to let the 12V battery absorb the energy from the solar pannel when the voltage of the solar pannel is under the actual battery's voltage
no, If the solar panel voltage is lower than the battery no current flows.
No amplification. It is just a switch.
Diode D2, (don't know what kind) but it will do nothing if the solar panel is connected right. The diode will (maybe) blow the fuse if the panel is connected backwards. I think this is why the diode is there but I think it will not function the way the designer thinks. If you have a 1 amp panel it can not produce more than 1A. If you connect backwards the panel will only make 1A. So you can't pop a 2A fuse with a 1A panel. I know there is on amp rating on the fuse.
F1 is for protection but I think it does nothing.
D1 is to keep the battery voltage off the panel, when the panel is not making power. Note it is a 1A high voltage diode. I think a 1N4001 will also work. (low voltage) Three are better diodes for this place.
R1&R2 allows the computer to measure the panels voltage. (there is 0.7V loss in D1) The computer can only measure 0 to 5 volts. A 3:1 divider will allow a measurement of 0 to 15 volts.
Q1. No amplification! If there is 0 volts across R3 then Q1 is "open" and no current will charge the battery. If R3 has -3 volts across it then charging current will flow. (assuming the panel is making more voltage then 12V)
There are other reasons why I do not like the circuit.
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What do you want the circuit to do?
What panel do you have? Volts and current and power?