A MOSFET does one of three things at any instant: it is off, it behaves like a voltage-controlled resistor, or it behaves like a voltage-controlled current source. Which of the three depends on two comparisons — gate-source voltage against threshold, and drain-source voltage against the overdrive. This calculator makes both comparisons, names the region, and computes the drain current with the small-signal parameters that follow from it.
It implements the classic square-law long-channel model, the one usually called Level 1. That model is the right tool for understanding what a MOSFET does and for a first-pass hand calculation, and Arb Digital has been explicit throughout this page about the places it stops being accurate — because a square-law prediction applied to a modern short-channel device can be wrong by a large factor, and knowing where the model breaks is more useful than the number itself.
What This MOSFET Calculator Does
Enter the gate-source and drain-source voltages, the threshold voltage, the process transconductance parameter and the width-to-length ratio, and the calculator returns the drain current with the operating region named underneath it. The grid gives the overdrive voltage, the small-signal transconductance, the output resistance produced by channel-length modulation, and the power the channel is dissipating.
All of those are the quantities you need to bias an amplifier stage or to check that a switch is genuinely fully on. The overdrive voltage is the one to watch: it sets the saturation boundary, it sets the transconductance, and it is the variable that circuit designers actually reason with, rather than VGS itself.
The device modelled here is an N-channel enhancement MOSFET, the common case. A P-channel device behaves identically with the signs of every voltage and the current reversed, so the same magnitudes apply if you work with absolute values. Depletion-mode devices, which conduct at zero gate voltage, are not covered by these expressions.
How to Use It
- Enter the gate-source voltage your bias network actually delivers. If a divider sets it, work it out first with the voltage divider calculator.
- Enter the drain-source voltage across the device. This decides whether the transistor is in triode or saturation and has almost no effect on current once it is saturated.
- Enter the threshold voltage from the data sheet, remembering that it is specified as a range and drifts with temperature.
- Set k′ and W/L, or set W/L to 1 and put the whole device transconductance parameter into k′ if that is how your figures are given.
- Read the region first, then the current. A current computed with the wrong region formula is meaningless, so the region label is the more important output.
The Formula: How It's Calculated
Everything starts with the overdrive voltage, VOV = VGS − VTH, and the device transconductance parameter k = k′ × W/L. There are then three cases.
Cutoff, when VOV ≤ 0: no channel forms and the ideal drain current is zero.
Triode, when 0 < VDS < VOV: the channel exists along the whole length and the device conducts like a resistor whose value the gate controls. ID = k (VOVVDS − VDS²÷2).
Saturation, when VDS ≥ VOV: the channel pinches off near the drain and the current becomes almost independent of VDS. ID = ½ k VOV² (1 + λVDS). The small-signal transconductance is gm = kVOV(1 + λVDS), equivalently 2ID÷VOV, and the output resistance is ro = 1 ÷ (λID) evaluated without the modulation term.
The physical basis for these expressions — how the gate field forms an inversion layer, why the current saturates when the channel pinches off, and how the characteristic curves follow — is set out in the HyperPhysics page on field effect transistors, and developed in full in MIT OpenCourseWare's 6.012 Microelectronic Devices and Circuits.
A worked example matching the defaults here: k′ = 200 µA/V² and W/L = 10 give k = 2 mA/V². With VGS = 3 V and VTH = 1 V the overdrive is 2 V, and since VDS = 5 V exceeds it the device is saturated. The ideal current is ½ × 2 × 2² = 4.0 mA, raised to 4.4 mA by the (1 + 0.02 × 5) modulation factor. Transconductance is 4.4 mS, output resistance is 1 ÷ (0.02 × 0.004) = 12.5 kΩ, and the channel dissipates 4.4 mA × 5 V = 22 mW.
The Saturation Boundary Is the Overdrive, Not the Gate Voltage
The most persistent error in MOSFET hand analysis is testing VDS against VGS instead of against VOV. With a 3 V gate, a 1 V threshold and 2.5 V across the device, testing against VGS says triode; testing correctly against the 2 V overdrive says saturation. The two formulas give different currents, and only one of them applies.
The reason the boundary sits at VOV is physical. The channel exists where the local gate-to-channel voltage exceeds the threshold. At the source end that voltage is VGS; at the drain end it is VGS − VDS. When VDS reaches VOV, the drain end of the channel is exactly at threshold and the inversion layer disappears there. Push VDS further and the pinch-off point moves back toward the source, but the voltage across the conducting part of the channel stays put — which is exactly why the current stops rising.
Why a Switch Wants Triode and an Amplifier Wants Saturation
These are opposite design goals using the same device. A power switch is driven with a large gate voltage so the overdrive is far above the drain voltage, putting it deep in triode where it behaves as a small resistance. In that region the useful parameter is RDS(on), and conduction loss is I²R. Data sheets quote RDS(on) at a specific gate voltage for exactly this reason, and running a switch at a lower gate drive than the one specified raises the resistance sharply.
An amplifier stage wants the opposite: saturation, where drain current is set by the gate and barely responds to the drain. That is what makes voltage gain possible, since the intrinsic gain of a stage is gmro and ro is large only in saturation. The bias point is chosen to keep VDS comfortably above VOV across the whole signal swing; letting a signal peak push the device into triode is what clips the output. Once you have a bias current and a load resistance, the stage gain calculation continues in the op-amp gain calculator for the feedback case, and the resistor network in the resistor combination calculator.
