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I'm puzzled in understanding how the P-channel enhancement MOSFET in figure is activated. Circuit

The role of the MOSFET is to disable the VBAT supply when VUSB is provided (a usb cable is plugged). I kind of understood that when VUSB is provided, 5V are both at gate and source, turning off the MOSFET, which is fine.

From the datasheet Vgs=-0.55V, so for the MOSFET to turn on and let VBAT power the U1 voltage regulator, there should be a difference in voltage between gate and source. Let's assume VUSB is disconnected (usb cable unplugged), the MOSFET is off and VBAT is connected to a 3.7V battery, where does the Vgs of at least -0.55V comes from?

Mosfet datasheet

Thanks!

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    \$\begingroup\$ Are you sure the MOSFET is not upside down? (The body diode will always let VBAT through.) \$\endgroup\$ – CL. Mar 3 '18 at 15:14
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    \$\begingroup\$ @CL. the body diode is just fine and the MOSFET is not upside down. \$\endgroup\$ – Andy aka Mar 3 '18 at 15:38
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I'm puzzled in understanding how the P-channel enhancement MOSFET in figure is activated.

When VUSB is not present, D1 is reverse biased so can be ignored. Under this situation the gate is connected to GND via R7 and the gate voltage settles at GND because the gate is insulated from the rest of the MOSFET. This is 50% of the criteria needed to turn on a P channel MOSFET. The other 50% comes from the source being a volt or so higher than the gate (MOSFET dependent).

When VBAT is considered, the obvious choice of conduction is through the MOSFET body diode thus raising the source voltage and, because the MOSFET is P channel (and the gate is now at GND), the MOSFET turns on and offers a high conduction path for current to be taken by the regulator from VBAT. This shorts out the previously conducting body diode and very little forward voltage is dropped across the MOSFET as it supplies current to the voltage regulator - the MOSFET is behaving as an "ideal diode".

If VUSB is high it will turn off the MOSFET and VUSB will source the current needed by the regulator through D1 losing about 0.7 volts in the process. The body diode should be reverse biased now so no current should be taken from the battery. This of course presumes that under normal circumstances VUSB is bigger than VBAT.

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