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Basically I am making a solar inverter without any battery or charge controller that will directly convert the dc output coming from solar panels (6 connected in parallel) into 220V AC.I am using solar panels 50W each, having an open circuit voltage of 20V and the voltage varies between 15-20V during the entire day provided a minimum amount of sunlight is there. Next, I am using a 12V buck converter circuit using an LM2576 and few more components to get a stable output voltage of 12V out of the panels. Now this 12V DC is fed to an inverter circuit which converts it into a square/modified sine wave 220V AC at approximately 50Hz. But, I am not getting desirable power output. From 6 panels, all I am able to power is a 45W LED bulb along with a small 3W LED bulb.

Probably, one problem is with LM2576 buck converter IC. This IC although providing a constant 12V output but it is rated at a 3A fixed output current. And I think probably this is the reason why we are unable to drive more loads. Is there a way to amplify current in this case? Or something else should be done which I am missing here ?

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closed as off-topic by Andy aka, Enric Blanco, Bruce Abbott, Bimpelrekkie, Dmitry Grigoryev Jul 10 '17 at 12:01

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  • \$\begingroup\$ You're dissipating a large amount of power in the buck converter. At the same time you want to amplify the output power. This is not engineering. \$\endgroup\$ – dirac16 Jul 9 '17 at 8:52
  • \$\begingroup\$ Yes, I understand but what else can be done here ? I mean the IC LM2576 is fixed at 3A output. So is there some other alternative to this IC for higher currents ? \$\endgroup\$ – Vishal Pant Jul 9 '17 at 8:55
  • \$\begingroup\$ Go look for switching converters with higher output - this isn't a shopping site because recommendations quickly becomes out of date and your question then becomes redundant over time. SE is looking for good questions with good answers that stand the test of time. \$\endgroup\$ – Andy aka Jul 9 '17 at 9:21
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    \$\begingroup\$ The average current a buck converter delivers is equal to the peak inductor current value minus half of the inductor ripple current. If the ripple is small (large inductor), then the average and the peak are close. To increase the output current, you have to select a controller driving an external MOSFET handling the necessary current. Why do you need so many intermediate stages and don't directly drive the LEDs via some constant-current dc-dc converter which does not care about input voltage variations? \$\endgroup\$ – Verbal Kint Jul 9 '17 at 9:24
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    \$\begingroup\$ @dirac16 "You're dissipating a large amount of power in the buck converter." - Why do you think he is? Are you confusing buck converters with linear regulators? \$\endgroup\$ – marcelm Jul 9 '17 at 10:08
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You are correct that your selection of the buck converter is the limiting factor. Since you would have up to 300 watts available from the panels, ideally you would need a converter that can handle at least 25 amps for a 12 volt output.

You may already be aware that without an MPPT (maximum power point tracking) mechanism you will typically realize less power from your panels than optimum. You should also consider that the buck converter has inefficiencies as well. An estimate of 90% efficiency is a starting point.

Then consider that your 220 volt inverter has a 90% or so efficiency. Under optimum conditions, you could power a 220 volt, 240 watt (~1 amp) load .

Do a web search for high power or high current buck converters. The LT1339 is one example you can study. Be prepared that at this power level the design will require external FETs and that the inductor design becomes more critical (and probably not off the shelf). The overall circuit is not a simple one like your current design. Much more engineering will be needed to obtain the desired results.

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1) Check if your inverter can accept 20V input. If it can work in a car, it will most likely have something like 11-15V but who knows?

2) If it accepts 20V, you're done.

Now, if your loads are all LED bulbs, then why the hell are you using an inverter? Use 12V LED bulbs. These work on 12VAC, which has a peak voltage of about 17V... if you pick bulbs with a proper switching driver, maybe they'll work on 20V too. Or just use a LED driver from banggood and some power LEDs. You don't need the efficiency penalty of an inverter for LED bulbs...

Anyway, depending on your loads, you might want to rethink the way you do things. For example, a car adapter for your laptop will most likely be more efficient from 12V than going through an inverter and then the laptop's power brick...

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