Best resistance for battery life

[dstin1]

The question i have is it better to use lower ohms and lower voltage or higher ohms and higher voltage to get the most out of a charged battery. Like most i want the most vape out of my battery but long life so more or less resistance is the question, i am still going to vape at about 8.5 watts.

 

[McGarnagle]

The question i have is it better to use lower ohms and lower voltage or higher ohms and higher voltage to get the most out of a charged battery. Like most i want the most vape out of my battery but long life so more or less resistance is the question, i am still going to vape at about 8.5 watts.

Assuming you keep wattage the same, a combination of high voltage and high resistance should (in theory) allow the battery to last longer. Batteries are rated in milliampere-hours (how many milliamps the battery can continuously supply in one hour), and amps will be what you can think of as what the battery stores. So the fewer amps you use over time the longer your battery would last. In your case, let's look at 2 different 8.5 watt configurations.

Case 1 (low resistance and low voltage) - For 8.5 Watts and a 1.8 Ohm resistance you would run at 3.9 Volts. The amps would be 2.2.
Case 2 (high resistance and high voltage) - For 8.5 Watts and a 3 Ohm resistance you would run at 5.0 or 5.1 Volts (the calculation is right at 5.05). The amps would be 1.7

This means the high resistance/volt combo should last longer.

Note: Batteries are tested at a given amperage to get their ratings and when changing the amperage the mah rating will not be the same. So there will not be a directly linear relationship in battery life increase with amperage decrease. Therefore I cannot tell you how much battery life you would gain with a simple calculation. The battery will also naturally drain over time, so of course the longer you use the battery the more you will lose due to that fact. Yet you should see an increase in battery life if you decrease amps.

 

[DavidOck]

Nicely put.

There is a current / time curve for batteries, but finding one for a specific battery is a bit of a problem!

The higher the current, the shorter the time the battery can provide it, so, as McGarnagle says, due to the laws of physics, the lower resistance, requiring higher voltage to maintain the same wattage, will deplete the battery quicker.

 

[Rader2146]

Output current is not the same as battery drain current.

The short answer is that resistance doesnt matter; battery life will be the same, no matter the resistance, as long as the power (watts) is the same.

The long answer is in this thread: http://www.e-cigarette-forum.com/fo...-question-about-variable-volt-real-power.html

 

[McGarnagle]

Output current is not the same as battery drain current.

The short answer is that resistance doesnt matter; battery life will be the same, no matter the resistance, as long as the power (watts) is the same.

The long answer is in this thread: http://www.e-cigarette-forum.com/fo...-question-about-variable-volt-real-power.html

I am going to edit this from my original response after looking at it some more.

I did make it to simplistic with the use of an ideal voltage source, which a mod is not. That is my mistake with that assumption.

Yet I also need to understand what is the power rating you are getting of the whole circuit and what a mod actually does. The power ratings used in that thread you mentioned only takes into account the power used at the carto and then does a simple p=iv for the whole device. That's not really true. The power used by the pv circuitry is ignored in that equation, and exactly how the boost converter works during use and not during use. We need to understand what find of current in flowing during the on and off states from the battery and exactly what the device circuitry is doing.

The amount of amperage flowing from the battery does matter for battery life. While I do understand conservation of energy, a battery will deliver that energy less or more efficiently depending on things like amps and it will also discharge more or less energy depending on that same variable.

 

[Rader2146]

When I get into these types of discussions (and there have been many) I usually assume an ideal circuit just for the fact that the efficiency is unknown for nearly all commercial PVs. However, I do usually includes the disclaimer that changes in efficiency will have an effect on the drain current. In the proof of concept for resistance vs battery drain current, taking efficiency into account does not change the results: at constant power, a change in resistance does not cause a change in battery drain current.

I have tested all of this on my OKR-T/6 mod when I wanted to compare the claimed efficiency vs actual efficiency and all results were as expected:

2.0Ω
8.2V input voltage
4.0V output voltage

Measured output current: 1.98A
Measured input current: 1.09A

Calculated output power: 7.92W
Calculated input power: 8.94W

Efficiency: 89%

3.0Ω
8.2V input voltage
4.9V output voltage

Measured output current: 1.61A
Measured input current: 1.12A

Calculated output power: 7.89W
Calculated input power: 9.2W

Efficiency: 86%

I was actually drawing slightly more current from the batteries with a higher resistance due to lower operating efficiency.

If the drain current is relatively equal no matter the resistance, then the battery's performance characteristics are moot. All the battery see's is equal current, what happens down stream in the converter has no effect on the battery.