Lithium Ion battery charging using a "pull-string" charger:
The charger is potentially usable but it would take a considerable time to charge a 1000 mAh LiIon battery - probably 8+ hours - see below.
Lion battery energy is > 4.2V x1000 Ma or about 4 Watt hour. Very approximate.
Combining batteries:
In a series string, voltages add and mAh is that of the lowest capacity battery. ALL batteries should be identical and initially at the same state of charge, and 'balancing' may be required during use.
In parallel: ALL batteries MUST have the same voltage and all should be identical. Voltage stays the same and mAh sums. While it is technically possible to combine batteries of different mAh capacity this will lead to sorrow of one sort or other, or several sorts at once. Combining batteries in parallel by direct connection provides no means to control the amount of current flow in or out of individual batteries and "bad things can happen".
Hand powered charging:
Based on measurements that I have made:
A good plastic* Chinese hand crank LED light charger will make 1 Watt continually with the user being the limiting factor. At that level it takes much effort. Operation for say 10 minutes is "very annoying" and beyond 10 minutes only the most enthusiastic would persist. (* plastic gears and case as found in many such products)
A properly built hand crank can return 5 x as much with less effort. Say 5 Watts continuous with relative ease or as much as 10 Watts continuous if a user is enthusiastic. 20 Watts for short period is doable by hand (or arm) but not for long for most people. This is with a direct drive alternator with no gearing and optimised for low speed use.
For comparison, a good quality foot / leg powered alternator can provide 50 Watts from most users over say 1 Hour without too much annoyance. 100 Watts for one hour is doable but you would preferably be fit. MUCH more than 100 Watts is possible for short periods if you are a fit athlete but for most people the above levels represent realistic levels.
"Real world" hand driven charger: I experimented with a hand driven charger using a Fisher & Paykel "Smartdrive" washing machine motor. These would probably make about as good a hand powered alternator energy source as anything that could be built. Speed of operation lies in the optimum range for comfortable use, efficiency is reasonable, the annoying force and load fluctuations and noise produced by small geared units is wholly absent. There is a small amount of low speed "cogging" or salience at extremely low speeds but this is essentially unnoticeable in use. Saliency can be reduced by pole profile modifications (with a grinder!) if desired.
These are 3 phase brushless dc motors rated at hundreds of Watts input power as a motor or output power as an alternator at design speeds. When operated by hand power at speeds in the one to few revolutions per second speed range outputs of say 5 to 30 Watts is possible at voltages of maybe 5 to 20 volts. Vout varies with wiring configuration and speed and available power at a given speed can be increased by changing the ferrite magnets to higher flux rare earth magnets. This is unlikely to be worthwhile in hand powered applications.
A pull string unit will make less power than a hand crank due to the intermittent nature of the power input. The awkward arm angle during part of the stroke unless you fasten it somehow and position it optimally will not help.
I have not measured one but guestimate upper sensible continuous output at say 500 mw.
- 4 Wh/0.5 W = 8 hours to charge :-(.
You'd need to place about 8 x 2450 LiIon in PARALLEL (not series) to get 1000 mAh but this would be "extremely unwise" (at best).
Better is to use 1 x 1000 mAh LiIon as you suggest.
Controlling charging: If a "real" charger such as the F&P alternator based one was used a small battery could easily be destroyed. Typical consumer goods LiIon batteries intended for phones and similar are usually rated at from C/2 to 1C charge rates. ie a 1000 mAh battery at 1000 mA charge = 1C, and at 500 mAh charge = C/2. At 10 Watts the F&P charger would provide about 2.5A if the voltage was matched to the cell. In practice Vout would probably be in the 5 - 20 Volt range. Use of a properly designed charger using a buck converter and proper battery management would be essentially essential. An F&P alternator and properly designed charging unit would be able to charge a suitably specified 4.2V, 3000 mAh 18650 cell in about an hour, hand-energy-inputter willing.
Charge regulation by existing equipment:
The existing charger MAY regulate the charge properly but this is unlikely.
Look to see if it has a charge control IC.
Look to see if it has a voltage regulator to limit overcharge.
You are unlikely to get overcharge with pull string charging and a 1000 mAh battery :-) - but the PV panel will charge the LiIon 1000 mAh fully if left long enough - which may take 1 to 2 weeks of all day sun based on what they say.
A LiIon battery MUST have its peak charge voltage limited to 4.2V or less.
4.1V is safer.
MUST!!!
If you do not do this the battery may die either with a whimper or with an enthusiastic fire show. REAL magic smoke.
There are other things you can do, but limiting Vmax is probably a very very good start in this case.
NEVER run a LiIon battery below about 3V when discharging.
Doing so may destroy it.
Slightly lower than that WILL destroy it.
Testing:
Disconnect battery.
Connect voltmeter across alternator (or rectified DC.)
Operate string charger enthusiastically and measure voltage.
Repest with solar panel in full sunlight.
If either gives > 4.3V you MUST regulate.
If neither gives > 4.2V then you MAY consider not regulating. But ... .
If you can prevent it discharging below 3V it will help it live long, if not actually prospering.
A 2.2F capacitor charged to 5V provides an energy storage of up to 27.5 joules: -
Energy stored limit = \$\dfrac{C\cdot V^2}{2}\$ = \$\dfrac{2.2\cdot 5^2}{2}\$ = 27.5 joules.
Clearly you'd need a lot of effort to get it charged up but I don't see why this would not work as a basic energy storage system. Once liberated of it's 10 joules (as per the question), the terminal voltage will be about 3.99 volts and hopefully sufficient to have powered the LEDs down to their limiting voltage but as the question doesn't contain this data I can't be sure.
Also, for the item shown above note that it has an ESR of 35 ohm and this may become a problem in utilizing the energy at a high rate because of volt-drop. For instance, a 1 watt output at 5 volts requires 200 mA and this current through 35 ohms is untenable with this device. You'd need to look for a much better device - something that is a couple of ohms maximum.
If this spec of supercap cannot be achieved then other means has to be found.
EDIT - following valiant work by @helloworld22, it looks like a suitable cap has been found. Here is the data sheet and note that it is on page 10 near the bottom - it is an EMHSR-0002C5-005R0 and it is a 2.5 farad, 5V device with an ESR less than 53 milli ohms. At 200 mA, the volt drop will be 11 mV and negligible and over ten seconds it will waste 21 milli joules to power the LEDs. Nice find!
Best Answer
The adapter does not connect to the laptop's battery directly, there is always a charging control circuit in between. In laptops this usually is a switching type charging circuit to avoid excessive power (heat) dissipation.
If the power adapter delivered 14.8 V then it would be difficult to impossible to fully charge the battery as there would be no "voltage headroom" left for the charging circuit.
The choice for 19.5 V instead of 24 V might be historical and practical. 19.5 V did the job and maybe 24 V might require components rated for a higher voltage.
The input rating of the power adapter is just that a rating which means that the current will not exceed the rated 1.6 A. Usually this maximum current is only reached when you plug in the adapter into the mains. During normal operation the current will be much less. How much will depend on how much power is required from the 19.5 V output.
If the adapter would actually continuously consume 1.6 A then it would get extremely hot as it would need to dissipate a lot of power. So no, you will not be paying a bill for 1.6 A.