Battery confusion.

[Waxxiii]

I'm getting my first device with removable batteries in the next couple of days so I started doing a little research into the various batteries, battery drain, draw, current, amp ..... *brain explodes*

Ok, so I got more than a bit lost. I would like to know how to work out which batteries are right for a particular mod and whilst I was looking I came across this:

W+(W•.1)/V = A

W =watts selected
V = the mods low voltage cut off under load
A = battery draw

Perhaps somebody could explain it?

Also, when using the battery drain calculator on Steam Engine, which results do I need to be looking at?

Could someone explain what I need to know regarding continuous maximum discharge current and maximum discharge current.


I would also like more specific advice. I'll be buying a 200w device. I will rarely be using it over 100w but I would like a battery that enables me to get the most out of the mod if I choose to. Should I buy two different types of batteries for different styles of vaping? This is probably a silly question. Do I need 20a or 30a batteries? The mod I'm looking at comes bundled with two Samsung 25R's at an insanely reduced price but I'm unsure they will be adequate being 20a. I was initially looking at Sony VTC4's.

Last but not least, flat top or button top?

 

[Izan]

 

[Waxxiii]

View attachment 548452

Thanks. I actually have this saved as a screenshot and I know the batteries I'm thinking of getting are both recommended here but I'm unsure whether I will need VTC4's (30A) or 25R's (20A) based on the mod I'll be buying and my vaping style.

 

[Susan~S]

Mooch gave the Sony VTC4's a 23A CDR.

• Sony VTC4 2000mAh 18650 Retest Results...a great 23A 2100mAh battery

If you are looking at true 30A CDR cells, look at LG HB2, HB4 or LG HB6.

• Which is the best performer: HB2 vs HB4 vs HB6?

As far as calculations when using a regulated mod. See below.

• Calculating Battery Current Draw for a Regulated Mod

 

[IMFire3605]

So if I am reading your OP right, we are dealing with a 200watt dual battery series mod. Also you are stating though it is 200watts, you'll be max 100watts (1/2 full throttle). So...Up to 100watts, any 20amp battery, best in class for long running time and CDR would be the LG HG2 (Chocolate Brown) and the Samsung 30Q (Lavender), up to 150watts the Samsung 25R, Sony VTC4, LG HD2C, up to 180watts Sony VTC4, at 200watts Sony VTC3, LG HB2, HB4, HB6 (this last category you will lose run time as true 30amp CDR batteries shed Mah to reach higher CDR).

hth

 

[Bunnykiller]

better brain explode than battery explode
the small round end is positive end

 

[Bunnykiller]

VTC4... been using the same 4 batteries for over 2.5 yrs now..

 

[sonicbomb]

The last link that Susan posted fully describes the calculations required. But in short you take the wattage you plan to vape at, calculate the amp draw when the battery discharged and make sure that the battery you use has a CDR that exceeds this value. How much of a safety margin you allow for is up to you. If you have multiple batteries then divide the total amp draw by the number of batteries.

If you don't like maths, use steam-engine, set regulated device, set wattage and the set 3.2v under 'battery voltage' on the top right as this is the usual cutoff point for most devices. Check the amp value under 'battery drain'.

 

[sparkky1]

I'm getting my first device with removable batteries in the next couple of days so I started doing a little research into the various batteries, battery drain, draw, current, amp ..... *brain explodes*

Ok, so I got more than a bit lost. I would like to know how to work out which batteries are right for a particular mod and whilst I was looking I came across this:

W+(W•.1)/V = A

W =watts selected
V = the mods low voltage cut off under load
A = battery draw

Perhaps somebody could explain it?

Also, when using the battery drain calculator on Steam Engine, which results do I need to be looking at?

Could someone explain what I need to know regarding continuous maximum discharge current and maximum discharge current.


I would also like more specific advice. I'll be buying a 200w device. I will rarely be using it over 100w but I would like a battery that enables me to get the most out of the mod if I choose to. Should I buy two different types of batteries for different styles of vaping? This is probably a silly question. Do I need 20a or 30a batteries? The mod I'm looking at comes bundled with two Samsung 25R's at an insanely reduced price but I'm unsure they will be adequate being 20a. I was initially looking at Sony VTC4's.

