MOSFET Threshold Voltage: Physical Derivation and Temperature Dependence#
Author: Tyler Mesko
1. Setup#
Derive $V_{TH}$’s dependence on temperature $T$, asserting $V_{SB}=0$ initially. Assuming fixed $V_{GS}$ and $V_{SB}$, the gate structure amounts to a single reverse-biased PN junction under an applied gate voltage $V_{GS}$, in which the negative charge in the p-type substrate is facilitated by the depletion region.
We derive an expression for $V_{TH}$ from the physics of a reverse-biased PN junction.
Assumptions: constant doping densities $N_A$ (p-substrate) and $N_D$ (n-well), with $N_D \gg N_A$, such that the depletion region exists almost entirely in the p-substrate.
2. Built-In Potential#
$$\psi_{bi} = \frac{kT}{q}\ln\left(\frac{N_A N_D}{n_i^2}\right)$$Charge neutrality requires the total charge per unit area to be equal and opposite on the p- and n-sides, imposing:
$$N_A W_1 = N_D W_2$$where $W_1, W_2$ are the depletion widths on the p- and n-sides respectively.
3. Solving Poisson’s Equation for the Potential Profile#
We want the voltage drop across the p-type region first, so we find $E_x$ for $-W_1 \le x \le 0$.
Gauss’s Law: $\dfrac{dE_x}{dx} = \dfrac{\rho(x)}{\varepsilon_{si}}$, with $\rho(x) = -qN_A$ (ionized acceptors) on the p-side, and $\varepsilon_{si}$ the permittivity of silicon.
Relate $E_x$ to $V_x$: $E_x = -\dfrac{dV}{dx}$
Combining, Poisson’s equation:
$$\frac{d^2V}{dx^2} = -\frac{\rho(x)}{\varepsilon_{si}} = \frac{qN_A}{\varepsilon_{si}}$$This tells us the charge density $\rho$ is proportional to the spatial curvature of the voltage — we integrate twice to recover $V(x)$.
First integration → $E(x)$#
$$\frac{dV}{dx} = \frac{qN_A}{\varepsilon_{si}}x + C_1 \;\Rightarrow\; E_x = -\frac{qN_A}{\varepsilon_{si}}x - C_1$$Boundary condition: $E_x = 0$ at $x=-W_1$ (field vanishes outside the depletion region):
$$0 = -\frac{qN_A}{\varepsilon_{si}}(-W_1) - C_1 \;\Rightarrow\; C_1 = \frac{qN_A W_1}{\varepsilon_{si}}$$$$\boxed{E_x = -\frac{qN_A}{\varepsilon_{si}}(x+W_1)}$$Second integration → $V(x)$#
$$V(x) = \frac{qN_A}{\varepsilon_{si}}\left(\frac{x^2}{2}+W_1 x\right) + C_2$$Boundary condition: $V(x)=0$ at $x=-W_1$ (reference — arbitrary, chosen to make the math clean):
$$0 = \frac{qN_A}{\varepsilon_{si}}\left(\frac{W_1^2}{2}-W_1^2\right) + C_2 \;\Rightarrow\; C_2 = \frac{qN_A W_1^2}{2\varepsilon_{si}}$$$$\boxed{V(x) = \frac{qN_A}{2\varepsilon_{si}}(x+W_1)^2}$$At $x=0$, we defined $V(0) = V_1$ (voltage drop across the p-side):
$$\boxed{V_1 = \frac{qN_A W_1^2}{2\varepsilon_{si}}}$$By symmetry, for the n-side ($0 \le x \le W_2$):
$$\boxed{V_2 = \frac{qN_D W_2^2}{2\varepsilon_{si}}}$$4. Total Depletion Width#
The total voltage drop across the depletion region is the built-in potential:
$$\psi_{bi} = V_1+V_2 = \frac{qN_A W_1^2}{2\varepsilon_{si}} + \frac{qN_D W_2^2}{2\varepsilon_{si}}$$Solving (using $N_AW_1=N_DW_2$) for the depletion widths:
$$W_1 = \sqrt{\frac{2\varepsilon_{si}\psi_{bi}}{qN_A}\cdot\frac{N_D}{N_A+N_D}}, \qquad W_2 = \sqrt{\frac{2\varepsilon_{si}\psi_{bi}}{qN_D}\cdot\frac{N_A}{N_A+N_D}}$$$$W = W_1+W_2$$In MOSFETs, we impose $N_D \gg N_A$, so the depletion region exists almost entirely within the p-substrate — meaning $W_1$ represents the depletion width under the oxide layer, and the $\frac{N_D}{N_A+N_D}$ factor $\to 1$:
$$\boxed{X \equiv W_1 \approx \sqrt{\frac{2\varepsilon_{si}\psi_{bi}}{qN_A}}}$$5. Depletion Charge and the Inversion Condition#
Charge per unit area in the depletion region:
$$Q = qN_A X = qN_A\sqrt{\frac{2\varepsilon_{si}\psi_{bi}}{qN_A}} = \sqrt{2qN_A\varepsilon_{si}\psi_{bi}}$$Inversion occurs when the surface potential $\psi_s$ reaches the critical value $2\phi_F$ (twice the Fermi potential), at which point no further increase in $V_{GS}$ changes the depletion width $X$ — the depletion charge saturates. In inversion, the depletion charge density becomes:
$$Q_b = \sqrt{2qN_A\varepsilon_{si}(2\phi_F)}$$where the Fermi potential is:
$$\phi_F = \frac{kT}{q}\ln\left(\frac{N_A}{n_i}\right)$$(from the law of mass action, $n_i$ is the intrinsic carrier concentration).
