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I am designing a system that uses a large amount of power. The MOSFET will be to turn off the batteries when mains is in use. The FET will therefore be on or off. No pulsing. The voltage will be between 34.5V and 20V with about 30A max of current draw.

Can someone please explain what I need to look at when choosing the correct FET or if there is a better solution. The driver IC I am planning to use is the LTC4414. Any help or recommendations are more than welcome. I know that I need a P-channel MOSFET But I am unsure of what specs I need to actually choose one.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ A schematic of the proposed design would help. Presumably this will be a P-channel MOSFET (based on the driver chosen) used as a high side switch, or is this a power source selector? \$\endgroup\$ – Anindo Ghosh Jan 16 '14 at 22:25
  • \$\begingroup\$ Well the LTC4414 is a high side driver... \$\endgroup\$ – user28910 Jan 16 '14 at 22:27
  • \$\begingroup\$ @user28910 "Automatic Switching Between DC Sources" is a key feature. \$\endgroup\$ – Anindo Ghosh Jan 16 '14 at 22:39
  • \$\begingroup\$ If the battery charger has sufficient capacity, you could just connect the load directly to the battery - no switching required. \$\endgroup\$ – Peter Bennett Jan 16 '14 at 23:07
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Three important requirements come to mind.

MOSFETs can withstand only so much drain-source voltage without breaking down. You need this to be at least your supply voltage, plus a healthy margin for robustness. This will be specified in the absolute maximums section of the datasheet, usually as \$V_{DS(max)}\$

Each MOSFET has a different gate voltage at which it is designed to operate. If the gate voltage is too low, the MOSFET isn't fully on, and it doesn't conduct well as you want it to. If it's too high, you break the gate insulation and the smoke gets out. You will need to pick a MOSFET with a gate voltage compatible with the driver you are using. From a brief reading of the LTC4414 datasheet, it looks like this is -8V to -9V.

Lastly, no MOSFET is a perfect conductor when on. The conducting channel has some resistance, and current through this resistance makes heat according to Joule's law: \$P=I^2 R\$. You need a MOSFET with a low enough resistance that the heat is manageable at the maximum current that will be drawn from your batteries. The MOSFET datasheet will usually specify a maximum current, but this usually assumes an unlimited heatsink. It's better to look at the resistance (\$R_{DS(on)}\$), calculate the losses, and perform an appropriate thermal design, given your room for a heatsink and allowable losses (which reduce battery time).

Tangentially relevant fact: electrons have higher mobility than holes, so an N-channel MOSFET will have a lower \$R_{DS(on)}\$ than a P-channel MOSFET of similar die size and cost. Depending on your cost and performance requirements, you might want to investigate drivers that can drive a high-side N-channel MOSFET, or re-arrange your circuit to disconnect the battery on the ground side, where you can use an N-channel MOSFET, if that's acceptable for your application.

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