# Battery charging via Solar pannel

Hey guys, I recently conducted an experiment which involved charging a battery via a PV module and also logging the I and V values to calculate conversion efficiencies. I also logged the I and V values from two other circuits with a fixed resistance. I understand that every irradiation has a Maximum Power Point (MPPT) so a fixed resistance will never draw the maximum energy yield from the PV module. But when I saw the results I saw that the power conversion for the battery was good under high and low irradiances. My question is why is this? Does a battery have some sort of control circuit to vary its resistance ? Thanks in advance.

mgb_phys
Homework Helper
No, but depending on the battery type the terminal voltage and so the current availabel from the PV will change as it charges.

For everythign you want to know about batteries see here http://www.batteryuniversity.com/

I've looked through that but I still don't understand what is actually happening.

As long as the PV output voltage is higher than the battery it will charge the battery. When the PV output drops, the battery will drain through the PV circuit. Many sailors have woken up to find drained batteries from either leaving their solar charger on at night or not installing diodes to keep the current flowing only to the battery.

Are you certain that durring the low irradiance test the battery was not what you measured for power conversion?

As long as the PV output voltage is higher than the battery it will charge the battery. When the PV output drops, the battery will drain through the PV circuit. Many sailors have woken up to find drained batteries from either leaving their solar charger on at night or not installing diodes to keep the current flowing only to the battery.

The forward voltage drop of a diode is roughly 0.6 volts, and is roughly 3 times the emitter-collector drop of a saturated PNP transistor, so using a simple regulator comprised of a simple control circuit and a PNP pass transistor on the + side of the charging voltage is better.