After reading this post (3rd answer):


I almost got it why the saturation regions of BJTs and MOSFETs are in different regions of their diagrams. The differences were explained, but it remained unclear for me why the same name is used for two quite different analysis.

Other posts I found focused on one of the transistor types so it remained obscure for me: why the saturation region of the MOSFETs has this name?

Since we put the load line in the "flat region" of both graphs for amplifier design, it's strange to call it active in BJTs and saturation for the MOSFETs.

Edit: in the mean time I've found a nice footnote in another book (Foundations.. from Agarwal and Lang) which I reproduce here.

"The saturation region in BJTs is completely unrelated to the saturation region in MOSFETS [...]. This duplication of terms [...] can be source of confusion, but, unfortunately, has become the norm in circuit parlance".

It's something like: I know, it's odd, live with it.

Like two answers below show very well: look at the internals, forget "inputs" and "outputs" and stick with the (quite different) physics of both devices.

Thanks to all!

  • 1
    \$\begingroup\$ I completely agree with you but what is your specific question? \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 11:59
  • \$\begingroup\$ Hi Andy. Since I understood that the BJTs preceded MOSFET in practice, why the name "saturation" was chosen differently than the current explanation at the time? \$\endgroup\$
    – Pedro
    Commented Oct 30, 2015 at 12:07
  • \$\begingroup\$ It is much more reasonable to think that if the output does not change any more when you change something at the input, the "thing" is saturated. \$\endgroup\$
    – Pedro
    Commented Oct 30, 2015 at 12:11
  • 3
    \$\begingroup\$ Maybe the term "saturation" refers more on the internal working conditions inside the MOSFET or BJT itself than on the properties we measure from the outside. So for a BJT the BE junction would saturate with carriers while in a MOSFET the channel would be saturated with carriers. Not sure if I'm making sense here though. \$\endgroup\$ Commented Oct 30, 2015 at 12:17
  • \$\begingroup\$ @FakeMoustache I like your explanation. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 12:20

2 Answers 2


why the saturation region of the MOSFETs has this name?

The post you link to does explain this, but in case it needs repeating and perhaps backing up with a textbook reference, the saturation region for a MOSFET is called so because the drain current saturates, i.e. basically stops increasing as Vds increases further.

You are correct that the active region of a BJT corresponds to the saturation region of a MOSFET when these devices are used as amplifiers.

The saturation region of a BJT (e.g. when turned on as a switch) corresponds to the triode/ohmic region of a MOSFET.

Some authors also call the saturation region of a MOSFET the "active mode", which does match the terminology used for BJTs. But they also call the triode/ohmic region the "linear mode" which perhaps doesn't help that much because "linear" suggests an amplifier rather than a switch. Linear here again refers to how the MOSFET characteristic looks like in that region rather than any external/use considerations. (Luckily, it seems nobody calls the BJT saturation region "linear mode".)

The only thing that's not confusing about this terminology is the cut-off region, which is the same for both. Here's a summary table for the correspondence (from an external/use viewpoint):

enter image description here

This summary also includes the reverse active region for BJTs, which is seldom used, but it doesn't include synonyms for the triode region; as I said "linear mode" or "ohmic region" are also used to denote the MOSFET triode region.

  • \$\begingroup\$ Thanks for all the references, specially the one from the 3rd author, who shares my pain... ;-) The summary table is also great. I'll edit my question to add another quote I've just found. \$\endgroup\$
    – Pedro
    Commented Oct 30, 2015 at 14:23
  • \$\begingroup\$ @Pedro: By the way, JEDEC has chosen "saturation region" and respectively "ohmic region" as their choice of standard terminology for MOSFETs. This is given in JESD77 on page "4-31". \$\endgroup\$ Commented Nov 4, 2015 at 1:49
  • \$\begingroup\$ Thanks! And for BJTs it states that saturation is: "a base-current and a collector-current condition resulting in a forward-biased collector junction". \$\endgroup\$
    – Pedro
    Commented Nov 5, 2015 at 10:26

I think there are better diagrams about this subject for JFETs than MOSFETs but the mechanism is the same - the channel is starting to pinch-off i.e. if you increase drain voltage then the "usable" region in the channel for current gets thinner - this basically means that the current stays the same because although voltage has increased (and is potentially able to "push" more current into the channel), the channel has pinched off more thus opposing the potential for current to flow: -

enter image description here

You should be able to see that pinch-off occurs more around the drain than the source and this is because the voltage difference between drain and gate is bigger than the voltage between source and gate.

So that's my explanation of why the saturation region is so called in a FET - you can't push any more current through it unless the device breaks down (sad). I will try and find a diagram of what it looks like specifically in a MOSFET. Here's one that should help: -

enter image description here

The blue area is the channel - note that the thickness of the channel (close to the source) is more than the thickness at the drain. This happens when the gate voltage exceeds the gate threshold voltage (\$V_{GS(THRESHOLD)}\$). Overcoming the threshold voltage is much easier around the source because the source is at a lower potential than the drain. Now it becomes the same story as the JFET - if the drain voltage rises then the pinch-off is more pronounced and current remains largely constant.

For a bipolar transistor, saturation means something else. Clearly a BJT doesn't have a channel so it has to mean something else! BJT saturation is when the collector voltage has fallen to such a low level that the collector-base region is becoming forward biased.

Normally, for an NPN transistor, the collector-base region is reverse biased as depicted in this diagram: -

enter image description here

If the collector voltage drops towards the emitter voltage there comes a point when the collector-base region becomes forward biased - in simple terms, the base cannot avoid driving current into the collector and, because of this, normal transistor action is starting to fail. Current gain (hFE) drops significantly and any increase in base current is NOT going to force the collector any lower - in fact increases in base current are starting to feed the collector with current thus preventing the collector voltage dropping further. This is what BJT saturation is about.

  • \$\begingroup\$ Thanks for all the effort. It is indeed much better to make the comparison without trying to mix internals with applications. Please take a look at the quote I've just added to the question. \$\endgroup\$
    – Pedro
    Commented Oct 30, 2015 at 15:14

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