Why the drain the source terminal of the MOSFET function differently while their physical structure is similar/symmetrical ?

This is a MOSFET:

You can see that the drain and source are similar.
So why do I need to connect one of them to VCC and the other to GND ?


3 Answers 3


Myth: manufactures conspire to put internal diodes in discrete components so only IC designers can do neat things with 4-terminal MOSFETs.

Truth: 4-terminal MOSFETs aren't very useful.

Any P-N junction is a diode (among other ways to make diodes). A MOSFET has two of them, right here:

MOSFET with diodes

That big chunk of P-doped silicon is the body or the substrate. Considering these diodes, one can see it's pretty important that the body is always at a lower voltage than the source or the drain. Otherwise, you forward-bias the diodes, and that's probably not what you wanted.

But wait, it gets worse! A BJT is a three layer sandwich of NPN materials, right? A MOSFET also contains a BJT:


If the drain current is high, then the voltage across the channel between the source and the drain can also be high, because \$R_{DS(on)}\$ is non-zero. If it's high enough to forward-bias the body-source diode, you don't have a MOSFET anymore: you have a BJT. That's also not what you wanted.

In CMOS devices, it gets even worse. In CMOS, you have PNPN structures, which make a parasitic thyristor. This is what causes latchup.

Solution: short the body to the source. This shorts the base-emitter of the parasitic BJT, holding it firmly off. Ideally you don't do this through external leads, because then the "short" would also have high parasitic inductance and resistance, making the "holding off" of the parasitic BJT not so strong. Instead, you short them right at the die.

This is why MOSFETs aren't symmetrical. It may be that some designs otherwise are symmetrical, but to make a MOSFET that behaves reliably like a MOSFET, you have to short one of those N regions to the body. To whichever one you do that, it's now the source, and the diode you didn't short out is the "body diode".

This isn't anything specific to discrete transistors, really. If you do have a 4-terminal MOSFET, then you need to make sure that the body is always at the lowest voltage (or highest, for P-channel devices). In ICs, the body is the substrate for the whole IC, and it's usually connected to ground. If the body is at a lower voltage than the source, then you must consider body effect. If you take a look at a CMOS circuit where there's a source not connected to ground (like the NAND gate below), it doesn't really matter, because if B is high, then the lower-most transistor is on, and the one above it actually does have its source connected to ground. Or, B is low, and the output is high, and there isn't any current in the lower two transistors.

CMOS NAND schematic

  • 2
    \$\begingroup\$ In an NFET it's clearly necessary that the source and drain potentials be no lower than the body potential, but that does not imply that the source and drain should have to have a fixed polarity relative to each other. It's hardly rare to have a situation where one wants to connect or disconnect two points, both of which will always be higher than some "ground" point, but either of which might be higher than the other. One could use two MOSFETs for that, but that would seem somewhat wasteful if a "four terminal MOSFET" could do the job. \$\endgroup\$
    – supercat
    Commented Jan 10, 2014 at 19:14
  • \$\begingroup\$ @supercat sure, but then you have to account for parasitic capacitances and inductances and analyze your circuit to guarantee that the source and drain remain at higher potentials than the body even in the presence of high dv/dt or di/dt. Given that these parasitics are highly dependent on layout and manufacturing variation, in many cases that seems more difficult than the alternative of designing a floating gate driver and using an ordinary 3-terminal MOSFET. \$\endgroup\$
    – Phil Frost
    Commented Jan 16, 2014 at 14:50
  • 1
    \$\begingroup\$ There are many circuits where three-terminal MOSFETs are just great. There are times, however, where it's necessary to switch current in two directions. One could use back-to-back MOSFETs, but that seems somewhat wasteful. It may be that a source/substrate connection is so advantageous to process geometry that a back-to-back pair with a given RDSon and current-handling capability can be made cheaper than could a single isolated-base MOSFET, in which case it wouldn't actually be wasteful, but I don't know if that's the case. \$\endgroup\$
    – supercat
    Commented Jan 16, 2014 at 15:54
  • 3
    \$\begingroup\$ @JasonS It's an NPN because that's how the silicon is doped. Look at the picture and you can read: "n", "p", "n". There is no asymmetry: I just arbitrarily chose to draw the symbol one way, but it doesn't matter because a BJT has some gain even if you flip it upside-down, especially when the BJT you are talking about is the parasitic one in a MOSFET and maximizing gain wasn't a design goal. \$\endgroup\$
    – Phil Frost
    Commented Mar 6, 2015 at 15:05
  • 1
    \$\begingroup\$ In typical integrated circuits many transistors share the body terminal. You make it that way rather than making a body for every transistor because it is smaller, cheaper, faster, and more power efficient. With SOI and modern technologies things are a bit different again. \$\endgroup\$
    – HKOB
    Commented Mar 8, 2015 at 14:19

Further to Phil's answer, occasionally you'll see a depiction of a MOSFET that gives more detail of the asymmetry

enter image description here

From electronics-tutorials.wa

The asymmetric link from substrate (body) to sources is shown as a dotted line.

  • 3
    \$\begingroup\$ The geometry of discrete MOSFETs is very different from that of integrated ones; while an integrated NFET will have a P-substrate, many discrete MOSFETs have an N-type substrate which is connected to the drain on one side of the transistor; the base (which behaves like the substrate of an integrated MOSFET) and source are connected on the other side of the transistor. \$\endgroup\$
    – supercat
    Commented Nov 13, 2014 at 18:32

From a physical device standpoint, they are the same. However, when discrete FETs are produced, there is an internal diode formed by the substrate which has its cathode at the drain and anode at the source, so you must use the marked drain terminal as the drain and marked source terminal as the source.


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