How this instrument works
A MOSFET switches on once the gate-source voltage clears the threshold, but how hard it conducts past that point is not a straight line. In saturation, drain current rises with the SQUARE of the overdrive voltage, Vgs − Vth — the portion of the gate voltage actually available to pull charge into the channel. That squared term comes from the gradual-channel model: the mobile charge at each point along the channel is proportional to the local overdrive, and integrating that charge against the drift velocity over the channel length produces a quadratic, not a linear, result.
The parameter k in this calculator folds together everything about the physical device that the square law does not otherwise capture: electron mobility in the channel, the gate oxide's capacitance per unit area, and the width-to-length ratio the layout gives it. Two transistors built from the same wafer with different channel widths will report different k values and different drain currents at an identical Vgs, even though the underlying square-law shape stays the same — which is exactly why textbooks split k into a process term and a geometry term.
This is the long-channel form fixed by the Shichman-Hodges model that most textbooks and SPICE's simplest level use, and it assumes the drain-source voltage is large enough to hold the transistor in saturation (Vds ≥ Vgs − Vth) with channel-length modulation ignored. Push a real, short-channel device hard enough and velocity saturation flattens the curve toward linear well before the square law predicts; treat this result as the clean textbook baseline, not a substitute for a datasheet at extreme bias.
- Enter the Transconductance parameter k in mA/V² — a fixed property of this transistor's fabrication and geometry, not something you tune per measurement.
- Enter Gate-source voltage, the voltage you are actually driving onto the gate relative to the source terminal.
- Enter Threshold voltage, the gate-source voltage below which the channel does not form and the device stays off.
- Read Drain current, the saturation current in mA. The result only holds while Vgs exceeds Vth and Vds is large enough to keep the device in saturation.
- To compare bias points, change only Vgs and watch how much faster current climbs than the voltage change itself.
Worked example — biasing a 0.5 mA/V² MOSFET 1.5 V past threshold
Take a transistor with k = 0.5 mA/V² and a threshold Vth = 1.5 V, driven with Vgs = 3 V at the gate. The overdrive voltage is Vgs − Vth = 1.5 V, so Id = 0.5 × 0.5 × 1.5² = 0.5 × 0.5 × 2.25 = 0.5625 mA. That is the current a designer would read off before setting the bias resistor or current-mirror ratio for an amplifier stage built around this device.
Now raise the gate to Vgs = 4.5 V, leaving k and Vth unchanged. The overdrive voltage doubles to 3.0 V, and Id = 0.5 × 0.5 × 3.0² = 2.25 mA — four times the original 0.5625 mA, not two. That fourfold jump from a twofold voltage change is the square law showing up directly in the numbers, and it is why small shifts in bias point move a MOSFET amplifier's operating current, and its gain, more than a linear intuition would predict.
Questions
Why does drain current depend on the square of the overdrive voltage instead of the voltage itself?
It falls out of the gradual-channel model: the mobile charge at any point along the channel is proportional to the local overdrive voltage, and drift current is that charge times velocity, integrated along the channel length. Carrying that integral through produces a (Vgs − Vth)² term rather than a linear one. A resistor's current is proportional to voltage; a saturated MOSFET's current is proportional to the square of overdrive voltage, which is the core difference between the two devices.
What is the difference between Vgs and the overdrive voltage?
Vgs is the total gate-source voltage you apply. Overdrive voltage, Vov = Vgs − Vth, is the portion of it actually available to pull charge into the channel once the threshold has been cleared. Only the overdrive enters the current equation — a device with Vth = 1.5 V and Vgs = 3 V behaves identically to one with Vth = 0.5 V and Vgs = 2 V, because both have a 1.5 V overdrive.
What happens if Vgs is below Vth?
The transistor is in cutoff and carries no drain current at all; the square-law formula does not apply because there is no conducting channel to model. This calculator's checks flag that condition directly, since Vgs must exceed Vth for the device to conduct in the first place. Raising Vgs above Vth reopens the channel and the square law takes over again.
What does the transconductance parameter k actually represent?
It is the device-specific constant that sets how strongly current responds to overdrive voltage: k = μn·Cox·(W/L), where μn is electron mobility in the channel, Cox is the gate oxide's capacitance per unit area, and W/L is the channel's width-to-length ratio. Fabrication process sets the first two factors; the circuit designer's layout sets the third, which is why widening a transistor is a direct way to raise its current at a fixed bias.
Is k the same thing as gm, the small-signal transconductance?
No, though they are related. gm is the slope of the Id-versus-Vgs curve at one bias point, gm = k·(Vgs − Vth) — it tells you how much the current changes for a small wiggle in gate voltage right there. k is the constant that shapes the whole square-law curve everywhere, not just at one point, so gm changes with bias even though k does not.
Why is it called saturation if the current keeps rising as Vgs increases?
Saturation describes the current's relationship with drain-source voltage, not gate voltage: once Vds climbs past Vgs − Vth, further increases in Vds barely move the current, and the channel is said to be pinched off and saturated. Current can still rise sharply with Vgs in this region — saturation just means the drain voltage has stopped being the limiting factor.