The trouble with solar chargers is that they only give you specs in the brightest sunlight so it always takes longer than you think it should. Lets try to do a "rough" (real rough) example:
Lets say the sun is at is directly overhead and its brightest for 10 hours a day. In this case your panel can put out 0.36 amps at 8.3 volts (3 watts). So lets say for the sake of easy math you got a real good one and it puts out 0.4 amps. So if your 4 AH battery is dead you could recharge it in one 10 hour day. (not quite true but close enough for an example
). This is the best you could do, no added circuits to waste power.
Now lets add the zener diode at 6.8 volts and look at the graph below. You have to divide the voltage by two to get from a 12 volt to a six volt battery. We will just look at 3 curves. The C/10, C/20 and C/40. These are the battery capacity (4AH) divide by the charge current, so in our case .4, .2, and .1 amps. So as long as the battery voltage is below the 6.8 volt zener all of the power goes into the battery. This goes on until the battery voltage reaches 6.8 volts at about 75% charged. At his point the zener starts to "steal current from the panel so we need to move over to the C/20 curve. Here the battery will only get .2 amps and the zener.2 amps until the battery reaches about 90% charged. Then we move over to the C/40 curve where it will reach 100% charged.
So now if we look at time. To get to 75% at .4 amps takes 7.5 hours. To get from 75 to 90% at .2 amps takes another 3 hours, and to get to 100% at .1 amps takes another 4 hours. So a total of 14.5 hours. Add a "fudge factor" for the time between the curves and maybe it is 17 hours. This is just an example to compare charge times. I real life it will take much longer due to the effects of the sun on the solar panel. In either case I think you might be better off using the simple method as the only losses are in the blocking diode there are no other circuits to waste power, so the charge time to 75% capacity is the best.
You might want to make sure your battery is not totally drained in between charges as they don't like that. So try to limit your lights or whatever keeping that in mind.
It is simple enough and inexpensive enough that you could build it and take some real world measurements to see how it would work. I would love to see your results.
I think the battery can stay at 6.8 volts based on this from battery university.
Once fully charged through saturation, the battery should not dwell at the topping voltage for more than 48 hours and must be reduced to the float voltage level. This is especially critical for sealed systems because these systems are less able to tolerate overcharge than the flooded type. Charging beyond what the battery can take turns the redundant energy into heat and the battery begins to gas. The recommended float voltage of most low-pressure lead acid batteries is 2.25 to 2.27V/cell. (Large stationary batteries float at 2.25V at 25°C (77°F.) Manufacturers recommend lowering the float charge at ambient temperatures above 29°C (85°F).
Lets say the sun is at is directly overhead and its brightest for 10 hours a day. In this case your panel can put out 0.36 amps at 8.3 volts (3 watts). So lets say for the sake of easy math you got a real good one and it puts out 0.4 amps. So if your 4 AH battery is dead you could recharge it in one 10 hour day. (not quite true but close enough for an example
). This is the best you could do, no added circuits to waste power.Now lets add the zener diode at 6.8 volts and look at the graph below. You have to divide the voltage by two to get from a 12 volt to a six volt battery. We will just look at 3 curves. The C/10, C/20 and C/40. These are the battery capacity (4AH) divide by the charge current, so in our case .4, .2, and .1 amps. So as long as the battery voltage is below the 6.8 volt zener all of the power goes into the battery. This goes on until the battery voltage reaches 6.8 volts at about 75% charged. At his point the zener starts to "steal current from the panel so we need to move over to the C/20 curve. Here the battery will only get .2 amps and the zener.2 amps until the battery reaches about 90% charged. Then we move over to the C/40 curve where it will reach 100% charged.
So now if we look at time. To get to 75% at .4 amps takes 7.5 hours. To get from 75 to 90% at .2 amps takes another 3 hours, and to get to 100% at .1 amps takes another 4 hours. So a total of 14.5 hours. Add a "fudge factor" for the time between the curves and maybe it is 17 hours. This is just an example to compare charge times. I real life it will take much longer due to the effects of the sun on the solar panel. In either case I think you might be better off using the simple method as the only losses are in the blocking diode there are no other circuits to waste power, so the charge time to 75% capacity is the best.
You might want to make sure your battery is not totally drained in between charges as they don't like that. So try to limit your lights or whatever keeping that in mind.
It is simple enough and inexpensive enough that you could build it and take some real world measurements to see how it would work. I would love to see your results.
I think the battery can stay at 6.8 volts based on this from battery university.
Once fully charged through saturation, the battery should not dwell at the topping voltage for more than 48 hours and must be reduced to the float voltage level. This is especially critical for sealed systems because these systems are less able to tolerate overcharge than the flooded type. Charging beyond what the battery can take turns the redundant energy into heat and the battery begins to gas. The recommended float voltage of most low-pressure lead acid batteries is 2.25 to 2.27V/cell. (Large stationary batteries float at 2.25V at 25°C (77°F.) Manufacturers recommend lowering the float charge at ambient temperatures above 29°C (85°F).
![BatteryVoltage-vs-StateOfCharge-at-charge-discharge[1].jpg](/data/attachments/55/55236-dd582f570d3217b71007b3d1fa33ff14.jpg)
