Bands, doping and the p-n junction

This chapter is optional. You know a MOSFET as a switch with a threshold voltage. Here we derive that threshold from the electrostatics of the gate stack:

and show what sets each term: gate work function and oxide charge, body doping through the Fermi level, and the depletion charge the gate has to support. Along the way we need carrier statistics (how doping positions the Fermi level), the p-n junction (which isolates every source and drain), and the MOS capacitor from accumulation to inversion.⁴

The treatment is the classic long-channel one. It is the reference the I-V curve and Shrinking chapters correct for short channels, quantum confinement and multi-gate geometries. Three numbers to carry through: silicon’s band gap of 1.12 eV, an intrinsic carrier density of about $10^{10}\,\mathrm{cm^{-3}}$, and the thermal voltage $kT/q \approx 25.9\,\mathrm{mV}$ at 300 K.¹

The occupancy of a state at energy $E$ is the Fermi function

More than a few $kT$ above $E_{\mathrm{F}}$ it reduces to a Boltzmann exponential, and integrating it against the density of states gives the carrier densities in compact form:¹

• $$\begin{aligned} n &= N_{\mathrm{c}}\, e^{-(E_{\mathrm{c}} - E_{\mathrm{F}})/kT} \\ p &= N_{\mathrm{v}}\, e^{-(E_{\mathrm{F}} - E_{\mathrm{v}})/kT} \end{aligned}$$
  with effective densities of states $N_{\mathrm{c}} = 2.8 \times 10^{19}$ and $N_{\mathrm{v}} = 1.04 \times 10^{19}\,\mathrm{cm^{-3}}$ for silicon at 300 K.
• Multiplying them removes $E_{\mathrm{F}}$:
  $$np = n_{\mathrm{i}}^2 = N_{\mathrm{c}} N_{\mathrm{v}}\, e^{-E_{\mathrm{g}}/kT}$$
  This mass-action law holds at equilibrium whether or not dopants are present.
• The intrinsic level $E_{\mathrm{i}}$, where $n = p = n_{\mathrm{i}}$, sits at midgap plus $(kT/2)\ln(N_{\mathrm{v}}/N_{\mathrm{c}})$, about 13 meV below midgap in silicon. Most hand analysis, and this chapter’s simulation, treats it as midgap.

Hu uses $n_{\mathrm{i}} \approx 10^{10}\,\mathrm{cm^{-3}}$;¹ the modern measured value at 300 K is $9.65 \times 10^9\,\mathrm{cm^{-3}}$.⁹ The difference shifts a built-in potential by only about 2 mV, but the temperature dependence is large: $n_{\mathrm{i}} \propto T^{3/2} e^{-E_{\mathrm{g}}/2kT}$, so it roughly doubles every 9 °C near room temperature (from the formula with $E_{\mathrm{g}} = 1.12\,\mathrm{eV}$). Anything proportional to $n_{\mathrm{i}}^2$, such as junction leakage, quadruples over the same step.

Two details matter later. First, silicon’s gap is indirect, so it absorbs and emits light poorly compared with direct-gap materials like GaAs.³ Second, hole mobility is about a third of electron mobility (470 vs 1,400 cm²/V·s in lightly doped silicon), partly because of the larger hole effective mass.² Both are bulk numbers; in a MOSFET’s inversion layer the mobilities are lower, because of surface-roughness scattering.⁵

With full ionization, charge neutrality is $n + N_{\mathrm{a}} = p + N_{\mathrm{d}}$. Combined with $np = n_{\mathrm{i}}^2$ it gives the exact majority density

which reduces to $n \approx N_{\mathrm{d}} - N_{\mathrm{a}}$ whenever the net doping is well above $n_{\mathrm{i}}$. Counter-doping (compensation) is routine: wells, threshold-adjust implants and halos are all layered on top of each other, and only the net doping sets the carriers, while the total doping sets impurity scattering.¹²

The Fermi level follows from the Boltzmann form: $E_{\mathrm{F}} - E_{\mathrm{i}} = kT \ln(n/n_{\mathrm{i}})$ for n-type, and $E_{\mathrm{i}} - E_{\mathrm{F}} = kT \ln(p/n_{\mathrm{i}}) \equiv q\phi_{\mathrm{B}}$ for p-type. Each decade of doping moves $E_{\mathrm{F}}$ by $kT \ln 10 \approx 60\,\mathrm{meV}$. That logarithm is why doping is such a gentle knob on band positions and such a strong knob on carrier densities.

