Charger question

[mattmc92]

to quote bwh79

Input amps drawn from the battery and output amps delivered to the atomizer are not the same thing. The only value that's (necessarily) the same on both input & output sides of the circuit board is watts.
a device is set to 64 watts, delivered to a .25Ω coil with a battery charge of 3.5 volts. To reach those 64 watts, at a charge state of 3.5 volts, the device pulls 18.2 amps from the battery. Over on the output side, however, it's a different story. To feed 64 watts through a .25 coil, it delivers 4.0 volts, and with 4 volts and a quarter of an ohm, that works out to an applied current of 16 amps. With a different coil at the same watts, you'll get a different value. That same 64 watts still draws 18.2A from the battery, but delivered to a .5Ω coil instead it requires 5.65 output volts and runs at only 11.3 amps applied current. But we don't really care about the applied current in either case. For purposes of battery safety, the amps we need to look at are the 18.2 that are coming out the battery, not the 11 or 16 that are going into the atomizer.

i don't know why you're being so aggressive.

 

[Bunnykiller]

just checked the voltage output across all terminals of a LUC V4..... there is voltage across all terminals, but it probably wouldnt control the current correctly when using different bays to charge a battery in the method you explained... it would charge but most likely continue to charge exposing the battery to excessive currents and potential overheating issues... seems that the negative ends of the charger are not all connected in series... so there isnt a common ground there... thus, each bay is independant in its capacity to sense the amount of voltage/charge each cell contains and adjusts current flow accordingly....

 

[Bunnykiller]

Its more like last night smoking i thought about it and was like "wonder if itll work. Well instead of trying it lets ask someone about it."

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curiosity is the sign of an active/open mind
I am constantly asking questions... and its kewl to learn the answers

 

[Eric LeClair]

Im going to first say that this question is purely for knowledge and wether it would work or not, im not going to try this. If you sodered a 2 inch long wire to the negative side of a battery and a seperate 2 inch long wire to the positive side, could you charge the battery by connecting the negative wire on one slot and the positive wire to the positive slot next to it?

haha only one way to find out - I think you should video tape what you are doing but in reality, I think it'll just short circuit.

 

[Bunnykiller]

Probably a lot more than you'd be able to comprehend.

I just can't figure what you're talking about here with "how many amps its taking the battery to fire".

Mininum amp before battery is dead?

A battery doesn't need amp to fire, it either has contained power or doesn't, this isn't an engine that you need to turn over... electricity/energy is stored, touching a positive to a negative and boom, it's like magic, it works. It's like opening a water system that's under pressure (such as gravity), pop that valve open, and it'll flow, you don't need to use a secondary system to pump it. Like popping a balloon, a controlled hole and the air pushes itself out, a bad ....., and it pops.

What you know that you're talking about might be clear in your head (and only your head), but care to explain your question instead of being arrogant?

a battery needs voltage and amperage to supply a current flow to the device being driven ( atty)
a hearing aid battery can have 3.7 volts but only carry .25 Ahr rating and our beloved 18650s have 3.7 volts at 20+ Ahr rating...
the 18650 will turn a .20 ohm coil into a glowing orange mass, but the hearing aid battery will do nothing near what we need from it... one can look at voltage as pressure in the hose and amperage as the amount of flow the hose can deliver... as in a 1/2" garden hose compared to a 6" fire hose both at the same pressure....

 

[Imfallen_Angel]

a battery needs voltage and amperage to supply a current flow to the device being driven ( atty)
a hearing aid battery can have 3.7 volts but only carry .25 Ahr rating and our beloved 18650s have 3.7 volts at 20+ Ahr rating...
the 18650 will turn a .20 ohm coil into a glowing orange mass, but the hearing aid battery will do nothing near what we need from it... one can look at voltage as pressure in the hose and amperage as the amount of flow the hose can deliver... as in a 1/2" garden hose compared to a 6" fire hose both at the same pressure....

Hi, thanks for your reply, but I actually know about this stuff with a good 30 years in electronics/electrical/computer work and repairs, so no worries.

The point was that the "question" was half baked, not clear, I didn't understand the point the person was getting at, and asking for clarification was too much for them and they needed to be nasty about it.

Talking about "firing a battery" without proper context was just too open to variances, especially with the added mention of "not to the atomiser", which at that point represented a battery just sitting in a mod doing nothing.

If your point is the correct one about the context of this guy's comment, then a simple "how much power does a battery need to have to provide a current that's sufficient to generate heat in a coil (via it's resistance) and not cause the battery to overload" would have been a simple clarification... but again, stating the absence of a coil/atomiser made it very unclear at that point.

And the answer isn't clear cut either, it's a balance of the resistance of the coil (or whatever). You can have a coil that only requires a fraction of power, compared to some monstrosities that almost require a car battery.

I could easily create a 0.2 ohm resistance with a thin enough wire and use a hearing aid battery, won't be able to create clouds, but it would manage to evaporate some sort of liquid for a few seconds.

It could have been about how much power for the reg. mod to activate, etc.

Anyways, the thread turned to crap over his ego, so I'm done.

 

[TheWalrus]

Hi, thanks for your reply, but I actually know about this stuff with a good 30 years in electronics/electrical/computer work and repairs, so no worries.

The point was that the "question" was half baked, not clear, I didn't understand the point the person was getting at, and asking for clarification was too much for them and they needed to be nasty about it.

Talking about "firing a battery" without proper context was just too open to variances, especially with the added mention of "not to the atomiser", which at that point represented a battery just sitting in a mod doing nothing.

If your point is the correct one about the context of this guy's comment, then a simple "how much power does a battery need to have to provide a current that's sufficient to generate heat in a coil (via it's resistance) and not cause the battery to overload" would have been a simple clarification... but again, stating the absence of a coil/atomiser made it very unclear at that point.

And the answer isn't clear cut either, it's a balance of the resistance of the coil (or whatever). You can have a coil that only requires a fraction of power, compared to some monstrosities that almost require a car battery.

I could easily create a 0.2 ohm resistance with a thin enough wire and use a hearing aid battery, won't be able to create clouds, but it would manage to evaporate some sort of liquid for a few seconds.

It could have been about how much power for the reg. mod to activate, etc.

Anyways, the thread turned to crap over his ego, so I'm done.

This was all very fun to watch from my perspective lol

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