One formula you will use often is:
Power (Watts) = Voltage (Volts) x Current (Amps).
The more power, the more energy it produces. If only amps and volts are specified, you can calculate the power from that by multiplying them.
The rating of a solar panel in Watts will be the maximum for full sunshine. So when it is cloudy, the power produced will drop significantly, down to approximately 10-20%.
You can use any battery, as long as you have a circuit which controls the charging of the battery. However, this circuit needs to be compatible with your battery and the solar panel. The larger the battery, the more energy it can store, for example to store energy from sunny days so it's available on cloudy days or at night. Too small a battery and you might not even have enough power at night.
I will go through one example for you, so you can follow the process for whatever system you want to use.
First, decide how much electricity you will use. Say you want to run a TV from it, for 4 hours a day. And say the TV uses 100 watts, and runs off 230V.
The total energy your TV will use per day is 4 x 100 = 400 watt-hours.
And say you have 12 hours of sunshine every day. To get 400 watt-hours, you would need a panel that has a power of 400 / 12 = 33.3 Watts. This could, for example, be a solar panel that produces 1 amp at 33.3V, or 66.6V at 0.5A, or whatever, as long as amps x volts is greater than or equal to 33.3W.
If you want to use the TV at night, you will want the battery to store 400 watt-hours. So if you use a 6V battery, it must have a usable capacity of 400 / 6 = 66.6Ah (Amp-hours). Or if you use a 24V battery, it must have a usable capacity of 400 / 24 = 16.6Ah. If you use a car battery, be aware that most car batteries need to maintain a charge of between 50% and 75% if they are to have a long life. So only a quarter of its stated amp-hour capacity can be used. So say you buy a 100Ah battery, you will only be able to use 25 amp-hours! It would be more cost-effective if you buy a so-called "deep cycle" battery, so you can typically use more like 80% of its capacity without causing premature ageing.
And say your solar panel produces 48V. You will need a system which solar panel controller that can charge a 6v, 67Ah battery from a 48v, 34W solar panel. This will be hard to find: you may need to start with the controller, then buy a solar panel and battery accordingly. The "controller" is made up of two parts, and you may want/need to buy them separately: The first part is a voltage converter, which takes the voltage of the solar panel, and converts it into a voltage for charging the battery. Then the battery charger is essentially a circuit that uses the supply from the voltage converter to charge the battery without overcharging it.
Finally you will need an inverter which produces at least 100 Watts, from a 6v supply, to 230V AC.
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:
- Don't hook up the panels in series to a boost converter.
- 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).
Best Answer
Solar panels act a little differently than you might think. For example, all "12 volt" solar panels actually output 18 volts, as high as 20 volts Voc in bright sunlight. You have to remember this is Voc and there really is no situation in which Voc is output when the panel is actually attached to a load and providing current.
I have read a couple of theories on why panels are made this way. One is that under lower light conditions you will still get the minimum voltage required for your load. This is the most common answer that I have read, but it doesn't make sense to me because under low light conditions amperage drops considerably making the panel not very useful.
The other more logical answer has to do with the power curve of the solar panel. If you graphed the current versus voltage curve of the panel you would find that for most cells, you get the maximum current (given the same sunlight) at about 75% of the max voltage of the panel or approx 14/18 volts.(when charging 12 volt lead acid batteries will rise to 14 volts and even a little higher)
So how does this apply to you? Well solar panels are interesting creatures and as soon as you hook it up to your 24 volt fan in bright sunlight, it will drop to 24 volt output and problem solved. No need for extra electronics. I am not sure of the correct terminology, but I believe that the panel's resistance is incredibly low, so that anything you hook it up to has a higher resistance. As a result the voltage of the panel drops to the voltage of the load.
For example my 12 volt panel that reads 20.5 volts in the bright sunlight drops immediately to 10.5 volts, to when I connect it to my depleted lead acid battery (which is 10.5 volts when completly depleted). I would think that the same should happen to your fans.
However, sunlight comes and go, birds, clouds, etc cause the panel's output to fluctuate and thus your fans would likely stop and start repeatedly. Have you considered hooking the panel to a battery first and the fan to a battery? It could be a relatively low capacity battery that could feed the fans a constant voltage and amperage and the panel could feed the battery. You would want a cheap charge controller for the battery. Two 12 volt lead acid batteries in series would work well. I do that with my wife's water fountain. It would shut down for a few seconds when a cloud flew over or the dog ran in front of the solar panel. I got a free lead acid battery that was essentially dead (measured capacity less than an alkaline C battery!). The panel charges the battery while the battery feeds the fountain. Works well even though the battery is almost dead, the battery only really providing full power for a few seconds at a time.
Hope this helps.