Where the simple model stops:

• Degenerate doping. The Boltzmann form is only valid when $E_{\mathrm{F}}$ is more than a few $kT$ inside the gap.¹ Since $N_{\mathrm{c}} = 2.8 \times 10^{19}\,\mathrm{cm^{-3}}$, donor levels of around $10^{19}$ and above, typical of source/drain regions, put $E_{\mathrm{F}}$ at or into the band. Full Fermi-Dirac statistics, incomplete ionization and band-gap narrowing then matter. The simulation stops at $10^{18}$ to stay inside the simple model.
• Temperature extremes. At cryogenic temperatures dopants freeze out ($n$ falls below $N_{\mathrm{d}}$); at high temperatures $n_{\mathrm{i}}$ catches up with the doping and the material turns effectively intrinsic.¹

In a real process, doping is the main knob for threshold flavors. SKY130’s low-$V_{\mathrm{T}}$ and high-$V_{\mathrm{T}}$ 1.8 V transistors have cross-sections identical to the standard devices except for the $V_{\mathrm{T}}$-adjust implants.¹³

Under the depletion approximation (neutral regions perfectly neutral, depletion region fully depleted, sharp edges at $-x_{\mathrm{p}}$ and $x_{\mathrm{n}}$), Poisson’s equation integrates to a triangular field and two parabolic potential segments.⁷ The results, for a junction with applied forward bias $V$:³

• $$\phi_{\mathrm{bi}} = \frac{kT}{q} \ln\frac{N_{\mathrm{a}} N_{\mathrm{d}}}{n_{\mathrm{i}}^2}$$
• $$\begin{aligned} W &= x_{\mathrm{n}} + x_{\mathrm{p}} \\ &= \sqrt{\frac{2\varepsilon_{\mathrm{s}}(\phi_{\mathrm{bi}} - V)}{q}\left(\frac{1}{N_{\mathrm{a}}} + \frac{1}{N_{\mathrm{d}}}\right)} \end{aligned}$$
  with $x_{\mathrm{n}}/x_{\mathrm{p}} = N_{\mathrm{a}}/N_{\mathrm{d}}$ from charge balance
• Peak field at the metallurgical junction
  $$E_{\mathrm{max}} = 2(\phi_{\mathrm{bi}} - V)/W$$
• Junction capacitance per area
  $$C = \varepsilon_{\mathrm{s}}/W$$

For the simulation’s junction ($N_{\mathrm{a}} = 10^{17}$, $N_{\mathrm{d}} = 10^{16}\,\mathrm{cm^{-3}}$, $\varepsilon_{\mathrm{r}} = 11.7$): $\phi_{\mathrm{bi}} = 0.774\,\mathrm{V}$, $W = 0.332\,\mu\mathrm{m}$ at zero bias with 91% of it (0.302 µm) on the n side. At 3 V reverse, $W = 0.733\,\mu\mathrm{m}$ and $E_{\mathrm{max}} \approx 103\,\mathrm{kV/cm}$.

Under bias the single Fermi level splits into quasi-Fermi levels $E_{\mathrm{Fn}}$ and $E_{\mathrm{Fp}}$, separated by $qV$ and flat across the depletion region. That gives the boundary condition $n(-x_{\mathrm{p}}) = n_{\mathrm{p0}}\, e^{qV/kT}$ for injected minority carriers, which then diffuse and recombine over a diffusion length $L = \sqrt{D\tau}$. Summing both sides gives the ideal diode law³

Three consequences worth keeping:

• Injection goes mostly into the lighter-doped side; in a one-sided junction one term of $I_0$ dominates.
• $I_{0} \propto n_{\mathrm{i}}^2$, so reverse leakage climbs steeply with temperature. Drain-to-body junction leakage is one of a MOSFET’s three off-state leakage paths, along with subthreshold conduction and gate tunneling.⁶
• The built-in potential can’t be measured at the terminals. The metal-semiconductor contacts carry their own contact potentials that cancel it, so a shorted diode at equilibrium carries no current.⁷

