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I have a 12 V 7 Ah battery. In order to charge a battery, we need to supply the voltage greater than the voltage of battery. Now lets say I have a power supply which is giving 14 V and battery current status is 12 V. When I will plug in the power supply pins to the battery, the battery voltage will reach upto 14 V and battery will start charging.

Now I have a solar charge controller PCB where I am using solar panel to charge the battery. Below is its schematic:

enter image description here

In the above image, X1 is the connector where will connect solar panel. A microcontroller is used to read the solar voltage through voltage divider circuit. CHARGEPIN is the pin which is used to enable the charging of battery.

So lets we have a full sunlight and solar output is 18 V. Now the problem I am facing is as soon as I connect the solar panel and start the charging, the voltage of solar panel reduces to 13 V but in the above case where I was using a power supply, the voltage didn't drop. It was same (power supply voltage was still 14 V). Because of this scenario, the battery is taking too long to charge.

Is the schematic wrong? Am I missing something?

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  • \$\begingroup\$ Is the battery charge done through a voltage divider? How does OP read voltage into his MCU? \$\endgroup\$
    – user400344
    Commented Jan 16, 2017 at 14:27
  • \$\begingroup\$ Funky... remember that this same circuit should prevent over-discharge (<10.4V), but safer to cut off at a higher value. \$\endgroup\$
    – user400344
    Commented Jan 16, 2017 at 14:30
  • \$\begingroup\$ No the schematic is incomplete. I am using two voltage divider circuit to read solar and batter voltages. So when battery voltage is below 10.8,I am cutting off the load and when the battery voltage is higher than 13.5, I stop charging. \$\endgroup\$
    – S Andrew
    Commented Jan 16, 2017 at 15:27
  • \$\begingroup\$ What are you reading these voltages with? \$\endgroup\$
    – user400344
    Commented Jan 16, 2017 at 15:28
  • \$\begingroup\$ I am reading Arduino to read the voltages. Its actually analog data. So then converting that analog data back to voltage using voltage divider formula \$\endgroup\$
    – S Andrew
    Commented Jan 16, 2017 at 15:29

2 Answers 2

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The thing you're missing is that your 14v power supply can output enough current to charge the battery and keep the terminal voltage at 14v. Your solar panel clearly does not output enough current, and the battery is taking all the charging current it can provide at 13v.

If you want to charge the battery with the panel, it will take as long as it takes. Get a bigger panel, or more light on the one you have, if you want it to charge faster, with that simple circuit. It may be possible to squeeze a higher charge rate if you used a MPPT converter, to operate the panel at its best power output point, which may not be 13v.

That said, the 1N4007 will drop a volt, a pretty wasteful form of protection, a MOSFET that can be switched on would be far better.

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    \$\begingroup\$ I'm sensing a lack of 't in your first sentence. \$\endgroup\$
    – Marcus Müller
    Commented Jan 16, 2017 at 9:27
  • \$\begingroup\$ @MarcusMüller I'm struggling to spot that, though if I reversed the order and added a 't it might be clearer? \$\endgroup\$
    – Neil_UK
    Commented Jan 16, 2017 at 9:29
  • \$\begingroup\$ "can output more current than the solar panel" \$\endgroup\$
    – Marcus Müller
    Commented Jan 16, 2017 at 9:30
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    \$\begingroup\$ @MarcusMüller yes, I'm guessing the PSU has enough current to charge it, whereas the panel doesn't, so no 't, as it's a positive assertion. \$\endgroup\$
    – Neil_UK
    Commented Jan 16, 2017 at 9:32
  • \$\begingroup\$ Ah, "the thing you're missing" as in the thing you're actually missing :) \$\endgroup\$
    – Marcus Müller
    Commented Jan 16, 2017 at 9:33
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You're trying to ignore the basic principle of conservation of energy. Your solar panel can only produce as much power as it can, and if that power isn't enough to quickly charge your battery, bad luck.

Notice that you could be charging for a longer period of time if you had a switch-mode power supply that converted from whichever your cell's actual output voltage to a suitable battery charging voltage.

Also, you'd typically use lower forward voltage diodes in your power path (wasted energy).

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