Electronic – Charging 12V SLA from Single Solar Cell

chargerlead-acidsolar cell

I'm interested in charging a 12V Sealed Lead-Acid battery from a single monocrystalline solar cell. The solar cell's voltage is only 0.5V at maximum power point, but it supplies 8A. I will need a 32X voltage multiplier to emulate a 12V panel to drive my SLA charger.

Could someone suggest an appropriate voltage multiplier design? I have read about Villard- and Dickson-type cascades, but they all will incur too much loss in the early-stage diodes.

Best Answer

Extracting power from such a low voltage is tricky because even otherwise small voltage drops are significant fractions of the input.

Fortunately, you have the 12 V battery available to power the control electronics of a boost converter. At its core, a basic boost converter will be a inductor, switch, and diode:

The switch closes, which makes the inductor current rise linearly with time. Once the inductor current gets to a decent level and a useful amount of energy is stored in it, the switch opens. At that point the inductor current can only go thru the diode, dumping current onto the output. The reverse voltage now on the inductor causes the current to decrease linearly over time, transferring the energy stored in the inductor to the output.

Since your input is 500-750 mV, the most critical part of this circuit is a switch that will only drop a small fraction of that. A bipolar transistor with maybe 200 mV saturation is not appropriate here. You want a FET with low on resistance (Rdson). Let's say you want to keep the drop accross the switch to 50 mV or less at 8 A. By Ohm's law, that means you need 6.3 mΩ Rdson or less. There are FETs that can do that, but for something like this paralleling a couple of them may be quite appropriate.

Fortunately, the drop accross the diode is relative to the output voltage, not the input voltage. A regular Schottky diode rated for the current should be good enough here. Let's say it drops 500 mV when dumping current onto the 12 V output. That's a 4% power loss, which in this case, especially considering you're asking here, I'd just swallow. To get around that you'd need to implement synchronous rectification, which is not a beginner concept unless you can find it included in a chip.

Since the battery is there, you can run the control circuit that turns the FET on and off from it. Many microcontrollers can do this. The micro would measure the input and output voltages, and produce the switch signal via built-in PWM hardware. The micro can power itself down when there is insufficient voltage from the solar cell to make it worth it to run the boost converter. With the right circuit, the micro can take less current than the internal leakage of the battery when powered down. It can then wake up every few seconds to check whether sufficient input voltage is available to run the converter, and go back to sleep when not.

You have to pay attention to low current design with the micro, and make sure things like the battery voltage measurement circuit don't take any current when off. Fortunately this is something a beginner can do, although it requires paying attention.

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