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Battery Life - Low Resistance, High Resistance, and Efficiency.

Published by Rader2146 in the blog Rader2146's blog. Views: 8195

I have often read posts on the subject of increasing a battery's run time (time between charging) through use of a higher resistance atomizer. The premise is that higher resistance, and thus higher voltage, will require less amperage to achieve the desired power (watts) than a lower resistance atomizer would require.

For Reference: Ohm's Law Calculator

Example:
4 Volts and 2.0Ω will produce 8 watts with a current of 2.0 amps.
5 Volts and 3.1Ω will produce 8 watts with a current of 1.6 amps.

On the surface it would appear that the premise would be correct, but we are missing a very important piece of the equation...

Now that we know the basis of this post...This is essentially "Part 2" of my previous post, Calculating Battery Drain Current. Please read "Part 1" if you have not done so already.

Now that we have read "Part 1" and know how to correctly calculate the battery drain current, lets look at the effects that atomizer resistance has on the battery drain current.

We now know that battery drain current is calculated by:
Input Current = Output Power / Input Voltage (A[SUB]I[/SUB] = P[SUB]O[/SUB] / V[SUB]I[/SUB])
_____________________________________________________
Low Resistance Example
  • 4.0V Output
  • 2.0Ω Atomizer
  • 3.7V Battery Voltage

Using Ohm's law (V[SUP]2[/SUP] / R) we find the output power:
4[SUP]2[/SUP] / 2.0Ω = 8W

Then we find the battery drain current (A[SUB]I[/SUB] = P[SUB]O[/SUB] / V[SUB]I[/SUB]):
8W / 3.7V = 2.16A battery drain current

High Resistance Example
  • 5.0V Output
  • 3.1Ω Atomizer
  • 3.7V Battery Voltage

Output Power:
5[SUP]2[/SUP] / 3.2Ω = 8W

8W / 3.7V = 2.16A battery drain current

Keep in mind that the figures above are for an ideal, zero loss, circuit. There is more to the puzzle.
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When I make this point in threads out in the forum, there is usually at least one person that will tell me "that can't be right, I get much better battery life with a high resistance atomizer". I dont doubt that to be true. It is very possible to get better life with high resistance, but it isn't directly attributed to the resistance. It is attributed to the efficiency of the voltage regulator.

Efficiency is not a constant value. Nearly any change in the circuit can affect the efficiency, even something as simple as the battery voltage dropping as it drains. So changing to a high resistance could put the APV into a range that is more efficient than low resistance, or vise versa.

So using an example of 8W output, lets look at the effect of two different efficiencies.

90% Efficiency:
8W / 90% = 8.88W total power
8.88 / 3.7V battery voltage = 2.4A drain current

70% Efficiency:
8W / 70% = 11.4W total power
11.4W / 3.7V = 3.1A drain current

Which one is the low resistance and which is the high resistance? What setup will put my APV into the best efficiency? There is no way to know unless you run tests on your specific APV to measure and calculate the input power and output power and calculate the efficiency yourself.
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It's also worth noting the effect that battery voltage has on the battery drain current. To compare we'll use the 90% efficiency example from above.

Fresh off the charger:
8.88W total power / 4.2V battery voltage = 2.11A

Just prior to hitting the cutoff voltage:
8.88W / 3.2V = 2.77A

Or the very far end of the spectrum for VV or VW APVs; 15W output power, 70% efficiency, and a low battery:
15W / 70% = 21.4W total power
21.4 / 3.2V = 6.7A battery drain current
  • Micgyver
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