18650 2000mah 10A (5C)
18650 1600mah 24A (15C)
18490 1100mah 16.5A (15C)
18350 700mah 6A (8.6C)
14500 600mah 4.0A (6.7C)
18650 2000mah 10A (5C)
18500 1100mah 8.8A (8C)
18350 800mah 6.4A (8C)
14650 950mah 9.5A (10C)
14500 700mah A (C) (No information found on this one)
18650 2000 mAh 16A (8C)
18500 1100 mAh 8.8A (8C)
18350 800 mAh 6.4A (8C)
14500 600mah 4.8A (8C)
IMR18650 1500mah 20A (13.3C)
CGR18650CH 2250mAh 10A (4.4C)
NCR18650PD 2900mAh 10A (3.4C)
INR18650-20R 2000mah 20A (10C)
us18650v3 IMR 2250mAh 10A (4.4C)
SE US18650VTC3 1600mAh 30A (18.75C)
Just to make sure everybody understands the difference between C and A
"A" is pretty simple it means amp so when you read the specs of a battery and see something like:
Max. continuous discharge rate: 15A
You know that the amp limit is 15 amps.
However if you see something like:
max. continuous discharge rate: 15C
This is not the amp limit of the battery, it is its "c rating". Now, you can figure out the amp limit of a battery by using the C rating pretty simply by doing some math: (C rating x mAh) / 1000 = amp limit or more simply C rating x Ah = Amp limit
So we can figure out that a 1100 mAh battery like this one with a C rating of 8C has an amp limit of 8.8 amps:
8C x 1100 mAh = 8 800
8800 / 1000 = 8.8 <- this is the amp limit
Note that it is only a coincidence that the C rating and amp limit of this perticular battery have almost the same value.
If we take a 700 mAh battery that as a C rating of 10 then:
10C x 700mAh = 7000
7000 / 1000 = 7 amps
ok lets do it this way
Using this tool Ohm's Law Calculator
on my gripper Mech I use an EFest 18350 battery (rated for 6.4 Amos by your chart)
If I put a 1.5 ohm coil on it
and a fully charged battery 4.2 volts
I put the ohms in the resistance line of the calculator 1.5
I put the 4.2 in the voltage line
Then hit the calculate button it tells me the current requirement is 2.8 Amps
the batterys have a 6.4 limit so 2.8 is fine with a little margin for safety
Now if I was to do a .5 ohm coil on the same battery and mech I would end up with an 8.4 Amp requirement which this battery can't even pretend to handle and probably major venting/explosive results
This is a list of IMR and hybrid "high drain" batteries with their continuous discharge rate in amps. Use the Ohm's Law Calculator to see how many amps your coil will pull from the battery.
PBusardo is now using my list on his Taste Your Juice website. tasteyourjuice.com/wordpress/battery-information/
18650 2000mah 10Amp CDR
18650 1600mah 24A
18490 (1100mah) 16.5A
18350 ?(700mah) 6A
18650 1500mah ?20A
Panasonic ?or Orbtronic hybrid
CGR18650CH (IMR/hybrid) 2250mAh 10A
NCR18650PF (LiNiCOMnO2) INR/ICR/IMR Hybrid 2900mAh 10A
NCR18650PD (LiNiCoAl) 2900mAh 10A
Orbtronic 18650 SX22 (hybrid) 2000mAh 22A
UR18650EX 2000mAh 20A
Samsung hybrid (LiNiCoMnP)
INR18650-22P 2200mAh 10A
INR18650-20R 2000mah 22A
us18650v3 IMR 2250mAh 10A
us18650vct3 1600mAh 30A
us18650vtc4 2100 mAh 30A
18650 2000 mAh 16A
18650 1500 mAh 22A
18500 1100 mAh 8.8A
18350 800 mAh 6.4A
18650 (IMR/hybrid) 2250mAh 10A
18650 2000mAh 10A
18650 1600mAh 30A
18490 1100mah 8.8A
18350 800mah 6.4A
(AWG) Average Diameter Heat
Treatment Resistance (68°F) Melting
Temp Max Operating Temp
24 0.51 mm Annealed
(soft) 0.17 ? / in. 1500°C 1400°C
28 0.32 mm 0.44 ? / in.
