25A wire

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You will absolutely need a gate driver... Chip or otherwise. If you attempt to switch the MOSFET gate directly from the optocoupler, it might work, but efficiency will be horrible. You want to drive the gate as hard as you can to minimize the time the MOSFET stands in between ON and OFF, because the majority of losses occurs during switching. Add to the that the fact that you are passing large amounts of current, and my prediction is that even if it "works" the MOSFETs will get very hot very fast without a proper gate drive.

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hugoboss wrote:
You will absolutely need a gate driver... Chip or otherwise. If you attempt to switch the MOSFET gate directly from the optocoupler, it might work, but efficiency will be horrible. You want to drive the gate as hard as you can to minimize the time the MOSFET stands in between ON and OFF, because the majority of losses occurs during switching. Add to the that the fact that you are passing large amounts of current, and my prediction is that even if it "works" the MOSFETs will get very hot very fast without a proper gate drive.

@kartman: That is great idea. Unfortunately i dont have access to a PV inverter.
I will get a gate driver to avoid the switching problems.
"High-speed gate drivers incorporate fast-reacting input circuits, short propagation delays, and powerful output stages delivering large current peaks for fast voltage transition times"*.
The mosfet RD(on) is 4.5mohm so the loss is 4.5e-3*20*20=1.8W. I think TO220 can handle this loss without a heatsink. Does it make sense?

*https://www.fairchildsemi.com/Assets/zSystem/documents/collateral/productOverview/High-Speed-Low-Side-Gate-Drivers-Product-Overview.pdf

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You calculated the static loss of the mosfet but guess as to the temperature rise. Not having looked at the datasheet for the mosfet, let's assume the junction to ambient thermal resistance is 70C/W. That means the junction temp is around 120C over ambient. Tj max is probably 150C so you're running close to the device melting down. You haven't considered switching losses so the picture is looking grim - i'd be thinking you'll need a heatsink.

If the U.S is anything like australia, you probably have had batches of crappy pv inverters that fail prematurely and get junked. Ask the right people and they would probably give you a few for nothing! You can then see how commercial designs work. For a 2kW inverter, they have big heatsinks and use igbts rather than mosfets - at 240vac, thats only 10Amps, 20Amps is going to be a lot worse. You also need big magnetics to filter the pwm - otherwise you're just making a 2kW rf noise transmitter.

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Quote:
The mosfet RD(on) is 4.5mohm so the loss is 4.5e-3*20*20=1.8W. I think TO220 can handle this loss without a heatsink. Does it make sense?

That is only the loss of passing a constant signal, like I said the majority of losses occur when the MOSFET is neither fully ON or fully OFF. The trick is to minimize the time it takes to go from one to the other, and you do that by ensuring you drive enough current into the gate to counteract the bigger MOSFET's gate charge and capacitance. If don't drive the gate hard enough, the actual switching time will be much longer than the advertised Ton Toff, and the losses from switching will add up fast.

Efficiency (thus losses) is a function of how fast you can switch the MOSFET and how often you switch it. A system with poorly driven gates that only switches once a second might be much more efficient than one where the gates are driven appropriately but switching at 1MHz for example. But even then, other considerations come into play, design wise. A system running at 60Hz might be more efficient than one at 1MHz, but it will need a lot more of much more expensive magnetics and capacitors to filter than the one at 1MHz.

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what does it high side and low side in gate driver mean?
This is a typical gate driver circuit.

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The low side are the ones at the bottom, the high side are the ones at the top. High as in higher voltage - they switch the positive side. Being N channel fets, the gate voltage needs to be greater than the source, so you need a means of boosting the gate voltage.

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