Calculating Battery Discharge Time

[Karnaj]

I'm just trying to estimate how many 5-second pulls I can take until my battery (mech mod) discharges from 4.2V to 3.9V. Things like battery charger cut-off voltage, voltage drop in rda/mech mod, etc. are not taken into account. Assuming that the mAh rating of most batteries is from 4.2V to 3.0V:

Sony VTC4 2100 mAh
0.2 ohm coil

(4.2V * 2) / 0.2Ω = 88.2W
4.2V / 0.2Ω = 21A

21A * 1000 = 21000mA
2100mAh / 21000mA = 0.1 hours
0.1 hr * 3600 = 360 seconds

4.2V - 3.9V = 0.3V (my selected voltage range)
4.2V - 3.0V = 1.2V (total voltage range used for most mAh measurements)
0.3V / 1.2V = 0.25 = 25%
360 seconds * 0.25 = 90 seconds
90 seconds / 5 second pulls = 18 pulls total from 4.2 to 3.9V?

Are my calculations correct and can I assume that I can get roughly 18 5-second pulls from 4.2 to 3.9V?

 

[State O' Flux]

Short of absorbing and confirming your math... I'll just point you to the Steam Engine battery drain calculator, so you can proof your calculations.

 

[Karnaj]

Short of absorbing and confirming your math... I'll just point you to the Steam Engine battery drain calculator, so you can proof your calculations.

Thanks for the link, but Steam Engine's Battery Drain page does not calculate the # of puffs between certain voltages. The page confirms that my math is correct up to the estimated total runtime (0.1 hr = 6 mins = 360 secs) and its associated # of puffs. However, I need to know whether I'm doing the right (or almost right) calculation for the # of puffs between 4.2V to 3.9V.

I have a feeling that this is quite complicated since, according to this 20A discharge chart, the voltage initially drops very sharply and then decreases at a more or less constant rate until 3~V:

Spoiler

Also, this chart only applies to continuous discharge rather than pulse, which is what I'm trying to calculate. Five second pulse, rest, five second pulse, etc. This leads me to another question. Why does the above chart show the voltage dropping instantaneously from 4.2V to 3.75~V when hardly any time has passed?

 

[State O' Flux]

Why does the above chart show the voltage dropping instantaneously from 4.2V to 3.75~V when hardly any time has passed?

You'll want to contact @Dampmaskin for an answer to that one... here through his username, or via lars.simonsen@ gmail.com.

 

[Mooch]

. This leads me to another question. Why does the above chart show the voltage dropping instantaneously from 4.2V to 3.75~V when hardly any time has passed?

A battery can be thought of just a large number of smaller batteries, each with different capacities and internal resistances. It's really all just the electrochemistry doing different things, at different times, in different ways, but the multiple battery analogy works to explain what is going on.

Some of those small batteries have very low capacities and discharge very quickly. This is what happens from 4.2V down to near the battery's nominal voltage.
Others of those small batteries have very large capacities. This is the bulk of the battery's capacity and the voltage stays near its nominal voltage for a long time.
And lastly, some of those batteries have small capacities and very high internal resistances which take a very long time to discharge. This is what happens when you discharge, seemingly, a cell down to almost zero but it rebounds back up to a much higher voltage after the discharge stops. The huge internal resistance causes a huge voltage drop and the cell seems to be at zero volts but it still has charge left in it. Once you stop the discharge the voltage drop disappears and the voltage rebounds.

An oversimplified description of why a battery does not have a linear, i.e., straight line, discharge curve but it works.

 

[Dampmaskin]

Hi. It is correct that it is not practical to try to calculate this, because the discharge curve is different for each battery. But with a discharge chart you can estimate it.

The area under the graph represents the stored energy. If you study the graph, you can see that between 4.2V and 3.9V, the area under the graph is very small, almost zero.

Different graphs show different curves (or the same curve in different ways), so you might want to check out a few more charts for the same battery, in order to be able to eyeball it more accurately.

With a slower discharge (a smaller current), the graph will look slightly different, with more "meat" in the 4.2-3.9V area.

 

[Karnaj]

Hey @Mooch and @Dampmaskin, thank you for your replies. I still don't understand why the voltage drops so much and so fast in these graphs from 4.2 to 3.7V. I've tried to search for answers to this, such as in this thread, but all that is mentioned is that the graph of the voltage's higher end (4.2 to 3.7V) is more visible during lower loads (10A and 5A). I don't doubt this, but I wish to know why this is the case. Does Mooch's multiple-smaller-batteries analogy apply specifically to this higher end of the curve or (I assume) does it apply to the whole curve from 4.2 to 2.5V? According to the 20A discharge graph above, does it mean that most of these batteries will only output 3.7V rather than 4.2V once the load is applied?