Last but not least, flat top or button top?

You need to be looking at battery drain
Enter 6.6V in the little window right above : battery voltage
You will notice as you raise the watts the amps will go up, your full charge will be 4.2 but mean current draw will be 3.7 - 3.6V BUT you want to put in the number when your battery is 80 % taxed ( dis-charged ) 3.3V because the amps will rise as the battery voltage charge decreases
Battery drain

 

[bwh79]

I came across this:

W+(W•.1)/V = A

W =watts selected
V = the mods low voltage cut off under load
A = battery draw

Perhaps somebody could explain it?

The most important thing is to know the maximum continuous amp draw your batteries can sustain, and not to exceed that limit. Amps, watts, volts, and ohms are all related to each other, and any change in one value will affect all the rest as well. With a wattage-regulated device, the battery powers the unit and the unit powers the coil -- the battery doesn't ever see the coil directly, and so the ohms are not particularly relevant in this case. The ohms only matter so that the device can know what voltage to apply, in order to provide the selected wattage. And that, the wattage, is what we really need to look at.

Remember how I said that amps, watts, volts, and ohms are all related? There are a couple of equations that demonstrate the relationship between those values. Here is the one you need to learn, regarding a wattage-regulated device:

A = W/V -- Amps = Watts/Volts

That states that the amperage at any given moment is equal to the wattage divided by the battery voltage. Note, this is the output voltage of the battery, not the voltage applied to the coil (remember how I said that the battery never sees the coil directly?) If you play around with the numbers, you will see that the amp draw is at its highest, when the voltage is at its lowest. For this reason, when doing safety calculations, you need to enter the lowest possible voltage your device will fire at. For most mods, this is around 3.1/3.2 volts. If you're not sure, you can just use a flat 3.0 to be on the safe side.

Next, realize that 100% of the power drawn from the batteries doesn't get applied to the coil. The device needs to "steal" a little power to power itself. So when you're at 50 watts, for example, the device is actually drawing slightly more than that from the batteries. That's where the extra (W*.1) comes in. "Watts, plus (watts time .1)" is the same as "1.1 times watts." Basically, it's saying to add an extra 10% onto the displayed wattage, to estimate the actual power being drawn from your batteries. For this, you will want to use the highest wattage you will be vaping at, or as high as the device will allow in case of accidental misadjustment. (Another way to do this is to take the final W/V value and divide by .9 for a similar, but not identical, result. As these numbers are all estimates anyway, either one should suffice.)

So in the end, what we're looking at, max amp draw is equal to max wattage (plus ten percent) divided by min cutoff voltage. So if you're looking at a 200-watt mod with 3 batteries and a cutoff voltage of 3.1v per battery, the equation looks like this:

A = (200 + (10%*200)) / (3* 3.1)
... = (200+20)/9.3
... = 220/9.3
... = 23.655... amps. So if you're using regular, 20-amp batteries (pretty much the industry standard -- don't believe anything that claims a 35-amp capability!) then they're not enough to safely run the full 200 watts. You will need to use a more powerful 25- or 30-amp battery like the LG HB2/HB6.

 

[Waxxiii]

The last link that Susan posted fully describes the calculations required. But in short you take the wattage you plan to vape at, calculate the amp draw when the battery discharged and make sure that the battery you use has a CDR that exceeds this value. How much of a safety margin you allow for is up to you. If you have multiple batteries then divide the total amp draw by the number of batteries.

If you don't like maths, use steam-engine, set regulated device, set wattage and the set 3.2v under 'battery voltage' on the top right as this is the usual cutoff point for most devices. Check the amp value under 'battery drain'.

Is that 3.2v per battery? So 6.4v for the two batteries that the Kbox uses?

 

[Waxxiii]

The most important thing is to know the maximum continuous amp draw your batteries can sustain, and not to exceed that limit. Amps, watts, volts, and ohms are all related to each other, and any change in one value will affect all the rest as well. With a wattage-regulated device, the battery powers the unit and the unit powers the coil -- the battery doesn't ever see the coil directly, and so the ohms are not particularly relevant in this case. The ohms only matter so that the device can know what voltage to apply, in order to provide the selected wattage. And that, the wattage, is what we really need to look at.