If a source-body bias $V_{SB}$ is applied, the depletion charge density becomes:
$$\boxed{Q_b = \sqrt{2qN_A\varepsilon_{si}(2\phi_F+V_{SB})}}$$6. Assembling the Threshold Voltage#
Some positive charge density $Q_{ss}$ always exists at the Si–SiO$2$ interface due to crystal discontinuities, and a work-function difference $\phi{ms}$ exists between the gate metal and the silicon substrate. These voltages and charge densities (via $Q = C_{ox}V$) add together to form the threshold voltage $V_{TH}$ — the voltage required to produce an inversion layer:
$$\boxed{V_{TH} = \phi_{ms} + 2\phi_F + \frac{Q_b}{C_{ox}} - \frac{Q_{ss}}{C_{ox}}}$$Substituting $Q_b$:
$$\boxed{V_{TH} = \phi_{ms} + 2\phi_F + \frac{\sqrt{2qN_A\varepsilon_{si}(2\phi_F+V_{SB})}}{C_{ox}} - \frac{Q_{ss}}{C_{ox}}}$$(This is consistent with the companion Subthreshold MOSFET Operation document’s body-effect formula $V_{TH}=V_{TH0}+\gamma(\sqrt{2\phi_F+V_{SB}}-\sqrt{2\phi_F})$, with $\gamma = \sqrt{2q\varepsilon_{si}N_A}/C_{ox}$.)
7. Temperature Dependence of $V_{TH}$#
Differentiate $V_{TH}$ with respect to temperature, assuming $N_A$, $N_C$, $N_V$ (conduction/valence-band effective density of states), $\phi_{ms}$, $Q_{ss}$, and $C_{ox}$ are all temperature-invariant. The temperature dependence enters entirely through $\phi_F$ (and hence $Q_b$), via $n_i$.
Expanding $\phi_F(T)$#
Using $n_i = \sqrt{N_C N_V},e^{-E_g/2kT}$ (mass-action law):
$$\phi_F = \frac{kT}{q}\ln\left(\frac{N_A}{n_i}\right) = \frac{kT}{q}\left[\ln N_A - \frac{1}{2}\ln(N_C N_V) + \frac{E_g}{2kT}\right]$$$$\boxed{\phi_F = \frac{E_g}{2q} + \frac{kT}{q}\ln\left(\frac{N_A}{\sqrt{N_C N_V}}\right)}$$Differentiating with respect to $T$ (treating $E_g$ as approximately temperature-invariant relative to the explicit $kT$ term):
$$\boxed{\frac{d\phi_F}{dT} = \frac{k}{q}\ln\left(\frac{N_A}{\sqrt{N_C N_V}}\right)}$$Because $N_A$ is typically much smaller than the geometric mean of the conduction- and valence-band effective density of states, $\sqrt{N_C N_V}$, the argument of the logarithm is less than 1:
$$\boxed{\frac{d\phi_F}{dT} < 0}$$Propagating to $dV_{TH}/dT$#
From $V_{TH} = \phi_{ms}+2\phi_F+Q_b/C_{ox}-Q_{ss}/C_{ox}$, with $Q_b \propto \sqrt{\phi_F}$:
$$\frac{dV_{TH}}{dT} = 2\frac{d\phi_F}{dT} + \frac{1}{C_{ox}}\frac{dQ_b}{dT} = \left[2 + \frac{\gamma}{\sqrt{2\phi_F}}\right]\frac{d\phi_F}{dT}$$Since the bracketed term is positive and $d\phi_F/dT < 0$:
$$\boxed{\frac{dV_{TH}}{dT} < 0}$$Result: because $N_A \ll \sqrt{N_C N_V}$, the threshold voltage has a negative temperature coefficient — $V_{TH}$ decreases as temperature increases. This matches the well-known experimental behavior of MOSFET threshold voltage (typically on the order of $-1$ to $-4\ \text{mV}/^\circ\text{C}$).