5
\$\begingroup\$

Could you please help me understand why in a reverse polarity circuit with P channel MOSFET Source of the MOSFET is connected to the LOAD and DRAIN to the power supply? I understand the concept of a body diode in the MOSFET, but still I do not quite understand is there any reason why not connect SOURCE to the power supply and DRAIN to the load? Let's neglect Zener diode and R1 in shown circuit. enter image description here

\$\endgroup\$
1

2 Answers 2

12
\$\begingroup\$

If you understand the body diode then you should see that if you apply significant negative voltage to the input the body diode needs to block or it will pass the incorrect polarity to the output, probably destroying the circuitry it is connected to.

In normal operation (positive input) the body diode is shunted by the Rds(on) channel resistance of the MOSFET. In blocking operation (negative input) both the body diode and MOSFET channel block.

\$\endgroup\$
11
\$\begingroup\$

MOSFETs have a neat party trick: they can conduct in both directions due to their symmetrical construction. 3-terminal FETs also have an issue: the body diode will conduct when the FET is connected and biased in ‘reverse’ (Vds negative for a pFET), even if the FET is ‘off’. The body diode is a side-effect of the internal source-substrate connection.

At any rate, here’s what the circuit does:

  • With +12 in, the FET conducts in ‘reverse’, through both the body diode and the main channel. But because the FET is turned ‘on’ into saturation there is little to no current through the diode: it’s shorted out. That’s the party trick.

  • With -12 in (swapped input), the main channel is shut off. Also, the body diode is reverse biased. So no current flows.

If you were to reverse the FET connection and have source to +12, that is, the ‘normal’ way, the body diode would conduct when the input is swapped, even though the FET gate-source is biased ‘off’. This would defeat the purpose of the circuit.

Another thing. It’s also possible to use an n-FET the same way, but on the GND side. It’s actually preferred since n-FETs typically have better Rds(on) characteristics than p-FETs.

Here's a sim showing how the p-FET works, and how to use an n-FET (simulate it here):

enter image description here

More here: What does this simple circuit do?

And here: Two directional FET switch. Is it possible?

The second link shows more party tricks… er, techniques for using FETs to pass or block signals, including back-to-back FETs and 4-terminal FETs that have a separate substrate connection.

Finally, for that particular FET (IRF5305) the Zener diode isn't needed as it has a Vgs rating of +/-20V, more than enough to deal with 12V.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ Minor point ; 3 terminal FETs on a conductive substrate have a body diode.eg Silicon on Sapphire doesn't. \$\endgroup\$
    – Russell McMahon
    Commented Oct 1, 2021 at 5:26

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