part13_1

Chapter 16 — MOS Fundamentals (continued)

16.2 Electrostatics — Mostly Qualitative

Before we crunch numbers, we want pictures. The MOS capacitor (metal–oxide–semiconductor, “MOS-C”) is a sandwich, and its behaviour is easiest to grasp with two kinds of cartoon: the energy band diagram (what the electrons’ energy landscape looks like) and the block charge diagram (where the charge sits).

16.2.1 Visualization Aids

Energy Band Diagram

The energy band diagram is an indispensable aid for visualizing the internal status of the MOS structure under static bias. Our first task is to build the diagram for the ideal MOS structure in equilibrium (zero bias).

Figure 16.2 shows the band diagrams of the three components on their own, each terminated abruptly by a vertical line that represents a surface. The “ledge” at the top of the vertical line is the vacuum level $E_0$: the minimum energy an electron must have to escape the material completely.

• Metal: the energy gap between the vacuum level and the Fermi level is the metal work function ${\Phi }_M$.
• Semiconductor: the surface barrier is described by the electron affinity $\chi$, the energy difference between the vacuum level and the conduction band edge at the surface. We use $\chi$ rather than “vacuum level minus $E_F$” because $E_C-E_F$ is not constant in a semiconductor: it changes with doping and with band bending near the surface. $\chi$, by contrast, is invariant with band bending. The quantity $(E_C-E_F{)}_{FB}$ is the $E_C$–$E_F$ separation in the flat-band (field-free) part of the semiconductor.
• Insulator: modelled essentially as an intrinsic, wide-gap semiconductor whose surface barrier is again specified through an electron affinity ${\chi }_i$.

Intuition: Think of each material as a lake whose water level is the Fermi energy $E_F$, sitting in a basin whose rim is the vacuum level. The work function says “how far below the rim is the water”; the electron affinity says “how far below the rim is the bottom of the empty upper floor (the conduction band)”.

Forming the zero-bias diagram is a two-step conceptual process:

1. Bring the metal and semiconductor together to a distance $x_o$ apart (the eventual oxide thickness) and let the two-component system equilibrate. In equilibrium the two Fermi levels line up (Subsection 3.2.4). Also, because the ideal MOS-C is defined with ${\Phi }_M={\Phi }_S$, the vacuum levels of M and S are automatically aligned too. Hence there is no charge and no electric field anywhere in the metal–gap–semiconductor system.
2. Insert the insulator of thickness $x_o$ into the empty gap. Because the field in the gap is zero, the only effect is to lower slightly the barrier between M and S.

The result is Fig. 16.3: flat $E_C$, $E_i$, $E_V$ in the semiconductor, a flat $E_F$ throughout, and a rectangular insulator barrier between the metal and the semiconductor.

Block Charge Diagrams

A block charge diagram is the complement of the band diagram: it shows the approximate charge distribution. In the ideal structure at equilibrium there is no charge anywhere. When a bias is applied, charge appears in the metal and in the semiconductor near the metal–oxide and oxide–semiconductor interfaces. A sample is Fig. 16.4: a $+Q$ block on the metal side and a $-Q$ block on the semiconductor side of the oxide (position $x$ along the horizontal axis; M, oxide, S labelled).

No attempt is made to depict exact charge distributions. A squared-off (“block”) approximation is used, so magnitudes and spatial extents are only qualitative. Nevertheless, because the electric field vanishes deep inside both the metal and the semiconductor (idealization 5), Gauss’s law demands that the total charge in the structure is zero.

Rule: in a block charge diagram, the area of the positive charge blocks always equals the area of the negative charge blocks.

16.2.2 Effect of an Applied Bias

General Observations

Assume normal operating conditions: the back of the MOS-C is grounded and $V_G$ is the d.c. bias on the gate. With $V_G\ne 0$:

1. The semiconductor Fermi level is unaffected and stays level (independent of position). This follows from the assumption of zero current under all static biasing: the semiconductor is always in equilibrium, whatever the gate voltage.
2. As in a pn junction, the applied bias separates the Fermi energies at the two ends of the structure by $q{V}_G$:

E_F(\text{metal})-E_F(\text{semiconductor})=-qV_G \tag{16.1}

Intuition (the handles): the metal and semiconductor Fermi levels are “handles connected to the outside world”. The back contact is grounded, so the semiconductor handle is bolted in place. The gate handle is moved down if $V_G>0$ and up if $V_G<0$. Because the barrier heights are fixed quantities, moving the metal handle bends everything in between, like twisting a rubber doll out of shape.

