7
\$\begingroup\$

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.

\$\endgroup\$
6
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Thanks, and thanks to @Szymon, this makes sense now. In my case the auxiliary power source will be 5V from a USB bus (when connected). I'm planning to use AA batteries. I have to be make sure the total voltage from the batteries isn't > 5V, or the LTC4412 will draw power from the batteries and not the USB VBUS. \$\endgroup\$ – David Aug 7 '12 at 20:35
5
\$\begingroup\$

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).

\$\endgroup\$
5
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.