0. Toolkit

The device lectures (2–5) move fast through a few tools they assume you already have. This chapter collects those tools in one place, so later chapters can say “integrate the charge to get the field” and you know exactly what that means.

After this chapter you can:

• work in the centimetre-based units of device physics and convert µm, nm and per-area quantities without slips;
• move between energies (eV) and voltages (V), and use the “60 mV per decade” rule;
• go from a charge profile to the electric field to the potential, in one dimension;
• read a band diagram: which way is up, what bending means, what a flat Fermi level says;
• tell capacitance $dQ/dV$ from the ratio $Q/V$Q/V, and a per-area (sheet) quantity from a per-volume one.

Everything in this chapter is a refresher the guide adds added. Lecture 2 itself teaches silicon, bands, carriers and doping; Chapter 2 covers that.

1. Units: centimetres everywhere

The idea. Device physicists measure lengths in centimetres, so concentrations come out per cubic centimetre and fields in volts per centimetre.

The lecturer says so at the start of Lecture 2: he writes “MKS” and then “cm” on the board, meaning SI units except that length is in cm (the old captions mishear “MKS” as “chaos”). Every formula in this guide uses these units.

Quantity	Symbol	Unit
doping, carrier concentration	$N_A, N_D, n, p$	$\mathrm{cm^{-3}}$
length, depletion width	$x, W$	$\mathrm{cm}$ (quote answers in µm or nm)
electric field	$\mathcal{E}$	$\mathrm{V/cm}$
permittivity	$\varepsilon = \kappa\varepsilon_0$	$\mathrm{F/cm}$
charge per area	$Q$	$\mathrm{C/cm^2}$
capacitance per area	$C_{ox}, C_j$	$\mathrm{F/cm^2}$

Conversions you will use constantly:

$$1\,\mu\mathrm{m} = 10^{-4}\,\mathrm{cm}, \qquad 1\,\mathrm{nm} = 10^{-7}\,\mathrm{cm},$$ $$1\,\mathrm{F/cm^2} = 10^{-8}\,\mathrm{F/\mu m^2} = 10^{7}\,\mathrm{fF/\mu m^2}.$$

Constants

Constant	Value used in this guide	Lecture’s rounded value
electron charge $q$	$1.602\times10^{-19}\,\mathrm{C}$
permittivity of free space $\varepsilon_0$	$8.854\times10^{-14}\,\mathrm{F/cm}$	$8.85\times10^{-14}\,\mathrm{F/cm}$
relative permittivity of silicon $\kappa_s$	$11.7$	$11.7$
relative permittivity of SiO₂ $\kappa_{ox}$	$3.9$	$3.9$
thermal energy $kT$ at 300 K	$0.02585\,\mathrm{eV}$	“26 meV”
thermal voltage $kT/q$ at 300 K	$0.02585\,\mathrm{V}$	“26 mV”
intrinsic concentration $n_i$ (Si, 300 K)	$10^{10}\,\mathrm{cm^{-3}}$	$10^{10}\,\mathrm{cm^{-3}}$
band gap $E_g$ (Si)	$1.1\,\mathrm{eV}$	$1.1\,\mathrm{eV}$

The lecturer writes $\kappa$ for relative permittivity (“dielectric constant”) and keeps $\varepsilon_0$ separate, so the permittivity of silicon is always written $\kappa_s\varepsilon_0$κs​ε0​. This guide does the same.

Worked example. What is $\kappa_s\varepsilon_0$κs​ε0​, and what is the capacitance per area of a 5 nm oxide?

1. Silicon: $\kappa_s\varepsilon_0 = 11.7 \times 8.854\times10^{-14} = 1.036\times10^{-12}\,\mathrm{F/cm}$10−12F/cm.
2. Oxide thickness in cm: $t_{ox} = 5\,\mathrm{nm} = 5\times10^{-7}\,\mathrm{cm}$10−7cm.
3. $C_{ox} = \kappa_{ox}\varepsilon_0/t_{ox} = (3.9 \times 8.854\times10^{-14})/(5\times10^{-7}) = 6.91\times10^{-7}\,\mathrm{F/cm^2}$10−7F/cm2.
4. Per square micron: $6.91\times10^{-7}\times10^{7} = 6.9\,\mathrm{fF/\mu m^2}$6.9fF/μm2. A gate of $1\,\mu\mathrm{m}\times0.1\,\mu\mathrm{m}$ therefore has about $0.7\,\mathrm{fF}$ of oxide capacitance.

