Battery Amp draw on vv series device

[ShaunSavage]

Good day, I recently got a tesla Invader 3, it's a variable voltage series mod, I'm running a 26guage kanthal dual parallel build ohming out to .18, vv potentiometer can go from 3.6v to 6.6v.. if I have it turned all the way down to 3.6volts across the two battries in series, will the amp draw from each battery still be 20amps each?

 

[93gc40]

Amp load is divided among the batteries.... so 10 amps each... but in reality the weakest battery will carry most of the load. So it may not always be a perfect 50/50 split.

Sent from my SCH-I605 using Tapatalk

 

[DaveP]

What 93gc40 said. batteries share the current load half and half (theoretically). You want to make sure that both batteries are new (purchased at the same time) to make sure that they both receive the same percentage of the load and discharge at similar rates.

Battery current is based on load. Each battery supplies half the current to operate the mod. 20 amps is the maximum current level the batteries can deliver in series and that's based on the coil resistance and the voltage/current setting you choose to vape at. As you turn the wattage up the current load on the batteries rises. At low wattage you might be working at 4A and at high wattage you could be pushing 20A. As the voltage rises across a given resistance, so does the amp draw.

When you take them out and put them on the charger they should both have the same or very similar voltage readings after they've been in the mod for a period of time. That indicates they are similar in charge and discharge characteristics. A weak battery in the chain will not discharge at the same rate and will probably heat up more than a strong, new cell and result in differing voltages as discharge continues.

 

[untar]

I'm no battery nor chemistry expert by any means but wouldn't one battery have the full current go through it?
It would need to provide only half of it by itself but the current from the other battery can't just teleport through and would have to go through the first battery, heating it up more and challenging the chemistry more.
They're in series so I don't see how it could happen in a different way, if someone knows exactly what happens please comment.
For true Amp split you'd need a parallel configuration imo.

My thinking would be while each battery only needs to provide half the peak current one of them still conducts the full load so you can't exceed the CDR of a single without taking more risk.

Edit: slight correction - since it's a closed circuit both batteries will have the full current go through them, the current doesn't disappear.

 

[Walee]

I'm no battery nor chemistry expert by any means but wouldn't one battery have the full current go through it?
It would need to provide only half of it by itself but the current from the other battery can't just teleport through and would have to go through the first battery, heating it up more and challenging the chemistry more.
They're in series so I don't see how it could happen in a different way, if someone knows exactly what happens please comment.
For true Amp split you'd need a parallel configuration imo.

My thinking would be while each battery only needs to provide half the peak current one of them still conducts the full load so you can't exceed the CDR of a single without taking more risk.

Edit: slight correction - since it's a closed circuit both batteries will have the full current go through them, the current doesn't disappear.

You're on the money. In a series circuit the current is the same at any point. The constant current you can safely draw is that of the weakest of the two batteries. The other measurement being mah or how much time the batteries will discharge is additive. It's sometimes difficult to discuss because the two indices are easily confused in verbiage used to describe them.

 

[Baditude]

 

[bwh79]

My understanding is that power (watts) is always the conserved quantity. Watts in == watts out. Watts at the atomizer == watts at the battery. So if you're running a .18 at 3.6v output voltage, that's equivalent to 72 watts (P = V*I = V*V/R since I = V/R). Divide that by six volts (input voltage aka battery charge state; using 3.0v which is about the lowest you should let your batts get before recharging -- does the mod have a low-voltage cutoff? -- and then times that by two 'cause they're in series) gives a 12-amp peak discharge at the battery. At higher input voltage (fuller charge) you divide by a larger number and so draw fewer amps. At higher output voltage (potentiometer) you start with a larger number and so draw more amps.

 

[ShaunSavage]

My understanding is that power (watts) is always the conserved quantity. Watts in == watts out. Watts at the atomizer == watts at the battery. So if you're running a .18 at 3.6v output voltage, that's equivalent to 72 watts (P = V*I = V*V/R since I = V/R). Divide that by six volts (input voltage aka battery charge state; using 3.0v which is about the lowest you should let your batts get before recharging -- does the mod have a low-voltage cutoff? -- and then times that by two 'cause they're in series) gives a 12-amp peak discharge at the battery. At higher input voltage (fuller charge) you divide by a larger number and so draw fewer amps. At higher output voltage (potentiometer) you start with a larger number and so draw more amps.

My understanding is that power (watts) is always the conserved quantity. Watts in == watts out. Watts at the atomizer == watts at the battery. So if you're running a .18 at 3.6v output voltage, that's equivalent to 72 watts (P = V*I = V*V/R since I = V/R). Divide that by six volts (input voltage aka battery charge state; using 3.0v which is about the lowest you should let your batts get before recharging -- does the mod have a low-voltage cutoff? -- and then times that by two 'cause they're in series) gives a 12-amp peak discharge at the battery. At higher input voltage (fuller charge) you divide by a larger number and so draw fewer amps. At higher output voltage (potentiometer) you start with a larger number and so draw more amps.

