Due to the symmetry in mosfet, source and drain are interchangeable. So we define the source of nmos is the terminal with the lower voltage. So if we force Vds to become negative the drain and source should flip. So Vds becomes positive again.

If that is true then the following is true?

If the overdrive voltage of an nmos is 0.2, for example, and I forced Vds to -0.3 will the mosfet operate in sat with source and drain flipped?

enter image description here Here is a circuit from CMOS analog design book by Razavi showing flipping of the drain and source.

  • \$\begingroup\$ some MOSFETs have special implants at the Drain, to handle high voltages; generic silicon FETs seem to have no special source or drain implants, at least in processes I've used; \$\endgroup\$ – analogsystemsrf Sep 10 '19 at 1:54

Yes, your understanding is correct. In reality, a MOSFET is a four-terminal device. The body is not necessarily connected to the source. For the planar MOSFETs used in VLSI design the source and drain are physically the same kind of structure.

So, the source of the NMOS transistor is the terminal with the lower voltage, out of the two terminals that could be either source or drain. This behavior is necessary for bidirectional analog multiplexers that use MOS pass transistors.

That means that the answer to your question is no, \$V_{DS}\$ cannot be negative, because the more negative terminal is always the source.


You assume the MOSFET is symmetric. This is not the case; real FETs (in discrete devices, anyway; internal to ICs is another matter but not one relevant to anyone not going deep into VLSI design) have the source terminal distinguished by being connected to the body terminal. A true symmetric MOSFET would have four terminals: source, gate, drain, and body/bulk/substrate (no one can seem to agree on the name of the fourth one).

This symbol, which seems to be considered the "most proper" symbol, makes that clearer:

The traditional MOSFET symbol, showing the body terminal connected to the source. This one is missing the body diode though.

  • \$\begingroup\$ And usually, there's a built-in diode in that body connection that causes current to flow from source to drain (in an NMOS) or drain to source (in a PMOS) when you get the connections flipped. \$\endgroup\$ – TimWescott Sep 9 '19 at 21:49
  • \$\begingroup\$ It is not always that case. These mosfets (source connected to bulk) require triple well technology (I don't know how much it is available now), but if you don't have access to triple well technology you have to connect the bulk to ground so in this case it will be symmetic (as you mentioned above). \$\endgroup\$ – Hassan Ibrahim Sep 9 '19 at 21:51
  • \$\begingroup\$ @HassanIbrahim Are we talking about internal to ICs? My experience with FETs is almost entirely using them as discrete devices for switching power converters, so I won't be able to give much help talking about VLSI design. \$\endgroup\$ – Hearth Sep 9 '19 at 21:52
  • \$\begingroup\$ @TimWescott Not usually, always. Except it's more proper to say it flows from body to drain or drain to body, if you want to be really specific. \$\endgroup\$ – Hearth Sep 9 '19 at 21:52
  • \$\begingroup\$ No, this is not true. In VLSI design the body is not necessarily connected to the drain. MOSFETs are four-terminal devices. \$\endgroup\$ – Elliot Alderson Sep 9 '19 at 22:03

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