Is this p-channel mosfet working backwards?

S

Speakerguy

Active Member

• Nov 5, 2009

• #1

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

https://www.electro-tech-online.com/custompdfs/2009/11/4412fa.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!

 

B

BrownOut

Banned

• Nov 5, 2009

• #2

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.

 

Last edited: Nov 5, 2009

S

Speakerguy

Active Member

• Nov 5, 2009

• #3

Then how does figure 4 work?

 

B

BrownOut

Banned

• Nov 5, 2009

• #4

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

 

S

Speakerguy

Active Member

• Nov 5, 2009

• #5

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

 

B

BrownOut

Banned

• Nov 5, 2009

• #6

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.

 

Last edited: Nov 5, 2009

crutschow

Well-Known Member

Most Helpful Member

• Nov 5, 2009

• #7

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.