Hey thanks for the info! and yes it does have low voltage protection, I just wanted to make Sure I was not in any danger, I really to try to stay within the range of my batteries, I'm using the LG HE2 3000MAH (brown wraps) batteries so I'm hoping these are also safe to use on a vv device

 

[bwh79]

I'm using the LG HE2 3000MAH (brown wraps)

Do you mean the HG2? HE2 is 2500mAh, and not sure about the wrap color, but I know HG2's are brown and 3000mAh from Mooch's "recommended batteries" chart. Either way, he's got both of them rated for 20A discharge which, in a dual-cell device, should be good up to about 120 watts max. You'll have to do your own calculations on resistance and voltage to determine your actual watts usage with any particular setup. On the .18 you should be good up to about 4.6v on the potentiometer (~117w).

 

[ShaunSavage]

Do you mean the HG2? HE2 is 2500mAh, and not sure about the wrap color, but I know HG2's are brown and 3000mAh from Mooch's "recommended batteries" chart. Either way, he's got both of them rated for 20A discharge which, in a dual-cell device, should be good up to about 120 watts max. You'll have to do your own calculations on resistance and voltage to determine your actual watts usage with any particular setup. On the .18 you should be good up to about 4.6v on the potentiometer (~117w).

Thanks alot! And sorry Yes I ment HG2, those are my favorite all around battries wanted to continue using them in this mod, didn't want to have to get Sony vtcs to have to use in vv, as I only use vtcs in my mech tube mods

 

[DaveP]

Battery University graphic for series and parallel connections below.

Voltage is additive and current capacity is the same through each battery in a series connection. At lower voltage levels the actual current draw will be less than the max for the batteries. As voltage rises, current delivery/draw across a given cell rises, but in series the max current is equal to the current capability of any single cell.

Serial and Parallel Battery Configurations and Information

for a parallel connection the total current capacity rises and voltage is the same as a single cell.

 

[Walee]

Battery University graphic for series connection. Voltage is additive and current is the same through each battery. At low voltage levels the current draw will be less. As voltage rises, current delivery across a given cell rises, but in series the max current is equal to the current capability of any single cell.

http://batteryuniversity.com/learn/article/serial_and_parallel_battery_configurations

for a parallel connection current capacity rises and voltage is the same as a single cell.

I think where some confusion creeps in for vapors is when they look at this to determine how long batteries will last or maybe better stated how many puffs they will get in different configurations. Both configurations deliver 48,960 mwh (milli watt hours) which is the indication of how many puffs each configuration will provide. There are really three values being represented here. Max voltage, max current, and power over time.

 

[DaveP]

I think where some confusion creeps in for vapors is when they look at this to determine how long batteries will last or maybe better stated how many puffs they will get in different configurations. Both configurations deliver 48,960 mwh (milli watt hours) which is the indication of how many puffs each configuration will provide. There are really three values being represented here. Max voltage, max current, and power over time.

Yes, vape time is affected by current load capacity, frequency of puffs, and condition of cells as well as the age of the cells being used. If you swap the same set(s) of cells in and out of a mod for a year all that becomes moot. New cells rule.

I'm using my Efest LUC 4 charger right now, but I really like my XTAR VC4 charger for looking at milliamps delivered to the cells during charging. Vape time is all about capacity and current draw. On the XTAR you see the MAH count while charging. If you discharge to the same voltage level each time the MAH reading is a telltale as the number reduces over time and lots of charge cycles.

 

[Walee]

Yes, vape time is affected by current load capacity, frequency of puffs, and condition of cells as well as the age of the cells being used. If you swap the same set(s) of cells in and out of a mod for a year all that becomes moot. New cells rule.

I'm using my Efest LUC 4 charger right now, but I really like my XTAR VC4 charger for looking at milliamps delivered to the cells during charging. Vape time is all about capacity and current draw.

Exactly. A lot going on behind the scene. Efficiency of the mod, voltage drop under load, efficiency of the battery at various loads, etc. etc. I've been watching voltage under load on my batteries to get a feel for their "condition". Not convenient, but it works.

 

[DaveP]

I have two sets of two Samsung 30Q's that I've been rotating in and out of my Eleaf Invoke for about a year now. They seem to be hitting discharge earlier in the day than they did when new.

I've been charging them with the LUC 4 charger, but I need to get the XTAR VC4 out and look at MAH numbers during charging to see if they've lost capacity. They used to hit somewhere around 2100 MAH on a charge from 3.2v to full charge. The printed MAH rating on a battery is based on charging from somewhere around 2.5v up to 4.2v.