Is this p-channel mosfet working backwards? - Electronic Circuits Projects Diagrams Free

speakerguy79 · 5th November 2009, 09:58 PM

Please reference page 9, figure 2 of the following data sheet:

http://cds.linear.com/docs/Datasheet/4412fa.pdf

Why do they have the P-channel FETs hooked up with the source towards the load? How are these devices working? I am more used to seeing things like Figure 7 for high-side P-channel power switches.

Thanks for any help!

BrownOut · 5th November 2009, 10:27 PM

It's not working backwards. Current flows from Drain to Source. Since it's a p-channel MOSFET, VDG > -VT, and the MOS is in the low power dissapation region. The Sense pin of the IC keeps the FET in that region.

speakerguy79 · 5th November 2009, 10:31 PM

Then how does figure 4 work?

BrownOut · 5th November 2009, 10:40 PM

Those MOSFET's ARE working backwards LOL! MOSFET's are allowed to work that way, so it's OK.

speakerguy79 · 5th November 2009, 10:42 PM

Really? Is this because the Vds is essentially zero in these cases?

BrownOut · 5th November 2009, 11:18 PM

VDS is nearly zero, but that's because in a P-channel MOSFET, when VGS< VT and VDG > -VT, a strong inversion layer is formed between the drain and source, which is bidirectional.

EDIT: I edited all my conditions for low power dissapation. I get those confused easily.

crutschow · 6th November 2009, 01:26 AM

The reason the source and drain is connected backward from what you might expect is so the parasitic MOSFET diode (which is connected anode to drain and cathode to source) does not conduct if the wall adapter voltage is significantly higher than the battery voltage, which would perform uncontrolled charging of the battery.

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5th November 2009, 09:58 PM	#1
speakerguy79	Is this p-channel mosfet working backwards? Please reference page 9, figure 2 of the following data sheet: http://cds.linear.com/docs/Datasheet/4412fa.pdf Why do they have the P-channel FETs hooked up with the source towards the load? How are these devices working? I am more used to seeing things like Figure 7 for high-side P-channel power switches. Thanks for any help!

5th November 2009, 10:27 PM	#2
BrownOut	It's not working backwards. Current flows from Drain to Source. Since it's a p-channel MOSFET, VDG > -VT, and the MOS is in the low power dissapation region. The Sense pin of the IC keeps the FET in that region. __________________ You don't need a quadraphonic Blaupunkt -- you need a curve ball. Last edited by BrownOut; 5th November 2009 at 11:25 PM.

5th November 2009, 10:31 PM	#3
speakerguy79	Then how does figure 4 work?

5th November 2009, 10:40 PM	#4
BrownOut	Those MOSFET's ARE working backwards LOL! MOSFET's are allowed to work that way, so it's OK. __________________ You don't need a quadraphonic Blaupunkt -- you need a curve ball.

5th November 2009, 10:42 PM	#5
speakerguy79	Really? Is this because the Vds is essentially zero in these cases?

5th November 2009, 11:18 PM	#6
BrownOut	VDS is nearly zero, but that's because in a P-channel MOSFET, when VGS< VT and VDG > -VT, a strong inversion layer is formed between the drain and source, which is bidirectional. EDIT: I edited all my conditions for low power dissapation. I get those confused easily. __________________ You don't need a quadraphonic Blaupunkt -- you need a curve ball. Last edited by BrownOut; 5th November 2009 at 11:25 PM.

6th November 2009, 01:26 AM	#7
crutschow	The reason the source and drain is connected backward from what you might expect is so the parasitic MOSFET diode (which is connected anode to drain and cathode to source) does not conduct if the wall adapter voltage is significantly higher than the battery voltage, which would perform uncontrolled charging of the battery. __________________ Carl Curmudgeon Elektroniker