current?
If the maximum allowed current is 50mA then they will survive longer if you operate them at 40mA or add a fan to cool them.
 
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