Equivalently, $V_G\ne 0$ causes potential drops and band bending inside the structure:

• No band bending in the metal (equipotential region).
• In the oxide and semiconductor the bands slope upward (increasing $E$ going from gate toward back contact) when $V_G>0$, and downward when $V_G<0$.
• Poisson’s equation in the oxide (an ideal insulator: no carriers, no charge centers) gives $d{\mathcal{E}}_{ox}/dx=0$, so ${\mathcal{E}}_{ox}$ is constant. The oxide band edges $E_C$, $E_V$ are therefore straight lines.
• In the semiconductor the bending has a more complex shape, but by idealization 5 it must vanish before the back contact is reached.

Specific Biasing Regions

Take an n-type Si substrate first.

Accumulation ($V_G>0$). Positive $V_G$ lowers $E_F$ in the metal relative to the semiconductor and makes the bands slope upward in insulator and semiconductor (Fig. 16.5a). The electron concentration $n=n_i{e}^{(E_F-E_i)/kT}$ increases as one approaches the interface (because $E_i$ moves farther below $E_F$ there). A situation where the majority-carrier concentration is greater near the interface than in the bulk is called accumulation. From the charge view: $V_G>0$ puts positive charge on the gate; charge balance draws negative electrons to the interface (Fig. 16.5b).

Depletion (small $V_G<0$). A small negative bias raises $E_F$ in the metal and gives the bands a small downward slope (Fig. 16.5c). Majority electrons are depleted near the interface. In charge language, the negative gate repels electrons from the interface and exposes positively charged donor sites (Fig. 16.5d). Here both $n$ and $p$ at the interface are less than the background doping ($N_D$ or $N_A$), hence “depletion”.

Inversion (large negative $V_G$). As $V_G$ becomes more and more negative, the semiconductor surface bands bend upward more and more. The surface hole concentration $p_s$ climbs: from less than $n_i$ when $E_i(\text{surface})<E_F$; to $n_i$ when $E_i(\text{surface})=E_F$; to greater than $n_i$ when $E_i(\text{surface})$ exceeds $E_F$. Eventually (Fig. 16.5e,f) we reach

E_i(\text{surface})-E_i(\text{bulk})=2\,[E_F-E_i(\text{bulk})] \tag{16.2}

and then

p_s=n_i\,e^{[E_i(\text{surface})-E_F]/kT}=n_i\,e^{[E_F-E_i(\text{bulk})]/kT}=n_{bulk}=N_D \tag{16.3}

So at this special bias $V_G=V_T$ the surface hole concentration equals the bulk electron concentration: the surface is no longer “depleted”. For $V_G<V_T$, $p_s$ exceeds $n_{bulk}=N_D$, and the surface region has changed character from n-type to p-type. The condition in which the minority-carrier concentration at the surface exceeds the bulk majority-carrier concentration is inversion (Fig. 16.5g,h).

Intuition (a see-saw of populations): Imagine a crowded dance floor (n-type: full of electron-dancers). A negative gate is a bouncer that first pushes dancers away from the door (depletion). Shove hard enough and the few “hole-guests” gather at the door until they outnumber the original crowd (inversion). The door region now behaves like p-type material sitting on n-type material, a pn junction created purely by an electric field!

For an ideal p-type device the results are those of Fig. 16.6, with reversed polarities: flat band ($V_G=0$), accumulation ($V_G<0$; a block of holes at the interface), depletion ($V_G>0$ small; exposed acceptors) and inversion ($V_G>V_T>0$; an electron sheet plus a depletion block of exposed acceptors).

