# how to calculate wattage



## Axkiker (Feb 20, 2009)

Goodness its been a LONG time since I had any electronics theory classes so maybe you all can help me out.

How do you calculate wattage. I have gotten on a few sites which have calculators but keep getting inconsistant numbers.

My questions revolves around 12 k2's which ill be running at either 700ma or 1000ma.

If I were to run them all in series what would the wattge be

also if I were to break them up into sets of 2 and run each set in series then run both sets off the same power supply in parallel what would the wattage be. 

I assume they would be the same but heck if I know

thanks


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## Tekno_Cowboy (Feb 20, 2009)

You have to multiply the forward voltage of the LED's by the current they are being driven at. Then that, IIRC, needs to be divided by the efficiency of the driver you are using.

Multiply the voltage of a single LED by the number of LED's in series, and multiply the current by the number of LED's in parallel.

For example:

12 LED's in series with a Vf of 2V would be 24V

12 of the same LED's in 6s2p would be 12V, but double the current.

The total wattage will be partly dependent on how efficient your driver is at the levels you are using. 

Example: you have a driver capable of running off 1V-10V, and you want to put out 5V, you should have better efficiency if your supply is close to 5V, unless the driver is optimized for a different input voltage.


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## baterija (Feb 20, 2009)

Tekno_Cowboy said:


> 12 LED's in series with a Vf of 2V would be 24V
> 
> 12 of the same LED's in 6s2p would be 12V, but double the current.



That works if your driver is constant voltage. Constant current drivers won't double the current though. If you are using a constant current driver each string in the parallel array see's part of the drive current. Assuming that Vf of each LED is 2, and a constant current driver pushing 1 amp:

12 Led's in series P = 24V * 1A = 24 watts

Same LED's in parallel 
P = Power in string1 + Power in string 2
= 12V * .5A + 12V * .5A 
= 12 watts


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## Tekno_Cowboy (Feb 20, 2009)

Sorry I wasn't clear. I meant that the driver would have to put out twice the current to provide each led with the desired current. For that example I should have said half the current to the led's.


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## Axkiker (Feb 20, 2009)

Tekno_Cowboy said:


> You have to multiply the forward voltage of the LED's by the current they are being driven at. Then that, IIRC, needs to be divided by the efficiency of the driver you are using.
> 
> Multiply the voltage of a single LED by the number of LED's in series, and multiply the current by the number of LED's in parallel.
> 
> ...


 

Okay I think im getting something messed up here.

If I have 12 leds in series. Say they have a forward voltage of 3.65V

would it be 3.65V x 12 = 43.8V
43.8V / 1A = 43.8watts ????????
43.8watts / 90% efficiency = 39.42 watts


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## baterija (Feb 20, 2009)

Axkiker said:


> Okay I think im getting something messed up here.
> 
> If I have 12 leds in series. Say they have a forward voltage of 3.65V
> 
> ...



First two lines are right on. If the LED is seeing 43.8 watts and the driver is 90% efficient total power is then
43.8 / .9 = 48.7 watts (you multiplied instead of divided)
48.7 - 43.8 = 4.9 watts used by the driver circuitry


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## Tekno_Cowboy (Feb 20, 2009)

baterija said:


> First two lines are right on. If the LED is seeing 43.8 watts and the driver is 90% efficient total power is then
> 43.8 / .9 = 48.7 watts (you multiplied instead of divided)
> 48.7 - 43.8 = 4.9 watts used by the driver circuitry


The second line was wrong, you should multiply by 1A, not divide.


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## Axkiker (Feb 20, 2009)

baterija said:


> First two lines are right on. If the LED is seeing 43.8 watts and the driver is 90% efficient total power is then
> 43.8 / .9 = 48.7 watts (you multiplied instead of divided)
> 48.7 - 43.8 = 4.9 watts used by the driver circuitry


 


HAHAHA whoooops.

I thought that looked wrong


thanks for the help


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## Tekno_Cowboy (Feb 20, 2009)

Happy to be of service


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## baterija (Feb 21, 2009)

Tekno_Cowboy said:


> The second line was wrong, you should multiply by 1A, not divide.



I'm not using current numbers in that calculation at all. That .9 is the power efficiency factor. If you are feeding 43.8 Watts to the LED through a 90 percent efficiency driver that's exactly how to calculate total wattage supplied to the driver. (Either that or we need to look at the Voltage of the cells and Current at the tailcap to directly calculate it.) The 43.8 watt number we computed is power at the LED (using LED Vf and If) not power provided to the driver. To be a little more clear about what I was computing

My second line:
Watts Total x driver efficiency = Watts to LED
With watts to the LED of 43.8 and efficiency of .9 we get
Watts Total x .9 = 43.8
Watts Total = 43.8/.9 = 48.7


So my third line is 
Watts to LED + Watts dissipated by the driver = Watts total provided to the driver
so 43.8 + watts dissipated = 48.7
solving that gives us 4.9 watts dissipated as heat by the driver


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## Tekno_Cowboy (Feb 22, 2009)

What I was referring to was this line:



Axkiker said:


> 43.8V / 1A = 43.8watts ????????



This should be Vf x If = W
Not Vf/If = W as it was written.


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## baterija (Feb 22, 2009)

Tekno_Cowboy said:


> What I was referring to was this line:



In the words of Homer Simpson, "Doh!" :naughty:


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