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I am developing a DC to DC converter(40V -> 12V) circuit using Mosfet. As I am using solar two panels rated 375 Watts (Max 10Amp,37.5V) each, so the convertor needs to handle 750 Watts at peak. However, most of the Mosfet which saw being available cant handle power dissipation up to 750 Watts.

Is there a Mosfet which can handle up to this load, as I don't want to put MOSFET's in parallel.

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    \$\begingroup\$ Should all the power of the solar panels go into the mosfet as heat? \$\endgroup\$
    – Jeroen3
    Commented Aug 28, 2020 at 11:42
  • \$\begingroup\$ Actually, looking at the datasheet of irf540link, it mentions max power dissipation as 150W, so I thought that 150W is the maximum power output. However, as per the answers provided, now I understand, that power dissipation mentioned in the datasheet is actually the maximum amount of energy that can be wasted as heat, and it not related to the maximum power which can be provided by the MOSFET. \$\endgroup\$
    – user228226
    Commented Aug 28, 2020 at 13:15
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    \$\begingroup\$ This shows a basic misunderstanding of DC-DC converter design, such that ... while an answer may help, I think a book is required. Look at "Switch-Mode Power Supplies" by C. Basso (or search here for "Verbal Kint".) \$\endgroup\$
    – user16324
    Commented Aug 28, 2020 at 13:15
  • \$\begingroup\$ You don't need MOSFETs that can handle 750 W .... That's the power the panels produce, not the power the MOSFETs need to sink. The power the MOSFETs will see is the current through them (squared) times their channel resistance (RDSon). P=I^2*R Hopefully the vast majority of the power they produce goes to your load ;) \$\endgroup\$
    – Kyle B
    Commented Aug 28, 2020 at 15:10
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    \$\begingroup\$ ^^^^ Plus some switching losses. But nowhere near 750W @BrianDrummond is right, you're best to study the topic and understand switchers before trying to design one. \$\endgroup\$
    – Kyle B
    Commented Aug 28, 2020 at 15:22

2 Answers 2

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A converter is used to transfer the energy from solar panel to battery or grid. If the converter has 90% efficiency, it means that 75W is lost as heat in the converter, some part of this lost is due to inductor heat, trace heat and also MOSFET. You don't want to have a such converter that converts 100% of the energy in pure heat.

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MOSFETs have Ron resistance. Effectively, open mosfet (conducting, I mean, there is confusion in terminology) is essentially a very small resistor. Thus, you can calculate heat dissipation as well as voltage drop across it just like you would do it for a resistor. I*R = V voltage drop actoss it, I^2 * R = Heat dissipation. This will allow you to predict the behavior of the circuit. Also, most heat dissipation ratings for mosfets assume you have top cooling thing, so in your case the mosfet will dissipate less unless you have half a ton heatsink. But nothing stops you from connecting more of them in parallel if you find no suitable part, it's a perfectly valid option, and mosfets self-regulate current distribution among them, you just throw them in and there will be no thermal runaway.

Take into account most mosfets have limited maximum voltage between gate and source. Pay attention at that, otherwise it can burn. If your drain/source is at 40V, your gate probably should only go as low as 30V (many mosfets have 12V maximum Vgs, and with 10V between gate and source it will already be fully open with minimum Ron resistance). Also, look at maximum Vds.

Remember, that you don't need to dissipate all the energy the panels provide. In fact, you want to lose as little as possible on the mosfets. If mosfets burn all the energy, you'll have no energy output :)

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