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When using MOSTEFS as a switch I always see the drain connected to the higher potential and and the load and the Source is always connected to ground. Can you switch those so that the Source pin connects to the higher potential and the drain is connected to ground?

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  • \$\begingroup\$ Why do you want to do that? \$\endgroup\$ – stevenvh Jun 7 '11 at 5:42
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    \$\begingroup\$ You can can do that with JFETs, but not MOSFETs. \$\endgroup\$ – Leon Heller Jun 7 '11 at 7:00
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    \$\begingroup\$ @Leon Why not? There's nothing preventing you from biasing the FET to the on-state and having current flowing from source to drain. Synchronous rectification is the 'killer app' for this function. \$\endgroup\$ – Adam Lawrence Jun 7 '11 at 14:08
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    \$\begingroup\$ I don't want to do it. I was just wondering if it would work if I did! \$\endgroup\$ – PICyourBrain Jun 7 '11 at 20:04
  • \$\begingroup\$ Related electronics.stackexchange.com/q/18884/17387 \$\endgroup\$ – try-catch-finally Feb 26 '17 at 17:25
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To clarify a bit what others have already said, a MOSFET has a internal diode that points from source to drain in N channel devices and drain to source in P channel devices. This is not something added deliberately by the manufacturer, but is a byproduct of the way MOSFETs are made. Most of the time this diode would prevent the MOSFET being useful when flipped around. There are some applications you can consider "advanced" where this diode is actually used deliberately. One example is to make a synchronous rectifier. That's basically a diode with a transistor accross it. The transistor is turned on when it is known the diode should be conducting. This lowers the voltage drop accross the diode and is sometimes used in switching power supplies to get a little more efficiency. A MOSFET with its internal diode can be thought of as the diode and transistor all nicely integrated into a single package.

Your observation of source being negative and drain positive is true for N channel FETs. Just like there are NPN and PNP bipolar transistors, there are N channel and P channel FETs that are mirror images of each other polarity-wise. A P channel FET would be connected with a positive source and negative drain. In the off state, the gate is held at the source voltage. To turn it on, the gate is lowered by 12-15V with respect to the source for most normal MOSFETs.

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    \$\begingroup\$ Another use is battery protection circuits, where you want current going in that direction anyway, but not the other. focus.ti.com/lit/an/slva139/slva139.pdf \$\endgroup\$ – endolith Jun 7 '11 at 16:13
  • \$\begingroup\$ @ Olin Lathrop: For an N type MOSFET, under normal usage the diode prevents current from flowing from the drain to the source, right? And then when a voltage is applied to the Gate that diode is essentially bypassed so that current is able to flow around the diode from the drain to the source? I guess my confusion is in the naming convention? It just seems like current should flow from a source to a drain (like a water faucet)! \$\endgroup\$ – PICyourBrain Jun 7 '11 at 20:15
  • \$\begingroup\$ Is it because the electron flow is in the opposite direction of current flow? \$\endgroup\$ – PICyourBrain Jun 7 '11 at 20:20
  • \$\begingroup\$ The diode is in parallel with the FET body, so never prevents current flow, only adds to what the FET would otherwise allow. In normal use, the diode is always reverse biased and therefore not there. "Drain" and "Source \ \$\endgroup\$ – Olin Lathrop Jun 8 '11 at 0:47
  • \$\begingroup\$ Argh, hit the wrong key in the middle of typing. Drain and Source are names relating to the semiconductor physics, not to current flow. Actually they refer to the flow of the minority carriers, which have different polarity in P and N type semiconductors. \$\endgroup\$ – Olin Lathrop Jun 8 '11 at 0:49
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If you want a ground-referenced load, you can use a P channel MOSFET. This will be a mirror-image of the circuit you describe, ie with the source connected to the higher voltage and the drain connected to 0V via the load. However, your gate drive will need to be reversed and will need to be close to your higher voltage to turn the load off.

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A mosfet is really a four terminal device. Drain, source, gate and body.

For a N channel mosfet the doping arrangements result in diodes that permit current flow from body to drain and from body to source.

If you have a mosfet with all four terminals brought out seperately then there is a symetry between drain and source. Provided body is kept at a potential that is less than or equal to both the drain and source voltage the mosfet can be used to switch currents in both directions.

However most discrete mosfets have body internally connected to source which effectively places a diode from source to drain. So the mosfet can only block current flow in one direction.

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The problem is internal diode, which will always conduct in reverse direction with 0.7V drop, so when you will switch MOSFET ON you will lower that drop down to 0V and that's it.

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You can do it if your application can cope with the reverse body diode - there are a few occasions where this can be useful, for example reverse-polarity protection with low voltage drop.

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