Remember how I said that amps, watts, volts, and ohms are all related? There are a couple of equations that demonstrate the relationship between those values. Here is the one you need to learn, regarding a wattage-regulated device:

A = W/V -- Amps = Watts/Volts

That states that the amperage at any given moment is equal to the wattage divided by the battery voltage. Note, this is the output voltage of the battery, not the voltage applied to the coil (remember how I said that the battery never sees the coil directly?) If you play around with the numbers, you will see that the amp draw is at its highest, when the voltage is at its lowest. For this reason, when doing safety calculations, you need to enter the lowest possible voltage your device will fire at. For most mods, this is around 3.1/3.2 volts. If you're not sure, you can just use a flat 3.0 to be on the safe side.

Next, realize that 100% of the power drawn from the batteries doesn't get applied to the coil. The device needs to "steal" a little power to power itself. So when you're at 50 watts, for example, the device is actually drawing slightly more than that from the batteries. That's where the extra (W*.1) comes in. "Watts, plus (watts time .1)" is the same as "1.1 times watts." Basically, it's saying to add an extra 10% onto the displayed wattage, to estimate the actual power being drawn from your batteries. For this, you will want to use the highest wattage you will be vaping at, or as high as the device will allow in case of accidental misadjustment. (Another way to do this is to take the final W/V value and divide by .9 for a similar, but not identical, result. As these numbers are all estimates anyway, either one should suffice.)

So in the end, what we're looking at, max amp draw is equal to max wattage (plus ten percent) divided by min cutoff voltage. So if you're looking at a 200-watt mod with 3 batteries and a cutoff voltage of 3.1v per battery, the equation looks like this:

A = (200 + (10%*200)) / (3* 3.1)
... = (200+20)/9.3
... = 220/9.3
... = 23.655... amps. So if you're using regular, 20-amp batteries (pretty much the industry standard -- don't believe anything that claims a 35-amp capability!) then they're not enough to safely run the full 200 watts. You will need to use a more powerful 25- or 30-amp battery like the LG HB2/HB6.

Thank you. I'll be saving this post to look over in more detail. I appreciate you taking the time to explain it so thoroughly .

 

[Waxxiii]

Thanks everyone. This info is a huge help. I decided to purchase the Samsung 25R's because at £6 for two it would be crazy not to. I'll use them with the Kbox 200 at a safe wattage and when I get some batteries needed for higher wattages I can always use the Samsungs in some of the 75w mods I intend to buy.

Quite honestly, I doubt I'll ever vape above 100w but I'd like to know I have the option to safely do so if I choose to.

 

[IMFire3605]

Is that 3.2v per battery? So 6.4v for the two batteries that the Kbox uses?

The Kbox 200watt has a battery cut off of 3.55v per battery is what I have witnessed with my own, so I guess they designed it to not tax the batteries very hard.

200watts/7.1volts low cut off=28.17amps so Sony VTC4's will work at that max wattage, not the VTC5 (the VTC5 is only 20 amps, unlike the VTC4 which is 23amps but can handle 28amps easily or the VTC3 which is a straight true 30amp battery). But at 100watts only or below, I still believe the LG HG2 and Samsung 30Q will serve you better at 3000mah at 20amps, extra 900mah over the VTC4 2100mah.

 

[Waxxiii]

The Kbox 200watt has a battery cut off of 3.55v per battery is what I have witnessed with my own, so I guess they designed it to not tax the batteries very hard.

200watts/7.1volts low cut off=28.17amps so Sony VTC4's will work at that max wattage, not the VTC5 (the VTC5 is only 20 amps, unlike the VTC4 which is 23amps but can handle 28amps easily or the VTC3 which is a straight true 30amp battery). But at 100watts only or below, I still believe the LG HG2 and Samsung 30Q will serve you better at 3000mah at 20amps, extra 900mah over the VTC4 2100mah.

Thank you. How do you find out a mods battery cut off point?

Thanks for the info. I always had it in my head to go for VTC4's because they ate the ones I hear about so often. I'll take a look at the others you mentioned.