Breakdown sets the reverse limit: avalanche multiplication when the peak field reaches a critical value, or band-to-band tunneling in heavily doped junctions, which is the basis of Zener diodes.³

Take a p-type body with doping $N_{\mathrm{a}}$, oxide capacitance $C_{\mathrm{ox}} = \varepsilon_{\mathrm{ox}}/T_{\mathrm{ox}}$ and surface potential $\phi_{\mathrm{s}}$ (band bending, positive downward). Kirchhoff around the gate stack and Gauss at the interface give⁴

where $Q_{\mathrm{s}}$ is all the charge in the silicon per unit area. The threshold computation then goes in three steps:⁸

1. Band bending at threshold. With $n(0) = n_{\mathrm{i}}\, e^{q\phi(0)/kT}$, setting the surface electron density to $N_{\mathrm{a}}$ gives $\phi_{\mathrm{s}} = 2\phi_{\mathrm{B}}$, where $\phi_{\mathrm{B}} = (kT/q)\ln(N_{\mathrm{a}}/n_{\mathrm{i}})$.
2. Depletion charge. The depletion width at that bending is $W_{\mathrm{dmax}} = \sqrt{2\varepsilon_{\mathrm{s}} \cdot 2\phi_{\mathrm{B}}/(qN_{\mathrm{a}})}$, holding $Q_{\mathrm{dep}} = -qN_{\mathrm{a}} W_{\mathrm{dmax}} = -\sqrt{2qN_{\mathrm{a}}\varepsilon_{\mathrm{s}} \cdot 2\phi_{\mathrm{B}}}$. Electrons are still negligible.
3. Sum the drops.
   $$V_{\mathrm{T}} = V_{\mathrm{fb}} + 2\phi_{\mathrm{B}} + \frac{\sqrt{2qN_{\mathrm{a}}\varepsilon_{\mathrm{s}} \cdot 2\phi_{\mathrm{B}}}}{C_{\mathrm{ox}}}$$

The flat-band voltage is the work-function difference less any oxide charge, $V_{\mathrm{fb}} = \psi_{\mathrm{g}} - \psi_{\mathrm{s}} - Q_{\mathrm{ox}}/C_{\mathrm{ox}}$. For an $\mathrm{n^+}$ polysilicon gate $\psi_{\mathrm{g}}$ equals silicon’s electron affinity, 4.05 V, and $\psi_{\mathrm{s}} = 4.05\,\mathrm{V} + (E_{\mathrm{c}} - E_{\mathrm{F}})/q$, so $V_{\mathrm{fb}} = -(E_{\mathrm{c}} - E_{\mathrm{F}})/q$ in the body.⁴¹²

A worked case (the simulation’s default): $N_{\mathrm{a}} = 10^{18}\,\mathrm{cm^{-3}}$, $T_{\mathrm{ox}} = 4\,\mathrm{nm}$ $\mathrm{SiO_2}$, $E_{\mathrm{i}}$ at midgap.

Above threshold the depletion charge is pinned near its threshold value, so the charge-control relation $Q_{\mathrm{inv}} = -C_{\mathrm{ox}}(V_{\mathrm{g}} - V_{\mathrm{T}})$ follows.⁸ Two refinements apply to real devices:

• Body effect. A source-to-body reverse bias $V_{\mathrm{sb}}$ adds depletion charge coupled through $C_{\mathrm{dep}} = \varepsilon_{\mathrm{s}}/W_{\mathrm{dmax}}$, giving $V_{\mathrm{T}} \approx V_{\mathrm{T0}} + \alpha V_{\mathrm{sb}}$ with $\alpha \approx 3T_{\mathrm{oxe}}/W_{\mathrm{dmax}}$ (3 being roughly 11.7/3.9, the ratio of dielectric constants). Steep retrograde doping, a light surface layer over a heavy one, pins $W_{\mathrm{dmax}}$ and reduces $\alpha$.⁵
• Electrical oxide thickness. Gate depletion in a poly gate and the finite inversion-layer thickness each add about a third of their thickness: $T_{\mathrm{oxe}} = T_{\mathrm{ox}} + W_{\mathrm{dpoly}}/3 + T_{\mathrm{inv}}/3$. Metal gates remove the first term; the second remains.⁴⁶

