Electronic – P-Channel MOSFET Drain-Source Polarity in Power Switch Application

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I'm looking at using the LTC4412 power switch in an application. From the datasheet, here is the typical application circuit: circuit diagram:

In the application circuit, an FDN306P p-channel MOSFET is used.

I'm kind of new to this, but my understanding is that for a p-channel MOSFET, current flows from the source to the drain, and that the source should be at a higher voltage than the drain. Looking at the application circuit, though, it appears that the drain is connected to the positive battery terminal.

What am I missing?

Thanks.

Best Answer

Note that this is not a power converter but effectively a chip that uses a PFET to implement active rectification. As such, the current is intended to flow from the battery to the load when the switch is on. The FET is being used like a diode, making use of the inherent body diode of the part. The chip then turns on the FET to make the diode look more ideal when it should already be on. The chip will turn off the FET quickly when is sees SENSE go higher than Vin. If it didn't, the "diode" would conduct backwards.

The advantage of such a design is that the effective diode has very low forward drop, which is useful when it's in series with a battery since less of the battery power will be wasted. The drawback is that the overall diode has slow reverse recovery time, since the chip has to sense the reverse voltage and then actively shut off the FET. In this case the diode is used for power ORing, so a few ┬Ás reverse on time won't matter much.