# Did I wire these solar panels correctly?

recently I bought 3 solar panels rated at 5V 200 mA each. I want to use them to charge a 5V battery bank to charge a phone. Thinking about the proper way to put them, I thought i can connect all in parallel to get maximum current, but realized that if the sun light was a little weak it will no generate full 5v thus preventing charging. So I decided to put 2 in parallel to give the equivalent of one 5V solar panel, connected in series with the 3rd panel to give the equivalent of 10v. sacrificing a little current to get higher voltage, to allow the charging to happen on a wider range of sun light power. The following picture shows the wiring and the schematic I intend to replicate.

Now that I have an equivalent of 10v, 400mA solar panel. I used a 7805 voltage regulator to cut down the excess to 5v.

Final step, I added a standard diode to prevent the panels from leaking the battery in the shade.

and now measuring:

Questions:

1. I didn't think about this before putting the diode, but is it okay to put the blocking diode on the ground wire? because I know some applications do not use the ground except for safety (i.e. 3 phase system). it would be helpful also to avoid the 0.7v drop across the diode before the regulator.
2. according to previous calculations, I'm supposed to get a maximum of 10v output before regulation, and considering that the sun was pretty shinny today, why was the reading I got not more then 6v? I have measured across each panel seperatly and got around 5.5V, are the connections right?

Putting a single panel in series with two other panels that are in parallel does not accomplish what you think it does. The overall current of such a setup is limited by the single panel to 200 mA, so the three panels will not produce any more power than you'd get by just putting two panels in series.

A single solar cell can be thought of as a current source in parallel with a silicon diode. The current source is driven by the incoming light. The diode "shorts out" the current source, which is why the voltage across a single cell can never be more than about 0.65 V, the forward drop of a silicon diode. A 5V panel is approximately 10 such cells connected in series. The current through all of the cells will be limited by the cell that is receiving the least amount of light.

Also, the blocking diode in your diagram is pointing the wrong way.

If your panels are rated at 5V, and your "battery bank" requires 5V to charge, then you don't need to do anything more than put all three panels in parallel and hook them directly to the battery. Forget about blocking diodes.

• Very useful information. Now it makes sense why the circuit is so much power inefficient. I will take your advice to connect in parallel, however I want to keep the blocking diode to prevent the panels from draining the battery when the voltage drops, (say I forget the charger in the shade...). Jan 25 '16 at 23:29
• for the diode direction I thought the symbol (arrow) should be pointing along the flow of current, i. e. positive terminal of the source along negative terminal orientation. right? and finally, each panel is measures at approx 5.3v. what's the worst and best case scenarios if I keep the regulator? Jan 25 '16 at 23:36

That's a mess, and your labels about what is + and what is - out of each panel seem inconsistant.

For the most effective use of the panels, wire them all in series. That will put out around 15 V under full sun. Now use a buck converter to make a regulated 5 V from that.

There are many buck converter chips available off the shelf at these low voltages. Finding one with a built-in switch and synchronous rectification shouldn't be too hard. You only have to supply the inductor, a few caps, and maybe a charge pump diode depending on the chip you chose.

The other advantage of a switching power supply is that you probably won't need heat sinking. The resulting overall circuit will be smaller, cheaper, and give better performance than just throwing a 7805 regulator at it. At 10 V in and 5 V 1 A out, the 7805 will dissipate 5 W. That's way beyond what it can do without a heat sink.