0
\$\begingroup\$

I have an unstable power source 0-10V (I don't actually know if it can drop to 0) from solar panels and I want to boost and stabilize it to charge 19V 2.1A laptop and a 12V 12Ah battery. I've already asked a question about this and got an answer that I need a boost converter. So I was thinking about getting a boost converter IC, then searched, found and bought TL499ACP. But I've made a foolish mistake... I forgot to check if it handles the current that I need...

When I bought solar cells for building the panels, per cell specification was 0.5V 8A 4W. So to get maximum power out of it, I need a current of 8 amps. As far as I know, to charge a 12V 12Ah battery I need current of 1 amp. The input of TL499ACP can be up to 500mA and output up to 100mA.

So is there any way to decrease the input current (8A from solar panels) for the booster and then increase its output externally (because more than 100mA is deadly for the booster) without loosing too much efficiency (balance between efficiency and letting the booster go)?

\$\endgroup\$
2
\$\begingroup\$

I'm sorry to say this, but sadness is in your future.

Please look again at the data sheet for your converter. The maximum recommended continuous current is 100 mA. This obviously will not do to run your laptop. You will need a much heftier converter.

That said, if you are going to roll your own system, you need to start thinking in terms of power, then current. Start with your laptop. If you need 19 V @ 2.1 A, the total power you need is 19 x 2.1, or 40 watts. Each of your panels is rated for 4 watts, so you need at least 10 panels. Note that this is on a cloudless day, with the panels pointed directly at the sun, and a perfectly efficient converter.

As a rule of thumb, for fixed panels, count on getting the equivalent of 4-6 hours of sunlight per day. In clear weather. If you want to run your laptop 6 hours per day total, and do it at night, you need to store all your power in a battery. Not only that, but battery charging is not terribly efficient. Let's assume overall efficiency of 50%.

This means that your battery needs to provide 40 watts for 6 hours per day, for a total of 240 watt-hours. A charging efficiency of 50% means that the panels have to produce a total of 480 watt-hours per day. Since they have to produce this power in 4 hours (equivalent), your array needs to produce a peak power of 480 / 4, or 120 watts. Since your panels are rated at 4 watts, you need 30 panels. If you are willing to assume a better environment, and figure on an equivalent of 6 hours per day, you need 480 / 6, or 80 watts peak, or 20 panels. Let's stick with the 4 hour number.

Knowing power, only now is it worthwhile to start talking about amps. Fortunately, you can connect your solar cells in series, so the output voltages add while the current remains the same. Then your array will have a peak output of 30 x .5 or 15 volts at 8 amps. Note that 15 x 8 = 120, which is the same as the 120 watts previously calculated, so no obvious boo-boo has occurred between then and now. However, this also assumes a 100% charger efficiency, and this does not happen. It would be a good idea to add 50% or so to the array capacity, so you actually need about 45 panels.

If you could count on peak output for 4 hours / day, you could make a very simple charger which drives your 15 volts into a 12-volt battery with very little loss. Unfortunately, this is not the case. I suggest you look into the workings of solar powered battery chargers. As for charger current, note that your peak charging current for a perfect battery/charger combination is 120 watts / 12 volts, or 10 amps. Again, assuming an imperfect charge process, add 50%, so you need a charger which will provide 15 amps peak.

As for the battery, as a first approximation it must provide 2.1 amps for 6 hours per day. Furthermore, it should not use more than 50% of its capacity to do so or the deep discharge will shorten its life. 2.1 amps x 6 hours is 12.6 amp-hours. Twice that is 25.2 amp-hours, which is the minimum battery capacity you need for each day of operation. The next question is: how many cloudy days in a row are you willing to plan for? Let's say the answer is 3 days. Then you need 75.6 amp-hours worth of battery.

So the bottom line is: if you need reasonable amounts of energy, you will need enough panels to be able to stack them and get higher voltage at the same current. This will ease the design of your battery charger. But you will still need a lot of panels.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.