Why the exact same transistor acts like a variable resistor in one region and a current source in the other — and why the name "saturation" is the opposite of what BJT experience trains you to expect.
A single MOSFET, with a single Vgs applied, can behave like two completely different components depending on nothing more than how much Vds is across it. Push it into one region and drain current climbs almost linearly with Vds, like a resistor whose value you set with the gate. Push it into the other region and drain current goes essentially flat — it stops caring about Vds at all and instead depends only on Vgs, like a current source. Nothing about the physical device changes between the two. Only the relationship between three voltages — Vgs, Vds, and the threshold voltage Vt — decides which behavior you get.
Once Vgs exceeds Vt, a conducting channel exists between source and drain, and the gate's overdrive — the amount Vgs sits above Vt, written Vgs − Vt — sets how "open" that channel is. What the device does with that open channel then comes down to a single comparison: is Vds smaller than Vgs − Vt, or larger? If Vds < Vgs − Vt, the channel stays open all the way from source to drain, and the device is in the triode region (also called linear or ohmic). If Vds ≥ Vgs − Vt, the channel "pinches off" near the drain end before it gets there, and the device is in the saturation region. That single boundary, Vds = Vgs − Vt, is the line every MOSFET output characteristic is built around.
The channel a MOSFET forms isn't uniformly thick along its length — its depth at any point depends on the local voltage difference between the gate and that point in the channel. Near the source, that difference is just Vgs, but near the drain it's reduced by whatever Vds has already been dropped getting there, so it's only Vgs − Vds. As long as Vds stays below Vgs − Vt, the channel stays open (if thinner near the drain) all the way across, and current through it behaves roughly like current through a resistor whose value is set by Vgs — rising with Vds, the way any resistor's current rises with the voltage across it. The moment Vds reaches Vgs − Vt, the channel depth at the drain end hits zero — it "pinches off." Beyond that point, adding more Vds doesn't widen the channel anywhere; it just moves the pinch-off point slightly back toward the source, and the current that does make it across is set almost entirely by how hard the gate is driving the channel (Vgs − Vt), not by how much extra voltage is waiting on the drain side. That's the whole reason saturation current is written as Id ≈ k(Vgs − Vt)² with no Vds term in the ideal first-order model.
This is one of the most common cross-over mix-ups for anyone who learned BJTs first. In a BJT, saturation is the low-resistance, fully-on state — the collector-emitter voltage collapses to a small fraction of a volt, and that's exactly the region you drive a switching BJT into. Carry that expectation over to a MOSFET and it leads you completely backward. MOSFET saturation is the constant-current region — the one where Id has flattened out and stopped depending on Vds, which is the region used for amplifiers and current sources, not the region you want for a fully-on digital switch. If you want a MOSFET to behave like a closed switch — low resistance, minimal voltage drop — you actually want the triode region, driven deep into it with Vds pushed as small as possible (a heavily overdriven gate, Vgs far above Vt, with Vds near zero). The names point in opposite directions between the two device families: BJT saturation = resistor-like "on." MOSFET saturation = current-source-like, and MOSFET triode is the one that's resistor-like.Mixing them up is a common source of real design errors, from picking the wrong bias point in an amplifier to misreading a datasheet's Ron (on-resistance) test conditions.
Explains why a MOSFET's behavior splits into two distinct regions — triode (voltage-controlled resistor) and saturation (voltage-controlled current source) — depending purely on the relationship between Vgs, Vds, and the threshold voltage Vt, and why the naming is a frequent source of confusion for anyone coming from BJTs.
BJT and MOSFET operating-region names look similar but describe opposite behaviors. BJT saturation is the low-resistance, fully-on state used for switching; MOSFET saturation is the constant-current, Vds-independent state used for amplification. Someone fluent in BJT terminology who assumes the same word means the same thing in a MOSFET datasheet or textbook chapter will draw exactly the wrong conclusion about which region to bias into for a given application.
Once Vgs exceeds the threshold voltage Vt, a MOSFET forms a conducting channel whose local depth depends on the gate-to-channel voltage at each point along its length. Near the drain, that voltage is reduced by whatever Vds has already been dropped, so it is effectively Vgs − Vds there. While Vds stays below the overdrive voltage (Vgs − Vt), the channel remains open end-to-end — the triode/linear/ohmic region — and drain current rises roughly with Vds for a fixed Vgs, much like a resistor. Once Vds reaches or exceeds Vgs − Vt, the channel pinches off at the drain end; additional Vds mostly just shifts the pinch-off point rather than adding more current, so Id plateaus and is set primarily by Vgs, following the first-order model Id ≈ k(Vgs − Vt)² with negligible Vds dependence — the saturation region.
Triode-region operation (with Vds driven very small via strong gate overdrive) is what makes a MOSFET a good low-resistance digital switch and a good voltage-controlled analog resistor for switches and choppers. Saturation-region operation is what makes a MOSFET useful as an amplifier or current source, since its output current stays nearly constant despite swings in the drain voltage — exactly the property a gain stage or current mirror needs. Choosing the wrong region for a given design intent — for example, biasing an amplifier stage into deep triode, or expecting a switch driven into MOSFET "saturation" to have near-zero resistance — is a real and recurring design mistake.
No — this is the single most common cross-over confusion between the two device families. BJT saturation is a low-resistance, fully-on state used for switching. MOSFET saturation is a constant-current, Vds-independent state used for amplification. The MOSFET region that behaves like a BJT switch in saturation (low resistance, small voltage drop) is actually the triode region, driven with Vds pushed close to zero.
The boundary is Vds = Vgs − Vt, the gate overdrive voltage. A higher Vgs opens a deeper channel and can sustain a larger Vds before that channel pinches off at the drain, so the triode-to-saturation transition point shifts to a larger Vds as Vgs increases — visible as each output curve's "knee" moving further right for higher Vgs.
Pinch-off does not stop current — it stops the channel from getting any wider or lower-resistance as Vds increases further. The pinched-off point moves slightly toward the source as Vds rises (channel-length modulation), which in real devices causes a small upward slope in the saturation plateau rather than a perfectly flat line, but the dominant behavior is still that Id is set mainly by Vgs, not Vds.
Deep triode region — drive Vgs well above Vt so the overdrive voltage is large, which keeps Vds small (often just millivolts) for a given load current and minimizes on-resistance (Ron) and power dissipation. This is the opposite of the saturation region, despite "saturation" sounding like it should mean "fully on" to anyone used to BJT terminology.
Yes, with polarities flipped. For a PMOS device, Vgs, Vds, and Vt are all typically negative, and the same comparison (in magnitude) between |Vds| and |Vgs − Vt| determines whether the device is in triode or saturation — the underlying pinch-off physics is identical, just mirrored in sign.
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