Summary. There are three physically distinct bias regions:

	Accumulation	Depletion	Inversion
ideal n-type	$V_G>0$	$V_T<V_G<0$	$V_G<V_T$
ideal p-type	$V_G<0$	$0<V_G<V_T$	$V_G>V_T$

Flat band ($V_G=0$, no band bending) marks the boundary between accumulation and depletion. The boundary $V_G=V_T$ is the depletion–inversion transition point, with Eq. (16.2) quantitatively specifying the onset of inversion for both n- and p-type devices.

Exercise 16.1

Problem. Construct line plots (with $V_G$ along the $x$-axis) that identify the voltage ranges for accumulation, depletion and inversion in ideal n- and p-type MOS devices.

Solution. The “plots” graphically restate the summary above. On a $V_G$ number line:

\text{n-type:}\quad & \underbrace{V_G<V_T}_{\text{INV}}\ \Big|\ \underbrace{V_T<V_G<0}_{\text{DEPL}}\ \Big|\ \underbrace{V_G>0}_{\text{ACC}}\\ \text{p-type:}\quad & \underbrace{V_G<0}_{\text{ACC}}\ \Big|\ \underbrace{0<V_G<V_T}_{\text{DEPL}}\ \Big|\ \underbrace{V_G>V_T}_{\text{INV}} \end{aligned}$$ The n-type axis reads (left to right) INV | $V_T$ | DEPL | $0$ | ACC; the p-type axis reads ACC | $0$ | DEPL | $V_T$ | INV. Note $V_T<0$ for n-type and $V_T>0$ for p-type. $$\boxed{\text{n: INV}<V_T<\text{DEPL}<0<\text{ACC};\qquad \text{p: ACC}<0<\text{DEPL}<V_T<\text{INV}}$$ --- ## 16.3 Electrostatics — Quantitative Formulation ### 16.3.1 Semiconductor Electrostatics #### Preparatory Considerations We now want analytic expressions for the charge density $\rho(x)$, electric field $\mathcal{E}(x)$ and electrostatic potential $\phi(x)$ inside an ideal MOS-C under static bias. The problem simplifies because: - The metal is an equipotential region. Charge near the metal–oxide interface sits only a few Ångstroms (1 Å $=10^{-8}$ cm) deep, so it is modelled as a **$\delta$-function of charge** at the M–O interface. Since the oxide holds no charge (idealization 3), the metal charge equals (in magnitude) the total charge in the semiconductor. - With no oxide charge, $\mathcal{E}_{ox}$ is constant and the oxide potential is linear in position. So solving the MOS-C essentially reduces to solving the semiconductor part. Starting from Poisson's equation we use the **depletion approximation** for a first-order closed form (closely paralleling the pn junction, Chapter 5). An exact solution exists (because the semiconductor is always in equilibrium regardless of bias); it is given in Appendix B. **Definitions.** $\phi(x)$ is the electrostatic potential at depth $x$ into the semiconductor, measured from the oxide–semiconductor interface (Fig. 16.7a). We use $\phi$ rather than $V$ to avoid confusion with externally applied voltages. The field $\mathcal{E}=-d\phi/dx$ vanishes deep in the substrate. By convention $\phi=0$ in the field-free bulk. The value of $\phi$ at $x=0$ is the **surface potential** $\phi_s$. Relation to the bands (Fig. 16.7b, consistent with Eq. 3.12): $$\phi(x)=\frac{1}{q}\left[E_i(\text{bulk})-E_i(x)\right] \tag{16.4}$$ $$\phi_s=\frac{1}{q}\left[E_i(\text{bulk})-E_i(\text{surface})\right] \tag{16.5}$$ and a key material parameter, the **Fermi potential**: $$\phi_F=\frac{1}{q}\left[E_i(\text{bulk})-E_F\right] \tag{16.6}$$ $\phi_F>0$ for p-type; $\phi_F<0$ for n-type; its magnitude is tied to the doping. For nondegenerate Si near room temperature (Chapter 2): $$p_{bulk}=n_i e^{[E_i(\text{bulk})-E_F]/kT}=N_A\quad(N_A\gg N_D) \tag{16.7a}$$ $$n_{bulk}=n_i e^{[E_F-E_i(\text{bulk})]/kT}=N_D\quad(N_D\gg N_A) \tag{16.7b}$$ Combining (16.6) and (16.7): $$\phi_F=+\frac{kT}{q}\ln\frac{N_A}{n_i}\quad\text{(p-type)} \tag{16.8a}$$ $$\phi_F=-\frac{kT}{q}\ln\frac{N_D}{n_i}\quad\text{(n-type)} \tag{16.8b}$$ > **Sign convention:** with the book's definitions, positive $\phi_s$ means the bands bend *downward* at the surface ($E_i$ at the surface is lower than in the bulk). In a p-type device positive $\phi_s$ therefore attracts electrons. Under flat band $\phi_s=0$. Substituting (16.5) and (16.6) into (16.2): $$\phi_s=2\phi_F\quad\text{at the depletion–inversion transition point} \tag{16.9}$$ For $\phi_F>0$ (p-type): $\phi_s<0$ is accumulation, $0<\phi_s<2\phi_F$ is depletion, and $\phi_s>2\phi_F$ is inversion. For an n-type semiconductor the inequalities are reversed (accumulation $\phi_s>0$, depletion $2\phi_F<\phi_s<0$, inversion $\phi_s<2\phi_F$). ### Exercise 16.2 **Problem.** (a) Construct line plots (with $\phi_s$ on the axis) identifying the surface-potential ranges for accumulation, depletion and inversion in ideal n- and p-type devices. (b) For each of the $(\phi_F,\phi_s)$ sets below (in units of $kT/q$), give the doping type and biasing condition, and draw the energy band and block charge diagrams. (i) $\phi_F=12,\ \phi_s=12$; (ii) $\phi_F=-9,\ \phi_s=3$; (iii) $\phi_F=-9,\ \phi_s=-18$; (iv) $\phi_F=15,\ \phi_s=36$; (v) $\phi_F=-15,\ \phi_s=0$. **Solution (a).** From (16.9) and the inequalities above: $$\begin{aligned} \text{p-type }(\phi_F>0):\quad & \underbrace{\phi_s<0}_{\text{ACC}}\ \Big|\ \underbrace{0<\phi_s<2\phi_F}_{\text{DEPL}}\ \Big|\ \underbrace{\phi_s>2\phi_F}_{\text{INV}}\\ \text{n-type }(\phi_F<0):\quad & \underbrace{\phi_s<2\phi_F}_{\text{INV}}\ \Big|\ \underbrace{2\phi_F<\phi_s<0}_{\text{DEPL}}\ \Big|\ \underbrace{\phi_s>0}_{\text{ACC}} \end{aligned}$$ **Solution (b).** The doping type follows from the sign of $\phi_F$. The condition follows from comparing $\phi_s$ with $0$ and $2\phi_F$: | Set | $\phi_F$ | $\phi_s$ | $2\phi_F$ | Doping | Condition | |---|---|---|---|---|---| | (i) | 12 | 12 | 24 | p | $0<12<24$: **Depletion** (middle, $\phi_s=\phi_F$) | | (ii) | $-9$ | 3 | $-18$ | n | $\phi_s>0$: **Accumulation** | | (iii) | $-9$ | $-18$ | $-18$ | n | $\phi_s=2\phi_F$: **Depletion/inversion transition** | | (iv) | 15 | 36 | 30 | p | $36>30$: **Inversion** | | (v) | $-15$ | 0 | $-30$ | n | $\phi_s=0$: **Flat band** | Diagrams: - **(i) p-type, depletion:** $E_F$ flat; $E_C,E_i,E_V$ bend downward toward the interface by $12\,kT$ (so $E_i$ is still above $E_F$ at the surface, i.e. the surface is still p-type but depleted of holes). Charge: $+Q$ on the metal; $-Q$ as a wide block of exposed acceptors. - **(ii) n-type, accumulation:** bands bend downward by only $3\,kT$ toward the interface, so $E_C$ moves closer to $E_F$ (more electrons). Charge: $+Q$ on metal; $-Q$ as a thin sheet of electrons at the interface. - **(iii) n-type, transition:** bands