Calculating Battery Drain Current

A common misconception is that you can just use Ohm's law with your output voltage and resistance to calculate the battery drain current. When you are using a variable voltage APV, this is the wrong way to calculate the current. The output is only half of the equation.

To accurately calculate the input current, you must use both sides of the equation; Input and Output.

The law of conservation of energy states that power in (watts) must equal power out. (The true statement of the law is power in equals power out + efficiency losses. But for simplicity sake, we'll get to efficiency below.)

Example of the wrong way:
Code:
For Reference:
[URL="http://www.ohmslawcalculator.com/ohms_law_calculator.php"]Ohm's Law Calculator[/URL]

Output = 5v and 2A
Input = 3.7v battery

Output Power: 5v * 2A = 10 watts
Input power [if we calculate the wrong way]: 3.7v * 2A = 7.4 watts

10 output power ≠ 7.4 input power

Now the right way:

Output Power = 5v * 2A = 10 watts

Since we know that output power must equal input power, we can do this...

Input Current = Output Power / Input Voltage = 10w / 3.7v = 2.70A

_____________________________________________________________

We are still missing one very important factor....efficiency losses. Efficiency losses are energy losses due to the resistance of doing work. Nothing runs a 100% (perpetual motion)

The first law of thermodynamics: said:
In all cases in which work is produced by the agency of heat, a quantity of heat is consumed which is proportional to the work done; and conversely, by the expenditure of an equal quantity of work an equal quantity of heat is produced.

Since heat is an output, we must add the efficiency losses to the output side of the equation.

Typical efficiency for a voltage converter is in the 75-95% range. Efficiency is not constant. It varies with the desired outputs. You can find the efficiency for certain [manufacturer chosen] situations in the regulators data sheet. Assuming a value of 80% efficiency we can figure our adjusted input current.

10w / 80% = 12.5w total output

And then figure the input current:

12.5w / 3.7v = 3.38A

Compare 3.38A input to the 2A output and you can see that there is the possibility for a very large error if you calculate the drain current using only the output values. This is especially important when talking about maximum safe discharge rates of batteries.

***Note for users of APVs that are calibrated for RMS voltage***

You cannot use the RMS voltage to accuratly calculate battery drain current. VRMS is not a "real" value that can be used in these calculations. You must uses the AVERAGE voltage.

Thanks for reading, feel free to leave questions or comments below.

Comments

So a 3 ohm atomizer uses the same amount of current as a 1.8 ohm atomizer if both are operating at 8 watts of power and the reason is due to efficiency losses primarily in the voltage converter? I always did think 8 watts was 8 watts regardless of how you arrived at it. I hadn't thought about those losses being the reason for battery drain remaining the same (hence: battery doesn't last longer with higher resistance at a higher voltage). Am I understanding this correctly? Thank you for your expertise on this... been wondering about it. So if it doesn't matter, then what is your suggestion for a good atomizer resistance?
 

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