The $2\phi_{\mathrm{B}}$ definition is a textbook landmark, not what a datasheet reports. In practice $V_{\mathrm{T}}$ is usually extracted as the gate voltage where the drain current reaches a fixed value such as $100\,\mathrm{nA} \times W/L$.⁶

Everything is analytic: Boltzmann statistics, full ionization, $E_{\mathrm{i}}$ at midgap, $n_{\mathrm{i}} = 10^{10}\,\mathrm{cm^{-3}}$, $\varepsilon_{\mathrm{r}}$ 11.7 (Si) and 3.9 ($\mathrm{SiO_2}$), 300 K. The junction uses the depletion approximation with $N_{\mathrm{a}} = 10^{17}$ and $N_{\mathrm{d}} = 10^{16}$. The MOS mode solves the exact one-dimensional charge equation for $\phi_{\mathrm{s}}$ (see Under the hood), so it is not built on the threshold formula it reports. Experiments:

• Compare the solved inversion charge with $C_{\mathrm{ox}}(V_{\mathrm{g}} - V_{\mathrm{T}})$. At the default and $V_{\mathrm{g}} = 2.5\,\mathrm{V}$ the model gives $Q_{\mathrm{inv}}/q \approx 9.3 \times 10^{12}\,\mathrm{cm^{-2}}$ against $1.04 \times 10^{13}$ from the formula: $\phi_{\mathrm{s}}$ keeps rising past $2\phi_{\mathrm{B}}$ by several $kT$, which the formula ignores.
• On the log plot, the slope below $V_{\mathrm{T}}$ is the subthreshold behavior: surface electron density rises by a decade per $m \times 60\,\mathrm{mV}$, where $m = 1 + C_{\mathrm{dep}}/C_{\mathrm{ox}}$. Thin the oxide and watch it steepen.
• Drop the body doping to $10^{16.5}$ with a thin oxide: $V_{\mathrm{T}}$ goes negative. An $\mathrm{n^+}$ gate on a lightly doped p body makes a depletion-mode (normally on) device, which is why the doping and the gate work function have to be chosen together.
• In the junction mode, check that the depletion width scales as $\sqrt{\phi_{\mathrm{bi}} - V}$ and that $x_{\mathrm{n}}/x_{\mathrm{p}}$ stays at $N_{\mathrm{a}}/N_{\mathrm{d}} = 10$.

The SKY130 documentation compares its model against e-test specs. For the 1.8 V NMOS (W/L = 7/8 µm) the extracted threshold is 0.538 V typical, 0.520 V fast and 0.557 V slow; the low-$V_{\mathrm{T}}$ device is 0.434 V typical. The two share a cross-section and differ only in the $V_{\mathrm{T}}$-adjust implant.¹³ The spread between corners (about ±20–25 mV) is smaller than the 100 mV gap between flavors, which is what makes multi-$V_{\mathrm{T}}$ libraries useful.

Computed with the simulation’s model (illustrative, uniform doping, 300 K):

Note the sensitivity: at $10^{18}\,\mathrm{cm^{-3}}$ the depletion term is the biggest lever on $V_{\mathrm{T}}$, and it scales linearly with $T_{\mathrm{ox}}$. That is why oxide-thickness control is a threshold-control problem as much as a capacitance one.

