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

Question

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

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.

Answer 1

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 µs reverse on time won't matter much.

Answer 2

This application uses the internal body diode of the transistor in the same way as 1N5819 diode is used in the schematic. Whichever source has higher voltage (wall adapter or battery) will supply the output and reverse bias the other input diode (1N5819 and body diode of FDN306P).

Additionally, when battery power is used, the MOSFET is turned on to lower the conduction losses of FDN306P (MOS structure is conducting, instead of its body diode).

Answer 3

The polarity of current flow is irrelevant. If the correct gate-source voltage is applied, the channel will conduct. This holds true for all MOSFETs - ON means ON, so to speak.

Consider this: when a MOSFET is used as a synchronous rectifier or an active OR-ing device, it is very easy for energy to travel backwards (from the output to the input) under some conditions, necessitating reverse-conduction detection and quick MOSFET turn-off.

Often, its the body diode polarity that determines where the source and gate are connected in the circuit.