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When I run a circuit simulation, I get a current in R1 which is around 16 mA. Isn't the current supposed to be zero, since this is an N-Channel EMOSFET, which I also think is symmetrical?

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    \$\begingroup\$ I remember dealing with a 4-terminal FET long ago when they were not really popular. The 4th terminal was the "body" of the transistor and this made the whole thing symmetrical. Most FETs have the body connected to the source and symmetrical they are not. \$\endgroup\$
    – fraxinus
    Aug 5 at 12:48

5 Answers 5

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16 mA is about right. Since you have a positive supply at the source, you are forward biasing the body diode of the FET, and you have 6 V - 0.7 V across 330 ohms, or 16.1 mA.

You can see this more clearly if you look at the device symbol with the body diode shown explicitly (from here):

enter image description here

Just about all commercially available power MOSFETs have a body diode as part of the structure, so you have to keep that in mind when applying them. If you need to block conduction in both directions one option is to use back-to-back FETs.

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  • \$\begingroup\$ I mean, it's not like you can manufacture a MOSFET without the body diode :) you can leave the body floating and hook it to a pin though. \$\endgroup\$ Aug 6 at 22:37
  • \$\begingroup\$ @VladimirCravero True- Years ago we made 4-terminal FETs like that with the body connected to a separate pin at the request of a particular customer. There wasn't much of a market for them though, so they were never offered as a catalog part. Don't know if anyone is making power MOSFETs like that these days. \$\endgroup\$
    – John D
    2 days ago
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Take a look at the datasheet for a 2N7000. Look at the symbol for the device on page 1, see the diode from source to drain? This is the body diode, and it is present in most discrete MOSFETs. On page 4, you'll see the "Drain-Source Diode Characteristics" and a typical Vf of 0.88V. (6V-0.88V)/330 Ohms is 16mA.

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You are forward biasing the bulk/body diode inside the MOSFET and, that will cause about 5 volts to appear across R1: -

enter image description here

Image from here.

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  • \$\begingroup\$ Given the substrate-to-channel junction symbol (triangle), isn't the body diode symbol somewhat redundant? I know of course that everyone draws these body diodes explicitly, so it's not a comment on your use of a common symbol, but rather on whether I understand the symbology correctly. \$\endgroup\$ Aug 6 at 17:58
  • \$\begingroup\$ @Kubahasn'tforgottenMonica a bit like how many loops you show on an inductor symbol or how many squiggles on some resistor symbols. \$\endgroup\$
    – Andy aka
    Aug 6 at 19:03
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The diagrams with piggy-back diodes always amused me; does no one teach kids anymore that symbology means anything? Well, apparently not. Matter of fact, I don't think anyone told me when I was in school, I had to figure it out myself. Well, go figure, then.

So, on that note, it's as simple as this:

Arrows denote P-N junctions, pointing in the direction of conventional current flow.

BJTs show the emitter; the collector can be drawn the same way (as a "double emitter" symbol), but one is usually left off to denote the asymmetrical junction, suggesting the direction of intended flow. (Transistors haven't been symmetrical since the germanium days!*)

*With some exceptions, or near exceptions. But by and large, diffused and epitaxial BJTs have low Vebo and much higher Vceo, among other effects.

JFETs, show the gate-channel junction. UJTs (back when they were relevant, heh) likewise; just drawn differently to imply the different function/behavior from JFETs (which they are otherwise fairly similar to, structurally speaking). The right-angle line segments to the FET channel suggest ohmic connections rather than rectifying (Schottky or PN) junctions.

And MOSFETs use an arrow, just as well. The substrate-channel junction, or "back gate", is the middle connection. The channel is the three line segments in a row (sometimes they are drawn as a solid line, particularly for depletion-mode FETs, which are "default on", hence a continuous line, eh?). Normally, substrate is tied with source, but this doesn't have to be (rarely, discrete parts are available this way, and it's the standard method in ICs).

You'll note sometimes a "MOSFET with emitter" sort of symbol is used. This newer style is all the more grating when you realize what the arrow is supposed to mean... Alas, that meaning has, it seems, been widely lost to time.

So, needless to say -- your diagram shows a diode from V6 to R1; but at least now it should be apparent. Cheers!

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The symbol of the device includes everything needed to figure this out. The various parts of the symbol represent functional features of the device:

A labeled description of the various parts of the NMOS enhancement mode device

There is a junction diode from the source+body terminal to the channel. This diode is forward-biased and conducts when the source+body terminal is above the drain.

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