Is it possible to have a high side MOSFET switch that will allow any direction of current? I was trying to design one but without the success. Or should I use a relay?

  • \$\begingroup\$ Maybe this: homemade-circuits.com/bidirectional-switch \$\endgroup\$
    – ErikR
    May 20, 2021 at 15:28
  • \$\begingroup\$ @ErikR I did try to simulate those configurations in the PSpice but for some reasons, the behaviour was not symmetrical. Mayme used models were not very good. \$\endgroup\$ May 20, 2021 at 15:31
  • \$\begingroup\$ Also look at the operation & design of analog switches, e.g. youtube.com/watch?v=z9GRiYPq7LM \$\endgroup\$
    – ErikR
    May 20, 2021 at 15:40
  • \$\begingroup\$ Here's ti.com/lit/pdf/slva948 how TI does it. \$\endgroup\$
    – Janka
    May 20, 2021 at 19:01
  • \$\begingroup\$ @Janka very interesting paper. \$\endgroup\$ May 20, 2021 at 19:30

1 Answer 1


Fun fact: MOSFETs (JFETs too!) will conduct in either direction when Vgs is above threshold.

The issue with MOSFETs (3-terminal ones, with the body tied to source) is the parasitic body diode formed by the FET structure and body-source tie also conducts when the drain-source voltage is in the reverse direction, even if the FET gate is off.

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

In other words:

  • FET 'on': conducts in both directions through drain-source channel with Rds(on) resistance
  • FET 'off': blocks forward drain-source, conducts reverse through body diode with one Vf drop.

Note that when 'on', the FET drain-source channel shorts out the body diode, so the current flows through the normal drain-source path. As the parasitic diode’s forward threshold isn’t met, there’s practically no diode current at all.

The workaround is to tie two FETs in series, back-to-back, source-to-source or drain-to-drain. This allows each FET to block the other's parasitic body diode when 'off'.

More here: Does MOSFET let current flow through source to drain as it allows it from drain to source?

BONUS: A simulation showing high-side switches built with n-FETs in three different ways: 4-terminal, 3-terminal with body diode, and back-to-back 3-terminal.

Simulate it here

enter image description here

  • 1
    \$\begingroup\$ A couple of caveats: It can sometimes be tricky to bias the gate if the voltage you are switching is greater than the allowed gate to source voltage of the FET. Also there will be a significant amount of capacitance across the switch when in the off state. \$\endgroup\$ May 20, 2021 at 17:44
  • \$\begingroup\$ Actually, it is very feasible to use a single MOSFET (e.g. n channel) to block bidirectional voltages. If the body is tied to VSS instead of the source, there will be two antiserial diodes in parallel to the channel and parasitic body conduction will not occur. Unfortunately 4-terminal MOSFETs are no longer sold as such, but some can still be found - labeled as NMOS analog switches. \$\endgroup\$
    – tobalt
    May 20, 2021 at 18:20

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