 

[Mooch]

Hey @Mooch and @Dampmaskin, thank you for your replies. I still don't understand why the voltage drops so much and so fast in these graphs from 4.2 to 3.7V. I've tried to search for answers to this, such as in this thread, but all that is mentioned is that the graph of the voltage's higher end (4.2 to 3.7V) is more visible during lower loads (10A and 5A). I don't doubt this, but I wish to know why this is the case. Does Mooch's multiple-smaller-batteries analogy apply specifically to this higher end of the curve or (I assume) does it apply to the whole curve from 4.2 to 2.5V? According to the 20A discharge graph above, does it mean that most of these batteries will only output 3.7V rather than 4.2V once the load is applied?

Yes, that's why the nominal voltage of these cells is 3.6V or 3.7V. It's the voltage, roughly, where the bulk of the discharge takes place.

The multiple battery analogy is over the entire discharge voltage range. That's why we need multiple batteries in parallel, in the analogy, to explain what happens at different voltages.

 

[Mooch]

It's best to think of these as 3.7V cells with a small "surface charge" that brings the voltage to 4.2V. There are negative ions at the cathode that are, literally, on or near the cathode's surface. This is in addition to the bulk of the ions spread fairly evenly across the cathode. When you start the discharge you quickly drain these surface ions and the voltage drops fast. Then the voltage holds fairly steady because the bulk of the electrons in the cathode start flowing out with a fairly even distribution across the cathode. When these ions "run out", the voltage plummets and the cell is discharged.

If you discharge at higher and higher rates, you increase the gradient in ion density across the cathode, causing a voltage drop. This is why capacity drops as discharge rates go up. The ions just can't get out fast enough. The voltage plunges and the cell seems empty. But, if you wait a few seconds for the ions to redistribute you find that there's still plenty of charge left to be used at lower discharge rates.

I don't know all the ugly electrochem tech details (always hated differential equations) but this is what I understand is going on. A lot easier to just use the multiple battery analogy sometimes.

 

[Karnaj]

It's best to think of these as 3.7V cells with a small "surface charge" that brings the voltage to 4.2V. There are negative ions at the cathode that are, literally, on or near the cathode's surface. This is in addition to the bulk of the ions spread fairly evenly across the cathode. When you start the discharge you quickly drain these surface ions and the voltage drops fast. Then the voltage holds fairly steady because the bulk of the electrons in the cathode start flowing out with a fairly even distribution across the cathode. When these ions "run out", the voltage plummets and the cell is discharged.

If you discharge at higher and higher rates, you increase the gradient in ion density across the cathode, causing a voltage drop. This is why capacity drops as discharge rates go up. The ions just can't get out fast enough. The voltage plunges and the cell seems empty. But, if you wait a few seconds for the ions to redistribute you find that there's still plenty of charge left to be used at lower discharge rates.

I don't know all the ugly electrochem tech details (always hated differential equations) but this is what I understand is going on. A lot easier to just use the multiple battery analogy sometimes.

Thank you so much for the detailed explanation. It's certainly easier to comprehend for the everyday vaper as opposed to electrochemical equations. Apparently the term "battery sag" can be used to describe this phenomenon, and I had not done much research on it before.

This raises another question that is quite important. Since the voltage quickly drops to the battery's nominal voltage under a high load, does this mean that the nominal voltage should be used to calculate the actual wattage delivered to the coils? For example, using a freshly charged 4.2V battery and a 0.2 ohm coil, the wattage quickly drops to (3.7 ^ 2) / 0.2 = 68.45W from the initial 88.2W (4.2V)? If so, then this would mean the current also drops quickly from 21A to 18.5A. Then again, this doesn't make sense since the current is constant (20A) in the discharge curves above.

 

[Mooch]

Your asumption about the current dropping when using the nominal voltage vs. the fully-charged voltage is correct, assuming you're using an unregulated mod. For the majority of the time you use that cell before recharging, it will be near its nominal voltage.

Discharge tests typically use constant-current (CC) because that's a standard for testing by the cell manufacturers. It doesn't emulate the way cells are used when vaping but it allows for easy comparison of the performance of various different cells. Whether against each other or against a particular manufacturer's discharge graphs.

 

[Mooch]

Wait...is your math right? Maybe I'm just too tired but those power calculations don't seem to work.

 

[Bunnykiller]

I would find out the mostly mathless way.... charge up a battery, verify voltage, pop it into the mod, take 25 pulls, remove battery and recheck voltage. find difference and divide by pulls for drop per pull

 

[Karnaj]

Your asumption about the current dropping when using the nominal voltage vs. the fully-charged voltage is correct, assuming you're using an unregulated mod. For the majority of the time you use that cell before recharging, it will be near its nominal voltage.

If this is true, then why does everyone use 4.2V for calculating the wattage and amperage of their mech mod setups even though it drops sharply to the nominal voltage? Is this for amperage safety concerns since a freshly charged battery does indeed output 4.2V at the start, no matter how brief it is? It seems to me that using the nominal voltage to calculate wattage and amperage is much more realistic. Therefore, the lowest "safe" coil resistance would be 3.7 / 20 = 0.185 ohms for my 20A batteries. Even at 0.185 ohms, I am confident that my batteries can take an initial 4.2 / 0.185 = 22.7A load for a very brief moment.

