2
\$\begingroup\$

I am probably being stupid and should probably go to bed but how is current flowing in this circuit below?

enter image description here

enter image description here

From my understanding, a P-channel MOSFET allows current to flow when its gate terminal is at a lower potential than its source terminal. Of course a certain threshold must be met to fully "turn on" the channel but regardless. What I don't understand with this circuit is how the capacitor is able to charge. As can be seen from the graphs, the VGS starts at 0V. VGS = 0V means that the gate and source are at the same potential, thus from the earlier criterion, current should not flow. However, it does flow and charges the capacitor to the source voltage. What is going on here?

\$\endgroup\$
0

3 Answers 3

Reset to default
4
\$\begingroup\$

The MOSFET has a body diode (not shown on your symbol) which will conduct if source is a lower voltage than drain, that is, the ‘reverse’ direction.

The body diode anode faces toward the drain, and the cathode faces toward the source. As connected, that diode is forward biased.

So, the cap is charging through the diode, even though your starting condition is Vgs = 0.

More about that pesky body diode here: How should I understand the intrinsic body diode inside a MOSFET?

\$\endgroup\$
4
  • \$\begingroup\$ Also, MOSFET's are pretty symmetrical devices -- so even if you ignored the body diode, a FET operated in "reverse" like that just has the drain acting like the source and the source acting like the drain. \$\endgroup\$
    – TimWescott
    Mar 25, 2021 at 5:56
  • 1
    \$\begingroup\$ Yes, that’s true when Vgd exceeds the threshold. Nevertheless The diode will conduct first in this case. \$\endgroup\$
    – hacktastical
    Mar 25, 2021 at 6:31
  • \$\begingroup\$ It would probably be helpful to the OP to point out that the FET is reverse polarised compared to usual use. You do mention "reverse direction" but I suspect that comment may not make the situation clear. (+1) \$\endgroup\$
    – Russell McMahon ♦
    Mar 25, 2021 at 8:37
  • \$\begingroup\$ I don’t think that drain and source are reversible usage like that. We known that body terminal of fet connected to source so the channel will create with voltage between gate and source only. But when it’s occur the current can flow in bidirectional as explained in my answer. \$\endgroup\$
    – M lab
    Mar 25, 2021 at 19:09
1
\$\begingroup\$

I found this kind if configuration in reverse polarity protection circuit as picture below. enter image description here First the current flow from drain terminal through body diode or intrinsic diode. After that source terminal got voltage as drain minus by diode’s forward voltage drop. Then the body that connect to source has some positive charge that build the channel along with gate plane. The channel is just N-type with charge density created by electric field from gate that allow current pass through in bidirectional. At this point, the voltage dropped across source and drain is lower than diode’s forward drop.

\$\endgroup\$
0
\$\begingroup\$

You are correcting using a P Mosfet as a high side switch.

Just invert source with drain by flipping horizontally the MOSFET and then your circuit is perfect.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ OP’s intent seems to be that the FET stays off when the power comes up. So swapping drain and source defeats that. Not that it matters - the FET still turns on even in reverse due to the Vgd coming on. \$\endgroup\$
    – hacktastical
    Mar 25, 2021 at 6:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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