• Choosing $V_{\mathrm{T}}$. Subthreshold current goes as $10^{(V_{\mathrm{gs}} - V_{\mathrm{T}})/S}$ with swing $S = \eta \cdot 60\,\mathrm{mV/decade}$ at 300 K, $\eta \ge 1$ set by $C_{\mathrm{dep}}/C_{\mathrm{oxe}}$ (and interface states). Every 100 mV of $V_{\mathrm{T}}$ reduction at $S = 100\,\mathrm{mV/decade}$ costs a decade of $I_{\mathrm{off}}$. Lowering $S$, by a thinner oxide or a wider depletion layer, is the preferred way to get both a low $V_{\mathrm{T}}$ and a low $I_{\mathrm{off}}$.⁶
• Doping profile. A uniform body forces a trade between $W_{\mathrm{dmax}}$ (short-channel control wants it small) and $V_{\mathrm{T}}$, mobility and body effect (which want light doping at the surface). Steep retrograde profiles, a light surface layer over a heavily doped one, decouple them: the depletion layer is basically the thickness of the light layer.⁵
• Random dopant fluctuation, back of the envelope. At $N_{\mathrm{a}} = 10^{18}\,\mathrm{cm^{-3}}$, $W_{\mathrm{dmax}} \approx 35\,\mathrm{nm}$. Under a 30 nm × 30 nm gate that volume holds $10^{18} \times (3 \times 10^{-6}\,\mathrm{cm})^2 \times 3.5 \times 10^{-6}\,\mathrm{cm} \approx 32$ acceptors. Poisson statistics give $\sigma = \sqrt{32} \approx 5.6$, an 18% spread in $Q_{\mathrm{dep}}$. With $|Q_{\mathrm{dep}}|/C_{\mathrm{ox}} \approx 0.65\,\mathrm{V}$, that is on the order of 0.1 V of $V_{\mathrm{T}}$ spread (an overestimate, since dopants deep in the depletion region count less, but the scale is right). The general point: $\sigma(V_{\mathrm{T}})$ from RDF grows as gate area shrinks.⁶
• Undoped channels and work-function $V_{\mathrm{T}}$. FinFETs were conceived with undoped channels, relying on fin dimensions for gate control, which removes RDF. The cost is that $V_{\mathrm{T}}$ must come from the gate work function, which adds process complexity. Light channel doping is an option, at some cost in mobility and RDF: in IBM and GlobalFoundries test wafers at $6 \times 10^{17}\,\mathrm{cm^{-3}}$ about a third of measured variability came from RDF, the rest from fin height, gate length and work-function variation.¹⁶ Metal gates for NMOS and PMOS generally need different metals with work functions near those of $\mathrm{n^+}$ and $\mathrm{p^+}$ poly.⁶
• Oxide scaling. $\mathrm{SiO_2}$ tunneling leakage rises exponentially as it thins ($10^3\,\mathrm{A/cm^2}$ at 1.2 nm). $\mathrm{HfO_2}$ with an equivalent oxide thickness of 1 nm leaks orders of magnitude less, at the price of chemical reactions with the silicon, lower mobility and more oxide charge, which is why a thin $\mathrm{SiO_2}$ interlayer remains under the high-k film.⁶

Equilibrium as drift balancing diffusion

Every band diagram in this chapter rests on one statement: at equilibrium the electron and hole currents vanish separately, so in every region drift exactly cancels diffusion.⁷ With the Einstein relation, that forces the Boltzmann relations

with $\phi$ measured from the intrinsic level, and $np = n_{\mathrm{i}}^2$ everywhere. Under bias, as long as the oxide or the depletion region keeps current small, the same relations hold in quasi-equilibrium.⁸ The remaining equation is Poisson’s, $dE/dx = \rho/\varepsilon_{\mathrm{s}}$.

The depletion approximation and its limits

Substituting the Boltzmann densities into Poisson gives a nonlinear equation. The depletion approximation replaces it with a box: $\rho = -qN_{\mathrm{a}}$ or $+qN_{\mathrm{d}}$ inside the depletion region and zero outside, with sharp edges. The field is then piecewise linear, the potential piecewise parabolic, and charge neutrality plus continuity of $\phi$ fix $x_{\mathrm{n}}$ and $x_{\mathrm{p}}$.⁷ The edges are really smeared over a few Debye lengths, $L_{\mathrm{D}} = \sqrt{\varepsilon_{\mathrm{s}} kT/(q^2 N)}$, about 4 nm at $10^{18}\,\mathrm{cm^{-3}}$ and 40 nm at $10^{16}$ (from the formula). The approximation is good when $W \gg L_{\mathrm{D}}$, which fails near flat band and in weak accumulation; that is exactly where the exact solution below is needed.

