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I have six solar panels but each are different from one another. The first is 4.5V. The second is 6V. And the last four are all equal up to 6V. It all adds up to 16.5V in series or on a good sunny day, 18V max. That's what my multimeter says.

I have a 1N1914 diode connecting to the positive lead connecting to my double pull switch. Now for my batteries, I have one 9V rechargeable battery and 4 AA (1.2V each) rechargeable batteries. The batteries total to 13.8V.

When I try to charge my batteries, it doesn't charge any of it. The switch actually works fine. But when I try to charge the batteries, the meter shows my battery voltage rather than my solar panel. What am I doing wrong?

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marked as duplicate by Kaz, placeholder, PeterJ, Keelan, Vladimir Cravero Sep 12 '14 at 7:29

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  • \$\begingroup\$ When the solar panel charges batteries, it is under load, and exhibiting voltage drop. From another point of view, the diode through which the solar panel is charging the batteries has an approximately fixed voltage drop (of say around 0.7V for a silicon diode like 1N1914). The voltage on one terminal of the diode won't be more than 0.7 off from the voltage taken on the other terminal. \$\endgroup\$ – Kaz Sep 12 '14 at 1:44
  • \$\begingroup\$ Your title is inconsistent; it says that your circuit shows battery voltage and not solar panel voltage, but the text says the opposite. \$\endgroup\$ – Kaz Sep 12 '14 at 1:45
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    \$\begingroup\$ Mixing different batteries is not such a great idea. \$\endgroup\$ – Kaz Sep 12 '14 at 1:47
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    \$\begingroup\$ The 1N914 is a signal diode, intended to carry low currents. In this application, you should use something like a 1N4001 - 1N4005 which are 1 amp rectifiers. \$\endgroup\$ – Peter Bennett Sep 12 '14 at 2:51
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Without more information, it is hard to know your problem. Using a different diode as suggested would help.

Also, your panel voltage might be = your battery voltage. What I mean is that the 1.2 volt Ni rechargeable batteries are rated 1.2 v nominal and quickly rise to 1.4 volts or higher depending on the brand when charging - this brings you up to 15.6 volts just for voltage rise for the AA with no voltage rise for the 9V, also you have the voltage drop of the panel from the diode - this brings them basically even!. I don't know what type of chemistry your 9 volt is, but if it rises when charging as well, then battery voltages will have risen to the panel voltage very quickly and then stopped charging. Try taking out a few batteries and see if it starts charging!

Good luck.

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  • \$\begingroup\$ I understand now thanks. But could it also be my blocking diode its a 1n914. \$\endgroup\$ – Janet Moreno Sep 14 '14 at 20:45
  • \$\begingroup\$ yes, honestly I would take out some batteries and the diode. I assume the diode is to primarily prevent the batteries from discharging into the panel in the night time. Just watch it without the diode and with fewer batteries. If you get charging, which you should, then re-insert the diode and see what happens. \$\endgroup\$ – Filek Sep 15 '14 at 5:53

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