P.S. In my previous post, I accidentally wrote (3.7 * 2) / 0.2 when the * sign should be a ^. The resulting calculation is correct, though.

I would find out the mostly mathless way.... charge up a battery, verify voltage, pop it into the mod, take 25 pulls, remove battery and recheck voltage. find difference and divide by pulls for drop per pull

Yup, I had the physical approach in mind as well, but it would be a hassle. I thought it would be nice if the number of pulls between voltages could be calculated easily instead, but it seems this is not feasible at the moment. I've moved on from this and I am now much more interested in the effect of battery sag on actual wattage in mech mods.

 

[Mooch]

If this is true, then why does everyone use 4.2V for calculating the wattage and amperage of their mech mod setups even though it drops sharply to the nominal voltage? Is this for amperage safety concerns since a freshly charged battery does indeed output 4.2V at the start, no matter how brief it is? It seems to me that using the nominal voltage to calculate wattage and amperage is much more realistic. Therefore, the lowest "safe" coil resistance would be 3.7 / 20 = 0.185 ohms for my 20A batteries. Even at 0.185 ohms, I am confident that my batteries can take an initial 4.2 / 0.185 = 22.7A load for a very brief moment.

P.S. In my previous post, I accidentally wrote (3.7 * 2) / 0.2 when the * sign should be a ^. The resulting calculation is correct, though.

Yes, it's for safety. And highly advisable for most users as it's a buffer against coils being lower in resistance than calculated, etc. But, you're right, using the nominal voltage is a more accurate way to calculate the wattage you'll be at (roughly) for most of the battery run time.

The voltage will drop instantly to less than 4.0V but can stay above the nominal voltage for several seconds, depending on the cell you're using and what current you're drawing. If you have true 20A batteries then, I agree, they'll easily handle 23A for several seconds without causing any more damage than use at 20A causes.

 

[State O' Flux]

If this is true, then why does everyone use 4.2V for calculating the wattage and amperage of their mech mod setups even though it drops sharply to the nominal voltage?

I don't know about anyone else... but I tend to use 3.8-4.0V, when running wattage/amperage calculations... the value is somewhat conditional, with battery brand, MCCD and age dependancy.

 

[bussdriver]

Close, but no cigar. Seems like math ain't quite right. 3.7 volts squared is equal to 3.7 times 3.7; not 3.7 times 2. Big difference.

Then when you get it all calculated, you realize your batteries have 150 charge/discharge cycles and it all goes out the window.

 

[Karnaj]

@Mooch and @State O' Flux, thank you guys so much for your help. I now have a much better understanding of batteries, battery sag, and unregulated mods. I am very likely going to buy a high wattage, dual 18650 regulated box mod due to the disadvantages of battery sag/life in mech mods when sub-ohming. It is my understanding that regulated box mods provide consistent vapor production by increasing the current as battery life decreases, something that is simply superior to mech mods.

Close, but no cigar. Seems like math ain't quite right. 3.7 volts squared is equal to 3.7 times 3.7; not 3.7 times 2. Big difference.

Then when you get it all calculated, you realize your batteries have 150 charge/discharge cycles and it all goes out the window.

I squared the volts in my original calculation, so the answer is actually correct: (3.7 ^ 2) / 0.2 = 68.45W. The only error I made was typing * instead of ^ in my post (not in my calculation), which was fixed. Good mention of the battery cycles, though. It only confirms that these calculations are just estimations due to the complexity of batteries and that measurements must be taken directly and correctly for true voltage, current, and wattage.

 

[bussdriver]

Good choice in reg over mechanical for your needs. Mech is good; I have plenty of 'em. There's a lot of variables in their use, more than most folks want to deal with. With a regulated mod you will gain consistency in your vape, no matter what the batteries are doing. And for high power, a dual battery is a MUST.

 

[State O' Flux]

@Mooch and @State O' Flux, thank you guys so much for your help. I now have a much better understanding of batteries, battery sag, and unregulated mods. I am very likely going to buy a high wattage, dual 18650 regulated box mod due to the disadvantages of battery sag/life in mech mods when sub-ohming. It is my understanding that regulated box mods provide consistent vapor production by increasing the current as battery life decreases, something that is simply superior to mech mods.

Although I presently use only tube mechs... there's much be said for a dual 18650 reg. mod. Aside from battery life, there's the option of going beyond Ohm's law parity, and "forcing" wattage on a build that may not be optimal for the limitations of good ol' V² ÷ Ω = P and V ÷ Ω = C.

Some day, I'll break down and get a 200-300 watt reg. box... just so I can see what change in performance another 10 or 20 watts above Ohm's law gets me.