The exact MOS charge equation (what the simulation solves)

For a uniformly doped p-type body, multiply Poisson by $d\phi/dx$ and integrate once from the bulk ($\phi = 0$, $E = 0$) to the surface. With $\beta = q/kT$ this gives the surface field, and by Gauss’s law the total silicon charge:

The first bracket is the hole and ionized-acceptor contribution and the second is electrons. Each regime appears as one term dominating:

• Accumulation ($\phi_{\mathrm{s}} < 0$): $e^{-\beta\phi}$ dominates and the hole charge grows as $e^{\beta|\phi|/2}$, so $\phi_{\mathrm{s}}$ barely moves past a few $kT$ negative.
• Depletion: $\beta\phi$ dominates, recovering $Q = -\sqrt{2qN_{\mathrm{a}}\varepsilon_{\mathrm{s}}\phi_{\mathrm{s}}}$, the depletion result.
• Inversion: the electron term catches up when $(n_{\mathrm{i}}/N_{\mathrm{a}})^2 e^{\beta\phi} \approx \beta\phi$, near $\phi_{\mathrm{s}} = 2\phi_{\mathrm{B}}$, and then grows as $e^{\beta\phi/2}$. That exponential is why $\phi_{\mathrm{s}}$ pins a few $kT$ above $2\phi_{\mathrm{B}}$ and why the depletion width saturates.

Since $V_{\mathrm{g}} = V_{\mathrm{fb}} + \phi_{\mathrm{s}} - Q_{\mathrm{s}}(\phi_{\mathrm{s}})/C_{\mathrm{ox}}$ is monotonic in $\phi_{\mathrm{s}}$, the simulation finds $\phi_{\mathrm{s}}$ for each gate voltage by bisection over −0.8 to 1.6 V (60 halvings, far below a microvolt). The inversion charge is the difference between $Q_{\mathrm{s}}$ with and without the electron term. Nothing in that loop uses the threshold formula; $V_{\mathrm{T}}$ is drawn on the plot only as the textbook landmark to compare against.

Below threshold

In weak inversion the electron sheet density is proportional to $e^{\beta\phi_{\mathrm{s}}}$, while $\phi_{\mathrm{s}}$ follows $V_{\mathrm{g}}$ through a capacitive divider: $d\phi_{\mathrm{s}}/dV_{\mathrm{g}} = C_{\mathrm{ox}}/(C_{\mathrm{ox}} + C_{\mathrm{dep}})$. So the current falls a decade for every $\eta \times 60\,\mathrm{mV}$ of gate voltage, with $\eta \approx 1 + C_{\mathrm{dep}}/C_{\mathrm{oxe}}$, larger when interface states add their own capacitance in parallel with $C_{\mathrm{dep}}$.⁶ The same 60 mV is the diode’s per-decade figure: both are $kT \ln 10/q$, the Boltzmann factor showing through. No conventional MOSFET at room temperature can switch more steeply than that.

How the parameters are measured

The MOS capacitor is also the main measurement structure. Its capacitance–voltage (C–V) curve, measured by adding a small AC signal to a DC gate sweep, is the usual way to determine oxide thickness, substrate doping, flat-band voltage and threshold voltage. In accumulation the structure is just $C_{\mathrm{ox}}$; in depletion it is $C_{\mathrm{ox}}$ in series with the depletion capacitance, falling as $W$ grows. In inversion a slow (quasi-static) sweep returns to $C_{\mathrm{ox}}$, while a high-frequency curve stays low because the inversion charge can’t follow the AC signal. A MOSFET’s source and drain junctions supply that charge quickly, which is why a transistor’s gate C–V follows the quasi-static curve at all frequencies.⁴

Assumptions the simulation makes

• Boltzmann statistics and full ionization, so doping is capped at $10^{18}\,\mathrm{cm^{-3}}$.¹
• $E_{\mathrm{i}}$ at midgap; $n_{\mathrm{i}} = 10^{10}\,\mathrm{cm^{-3}}$; $T = 300\,\mathrm{K}$.
• Uniform doping, one dimension, no oxide or interface charge, an ideal $\mathrm{n^+}$ gate with no depletion.
• No quantum confinement: a real inversion layer’s electrons sit effectively 1.5–3 nm below the interface, which adds $T_{\mathrm{inv}}/3$ to the electrical oxide thickness.⁴
• The junction’s quasi-Fermi levels are drawn flat on each side; diffusion and recombination of the injected carriers are only indicated by dots.