Hello all,

I accidentally wired a p-mosfet with power at the drain, and default same voltage at the gate. If the power we're at the source I would expect the led load to be normally off when the gate is high, and on when the gate is pulled low. Right now the led is normally on. I thought this could be because the resulting Vgs is less than zero, but I am wondering why the drain-source is conducting. Since it is conducting, is the body diode not connected to the source, and the mosfet is symmetrical, meaning what is labeled drain in the data sheet is actually the source by definition of being attached to power? This would mean the gate is low? I just realized I could measure this, but I'll post anyway as it is an interesting question and I am out of the office until tomorrow...

 

Draw your scheme please. Then I will understand your problem.

 

I echo Bordodynov's request in the post above. There are too many ambiguities in your post to be able to be sure of providing a correct definitive answer.

You can treat P-channel MOSFETs the same as N-channel MOSFETs if you reverse the voltages.

Generally speaking (writing) driving the gate the same polarity as the drain will tend to turn the device on.

 

You know there is a diode in parallel with the mosfet? If you applied reverse voltage it will just conduct regardless of what you do to the gate.

 

You know there is a diode in parallel with the mosfet?

Do you know why it's there?

 

Do you know why it's there?

Its inherent in the die fabrication.

 

Hello,

Have a look at page 2 of the attached PDF.
There the bodydiode is shown.

Bertus