DIY 5v Solar Panels

It is a long question, but better than a short one, as you've shown your own research.

1) Solar cells. If you're stacking your own ones, stack 9 of them and get the 4.5V of the original circuit.

2) Battery charging. Batteries are the only thing you've left out of your spec. This is an area where the circuit design relies on cutting a lot of corners. In theory it might be out of spec, if you were to put 4.5V at 280ma through AA NiMH cells indefinitely. In practice, you don't get full sun all day, you'll be using it indoors, and you're not going to get optimal power transfer from the cells, so this isn't going to cause problems.

3) Diode. It's just a regular diode, not a zener. Current through it is actually determined by the battery and right hand side circuit, not the solar panel - the transistor is off when the panel is generating electricity. The original 1N914 will be fine. 1N4004 will also be fine.

4) Resistors: not a precision component here, use whatever meets your cost constraint. 5.1k for 5k is fine.

5) Wire: not critical. Your ebay link looks suitable. Thinner is better for the toroid.

6) Transistors: stick with the exact part numbers. Design may rely on specific parameters.

7) LED: again, this circuit relies on cheating. Normally a white LED won't run from two NiMH cells. The joule thief part provides a boost converter that gives small pulses of higher voltage. It doesn't have the capacity to provide a lot of current at that voltage. In combination with the pulsing this means there should be no risk of damaging it.

(A proper analysis of this circuit would be good, if nobody else supplies one I'll do it in a few days).

I dont understand in which irradiance does the panel give those values. This is my first question.

Not too hard to calculate from the data provided, given a few assumptions. The "peak voltage" they give is less than the open circuit voltage, so ostensibly they mean to specify the peak voltage and current at the maximum power point. The "mp" after "Imp" and "Vmp" would also conventionally indicate "maximum power". So they are saying this panel can produce (under optimal conditions):

$$ 3 \mathrm V \cdot 0.2 \mathrm A = 0.6 \mathrm W $$

The area of the panel is:

$$ 92 \mathrm{mm} \cdot 61 \mathrm{mm} = 0.005612 \mathrm m^2 $$

They claim an efficiency of 15%, so the irradiance \$I\$ must have been:

$$ 0.15 \cdot I = \frac{0.6 \mathrm W}{0.005612 \mathrm m^2} \\ I = 713 \mathrm W / \mathrm m^2 $$

Irradiance on a very clear day at Earth's surface with the sun directly overhead is on the order of 1kW per square meter, so this sounds reasonable enough, considering that the efficiency number may not account for the fact that this device has some margins around the panels.

My second question is why there is no datasheet? I guess that depending on the type of panel (for example polycrystaline) the I-V curves are same for every panel. Am I right?

Some products just suck. That said, silicon junction photovoltaics pretty much have a common shape to their I-V curves, which is more or less a top-for-bottom flip of the I-V curve of a silicon diode. A more expensive product would probably have more detailed specifications; for something cheap like this you will just have to infer or measure for yourself.