30 0.25 mm 0.70 ? / in.
32 0.20 mm 1.09 ? / in.
34 0.16 mm 1.76 ? / in.
I came across this, written by Manu, it really helps clarify things for anyone trying to gauge a mech mod and how it performs and if after modding it you have a better device.
MANU, 06/02/2013, 1:27 PM:
A lower atomizer resistance translates into higher current = more power = more vapour. However, the higher the current, the higher the voltage drop on the internal battery resistance and mod resistance.
Remember that voltage drops at high currents cannot be completely eliminated. The presence of voltage drop only means that the efficiency is decreased. With lower resistance coils, the power on the atomizer increases, but a higher percentage of it is lost on the way.
Let’s get to an example:
Joe has an 18500 non-IMR battery. He vapes with an RBA that has a 1.2 Ohm coil. His battery is fully charged, so he decides to measure the voltage drop of his Roller.
Joe measures the battery voltage at open-circuit, and finds it to be 4.2 Volts.
Then, Joe measures the voltage across the atomizer terminals under load, and finds it to be 3.79 Volts. He assumes that his mod is giving him a voltage drop of 0.41 Volts and frowns unhappily. He is convinced that his mod is not giving him a good hit, although it vapes like a train.
Jay has a fresh 18500 IMR battery. He vapes with an RBA that has a 2.2 Ohm coil. His battery is fully charged, so he decides to measure the voltage drop of his Roller.
Jay measures the battery voltage at open-circuit, and finds it to be 4.2 Volts.
Then, Jay measures the voltage across the atomizer terminals under load, and finds it to be 4.02 Volts. He assumes that his Roller is giving him a voltage drop of 0.18 Volts and goes on to say how amazing it is and how great it vapes.
One minor detail I forgot to mention is that Joe sold his Roller to Jay because of the “voltage drop issue”, so it’s the *same* device.
On to the explanation:
The voltage across a resistance is given by the rather simplified formula V = I * R, where I is the current supplied by the battery and R the resistance we are examining.
The current I flowing through a mod is roughly equal to: I = Voc / ( Ra + Rm + Ri ), where Voc is the open circuit voltage of the battery, Rm is the equivalent mod resistance, Ri is the internal battery resistance and Ra is the atomizer resistance.
A typical value for Ri would be around 0.08 Ohms for a non-IMR battery that is in *OK* condition, while a newer, larger capacity, non stressed, high-drain battery might be better. Take this value with a grain of salt, since each battery is different.
Now, the mod’s equivalent resistance is again not a static quantity, since it depends on how tight the different components are screwed, how clean they are, and many other variable factors. A typical equivalent value would be around 0.05 Ohms, perhaps even less.
So, with the same mod (Rm = 0.05) in the same condition and configuration, Joe was vaping at a current of:
I_joe = 4.2 / (0.08 + 0.05 + 1.2) = 3.16 Amps,
which gives an atomizer voltage of
Va_joe = 3.16 * 1.2 = 3.79 Volts.
In the case of Jay, these values were:
I_jay = 4.2 / (0.05 + 0.05 + 2.2) = 1.83 Amps
Va_jay = 1.83 * 2.2 = 4.02 Volts
So – the voltage drop says nothing. In fact, Joe’s kit was putting out many more watts than Jay’s, because of the low atomizer resistance. Joe was vaping at:
P_joe = Va_joe * I_joe = 3.16 * 3.79 = 12 Watts
while Jay is vaping at:
P_jay = Va_jay * I_jay = 1.83 * 4.02 = 7.36 Watts
If you want to vape with a low resistance coil, the best thing to do in order to maximize efficiency is to use a high-drain, high-energy battery (18500/18650 IMR).
If Joe had a good IMR battery, his current and voltage would be:
I_joe’ = 4.2 / (0.04 + 0.05 + 1.2) = 3.26 Amps,
which would give an atomizer voltage of
Va_joe’ = 3.16 * 1.2 = 3.91 Volts
and a power of
P_joe’ = 12.73 Watts,
which is much better than the 3.79 Volts and 12 Watts he got with the non-IMR battery.
You need to be logged in to comment