MOSFET Output Characteristics Calculator
MOSFET Output Characteristics Calculator
Drain current against drain-source voltage for a MOSFET at two gate voltages, with a resistive load line drawn across them and the operating point where they cross. The square-law model, stated with its limits — a modern power MOSFET is not a square-law device and its datasheet curve is the truth — plus the dissipation at the operating point and whether it sits in the constant-current region, which is where MOSFETs die.
MOSFET output curves and load line
a device with a 2.0 V threshold and K = 0.5 A/V², driven at 4.0 V and at 3.0 V, switching a 10 Ω load on 12 V
Square law above pinch-off, triode below it
saturation (Vds ≥ Vov): Id = K·Vov²
triode (Vds < Vov): Id = K·(2VovVds − Vds²)
load line: Id = (Vdd − Vds) ÷ RL
small-Vds resistance = 1 ÷ (2·K·Vov)
- K
- the transconductance parameter, in A/V². It bundles mobility, oxide capacitance and the channel’s width-to-length ratio. Sedra and Smith write the same law as ½k′(W/L)Vov², so their k′(W/L) is twice this K
- Vov
- overdrive. It is also the knee voltage: the device leaves the ohmic region when Vds reaches it
- RL
- the load. The line runs from (0, Vdd/RL) to (Vdd, 0), and the operating point is wherever it meets the device curve
Worked example
a device with a 2.0 V threshold and K = 0.5 A/V², driven at 4.0 V and at 3.0 V, switching a 10 Ω load on 12 V
At 4.0 V the overdrive is 2 V, so the saturation current would be K·Vov² = 2 A — well above the 1.2 A the load line can supply, so the crossing must be in the triode region
There, K·RL·Vds² − (2K·RL·Vov + 1)·Vds + Vdd = 0, whose lower root is 682.3 mV; the current is (12 − 0.6823) ÷ 10 = 1.132 A
So the device dissipates 682.3 mV × 1.132 A = 772.2 mW, behaving like a 602.8 mΩ resistor. The small-signal figure 1 ÷ (2K·Vov) = 500 mΩ is the limit as Vds goes to zero
Drop the gate to 3.0 V and the overdrive halves to 1 V, so the saturation current falls by four to 500 mA — now BELOW what the load line can supply. The crossing moves into the constant-current region at 7 V, and the dissipation rises to 3.5 W
That is a factor of 4.5 for a 1.0 V change in gate drive, and it is the whole argument for driving a switching MOSFET hard and fast. The worst any device could dissipate on this load line is Vdd² ÷ 4RL = 3.6 W, halfway along it
Two meanings of linear, and which one kills devices
| Region | Textbook name | Power-datasheet name | What happens there |
|---|---|---|---|
| Vds below Vgs − Vth | triode, or linear region | ohmic region, fully on | the device behaves like a resistor. Dissipation is I²R. A hot spot has HIGHER resistance, so current moves away from it — negative feedback, and safe |
| Vds above Vgs − Vth | saturation, or active region | linear mode, or constant-current | the device holds voltage and current at the same time. Below the zero-temperature-coefficient gate voltage a hot spot conducts MORE, so it gets hotter — positive feedback, thermal instability, and the reason the SOA chart has a leg steeper than constant power |
| Gate below threshold | cutoff, or subthreshold | off, with leakage | the square law says zero; the real device conducts an exponentially small current that doubles every few degrees |
What the square law gets right, and where it stops
An output characteristic is drain current plotted against drain-source voltage, one curve per gate voltage. Each curve rises steeply from the origin, bends over at a knee, and then runs nearly flat: below the knee the channel is a resistor whose value the gate sets, and above it the channel is pinched off at the drain end and the current stops caring about the drain voltage. The knee sits at Vds = Vgs − Vth, which is also the overdrive, and the flat current is K times the overdrive squared. Those two expressions and the requirement that they meet smoothly are the entire square-law model.
The load line turns two curves into one answer. Put a resistor between the supply and the drain and the circuit constrains Vds and Id to a straight line from (0, Vdd/RL) to (Vdd, 0). The device constrains them to its own curve. Both must hold, so the operating point is where they cross — and that is why this page draws two device curves and the load line rather than three device curves. Without the load line there is no operating point to find; with it, the picture tells you the whole story, including what happens when the gate drive changes.
