How this instrument works
A MOSFET conducts once its gate-source voltage clears a threshold, Vth, and beyond that point the drain current in saturation follows a square law: Id = (k⁄2)(Vgs − Vth)². The square appears because channel charge grows in proportion to the overdrive voltage, Vgs − Vth, and so does the electric field driving that charge along the channel — multiply two quantities that both scale with overdrive and the result scales with its square. Undoing that square is why this calculator's formula carries a square root: Vth = Vgs − √(2Id ⁄ k).
This is not a design equation, it is a verification equation. A device physicist sizing a new transistor picks Vth as a target and derives everything else from it; a test engineer holding a finished part does the opposite. Bias the gate to a known Vgs, measure the current that actually flows, and solve backward for the Vth the silicon delivers, which may drift from the datasheet's typical figure with temperature, process lot, or age. The transconductance parameter k = μnCox(W ⁄ L) has to be known first, usually pinned down from a separate curve fit or the manufacturer's SPICE model, before this equation can isolate Vth alone.
The result only means what it claims while the transistor sits in saturation, that is while Vds is at least Vgs − Vth; below that, in the linear region, drain current depends on Vds too and this formula no longer applies. The square-law model is itself an approximation: real datasheets usually define Vgs(th) as the gate voltage that produces one specific, deliberately small current, often 250 µA or 1 mA, not by this general square law, and short-channel devices depart from the square law further still because of velocity saturation and mobility degradation.
- Enter the current you measured on the bench into Measured drain current, in mA.
- Enter the device's known transconductance parameter into Transconductance parameter k, in mA/V² — from a datasheet curve fit or prior characterization, not a guess.
- Enter the gate-source voltage the current was measured at into Gate-source voltage at measurement, in volts.
- Read Threshold voltage — the calculator solves Vth = Vgs − √(2Id ⁄ k) for you.
- If Vth comes out negative or larger than Vgs, check that the device was actually biased in saturation before trusting the figure.
Worked example — backing out Vth from a 2 mA bench reading
A bench technician biases a small-signal MOSFET at Vgs = 3 V and measures 2 mA of drain current. The device's transconductance parameter, already pinned down from an earlier curve fit, is k = 0.5 mA/V². These three figures go straight into Measured drain current, Transconductance parameter k, and Gate-source voltage at measurement.
The saturation equation runs backward: 2Id ⁄ k = 2 × 2 ⁄ 0.5 = 8, and √8 = 2.828427. So Vth = 3 − 2.828427 = 0.171573 V. The calculator reports 0.171573 V to six figures, and that is the actual threshold voltage this particular part is exhibiting on the bench right now, not the datasheet's typical value, which was measured on a different sample at a different test current.
Two more readings on the same setup show why the direction of each term matters. Push the same Vgs = 3 V to a higher measured current, 8 mA instead of 2 mA, with the same k, and Vth drops to about −2.66 V, because more current at an unchanged gate bias only happens if the device turns on more easily than the first reading implied. Double k instead, to 1.0 mA/V² at the original 2 mA, and Vth rises to exactly 1.0 V: a more sensitive device needs less overdrive to pass the same current, so more of the 3 V gate bias must have gone toward simply clearing threshold.
Questions
Why does the formula use a square root instead of just subtracting Vgs and Vth?
Because the relationship it is undoing is not linear — saturation drain current is proportional to the square of the overdrive voltage, Id = (k⁄2)(Vgs − Vth)². Solving for the overdrive first, then Vth, means dividing out that square with a square root, √(2Id ⁄ k). Skip the root and the answer has the wrong units and roughly the wrong size.
Why does a higher measured current give a lower calculated Vth?
Because the two are inversely linked at a fixed gate bias: more current at the same Vgs can only occur if less of that Vgs was needed to clear threshold, meaning Vth itself is smaller. Raising Id from 2 mA to 8 mA at Vgs = 3 V and k = 0.5 mA/V² pushes Vth from about 0.17 V down to about −2.66 V, confirming the device turns on more readily than the first reading implied.
Where does the transconductance parameter k come from?
It is measured separately, not guessed. k = μnCox(W ⁄ L) combines electron mobility, gate oxide capacitance per area, and the transistor's width-to-length ratio, values a fabrication process fixes and a SPICE model file usually reports directly. In practice, k is pinned down by fitting the square-law equation to several Id–Vgs points before this calculator's single backward step for Vth becomes meaningful.
Is the Vth this calculator returns the same number printed on a datasheet?
Not necessarily. A datasheet's Vgs(th) is defined at one specific, deliberately small drain current, often 250 µA or 1 mA, chosen so the figure is repeatable across production lots. This calculator returns whatever Vth makes the square-law equation consistent with your own Id, Vgs, and k, which can differ from the datasheet spec if your test current, temperature, or sample differs from theirs.
Does this equation still work if the MOSFET is in the linear, or triode, region?
No. The square-law relationship this formula inverts, Id = (k⁄2)(Vgs − Vth)², only holds once Vds is at least Vgs − Vth, the saturation condition. Below that, in the linear region, drain current depends on Vds as well, and solving with the saturation formula returns a number that does not correspond to the physical threshold.
Can this be used for a PMOS device, not just NMOS?
Yes, with sign conventions handled by how the numbers are entered. PMOS conducts once Vgs falls below a negative Vth, so measured current, Vgs, and the resulting Vth are all treated as negative relative to source; enter the signed values consistently and the same Vth = Vgs − √(2Id ⁄ k) relationship holds, since it is derived from the identical square-law current equation.