I am seeking clarification on the application of a p-channel MOSFET as illustrated in the diagram below. Could someone please provide an explanation?


In the diagram, legs 1, 2, 6, and 5 serve as drains, while leg 3 functions as the gate and leg 4 as the source. A 12V input is directed into the voltage regulator.

Before delving into my queries, I'd like to establish what I understand to be correct, and I welcome corrections if any inaccuracies are present:

The standard current flow in a p-channel MOSFET is from source to drain when a negative voltage is applied to the gate (Vgs). However, in this scenario, we aim for the current to flow from drain to source, as the positive terminal of the battery is connected to the drain.

The gate is connected to ground, indicating a zero voltage applied to the gate.

Now, onto my questions:

Given our objective to reverse the current flow (from drain to source instead of source to drain), should we apply a positive voltage to the gate rather than a negative one?

Whether the answer to the previous question is affirmative or negative, the following question persists: Since the gate voltage is already zero and the drain possesses 12V with no current flowing, how will the MOSFET be activated?

I appreciate any insights or clarifications on these matters. Thank you.

  • \$\begingroup\$ \$V_{GS}\$ is not gate potential. \$V_{GS}=V_G - V_S\$, the potential difference between gate and source. Since sources are permanently connected to +12V, and gates are at 0V, \$V_{GS}=(0V)-(+12V)=-12V\$, and the MOSFETs are permanently on. \$\endgroup\$ Commented Mar 31 at 3:53
  • \$\begingroup\$ +12V is connected to Drain not Source. \$\endgroup\$ Commented Mar 31 at 3:55
  • \$\begingroup\$ Your schematic clearly shows 12V connected to sources. \$\endgroup\$ Commented Mar 31 at 3:57
  • \$\begingroup\$ here is the datasheet. It shows those legs are drains and not source. diodes.com/assets/Datasheets/DMP3056LDM.pdf \$\endgroup\$ Commented Mar 31 at 3:58
  • 1
    \$\begingroup\$ @user23910965 IMPORTANT I'm a moderator. (1) It appears that you have used AI to improve your question. As long as it's your question and you are using the AI as a tool to improve the format then this is (probably) OK. BUT the 1st paragraph (Certainly ..." ) did not read like human input, was wholly unnecessary, and detracted from your question. If using tools to assist you you should read the result carefully and ensure that it is "speaking in your voice". (2) To meet site rules, please add the source of the diagram to your question. eg "Circuit created by me" or "from http:// ... " etc \$\endgroup\$
    – Russell McMahon
    Commented Mar 31 at 6:39

1 Answer 1


If this is a reverse polarity protection system, then the node labelled "12V" is misleading. I'll redraw the design with some label changes to reflect the input and output roles less ambiguously:


simulate this circuit – Schematic created using CircuitLab

For negative inputs, \$V_{IN} < 0\$, the body diodes are reverse biased, and the load keeps \$V_{OUT}\$ at 0V. In this condition, source potential \$V_S=0V\$:

$$ V_{GS} = V_G - V_S = (0V) - (0V) = 0V $$

So, for \$V_{IN} < 0\$, the MOSFETs are off, and their body diodes do not conduct. Current \$I=0A\$.

When \$V_{IN}\$ rises above 0V (becoming positive), the body diodes are forward biased, ensuring that source potential \$V_S\$ is never lower than \$V_S = V_{IN} - 0.7\$. Since source potential also rises, \$V_{GS}\$ is:

$$ V_{GS} = V_G - V_S = (0V) - (V_{IN} - 0.7) = -(V_{IN} - 0.7) $$

This will be strongly negative, if \$V_{IN} >> 0\$, the condition required for the MOSFETs to be on, and they conduct strongly. Their impedance drops enough to bypass the body diodes altogether, raising \$V_S\$ to become \$V_S = V_{OUT} = V_{IN}\$.

  • \$\begingroup\$ Where did you get the 0.7 ? \$\endgroup\$ Commented Mar 31 at 16:32
  • \$\begingroup\$ @user23910965 It's the forward voltage of a silicon diode, 0.7V \$\endgroup\$ Commented Mar 31 at 17:11

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