Where the Square-Law Model Stops Being True
This model was built for long-channel devices and it describes them well. Modern short-channel MOSFETs violate several of its assumptions, and the errors go in a consistent direction.
Velocity saturation is the big one. In a short channel the lateral electric field is strong enough that carrier velocity stops increasing with field. Drain current then rises roughly linearly with overdrive rather than as its square, so the square-law formula overestimates current at high overdrive, sometimes by a factor of two or more. Transconductance becomes roughly constant instead of proportional to overdrive.
Subthreshold conduction is the other. Below threshold the ideal model says zero current; the real device conducts an exponentially decaying leakage that is negligible for one transistor and dominant for a billion of them. That leakage is why standby power is a first-order design constraint in modern chips and why the model's clean "cutoff" is a simplification.
The body effect raises the threshold when the source is not tied to the bulk, which happens constantly in stacked circuits and source followers. Temperature pulls in two directions at once: threshold falls as the device warms, which raises current, while mobility falls, which lowers it. At high current the mobility term wins, so RDS(on) rises with temperature — the property that lets power MOSFETs share current in parallel without thermal runaway, and the reason real dissipation calculations use the hot value rather than the 25 °C one. The power dissipation calculator and the Joule heating calculator handle that side of the problem.
Reading a Data Sheet Against This Model
Data sheets rarely publish k′ or W/L, because they describe a packaged part rather than a process. What they give instead is a transfer characteristic — drain current against gate voltage at a fixed drain voltage — from which you can extract an effective k by fitting two points in saturation, and a threshold voltage specified over a range that is often two to one between minimum and maximum.
That range is not sloppiness; threshold varies between parts from the same reel and drifts with temperature at a few millivolts per kelvin. Any design that depends on a precise threshold is fragile. This is why analogue designers bias with a current source and let the gate voltage settle where it must, rather than setting a gate voltage and hoping for a current, and why switching circuits use a gate drive far above the worst-case threshold rather than one just clear of the typical value. Gate-drive current for the switching edge itself is a capacitance problem, covered by the capacitor charge time calculator.
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Browse All Free Tools Talk To Arb DigitalCommon Mistakes to Avoid
- Testing VDS against VGS to find the region — the saturation boundary is at the overdrive voltage, VGS − VTH.
- Applying the square law to a short-channel device at high overdrive — velocity saturation makes current rise roughly linearly, so the formula overestimates it.
- Designing around a typical threshold voltage — it is specified as a range and moves with temperature, so a design that needs a precise value will not hold across parts.
- Using the 25 °C RDS(on) for a loss calculation — on-resistance rises substantially at operating temperature, and the hot value is the one that matters.
- Assuming zero current below threshold — subthreshold leakage is exponential, not absent, and it dominates standby power in large designs.
Related Free Tools From Arb Digital
Set a gate bias with the voltage divider calculator, combine bias resistors with the resistor combination calculator, and check basic branch currents with the Ohm's law calculator. For thermal work use the power dissipation calculator and the Joule heating calculator, and for gate-drive timing the capacitor charge time calculator. The semiconductor physics underneath sits in the Fermi level calculator and the intrinsic carrier concentration calculator. Everything else is in the free online tools hub.
Frequently Asked Questions
Compare the drain-source voltage with the overdrive voltage, which is the gate-source voltage minus the threshold. Below the overdrive the device is in triode and behaves like a resistor; at or above it the device is saturated and behaves like a current source. Comparing against the gate voltage instead is the most common error in the topic.
It is the gate-source voltage minus the threshold voltage, and it is the variable that actually controls the device. It sets the saturation boundary, the drain current through its square, and the transconductance. Two devices with different thresholds but the same overdrive behave the same way.
Most likely velocity saturation. The square-law model assumes carrier velocity keeps rising with electric field, which stops being true in short channels. Real devices then show current rising roughly linearly with overdrive rather than as its square, so the square-law result runs high at large overdrive.
In saturation the pinch-off point moves slightly toward the source as drain voltage rises, shortening the effective channel and letting a little more current through. It is modelled by the factor one plus lambda times drain-source voltage, and it is what gives a saturated MOSFET a finite output resistance instead of an infinite one.
No. The ideal model says zero, but real devices carry a subthreshold current that falls exponentially as gate voltage drops. It is irrelevant for a single switching transistor and decisive in large integrated circuits, where the sum of that leakage across millions of devices sets the standby power.
Because carrier mobility falls with temperature faster than the threshold voltage drops. The net effect at high current is a positive temperature coefficient, which is useful: paralleled MOSFETs naturally share current, since whichever device carries most heats up and pushes current to the others.
Yes, if you work in magnitudes. A P-channel device obeys the same relations with every voltage and the current reversed in sign, so entering the absolute values of the gate-source, drain-source and threshold voltages gives the correct magnitudes. Depletion-mode devices are a different case and are not covered by these expressions.
This tool is provided for educational and design-estimate use. It applies the long-channel square-law model and does not account for velocity saturation, subthreshold conduction, the body effect, gate leakage, temperature coefficients, safe operating area or avalanche ratings. Component selection, thermal design and any mains-connected or high-power circuit should be verified against the manufacturer's data by a qualified engineer.