Common mistakes.

• Leaving a thickness in nm while $\varepsilon_0$ is in F/cm. The answer is then off by $10^7$107.
• Forgetting $\varepsilon_0$ and using $\kappa$ alone as the permittivity.

Check yourself A depletion width comes out as 3.3×10⁻⁵. In what unit, and how many µm is it?

In cm, because every input was in cm units. $3.3\times10^{-5}\,\mathrm{cm} = 0.33\,\mu\mathrm{m}$ (divide by $10^{-4}$10−4).

2. Energy and voltage

The idea. An electron’s energy goes down when the electrostatic potential goes up, because its charge is negative.

Band diagrams plot electron energy in electron-volts (eV). Circuits use potential in volts (V). The bridge between them is

$$E = -q\psi .$$

So a region at higher potential $\psi$ sits lower on a band diagram. Moving an electron through a potential difference of 1 V changes its energy by 1 eV. Numerically, an energy in eV and the matching voltage in V are the same number, but they are different quantities. Keep the $q$ when you switch between them:

• $kT$ is an energy (26 meV at room temperature);
• $kT/q$ is a voltage (26 mV), called the thermal voltage;
• a band-diagram spacing such as $E_i - E_F$ becomes a voltage once you divide by $q$q, for example $\phi_F = (E_i - E_F)/q$ in Chapter 3.

Check yourself The n side of a junction sits at a higher potential than the p side. On the band diagram, which side's bands are lower?

The n side’s. Higher potential means lower electron energy ($E=-q\psi$E=−qψ), so the bands bend down going from p to n.

3. Exponentials, logarithms and “60 mV per decade”

The idea. Carrier concentrations depend exponentially on energy, so a small energy change produces a large concentration change, and logarithms undo that.

In Lecture 2 the carrier concentrations are written as

$$n = n_i\, e^{(E_F - E_i)/kT}, \qquad p = n_i\, e^{(E_i - E_F)/kT}.$$

To find where the Fermi level sits for a given concentration, take the natural log of both sides:

$$E_F - E_i = kT \ln\frac{n}{n_i}.$$

Facts worth memorising:

• $\ln(ab) = \ln a + \ln b$ and $\ln(a/b) = \ln a - \ln b$lnb. This is why a built-in voltage $\ln(N_AN_D/n_i^2)$ splits into a p-side part plus an n-side part.
• $\ln 10 = 2.303$2.303, so $\ln(10^k) = 2.303\,k$2.303k.
• $kT\ln 10 = 0.02585 \times 2.303 = 0.0595\,\mathrm{eV}$0.0595eV: each factor of 10 in concentration moves the Fermi level by about 60 meV. For voltages it is about 60 mV per decade.

Worked example. Silicon doped with $N_D = 10^{16}\,\mathrm{cm^{-3}}$1016cm−3. Where is $E_F$ relative to $E_i$Ei​?

1. Majority electrons: $n \approx N_D = 10^{16}\,\mathrm{cm^{-3}}$1016cm−3.
2. Ratio: $n/n_i = 10^{16}/10^{10} = 10^{6}$106, which is 6 decades.
3. Exactly: $E_F - E_i = 0.02585 \times \ln(10^6) = 0.02585 \times 13.82 = 0.357\,\mathrm{eV}$0.357eV.
4. With the rule of thumb: $6 \times 60\,\mathrm{meV} = 0.36\,\mathrm{eV}$0.36eV. Same answer.

Common mistakes. Using $\log_{10}$ where the formula has $\ln$ln. That makes the answer 2.3 times too small.

Check yourself Doping goes up by a factor of 100. How far does the Fermi level move?

Two decades, so about $2\times60 = 120\,\mathrm{meV}$ (exactly $kT\ln 100 = 0.119\,\mathrm{eV}$0.119eV), toward the band of the majority carriers.

