1. Setup and Key Relations#
For an NMOS device, the basic (strong-inversion) drain current relation and overdrive voltage are:
$$I_D = K\left(\frac{W}{L}\right)\left(V_{OV}\right)^2, \qquad K = \mu_n C_{ox}, \qquad V_{OV} = V_{GS}-V_{TH}$$$V_{TH}$ itself depends on bandgap energy, the body-effect coefficient, and the Fermi potential — its full physical derivation (including temperature dependence) is treated separately. For this document, the relevant threshold-voltage parameters are:
$$\varepsilon_{si} = \varepsilon_r\varepsilon_0 \quad(\text{permittivity of silicon}), \qquad N_A \quad(\text{acceptor/p-substrate doping concentration}), \qquad n_i \quad(\text{intrinsic carrier concentration})$$$$\phi_F = \frac{kT}{q}\ln\left(\frac{N_A}{n_i}\right) \quad (\text{Fermi potential})$$2. Weak Inversion: $V_{GS} < V_{TH}$#
In weak inversion, drain current is the sum of drift and diffusion components — but diffusion dominates, arising from the gradient in minority carrier concentration across the channel.
In this regime, the NMOS structure behaves analogously to an NPN BJT: source $\approx$ emitter, p-substrate $\approx$ base, drain $\approx$ collector. The dominant contribution to drain current comes through this body transconductance path.
Deriving the drain current#
With a grounded substrate, increasing $V_{GS}$ increases the surface potential $\psi_s$, which decreases the reverse bias across the source-body (“B-E”) junction and exponentially increases the electron concentration in the p-substrate at the source (“emitter”) junction, $x=0$:
$$n_p(0) = n_{p0}\,e^{\,q\psi_s/kT}$$where $n_{p0}$ is the equilibrium electron concentration in the p-substrate (“base”).
At the drain (“collector”), $x=L$, the B-C junction is reverse-biased by $V_{DS}$, giving an offset:
$$n_p(L) = n_{p0}\,e^{\,q(\psi_s-V_{DS})/kT}$$The diffusion current density is $J_n = qD_n\dfrac{dn}{dx}$, where $D_n$ is the electron diffusion constant. Approximating a linear minority-carrier profile across the channel (short-base approximation, same as BJT base transport):
$$I_D = qAD_n\,\frac{n_p(0)-n_p(L)}{L}$$where $A$ is the effective cross-sectional area of the channel (channel width $\times$ effective channel depth). Substituting:
$$I_D = \frac{qAD_n\,n_{p0}}{L}\left[e^{q\psi_s/kT} - e^{q(\psi_s-V_{DS})/kT}\right] = \frac{qAD_n\,n_{p0}}{L}\,e^{q\psi_s/kT}\left(1-e^{-qV_{DS}/kT}\right)$$(Conventional current flows opposite to electron flow into the device, so the sign convention is flipped to reflect current flowing into the drain terminal — consistent with the expression above.)
Simplifying with the Einstein relation and mass-action law#
Using the Einstein relation, $D_n = \mu_n V_T$ (where $V_T = kT/q$ is the thermal voltage), and the mass-action law, $n_{p0} = n_i^2/N_A$:
$$I_D = \frac{qA\mu_n V_T}{L}\cdot\frac{n_i^2}{N_A}\,e^{\psi_s/V_T}\left(1-e^{-V_{DS}/V_T}\right)$$Define the subthreshold current parameter $I_0$ (a function only of process parameters):
$$I_0 = \frac{qA\mu_n n_i^2}{L\,N_A}$$giving the compact form:
$$\boxed{I_D = I_0\,e^{\psi_s/V_T}\left(1-e^{-V_{DS}/V_T}\right)}$$3. Relating Surface Potential $\psi_s$ to $V_{GS}$#
$\psi_s$ is approximately linear in $V_{GS}$ in weak inversion. Neglecting charge stored at the Si–oxide interface, $\psi_s$ is controlled by $V_{GS}$ through a capacitive divider between the oxide capacitance $C_{ox}$ and the depletion-region (“junction”) capacitance $C_{js}$:
$$\frac{d\psi_s}{dV_{GS}} = \frac{C_{ox}}{C_{ox}+C_{js}} \equiv \frac{1}{n}$$where $n$ is the subthreshold slope factor. Integrating:
$$\psi_s = \frac{V_{GS}}{n} + C$$Initial condition: define $V_{TH}$ as the value of $V_{GS}$ at which $\psi_s = 0$:
$$0 = \frac{V_{TH}}{n} + C \;\Rightarrow\; C = -\frac{V_{TH}}{n}$$$$\boxed{\psi_s = \frac{V_{GS}-V_{TH}}{n}}$$Final subthreshold current expression#
Substituting into the drain current expression:
$$\boxed{I_D = I_0\,e^{\,(V_{GS}-V_{TH})/nV_T}\left(1-e^{-V_{DS}/V_T}\right)}$$This is the standard weak-inversion (subthreshold) MOSFET current equation.
4. Body Transconductance Effect on $V_{TH}$#
Now consider the body effect: how the substrate bias $V_{SB}$ modifies $V_{TH}$.
Define the rate of change:
$$X \equiv \frac{dV_{TH}}{dV_{SB}} = \frac{\gamma}{2\sqrt{2\phi_F+V_{SB}}}$$Integrating:
$$\int dV_{TH} = \int \frac{\gamma}{2\sqrt{2\phi_F+V_{SB}}}\,dV_{SB}$$$$V_{TH} - V_{TH0} = \gamma\left(\sqrt{2\phi_F+V_{SB}} - \sqrt{2\phi_F}\right)$$$$\boxed{V_{TH} = V_{TH0} + \gamma\left(\sqrt{2\phi_F+V_{SB}}-\sqrt{2\phi_F}\right)}, \qquad \gamma = \frac{\sqrt{2q\varepsilon_{si}N_A}}{C_{ox}}$$($\gamma$ is the body-effect coefficient.)
Complete $I_D$–$V_{GS}$ relationship with body effect#
$$\boxed{I_D = I_0\,\exp\!\left[\frac{V_{GS}-V_{TH0}-\gamma\left(\sqrt{2\phi_F+V_{SB}}-\sqrt{2\phi_F}\right)}{nV_T}\right]\left(1-e^{-V_{DS}/V_T}\right)}$$5. Transconductance Efficiency: $g_m/I_D$#
This is the key figure of merit for analog IC design (gm/Id methodology).
Assert $V_{DS} \gg V_T$ so that $e^{-V_{DS}/V_T} \to 0$ (saturation-region approximation), and hold $V_{SB}$ fixed while differentiating with respect to $V_{GS}$:
$$I_D \approx I_0\,e^{(V_{GS}-V_{TH})/nV_T}$$$$\ln I_D = \ln I_0 + \frac{V_{GS}-V_{TH}}{nV_T}$$Differentiating both sides with respect to $V_{GS}$:
$$\frac{d(\ln I_D)}{dV_{GS}} = \frac{1}{I_D}\frac{dI_D}{dV_{GS}} = \frac{g_m}{I_D}$$(using the chain rule identity $\dfrac{d\ln f}{dx} = \dfrac{1}{f}\dfrac{df}{dx}$ on the left, and direct differentiation of the linear exponent on the right)
$$\boxed{\frac{g_m}{I_D} = \frac{1}{nV_T}}$$This is the maximum achievable $g_m/I_D$ — the transistor is at its most efficient in weak inversion, which is exactly why the subthreshold region is so valuable for low-power analog design.
Impedance looking into the source follows directly:
$$r_s = \frac{1}{g_m} = \frac{nV_T}{I_D}$$