Where the model stops being true. A real power MOSFET is not a square-law device. Short-channel effects — velocity saturation above all — make the transfer characteristic closer to linear than quadratic over most of the useful range, so a fit that matches one datasheet point will be wrong at another. The flat region is not flat either: channel-length modulation gives it a real slope, and the output conductance matters if you are using the device as an amplifier. The threshold falls by a few millivolts per degree, the mobility falls with temperature too, and the two cancel at the zero-temperature-coefficient point — which is why a device operated above that point is thermally stable and one operated below it is not. Use this page to understand the shape and to size things roughly. For the actual number, use the datasheet’s own output and transfer curves, and its SOA chart at your pulse width.
The dissipation number is the one to look at. A MOSFET used as a switch spends its life deep in the triode region, where the dissipation is I²·RDS(on) and a hot spot pushes current away from itself. A MOSFET held part-way on sits in the constant-current region with both a large voltage and a large current, and there a hot spot pulls more current towards itself. That positive feedback is thermal instability, and it is why a power MOSFET’s SOA chart has a leg steeper than the constant-power hyperbola. Modern low-RDS(on) parts are markedly worse at it than the older, slower ones they replaced. The page reports the dissipation at the operating point and says which region it is in. For the loss of a device that is switching rather than sitting still, use the MOSFET loss calculator; for the path it takes through this same plane during a transition, the switching load locus calculator; and for getting the heat out, the heatsink calculator.
Frequently asked questions
What value of K should I use?
If the datasheet gives a transfer curve, pick a point on it and use K = Id ÷ (Vgs − Vth)². If it only gives RDS(on) at a stated Vgs, use K ≈ 1 ÷ (2·RDS(on)·(Vgs − Vth)), because the on-resistance in the ohmic region is 1 ÷ (2K·Vov). The two will not agree well on a modern power part, and that disagreement is the square-law model’s error, not a mistake in your arithmetic.
Why does the same device dissipate ten times more with a slightly lower gate voltage?
Because it changes region. With enough overdrive the saturation current is far above what the load line can deliver, so the crossing is in the ohmic region at a low voltage. Drop the overdrive and the saturation current falls as the square; once it drops below the load line’s current the crossing jumps into the constant-current region, where the device holds most of the supply voltage. That is why a gate driver’s job is to get through the middle fast.
Is the linear region dangerous or safe?
It depends whose word you are using. The textbook linear region is the ohmic one, where a switch belongs and which is safe. Power-device linear mode is the constant-current region, where the device holds voltage and current together and thermal instability lives, and which is dangerous. This page says triode and saturation to avoid the collision, and flags the constant-current case explicitly.
Why is my operating point not where the datasheet curve says?
Threshold voltage spread is the usual reason — the same part number is often specified over a 2:1 range — followed by temperature, which lowers the threshold by a few millivolts per degree and lowers the mobility as well. If you are anywhere near the constant-current region, both of those move the answer a long way.
Can I use this for a load that is not a resistor?
Only as a first look. A resistive load gives a straight load line; an inductive load gives a trajectory that leaves the straight line entirely during switching, which is what the load locus page is about. A constant-current load is a horizontal line and a lamp is a curve, because its resistance rises with temperature.
Related calculators
References
- Sedra A S, Smith K C. Microelectronic Circuits, 7th ed. Oxford University Press, 2015. The MOSFET chapter: the i-v characteristics, the triode and saturation expressions, the knee at Vov, and the caveat that short-channel devices depart from the square law. Edition verified by search; the chapter number is not quoted because it was not.
- Infineon Technologies. Linear Mode Operation and Safe Operating Diagram of Power-MOSFETs, application note AP99007, V1.1, May 2017 (J. Schoiswohl). The source for the power-device meaning of linear mode, the zero-temperature-coefficient point, and why thermal instability gives the SOA a leg steeper than constant power.
- Spirito P, Breglio G and colleagues. Thermal instabilities in high current power MOS devices: experimental evidence, electro-thermal simulations and analytical modeling, 2002. The primary-literature account of the mechanism behind the steep leg of a power MOSFET’s SOA. Locator unverified: the title and lead authors were confirmed by search, the proceedings volume and page numbers were not. The same mechanism, with the same conclusion, is set out in the Infineon note above, which was fetched and checked.