Regulated Mod -Ohms law and Battery safety

[Koutroulis87]

I just want to make a guestion and hopefully I get an answer to finally solve my confusion that goes on in my head

I know how ohm law works and what u need to take care when running a mechanical mode. Basically u need to know what's your battery CDR (LETS SAY 25A) and then divide the voltages you running ur device with the resistance u have on .. if its bigger than your batteries CDR Then maybe boom comes .

But where my confusion comes is when using a regulated mode. I know they have build in chips for protection and actually the modem have it's own voltage out put.

I run for example the smok R-kiss kit and it has voltage out put from 0.5V-8V and power range from 6W -200. Resistances from 0.1 to 2.5 Ohm. The batteries I'm running is the MURATA VTC5A (25A).

The coil I use is 0.16 and I work it at 110 watts .That give us 4.19 V and current I=26.2 A... do I already push the batteries in the limits cause my batteries has limit 25A. ? The coil factory suggest to vape from 90-140 WATTS with best between 110-130..

My mode has the batteries in series connection, so I know that both batteries together output 4.2V+4.2V=8.4 V when full charge! And in worst case senario 3.2V+3.2V=6.4V , but when running in series the battery CDR is still as one battery so it still stays at 25A!

So what's the maths to calculate when running regulated mod..? I know I am not in danger and I shouldn't worry at that levels of watts (110) with my MURATA VTC5A but I still want to know how things work

Thank u everybody in advantage and sorry if my English are not so good

 

[Koutroulis87]

I just want to make a guestion and hopefully I get an answer to finally solve my confusion that goes on in my head

I know how ohm law works and what u need to take care when running a mechanical mode. Basically u need to know what's your battery CDR (LETS SAY 25A) and then divide the voltages you running ur device with the resistance u have on .. if its bigger than your batteries CDR Then maybe boom comes .

But where my confusion comes is when using a regulated mode. I know they have build in chips for protection and actually the modem have it's own voltage out put.

I run for example the smok R-kiss kit and it has voltage out put from 0.5V-8V and power range from 6W -200. Resistances from 0.1 to 2.5 Ohm. The batteries I'm running is the MURATA VTC5A (25A).

The coil I use is 0.16 and I work it at 110 watts .That give us 4.19 V and current I=26.2 A... do I already push the batteries in the limits cause my batteries has limit 25A. ? The coil factory suggest to vape from 90-140 WATTS with best between 110-130..

My mode has the batteries in series connection, so I know that both batteries together output 4.2V+4.2V=8.4 V when full charge! And in worst case senario 3.2V+3.2V=6.4V , but when running in series the battery CDR is still as one battery so it still stays at 25A!

So what's the maths to calculate when running regulated mod..? I know I am not in danger and I shouldn't worry at that levels of watts (110) with my MURATA VTC5A but I still want to know how things work

Thank u everybody in advantage and sorry if my English are not so good

 

[DeloresRose]

The whole point of regulated mods is you don’t have to do the math, because the mod does. If the batteries aren’t capable of firing safely the mod won’t fire. Easy peasy.

All you have to be concerned about is that they are not rewraps and have no damage. Use married batteries in two battery mods, and charge with a good charger.

 

[bombastinator]

No maths at all I think anyway. I’m not familiar with any of that particular gear, so I could be wrong here, but generally, If a VV mod is working correctly it simply won’t fire if you exceed its capacity, and if the electronics do fail they will most likely fail into a non working state. This is part of the point of VV mods. They’re vastly more idiot resistant. Being an idiot myself I very much appreciate this.

 

[Topwater Elvis]

---> Calculating battery current draw for a regulated mod | E-Cigarette Forum

Sometimes folks get a false sence of security when using a regulated power device.
Proper battery selection for the power range you use is still an important safety consideration.

Near & at the power device's low voltage cutoff amp demand is highest / as the cell voltage decreases amp demand increases.

 

[DeloresRose]

Being an idiot myself I very much appreciate this

I resemble that remark. I can’t even balance my stupid check book. But at least that won’t blow up on me if it’s off a few bucks.

 

[Topwater Elvis]

This represents why starting two threads with the exact same question is detrimental.

Proper battery selection for a regulated power device is important.
As cell voltage decreases amp demand increases.
As cells age capacity diminishes which results in a reduction of the amount & rate amps can be demanded from the cell safely.

 

[bombastinator]

This represents why starting two threads with the exact same question is detrimental.

Proper battery selection for a regulated power device is important.
As cell voltage decreases amp demand increases.
As cells age capacity diminishes which results in a reduction of the amount & rate amps can be demanded from the cell safely.

A good point. Which is the other thread? Maybe a mod can do that merging thing.

 

[Topwater Elvis]

Same subforum, same page as this one, same exact thread title, same member posting the thread.

I already 'notified' in hopes of combining the two.

 

[Koutroulis87]

Sorry guys for the double thread , I sign up from my mobile phone and use it as well to write so I didnt know it has upload it at first place and came twice

I agree on that having a regulated mod doesn't mean we should not care about the power of CDR in our cells .That's why I try to find the maths behind how to calculate when using a regulated mod .

BTW I will try to delete the other thread and sorry again

 

[Topwater Elvis]

I provided a link to an easy to understand 100% factual answer to your question in your other thread.

Having it deleted also deletes the easy to understand factual answer.
Combining the two threads is the best option.

 

[Koutroulis87]

I provided a link to an easy to understand 100% factual answer to your question in your other thread.

Having it deleted also deletes the easy to understand factual answer.
Combining the two threads is the best option.

Hiw can I combine them together? Of course it's better to do so ,if u tell me the way
Sorry but I'm not used to forums and all this and I'm new here as well and try to find out

 

[sonicbomb]

Ohms/Watts Law - Calculating safe amp usage | E-Cigarette Forum

 

[Koutroulis87]

Thanks Topwater Elvis and sonicbomb for your help , your links helped me a lot !! Basically the confusion came to the factor that in regulated mods doesn't matter if the batteries are connected in series or parallel ..

So in my case , if I run my device at 110Watt let's say , each battery has to give 55 watts. 55 watts / 3.2 V when battery is exhausted (worst case scenario) that's 17.1 A for each battery ! We can add 10% room of mistake that's 18.8A .Pretty safe for my MURATA VTC5A 25A

 

[Ablonz]

Here is a excellent post to figure it out for you @Koutroulis87 Max Amps Per Battery = (75W / 3.2V) / 0.95 = 24.7A That would be for 150 watts, yours at 110 max would be 18.10 according to Mooch's post. Also, you may want to check out his blogs as he is our site battery guru.

 

[greek mule]

Welcome to the forums.
Others have answered all your questions pretty well.

 

[sonicbomb]

So in my case , if I run my device at 110Watt let's say , each battery has to give 55 watts. 55 watts / 3.2 V when battery is exhausted (worst case scenario) that's 17.1 A for each battery ! We can add 10% room of mistake that's 18.8A .Pretty safe for my MURATA VTC5A 25A

Steam-engine is the go-to site for all your vaping calculations

Battery drain | Steam Engine | free vaping calculators

 

[classwife]

Merged duplicate threads

 

[BrotherBob]

Thank u everybody

Welcome and glad you joined.

 

[greek mule]

You can add your country location to your profile so that people can answer your questions more easily.
Mouse over your username in the upper right corner and select Personal Details. You'll find the Location field on that page. Enter your country there and then click "Save" at the bottom of the page.