bend upward by $18\,kT$; at the surface $E_i$ lies as far *above* $E_F$ as it lies below it in the bulk, so $p_s=N_D$. Charge: $-Q$ on the metal; $+Q$ as a wide block of ionized donors. - **(iv) p-type, inversion:** bands bend downward by $36\,kT$, beyond $2\phi_F=30\,kT$. Charge: $+Q$ on metal; $-Q$ as a thin electron sheet at the interface plus a block of exposed acceptors of width about $W_T$. - **(v) n-type, flat band:** all bands flat, $E_F$ level across the structure, no charge anywhere. --- ### Delta-Depletion Solution The approximate closed-form solution based on the depletion approximation is divided into three segments matching the three bias regions. The exact-solution plots of Fig. 16.8 (a p-type device with $\phi_F=12\,kT/q$, $T=300$ K, $kT/q=0.0259$ V) guide the approximation. - **Fig. 16.8(a), accumulation** ($\phi_s=-6kT/q$): the accumulation-layer charge is squeezed into an extremely narrow region at the interface. - **Fig. 16.8(b), middle of depletion** ($\phi_s=\phi_F=12kT/q$): the depletion region reaches deep (a fraction of a micron). - **Fig. 16.8(c), onset of inversion** ($\phi_s=2\phi_F=24kT/q$): wider depletion region with charge density $-qN_A$. - **Fig. 16.8(d), deep inversion** ($\phi_s=2\phi_F+6kT/q=30kT/q$): a very thin, very dense electron layer at the interface (hundreds of $qN_A$) on top of a depletion region only slightly wider than in (c). (The $\rho$ plots use a linear scale with the $x$ axis pointing downward to match Fig. 16.6; dashed lines in (b)–(d) outline the depletion-approximation charge.) **Accumulation.** Since the accumulation charge sits in an extremely narrow region while the depleted region is far deeper, replace the accumulation charge by a $\delta$-function of equal charge at $x=0$. Then automatically $\mathcal{E}=0$ and $\phi=0$ for all $x>0$. This is somewhat inaccurate but acceptable to first order. **Inversion.** Likewise the inversion charge is confined to an extremely thin layer, so it is modelled by a $\delta$-function at the interface. In addition, comparing $\phi_s=\phi_F$, $2\phi_F$ and $2\phi_F+6kT/q$, the depletion width grows a lot up to $2\phi_F$ but only slightly afterwards. So in the delta-depletion model the $\delta$-function added in inversion exactly balances the extra gate charge; the depletion width, the field and the potential for $x>0$ stay frozen at their $\phi_s=2\phi_F$ values. The inversion solution = end-of-depletion solution + a $\delta$-function surface charge. > **Intuition (the pinned surface):** once inversion begins, electrons are so cheap to supply that any extra gate charge is answered by extra electrons in a razor-thin layer. It is like a reservoir with an overflow pipe: the water level ($\phi_s$) stops rising, and extra input spills into the inversion layer. **Depletion.