# Yet another ohms spreadsheet



## Rellik (15/11/14)

Hi Guys
I made this little spreadsheet for my own usage. Before I start referencing and trusting it, can anyone please check that my logic (and Maths of course !) is correct.

I calculated the amps on each of the coils I estimate that I would be building. I have a Hana DNA 30 with the 2500mAh eFest IMR 18650 (35A) battery. These all look very much in safe limits hey?
I used the Formula of I=SQRT(P/R)

Reactions: Like 2 | Informative 2


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## Andre (15/11/14)

The continuous discharge rating on those Efests is 20A, but you are still more than safe. And your calculations look correct to me....checked a few, not all.


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## Silver (15/11/14)

Nice spreadsheet @Rellik - looks good
Nice way of checking the amp draw for safety
Looks like you are nowhere near the limits of that battery of yours on any of those coils and power settings


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## Johnny2Puffs (16/11/14)

Your formula is correct and is derived from the mixture of the 2 most basic Ohms law formulas as follows;
P=I*V and V=I*R
Therefore,
P=I*I*R
I Squared =P/R
I=Square root of P/R


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## zadiac (16/11/14)

Rellik said:


> Hi Guys
> I made this little spreadsheet for my own usage. Before I start referencing and trusting it, can anyone please check that my logic (and Maths of course !) is correct.
> 
> I calculated the amps on each of the coils I estimate that I would be building. I have a Hana DNA 30 with the 2500mAh eFest IMR 18650 (35A) battery. These all look very much in safe limits hey?
> ...



Thanks for the post, but could you add some higher wattage and lower resistanse too please, for us sub-ohm vapers living on the edge 
I am terrible at math, so much so that my 9th grade math teacher begged me to take another subject...soooo...


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## Rellik (17/11/14)

zadiac said:


> Thanks for the post, but could you add some higher wattage and lower resistanse too please, for us sub-ohm vapers living on the edge
> I am terrible at math, so much so that my 9th grade math teacher begged me to take another subject...soooo...



@zadiac I have updated the Original Post to include higher Wattage and lower resistance.


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## Andre (17/11/14)

Rellik said:


> @zadiac I have updated the Original Post to include higher Wattage and lower resistance.


Ah, now it starts being more difficult. In the spreadsheet your constant is the Wattage. So, for a regulated mod, using wattage mode, your calculations will be correct, but not for a mechanical mod on a fresh battery (4.2V). For example, at 0.2 ohms on a fresh battery in a mechanical mod you will get 21A and 88.2W. At 0.1 ohms the result will be 42A and 176.4W. 
@zadiac a good online calculator: http://www.onlineconversion.com/ohms_law.htm


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## Johnny2Puffs (17/11/14)

Andre said:


> Ah, now it starts being more difficult. In the spreadsheet your constant is the Wattage. So, for a regulated mod, using wattage mode, your calculations will be correct, but not for a mechanical mod on a fresh battery (4.2V). For example, at 0.2 ohms on a fresh battery in a mechanical mod you will get 21A and 88.2W. At 0.1 ohms the result will be 42A and 176.4W.
> @zadiac a good online calculator: http://www.onlineconversion.com/ohms_law.htm



Not quite correct. Your bat is rated at 3,7v and you should use this value in your calculation to get actual figures. Not 4,2v as this is "open circuit" voltage. When you close the circuit by firing it, it drops from 4,2 to 3,7v. 
When using a VV ciggie, the voltage set is the actual voltage when fired.

Reactions: Thanks 1


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## zadiac (17/11/14)

Rellik said:


> @zadiac I have updated the Original Post to include higher Wattage and lower resistance.



I thank you sir 



Andre said:


> Ah, now it starts being more difficult. In the spreadsheet your constant is the Wattage. So, for a regulated mod, using wattage mode, your calculations will be correct, but not for a mechanical mod on a fresh battery (4.2V). For example, at 0.2 ohms on a fresh battery in a mechanical mod you will get 21A and 88.2W. At 0.1 ohms the result will be 42A and 176.4W.
> @zadiac a good online calculator: http://www.onlineconversion.com/ohms_law.htm



Thanks @Andre , I'll check it out. I only use the Reo and my regulated mod now, so will still need it for the Reo as I sub-ohm with her too (if you know what I mean )


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## Andre (17/11/14)

Johnny2Puffs said:


> Not quite correct. Your bat is rated at 3,7v and you should use this value in your calculation to get actual figures. Not 4,2v as this is "open circuit" voltage. When you close the circuit by firing it, it drops from 4,2 to 3,7v.
> When using a VV ciggie, the voltage set is the actual voltage when fired.


Yes, of course there is voltage drop, but that is dependent on the equipment you use and, as I have it, it does not go down to 3.7V as a matter of course. I have always understood that the standard for mechanical mods is to calculate at 4.2V.
Certainly not my field of expertise, maybe @johan can be of help.
In any event, whatever voltage is used, the calculations in the OP's table is not correct for mechanical mods.


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## johan (17/11/14)

Andre said:


> Yes, of course there is voltage drop, but that is dependent on the equipment you use and, as I have it, it does not go down to 3.7V as a matter of course. I have always understood that the standard for mechanical mods is to calculate at 4.2V.
> Certainly not my field of expertise, maybe @johan can be of help.
> In any event, whatever voltage is used, the calculations in the OP's table is not correct for mechanical mods.



