Read my just posted answer to this question. While not identical it covers aspects which will answer some of your questions.
3 x 18650 LiIons (or any 3 LiIons) will have a fully charged voltage of 3 x 4.2V = 12.6V and a fully discharged voltage of ABOUT 3 x 3 = 9V. How low low goes is up to you. Too low and battery dies.
Read my answer above re balancing. It is not NECESSARY as long as you are CERTAIN that no cell is ever deep discharged AND if charging in series, as long as no cell is in constant voltage tail off mode while you are attempting to inject full constant current at 1C. 'Attempting to" period may be short.
IF you charge this off the bike and if all 3 cells are isolated from the world (but connected to each other) then my answers above re charging one at a time apply. You can charge 3 at a time with 3 chargers ** as long as** all charger outputs are truly isolated.
An easy way to get 12V is to use one of the many many available switch mode power supplies. You can get 1 or 2 or 3 cell LiIon to 12V capable supplies.
An 18650 LiIon cell is has a capacity of about 2000 mAH x 3.6V nominal =~~ 7 Watt hours. IF your flasher worked at 1 Watt average and was anything like serious it would blow following motorists off the road. Depends on design. 1 Watt at 10% duty cycle = 10 Watts when one. 1 Watt at 1% duty cycle = 100 Watts when on. Properly collimated a 1 att red LED willl do a very very very very good job. So a single 18650 cell with inverter of say 7% efficient (low) will run for 7 Wh/1 Watt x 70% = 5 hours. Ample for most people.
ADDED:
- OK, so some clarifying questions.
1) how can I be certain that no cell is ever "deep discharged"?
No cell ever under 3 Volt.
- Monitor voltage and prevent this happening
OR
- Never discharge beyond known capacity to ensure this is true.
Murphy says you will fail if you take the 2nd choice.
- 2) if I'm using a COTS charger (and charging each cell separately), what do you mean by the charger outputs must be truly isolated? If I'm using a physically separate charger for each cell,
If the cells are not connected in any way this is irrelevant.
This is an issue only if the cells are connected as in a battery holder.
Outputs are fully isolated from the charger input.
If you operate 2 chargers from mains simultaneously you must get no sensible voltage readings when measuring from eg V+ out of one to V+ out of other.
If you plav a resistor from V+ out of oneto Ground out of th eother no current flows.
Well under 1 mA would be acceptable.
I'd expect good isolation. What is to be avoided is having ground out hard connected to ground pin in.
3 truly isolated chargers will work happily on 3 cells in series if there is no closed current path apart from the cell interconnections.
- will it be safe if when they're done charging separately I put all three cells back into the same system together
Yes.
You MUST look at the diagrams in the data sheet and understand what they show.
You must read the answers that people give and actually take note of what they say.
apalopohapa have you a completely clear answer.
You ignored what he said and just asked the question again.
He said "The battery goes both to the charger and the circuit."
This means
- The battery goes to the charger
and
- the battery at the same time goes to the circuit.
If you read a sentence like that and it makes no sense then read it again (and again and again) - maybe draw a diagram - or simply look at the diagram in the datasheet.
It shows the output going to the battery
AND AT THE SAME TIME to the system load.
And apalopohapa also said "When you are charging the battery, you are still powering the circuit."
This means
If that does not make sense then reread, rereread, rererereread.
Look at this diagram.
The charger IC output is connected to the battery
AND
AT THE SAME TIME
The charger IC output is connected to the load.
It connects to the load and battery at the same time.
It connects to the battery and the load at the same time.
All 3 are connected together at once.
Does that make sense?
Best Answer
USB supplies 5V at maximum 500mA. Power-wise this is 5 x 0.5 = 2.5W.
Your charger supplies 4.2V at 700mA. Power wise this is 4.2 * 0.7 = 2.94W.
You cannot get more power out than goes in, so you cannot charge at 4.2V @ 700mA. To increase the current to 700mA you would need to lower the voltage below 4V, which is not really an option for Li-Ion charging. The maximum current you could achieve (assuming 100% conversion efficiency) would be 2.5W / 4.2V = 0.595A. In reality due to losses it would be more like 595mA * 0.9 = 535mA.
So you might as well skip the conversion and just use the 500mA directly.
At 500mA, it will just take a bit longer to charge.
Li-Ion charging can get quite complicated, but to get to started the MCP73831 is a very simple/cheap charge control IC from Microchip, who make quite a few decent charge control ICs of varying complexity. It is selectable between 100mA and 500mA, and has a pin for LED indicator.
The MCP73837 can switch between a DC apadtor and USB, has more control over charge current, and a thermistor input.
Read both datasheets thoroughly, they give plenty of good information and example schematics.