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Threshold Voltage
-0.1707
volts (V)
2φF (surface inversion) 0.6 V
Depletion / body charge term 0.1293 V

What is the MOSFET Threshold Voltage?

The threshold voltage (Vth) of a MOSFET is the gate-to-source voltage at which a conducting channel begins to form between the source and drain. Below this voltage the transistor is essentially off; above it, the device turns on. Vth is one of the most important parameters in transistor and analog/digital circuit design, setting bias points, logic levels, and leakage behavior.

Cross-section of an n-channel MOSFET showing gate, oxide, source, drain and substrate with an inversion channel
Cross-section of a MOSFET: the gate voltage forms an inversion channel between source and drain once Vth is reached.

The Formula

For an n-channel (NMOS) device built on a p-type substrate, the long-channel threshold voltage is:

$$V_{th} = V_{fb} + 2\phi_F + \frac{\sqrt{2\,\varepsilon_s\, q\, N_a \cdot 2\phi_F}}{C_{ox}}$$

where \(V_{fb}\) is the flat-band voltage, \(\phi_F\) is the Fermi potential, \(\varepsilon_s = \varepsilon_r \cdot \varepsilon_0\) is the silicon permittivity (\(\varepsilon_0 \approx 8.854\times10^{-14}\) F/cm), \(q = 1.602\times10^{-19}\) C is the electron charge, \(N_a\) is the substrate acceptor doping concentration (cm⁻³), and \(C_{ox}\) is the gate-oxide capacitance per unit area (F/cm²). The \(2\phi_F\) term sets the strong-inversion surface potential, and the square-root term is the voltage needed to support the depletion-region (body) charge.

Band-diagram style illustration of MOSFET threshold voltage as sum of flat-band voltage, Fermi term and depletion charge term
Vth is built from the flat-band voltage, twice the Fermi potential, and the depletion-charge term divided by oxide capacitance.

How to Use It

Enter the flat-band voltage, Fermi potential, substrate doping, oxide capacitance, and the relative permittivity of silicon (default 11.7). Keep all quantities in CGS-compatible units: doping in cm⁻³ and capacitance/permittivity per cm². The calculator returns \(V_{th}\) along with the surface-inversion term and the depletion-charge term so you can see how each contributes.

Worked Example

Take \(V_{fb} = -0.9\) V, \(\phi_F = 0.3\) V, \(N_a = 1\times10^{16}\) cm⁻³, \(C_{ox} = 3.45\times10^{-7}\) F/cm², \(\varepsilon_r = 11.7\). Then \(\varepsilon_s = 1.036\times10^{-12}\) F/cm. The inside of the root is $$2 \cdot \varepsilon_s \cdot q \cdot N_a \cdot 0.6 = 1.992\times10^{-15},$$ whose square root is \(\approx 4.463\times10^{-8}\) C/cm². Dividing by \(C_{ox}\) gives \(\approx 0.1294\) V. So $$V_{th} = -0.9 + 0.6 + 0.1294 \approx -0.171 \text{ V}.$$

FAQ

Does this work for PMOS? The same structure applies, but signs of \(V_{fb}\), \(\phi_F\), and doping type flip; this tool models the NMOS case.

What units should I use? Use cm-based units (cm⁻³ for doping, F/cm² for capacitance) so that the embedded constants \(q\) and \(\varepsilon_0\) are consistent.

Why is Vth sometimes negative? A negative \(V_{th}\) (depletion-mode device) can result when the flat-band voltage is strongly negative relative to the inversion and body terms.

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