Current output of a solar panel when connected to battery and load

Don't even think about creating the setup you just described. It is bloody dangerous.

If you wire the "solar cell pack" and the two battery packs in parallel without connecting the Raspberry Pi, you'll get a loop. Kirchhoff's second law explains that the sum of voltages around a loop must be zero. In this case, if you start going around the loop in one direction, you'll encounter the two power sources with opposite directions, so now their difference must be zero - so they must be at an equal voltage. Will this be 6V (dictated by the solar panels) or 3.7V (dictated by the battery packs)? The following will happen:

• Initially, without sunlight, the common voltage will be 3.7V. No current flows, since the solar panel does not let current flow backwards (its resistance goes near infinity). All is well for now.
• Then you apply sunlight. The solar panels try to increase the voltage to 6V, but at this voltage the batteries would allow through much more current than they can supply. So the panels drop their voltage to 3.7V, but still begin to charge the batteries with the couple hundred milliamps they can supply, until the voltage in the battery packs reaches 6V. And there's a pretty good reason the batteries are rated at 3.7V.
• If I learned anything about Li-ion and Li-polymer batteries is that they are very easy to upset. And they especially don't like being overcharged. If they are indoors, they will blow up your desk and burn your house down. If they are oudoors, they'll happily ignite the grass around them. Then burn your house down. Li-ion and Li-polymer batteries are not toys. Don't even think about putting them in a circuit where there's even a slight chance they'll get overcharged.

Connecting the Raspberry Pi before the detonation wouldn't work out well either. The 3.7V combined power supply is not enough for the Pi, which will then do one of the following (I'm not familiar with the Pi's power supply circuit):

• Pull a lot of current overheating the battery, the solar panel and maybe even its own on-board voltage regulator. It will not boot, or even if it does, it will frequently crash and reboot because of the inadequate voltage. This goes on until one of the components fail: if it's the solar panel, you're safe. If it's the Pi, it's the time bomb scenario all over again. The battery packs also don't like being over-discharged, but as far as I know, they don't burst out in flames then. They just don't work anymore.
• Don't pull any current at all. Then it's like you didn't connect it at all. Time bomb again.

Creating a circuit which safely combines solar and battery power requires advanced electronics skills and dedicated circuitry. In your case, I would follow S.J. Becker's advice (+1) and buy a (solar powered) power bank from eBay. The circuitry is there, pre-made for you and it can power your Raspberry Pi longer than your setup would have even if it worked. I know they are not as cheap as using things from your parts bin, but they are definitely cheaper than replacing your burnt furniture.

Additionally, does the thing have to be solar powered?

Edit: Some battery packs have built-in protection circuitry that shuts the power off if the battery is overcharged, so there's a chance your setup won't actually ignite but just not work at all.

I'm confused -- if you have two 4V solar panels in series, your open-circuit solar cell voltage is going to be 8V, and you want to have a buck converter to get 5V for charging a phone. Hooking up 8V to that Amazon module will likely damage it.

If you have two 4V solar panels in parallel, now a boost converter would make sense to get 5V, and that module could potentially work for you.

Regardless, think about your problem in terms of power. A solar panel with 4V open-circuit voltage, and let's say your 50mA number is short-circuit current, we can back-of-the-envelope as a 200mW device. If you put these both in parallel, you have about 400mW (theoretical) you can deliver to your device, assuming perfect power transfer (which of course we cannot do).

Now, assuming that you are not doing any peak-power tracking and are applying the output of the cells straight into a boost converter to get 5V, assuming perfect everything (which again, is not true), your best case delivery to a phone is 5V @ 80mA...which is not very much. I doubt that's enough to even trickle charge a modern smartphone. If we assume that your energy conversion process is 80% efficient, that's even less transfer -- only 64mA delivered ((0.8 * 400mW) / 5V).

In the scenario where you have placed them in series, you would have the same amount of power available, but you would use a buck converter instead of a boost converter.

So:

1. Don't hook up the panels in series to a boost converter.
2. You don't have enough power in either scheme to do what you want and charge a phone sufficiently (best case, you charge a phone that is off very, very slowly).