Hi, pls suggest me for the following A 20Watt ( 20V , 1000 ma) solar panel is connected to charge a 12V, 80AH battery . What is the value and wattage of a Current limiting resistor to be added?

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I think the charging current from this panel is small enough, so that no charge controller is needed.

A single diode in series would be a good idea, to keep the battery from pushing current backwards through the panel, whenever there is low light; e.g. night time, or clouds.

By the way, for charging batteries, the charging current is often scaled to the ampere*hour capacity of the battery, and you will see numbers like "C/3", "C/10", "C/100", and so forth. I think this convention might be called, "C-rate"

https://batteryuniversity.com/learn/article/what_i...

The way it works is, the number in the denominator is a number of hours, and C is the battery capacity in ampere*hours, so this division problem gives an answer in units of current.

As an example, the amount of current needed to charge 80 A*hour battery, in 10 hours, is (80 A*hour)/(10 hour) = 8 A

Equivalently, the C/10 rate for this battery is 8 A.

Also for this battery, a charging current of 1 A, is C/80.

And that charging rate might be low enough to be considered, "trickle charging", which is essentially charging so slowly there is no danger of overcharging the battery, even if that current were connected forever.

Of course that is not possible with solar power, since that is naturally turned off at night. So that might actually be helpful, to prevent overcharging the battery.

By the way, you did not mention the chemistry of your battery, but I am guessing it is lead acid.

Also by the way, I think the way the professionals do this, is to buy a charge controller matched to the battery being charged, and maybe also matched to the power source, like solar or mains power. Then the charge controller uses some kind of logic, or algorithm, to make decisions about when to turn on, turn off, charging current.

I linked to batteryuniversity, and I should probably also link to powerstream, because they have some good tutorials also.

https://www.powerstream.com/SLA.htm

If I follow what you are saying, the (dis)charge, per day, used by your load is approximately:

(24 hour)*(0.05A) = 1.2 A*hour

while the charge, per day, from the solar panel is approximately:

(8 hour)*(1.5 A) = 12 A*hour

And this is a problem because one is bigger than the other by a factor of about 10 to 1.

So you want to reduce the charging current by a factor of around 10? Or maybe that is too much? Maybe only by a factor of 5?

Can we imagine the present charging current is due to some kind of internal resistance? But how big?

Guess around 4 ohms? Or whatever would solve I = (Vpanel-Vbattery)/R.

E.g. Vpanel = 18V, Vbattery = 14V, I = 1.0 A, R = 4 ohm.

So increasing that resistance to 20 ohm (by adding 16 ohm), or increasing it to 40 ohm (by adding 36 ohm), would make resistance between the panel and battery larger by a factor of about 5 to 10, and decrease charging current by the same amount.

So that is my guess: a few 10s of ohms; e.g. in the range from 10 ohm to 40 ohm, required to limit current in this way, using an external resistor.

I am guessing the power dissipated (V^2/R) by this resistor would be around 4 to 5 volts squared, i.e. 16 V^2 to 25 V^2, divided by R; e.g. 2.5 watt for a 10 ohm resistor with 5 volts across it.

Another trick is to buy, or build, a charge controller that watches the battery voltage, and switches on the when battery voltage is too low (e.g. around 13 volts), and switches off the when battery voltage is too high (e.g. around 15 volts)