** In the depletion approximation the true charge is replaced by a squared-off distribution ending abruptly at $x=W$. For p-type: $$\rho=q(p-n+N_D^+-N_A^-)=-qN_A\qquad(0\le x\le W) \tag{16.10}$$ Poisson's equation reduces to $$\frac{d\mathcal{E}}{dx}=\frac{\rho}{K_s\varepsilon_0}=-\frac{qN_A}{K_s\varepsilon_0}\qquad(0\le x\le W) \tag{16.11}$$ Integrating with $\mathcal{E}=0$ at $x=W$: $$\mathcal{E}(x)=-\frac{d\phi}{dx}=\frac{qN_A}{K_s\varepsilon_0}(W-x)\qquad(0\le x\le W) \tag{16.12}$$ A second integration with $\phi=0$ at $x=W$: $$\phi(x)=\frac{qN_A}{2K_s\varepsilon_0}(W-x)^2\qquad(0\le x\le W) \tag{16.13}$$ Applying $\phi(0)=\phi_s$ determines $W$: $$\phi_s=\frac{qN_AW^2}{2K_s\varepsilon_0} \tag{16.14}$$ $$W=\left[\frac{2K_s\varepsilon_0}{qN_A}\,\phi_s\right]^{1/2} \tag{16.15}$$ Equations (16.10), (16.12), (16.13), (16.15) constitute the depletion-bias solution. For an n-bulk device replace $N_A$ by $-N_D$ (and use the magnitude of $\phi_s$ inside the square root). **Maximum depletion width.** In the delta-depletion picture $W$ stops growing once the surface inverts ($\phi_s=2\phi_F$), so it reaches its maximum equilibrium value $$W_T=\left[\frac{2K_s\varepsilon_0}{qN_A}\,(2\phi_F)\right]^{1/2} \tag{16.16}$$ Fig. 16.9 plots $W_T$ against doping for Si at 300 K over the typical MOS range (log–log): it falls monotonically with doping, from several micrometres near $N_A\sim10^{14}\ \text{cm}^{-3}$ to a few hundredths of a micrometre near $10^{18}\ \text{cm}^{-3}$. > **Sanity check:** for $N_A=10^{16}\ \text{cm}^{-3}$, $\phi_F=0.0259\ln(10^{6})=0.358$ V and $W_T=\sqrt{2(11.8)(8.854\times10^{-14})(0.716)/[(1.6\times10^{-19})(10^{16})]}\approx0.31\ \mu\text{m}$. ### 16.3.2 Gate Voltage Relationship The results above are cast in terms of the surface potential $\phi_s$, an internal quantity. The knob we actually turn is $V_G$, so we need a relation between $V_G$ and $\phi_s$. The gate voltage in the ideal structure is dropped partly across the oxide and partly across the semiconductor: $$V_G=\Delta\phi_{semi}+\Delta\phi_{ox} \tag{16.17}$$ Since $\phi=0$ in the bulk, the semiconductor drop is just $$\Delta\phi_{semi}=\phi(x=0)=\phi_s \tag{16.18}$$ So we must express $\Delta\phi_{ox}$ in terms of $\phi_s$. In the ideal insulator $$\frac{d\mathcal{E}_{ox}}{dx}=0 \tag{16.19}$$ $$\mathcal{E}_{ox}=-\frac{d\phi_{ox}}{dx}=\text{constant} \tag{16.20}$$ so integrating across the oxide of thickness $x_o$, $$\Delta\phi_{ox}=\int_{-x_o}^{0}\mathcal{E}_{ox}\,dx=\mathcal{E}_{ox}x_o \tag{16.21}$$ Relating $\mathcal{E}_{ox}$ to the semiconductor field uses the boundary condition on normal displacement at an interface: $$(D_{semi}-D_{ox})\big|_{O\text{-}S}=Q_{O\text{-}S} \tag{16.22}$$ where $D=K\varepsilon_0\mathcal{E}$ and $Q_{O\text{-}S}$ is the interface charge per unit area. Since $Q_{O\text{-}S}=0$ in the ideal structure (idealization 3): $$D_{ox}=D_{semi}\big|_{x=0} \tag{16.23}$$ $$\mathcal{E}_{ox}=\frac{K_s}{K_o}\mathcal{E}_s \tag{16.24}$$ $$\Delta\phi_{ox}=\frac{K_s}{K_o}\mathcal{E}_sx_o \tag{16.25}$$ Here $K_s$ is the semiconductor dielectric constant, $K_o$ that of the oxide, and $\mathcal{E}_s$ the semiconductor field at the interface. Substituting (16.18) and (16.25) into (16.17): $$V_G=\phi_s+\frac{K_s}{K_o}\mathcal{E}_sx_o \tag{16.26}$$ with $\mathcal{E}_s$ a known (or readily computed) function of $\phi_s$. In the delta-depletion solution, (16.12) at $x=0$ combined with (16.15) gives $$\mathcal{E}_s=\left[\frac{2qN_A\phi_s}{K_s\varepsilon_0}\right]^{1/2} \tag{16.27}$$