Ipv v3 question about amps

[BrennanElectric]

Ipv v3 question. I wanna get one. Using steam engine calculator but it's not making sense. V3 runs in series, so double volts, same amps. So if I have two 20 amp batteries, what are my limits how does it work? According to the battery drain calculator if I put say, 0.3 ohms and use 70 watts but put the battery at 4.2 it gives me the first picture. Now since there's two batteries, do I change the voltage to 8.4, because when I do that it cuts the amps to the battery in half, and do I base the amps to the atty into my 20 amp battery or just the battery amps part of the column? And so all I'm asking is how you'd do the math running a series as to how low you can go without going over your amps, but in the example of using 20 amp batteries, I'm not sure how to use the calculator

 

[tehdarkaura]

The ipv3 has the batteries stacked "in series". So the voltage is doubled but the amperage capabilities of the batteries remains the same. If they were side-by-side "in paralell" then the voltage output would stay the same but they would be able to provide twice the output of amps (current) as one of them alone could of.

 

[BrennanElectric]

The ipv3 has the batteries stacked "in series". So the voltage is doubled but the amperage capabilities of the batteries remains the same. If they were side-by-side "in paralell" then the voltage output would stay the same but they would be able to provide twice the output of amps (current) as one of them alone could of.

Right I understand that, here let me post something that explains better

Ipv v3 question. I wanna get one. Using steam engine calculator but it's not making sense. V3 runs in series, so double volts, same amps. So if I have two 20 amp batteries, what are my limits how does it work? According to the battery drain calculator if I put say, 0.3 ohms and use 70 watts but put the battery at 4.2 it gives me the first picture. Now since there's two batteries, do I change the voltage to 8.4, because when I do that it cuts the amps to the battery in half, and do I base the amps to the atty into my 20 amp battery or just the battery amps part of the column? And so all I'm asking is how you'd do the math running a series as to how low you can go without going over your amps, but in the example of using 20 amp batteries, I'm not sure how to use the calculator

 

[BrennanElectric]

So all in all, changing the volts on the steam engine calculator to 8.4 (Dual batteries) is producing the correct information? Well I know with a regulated box like my ipv v1 that I have, it has one battery, 4.2 fully charged, but to get the full wattage at my resistance the volts are 4.8, which makes the amps higher as the battery drains, due to it running a higher volt. So I'm guessing since the ipv v3, actually has two batteries, it cuts the work in half, and allows your atty to get double the amps, while splitting the work between both batteries instead of using one. I think it makes sense in my head, but I figured I'd ask people who are more experienced in the science behind it.

 

[tehdarkaura]

well I installed steam engine and I see they have a preset for the ipv3 -- under the pioneer for you section.. You could use that i suppose....

You can change watts to 165 and lowest resistance to 0.03 and see the difference between stock and after the firmware update is applied too.

 

[BrennanElectric]

Anybody got an answer?

 

[Dampmaskin]

Oh, hi, I just got this question via the Steam Engine feedback form, but I didn't have a sender address to reply to, so I wrote an answer on Steam Engine's Facebook page instead.

Here's a copypasta of my answer:

First of all: The IPV2 is a regulated mod, so remember to select this option.

In a mechanical mod, the current going through the atomizer is identical to the current going through the battery, but in a regulated mod it is not.

This confuses a lot of people, especially those who are used to thinking in mech mod terms. I wrote a text about it a while ago, but I'm not going to repeat everything here. Instead, you can click the "How it works" link at the bottom left of the battery drain calculator. The text will appear on the bottom of the page. I hope this clears things up for you.

Another thing to take note of: You're using 8.4 volts to calculate the current, probably because you want to figure out the worst case scenario, right?

Remember that in regulated mods, the battery current increases as the batteries are drained. This is the opposite of what happens with mech mods. So with regulated mods, using the fully charged voltage is the best case scenario, not the worst case scenario.

To calculate the current in the worst case scenario, assume that your batteries are nearly empty. Yes, this sounds counter-intuitive when you're used to mechs, but that is how it is.

Anyway, regarding the IPV3: If we trust YiHi's assertion that the SX330v3 chip is 95% efficient, you'll reach 20A at the battery side at 120W output power, when your batteries are low (6.4V combined). However, it is not wise to push batteries to the limit, not even the continuous one.

An output of 85W will give you somewhere around 30% headroom, which hopefully is OK for stacked batteries.

When your batteries are full, you can get 160W from them, with zero headroom. Again, you probably want a certain safety margin, so 110W will give you around 30% headroom, with full batteries.

I'm not making any guarantees though. Use your own judgement. The battery safety of your mod is ultimately - and literally - in your own hands.