It all depends on the quality of battery as well as the resistance of coil in use; at a breakfast meet beginning of the year it was illustrated that a freshly charged battery (4.2V), and a well setup Reo, the voltage drop measured over various coils, were on average 200mV (0.2V). Can't remember all the different readings as such. The best I could achieve was 180mV (0.18V), voltage drop on a 0.93 Ohm coil with a new Efest battery fully charged to 4.23V. (4.23 - 0.18 = 4.05V).

Again here's an Ohms Law wheel for easy formula checkup:

Reactions: Thanks 2


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## Ccoetzee (17/11/14)

Andre said:


> The continuous discharge rating on those Efests is 20A, but you are still more than safe. And your calculations look correct to me....checked a few, not all.



Howsit @Andre,

Please tell me if I'm mistaken, but I have a few eFest's and the 3100mAh batteries are 20A continuous and the 2500mAh are 35A continous. Really good quality batteries

Regards

Chris


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## Stochastic (17/11/14)

Is the table for a mech or regulated mod? From what I understand that is a mech table and would be different for a regulated mod due to boosting the volts to get the desired wattage.


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## Andre (17/11/14)

Ccoetzee said:


> Howsit @Andre,
> 
> Please tell me if I'm mistaken, but I have a few eFest's and the 3100mAh batteries are 20A continuous and the 2500mAh are 35A continous. Really good quality batteries
> 
> ...


Unfortunately, you are mistaken. The 20A and the 35A are the burst or pulse discharge rating. Efest has publicly admitted (copy of their letter was posted somewhere on the forum) that the continuous discharge rating of these are 10A and 20A, respectively.


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## Andre (17/11/14)

Stochastic said:


> Is the table for a mech or regulated mod? From what I understand that is a mech table and would be different for a regulated mod due to boosting the volts to get the desired wattage.


As per the first post, the OP designed the table for his Hana regulated mod. As I tried to show above the table is accurate for a regulated mod if using the Watts at the top of the table. It is not accurate for mechanical mods.


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## Rellik (18/11/14)

Thanks for the reply @Andre . I was a bit busy yesterday and couldn't post here. As @Andre mentioned, this table is only accurate for regulated mods. Please always recheck your calculations and don't just trust a spreadsheet of the internet

Reactions: Like 1


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## Stochastic (18/11/14)

I still don't get the resistance part though, is it the coil?


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## JW Flynn (18/11/14)

Stochastic said:


> I still don't get the resistance part though, is it the coil?


your coils that provides the heat has a resistance and works much like one of those electric heaters... it is built with resistance wire (made up of a couple of elements) so your coil has the resistance, your battery provides the voltage and also depending on your battery it can handle a set amount of amps. wikipedia ohms law, it should provide all the info you need 

I have recently seen a instagram post where the guy recons he has learnt more about electricity since he started vaping than he ever did while he was at school, LOL...

Reactions: Informative 1


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## Stochastic (18/11/14)

I get that, but I don't get calculating amp draw on the battery using the coil's resistance. Here is a link to an explanation on how I also understand it. http://www.e-cigarette-forum.com/forum/blogs/dampmaskin/5901-battery-drain.html


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## Andre (18/11/14)

Stochastic said:


> I get that, but I don't get calculating amp draw on the battery using the coil's resistance. Here is a link to an explanation on how I also understand it. http://www.e-cigarette-forum.com/forum/blogs/dampmaskin/5901-battery-drain.html


That is the only way you can do it for a mechanical mod as you only have the resistance and battery voltage available. That is also how I understand the article you linked.


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## Stochastic (18/11/14)

Andre said:


> That is the only way you can do it for a mechanical mod as you only have the resistance and battery voltage available. That is also how I understand the article you linked.



But we are talking about a regulated mod as you said, for which the author argues using coil resistance is nonsensical in calculating amp draw on the battery.


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## johan (18/11/14)

Before you can start doing calcs for a regulated mod you need to know the following from the specification of the regulated mod: (1) minimum and maximum output voltage, (2) maximum deliverable current and obviously; (3) maximum output wattage, then only can calculations be made to fit into the limiting set parameters.

Reactions: Winner 1


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## Stochastic (18/11/14)

So the current drain on the battery is not I = P/V? (V being the battery's voltage and P being the desired wattage setting and ignoring efficiency)


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## johan (18/11/14)

Stochastic said:


> So the current drain on the battery is not I = P/V? (V being the battery's voltage and P being the desired wattage setting and ignoring efficiency)



Don't understand your question properly but assume you're looking for I=V/R - anyhow here's the wheel again with all the formulas (Ohms law):

​


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## Andre (18/11/14)

Stochastic said:


> So the current drain on the battery is not I = P/V? (V being the battery's voltage and P being the desired wattage setting and ignoring efficiency)


Yes, it is, if not limited by one of the factors listed by @johan above. If I understand you correctly.

Reactions: Like 1


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## Rellik (18/11/14)

Sheesh, I posted the initial spreadsheet to try and clear up some confusion. Seems it has caused more confusion


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## Stochastic (18/11/14)

No way, it's a good thing IMO, I learned to look at it from your perspective. Thanks for posting it @Rellik and the input from the others.

Reactions: Agree 1


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## Andre (19/11/14)

Rellik said:


> Sheesh, I posted the initial spreadsheet to try and clear up some confusion. Seems it has caused more confusion





Stochastic said:


> No way, it's a good thing IMO, I learned to look at it from your perspective. Thanks for posting it @Rellik and the input from the others.


Totally agree, we hopefully all learnt something, which would not have happened did you not post. And, as they say, knowledge is power!

Reactions: Like 2 | Agree 1 | Thanks 1


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