This calculator computes nMOSFET drain current with the long-channel square-law model plus channel-length modulation and plots it as a family of ID–VDS curves. It shows at a glance whether your operating point is in cutoff, the linear region or saturation, along with the drain current and small-signal parameters there.
Operating point
Region
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Drain current ID
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Saturation voltage VDS,sat = VGS − Vth
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Transconductance gm
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Output resistance ro
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The thick line is your VGS; thinner lines are lower gate voltages. The dashed line is the boundary between the linear and saturation regions (VDS = VGS − Vth). Subthreshold current and short-channel velocity saturation are not included.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
The square-law model
In a long-channel nMOSFET, a channel forms once VGS exceeds the threshold voltage Vth. In the linear (triode) region at small VDS, ID = µnCox(W/L)[(VGS − Vth)VDS − VDS²/2], so the transistor behaves like a voltage-controlled resistor. When VDS exceeds the overdrive voltage VGS − Vth, the channel pinches off at the drain end, and in saturation ID = (1/2)µnCox(W/L)(VGS − Vth)², nearly independent of VDS.
Channel-length modulation and output resistance
In a real device, raising VDS in saturation moves the pinch-off point toward the source and shortens the effective channel, so the current still rises slightly. This channel-length modulation is modeled by multiplying ID by (1 + λVDS). The slope gives the small-signal output resistance ro ≈ 1/(λID), which sets an amplifier's intrinsic gain gm·ro. Shorter channels have larger λ and smaller ro, which is why analog circuits often use longer-than-minimum channels.
Finding the transconductance
In saturation, the change in current per change in gate voltage is gm = µnCox(W/L)(VGS − Vth) = 2ID/(VGS − Vth). At a fixed current, a smaller overdrive gives a larger gm but requires a larger device (larger W/L) and changes the voltage headroom. With the default values of Vth = 0.5 V, µnCox = 200 µA/V², W/L = 10, λ = 0.05 /V, VGS = 1.5 V and VDS = 1.2 V, ID ≈ 1.06 mA, gm ≈ 2.1 mS and ro ≈ 19 kΩ, giving gm·ro of about 40.
Limits of the model
The square law is an approximation for long-channel devices. In modern short-channel transistors, carriers reach velocity saturation in strong fields, so the current depends on (VGS − Vth) closer to linearly than quadratically. Below threshold the current is not zero either: subthreshold current changes exponentially with gate voltage, needing at least about 60 mV per decade at room temperature. Circuit design therefore relies on detailed compact models such as BSIM in simulation.
In saturation, a long-channel MOSFET follows ID = (1/2)μCox(W/L)(VGS − Vth)², so drain current is proportional to the square of (VGS − Vth) (the square law). VGS − Vth, the overdrive voltage, is a basic analog design parameter. In short-channel devices velocity saturation pushes the dependence toward linear, and mobility degradation and parasitic resistance add further deviations, so real designs use detailed models such as BSIM.
ALike a variable resistor whose resistance changes with gate voltage
BLike a constant-current source independent of VDS
CLike a diode that conducts in only one direction
Show answer and explanation
Answer: A. Like a variable resistor whose resistance changes with gate voltage
In the linear region drain current is nearly proportional to VDS, and the slope—conductance μCox(W/L)(VGS − Vth)—changes with gate voltage, so the device acts as a gate-controlled variable resistor. A switch's on state lies in this region, and on-resistance falls as gate voltage rises. The property is also used in analog switches and variable-gain circuits. When VDS exceeds VGS − Vth, the device enters saturation.
With VGS > Vth and a channel present, once VDS ≥ VGS − Vth the gate-to-channel voltage at the drain end falls below Vth, the channel pinches off, and drain current becomes nearly independent of VDS—the saturation region. Amplifiers operate here, and the minimum VDS (= VGS − Vth) sets the lower limit of output swing. In short-channel devices velocity saturation makes current saturate at a lower VDS.
Including channel-length modulation, saturation drain current rises slightly with VDS as ID(1 + λVDS), and the inverse of this slope is the output resistance ro ≈ 1/(λID). λ is roughly inversely proportional to channel length, so doubling L about doubles ro. The intrinsic gain gm·ro is only about 10–20 in highly scaled transistors, so cascodes and multistage designs are used to obtain enough gain.
FAQ
When is a MOSFET in saturation?
For an nMOSFET, a channel exists when VGS > Vth, and the device is in saturation when VDS ≥ VGS − Vth. With smaller VDS it is in the linear (triode) region.
Does saturation mean the same thing for a MOSFET and a BJT?
No. MOSFET saturation is the region where current is nearly independent of VDS and is used for amplification. BJT saturation means the transistor is fully switched on, which corresponds more closely to the MOSFET's linear region.
How do you calculate MOSFET output resistance?
In saturation, estimate it from the channel-length modulation coefficient and drain current as ro ≈ 1/(λID). For λ = 0.05 /V and ID = 1 mA, ro is about 20 kΩ.