 

[BrennanElectric]

That was actually me that sent the feedback as well, I also just posted on your blog, here is what I said, just want to make sure my calculations are correct if I'm running at the lowest ohm build ohms law will allow me to to be safe

Hey Dampmaskin, just wanted to double check and clarify, if you're running at dual 18650 regulated mod, to figure out the battery drain, do you just change the voltage to 8.4, and is that calculating in series? I plan on getting the ipv v3, and just want to know if the calculations are correct and safe. According to ohms law, if I had two 20 amp batteries at continuous, in a series that would be 8.4/20=0.42 ohms, so that would be the lowest safe resistance I could go to with those batteries on a series box mod correct? and then 8.4820=168 watts, which calculates to how much power I can give it to hit 20 amps correct? Well I went ahead and used your calculator, and of course there is efficiency involved, does that account for voltage drop as well? With regulated mods, it uses more amps as the battery drains to hit that wattage..so setting it at 168 watts, actually uses 186.67 W on the battery, so I went ahead and lowered the wattage on the device part, all the way down to 150 watts which gave me 1 percent head room on the battery, but that's at a full charge at 8.4 with two batteries, so I lowered it down to a pretty much drained battery at 7.2 which made the amps go up, so then I lowered the wattage even more to 129 watts which gave me 0% headroom right below 20 amps, so basically I was wondering if I'm using your calculator correctly and if so I could basically run two 20 amp batteries at a .42 ohm build at about 120 watts safely? I am new to this and hope I understood it correctly and also was wondering how accurate this was? And is efficiency the same as voltage drop, or do you have to add in voltage drop as well to see how many watts you are truly getting?

I also set it at 90 percent efficiency, just to give it a little more margin for error, I Probably wouldnt ever use watts that high, but just wanted to know how far I could push it, I currently own an IPV v1 right now, so I'm somewhat familliar with how to use the calculator just didn't know if all you had to do for 2 batteries was double the voltage, and then account for a drained battery and watch the amps raise. Thanks for responding though, I love how close the vaping community is, to me your calculator is legendary and I praise you for programming that and making it easier!

 

[Dampmaskin]

if you're running at dual 18650 regulated mod, to figure out the battery drain, do you just change the voltage to 8.4, and is that calculating in series?

Yes. The run time result will be off though. You can double the mAh number to fix that, but then you'll also have to cut the C rating in half in order to compensate for that. (Because batteries in series don't have double the amp limit as a single battery.)

And is efficiency the same as voltage drop, or do you have to add in voltage drop as well to see how many watts you are truly getting?

Voltage drop is not exactly the same as efficiency, but for all practical purposes you can regard them as analogous. You can say that efficiency is "wattage drop", if that makes sense. There's no point in calculating both for everyday use.

and if so I could basically run two 20 amp batteries at a .42 ohm build at about 120 watts safely?

Since the mod is regulated, and you're adjusting the watts, the resistance of the atomizer is not relevant to the calculation.

You can run 110 or 120W relatively safely with full batteries, but as the batteries drain, you need to lower the watts accordingly. As long as your batteries are good, and you're following stacking protocols, 85W should be safe all day long.

 

[BrennanElectric]

Alright thanks so much, also what is considered the cut off for drained batteries on the ipv v3? Is that like you said 6.4 volts? I'll play around some more with the calculator! I appreciate all of your help, you've been the one person to skillfully and willfully help me understand what's going on!

 

[Dampmaskin]

IMR batteries can survive going as low as 2.5V, but it is recommended to not let them go below 3.0V if you can avoid it. The lower you drain them, the faster they will wear out. Most mods cut out when the battery reaches 3.4 or 3.2 V. What the cutoff is with the IPV3 (or other stacked mods for that matter), I don't know.

 

[BrennanElectric]

IMR batteries can survive going as low as 2.5V, but it is recommended to not let them go below 3.0V if you can avoid it. The lower you drain them, the faster they will wear out. Most mods cut out when the battery reaches 3.4 or 3.2 V. What the cutoff is with the IPV3 (or other stacked mods for that matter), I don't know.

Also I don't know if you know or not, but that list you got from baditude, it lists the max continuous discharge currents, but no c ratings. How would you calculate the c rating, since your calculator doesn't have an amp rating?

 

[BrennanElectric]

IMR batteries can survive going as low as 2.5V, but it is recommended to not let them go below 3.0V if you can avoid it. The lower you drain them, the faster they will wear out. Most mods cut out when the battery reaches 3.4 or 3.2 V. What the cutoff is with the IPV3 (or other stacked mods for that matter), I don't know.

Or did you already calculate the C rating, from that list? I see the samsung 25r has an 8.8 C rating but has 20 amps and 2500 mah, but my lg he 2's have the same specs, just not the same C rating, so that is where I'm confused, as to how that C rating is calculated.

 

[Dampmaskin]

Amp = C * Ah

So

C = Amp / Ah

or

C = Amp / (mAh/1000)

Hmm.... it's well past midnight here and I'm getting sleepy, so I hope I got that right.

I've got to get me some sleep now. See you around.

 

[BrennanElectric]

Amp = C * Ah

So

C = Amp / Ah

or

C = Amp / (mAh/1000)

Hmm.... it's well past midnight here and I'm getting sleepy, so I hope I got that right.

I've got to get me some sleep now. See you around.

Yes you did get it right, your list on batteries and the C ratings are a little off, but doing the math myself and tweaking the calculator to make it 20 amps made it work correctly, thank you for helping me use your program confidently! I won't ever push it too far, and you've helped take a lot of stress off of me for having a mental block, it makes a lot more sense now. You have a good night sir, and I will see you around!