# Calculating Power

Discussion in 'General Electronics Chat' started by RRITESH KAKKAR, Mar 11, 2015.

1. ### RRITESH KAKKAR Thread Starter Senior Member

Jun 29, 2010
2,831
91
Hello,
There are 3 white leds in series.
total led 15x6 = 90.
30 string with 120 ohms 1/4watt two parallel Resistance so, ~60ohms.
at 12V taking 800mA. I want to calculate the Led power of lights intensity!
How to calculate this??

Jan 15, 2015
2,101
891
Unless you accurately know the true LED Lumens per Watt of the LEDs you can't.

All you really know is the entire circuit, including LEDs and resistors consumes 12 Volts @ 800 mA which is total circuit power of about 9.6 Watts. The 60 Ohm resistors are consuming about 1.2 Watts of the 9.6 Watts leaving 8.4 Watts consumed by the LEDs. However, you still have no way to get back to the Lumens per Watt.

Ron

3. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
3,080
1,839
As I understand your question, you wish to determine the optical output power...

Neglecting loss in the resistive ballasts, conductors, etc...
Optical power (Watts) = Power consumption (Watts)*LED efficiency (lm/W)

Note that (with some qualifications) ~682.779 lm = 1 watt optical power (hence ~682.779 lm/w represents 100% efficiency)

Best regards
HP

4. ### mcgyvr AAC Fanatic!

Oct 15, 2009
5,082
1,104
As stated above the "light intensity" cannot be calculated without the specs/datasheet of that specific LEDs..

5. ### RRITESH KAKKAR Thread Starter Senior Member

Jun 29, 2010
2,831
91
But there is no data sheet of these local Chinese leds!
they are drum types ! so, we have to calculate total voltage across string x current or per led in series?

6. ### MrAl Distinguished Member

Jun 17, 2014
3,754
791
Hi,

You can calculate the total electrical power by I*E. You can calculate each LED by I*E also where E is the voltage across that one LED. If you can find the LED part number then you can use the efficacy to estimate the light output, but you have to realize that the data sheet will only show a range of typical values not any exact value.
If you want to know the light output for an unknown part instead then you have to use an integrating light sphere.