4. One-dimensional electrostatics: charge → field → potential

The idea. Integrate the charge density once to get the electric field, and integrate the field (with a minus sign) to get the potential.

Every device in these lectures is treated as one-dimensional: things vary with depth $x$ only. Two equations then carry all the electrostatics.

Gauss's law in one dimensionadded

$$\frac{d\mathcal{E}}{dx} = \frac{\rho(x)}{\kappa\varepsilon_0}$$

The slope of the field equals the local charge density divided by the permittivity. Where there is no charge, the field is constant.

Holds when: a charge density ρ(x) in a material of uniform permittivity κε₀

Field and potentialadded

$$\mathcal{E} = -\frac{d\psi}{dx},$$$$\psi(b) - \psi(a) = -\int_a^b \mathcal{E}\,dx$$

The field points downhill in potential. The potential difference across a region is the (negative) area under the field curve.

Holds when: always (electrostatics)

Combining them gives Poisson’s equation, $d^2\psi/dx^2 = -\rho/(\kappa\varepsilon_0)$−ρ/(κε0​), but in practice you integrate one step at a time, as the lecturer does on the board.

The pattern to remember (the lecturer says it in Lecture 3 at 5:25): “I integrate a constant, I get a linear; I integrate a linear, I get a quadratic.”

Charge density $\rho$	Field $\mathcal{E}$	Potential $\psi$
zero	constant	linear
constant	linear	quadratic

Why neutral objects have no field outside. Start to the left of all the charge, where $\mathcal{E}=0$0. Integrating Gauss’s law across the whole object gives $\mathcal{E}_{\text{right}} = \frac{1}{\kappa\varepsilon_0}\int\rho\,dx$κε0​1​∫ρdx, the total charge per area divided by the permittivity. If the object is neutral, the field returns to zero on the far side. This is how the pn junction and the MOS capacitor are solved: the field is zero in the neutral regions on both ends.

Try it below. There are two slabs of fixed charge, negative on the left and positive on the right, the same arrangement as a pn junction. Make the two areas $N\times\text{width}$ equal and the field closes back to zero. Make them unequal and a field is left over on the right.

negative slab density

2.0×10¹⁶ cm⁻³

negative slab width a

0.10 µm

positive slab density

1.0×10¹⁶ cm⁻³

positive slab width b

0.20 µm

Net charge per area: 0.00 nC/cm². Neutral, so the field returns to zero on the right.

Takeaway. The field is a triangle when two neighbouring slabs balance. Its peak sits at the boundary between them, and the total potential step is the triangle’s area.

Worked example. A slab of negative charge $\rho = -qN$ with $N = 10^{16}\,\mathrm{cm^{-3}}$1016cm−3, $0.1\,\mu\mathrm{m}$ wide, in silicon. How much does the field change across it?

1. Integrate Gauss’s law across the slab: $\Delta\mathcal{E} = \rho\, a/(\kappa_s\varepsilon_0)$ρa/(κs​ε0​).
2. Numbers: $\rho = -1.602\times10^{-19}\times10^{16} = -1.602\times10^{-3}\,\mathrm{C/cm^3}$10−3C/cm3, and $a = 10^{-5}\,\mathrm{cm}$10−5cm.
3. $\Delta\mathcal{E} = (-1.602\times10^{-3})(10^{-5})/(1.036\times10^{-12}) = -1.55\times10^{4}\,\mathrm{V/cm}$104V/cm, about $-15.5\,\mathrm{kV/cm}$−15.5kV/cm.
4. If the field also starts at zero, the potential changes by the triangle’s area: $\tfrac12 \times 10^{-5} \times 1.55\times10^{4} = 0.077\,\mathrm{V}$0.077V.

Common mistakes.

• Dropping the minus sign in $\mathcal{E} = -d\psi/dx$−dψ/dx. Check signs at the end with physics instead: positive charge on the right and negative on the left make a field pointing left, which is negative.
• Thinking zero charge means zero field. A charge-free oxide between charged plates has a constant, nonzero field.

Check yourself A region has constant positive charge density. Is its field curve rising, falling or flat? Its potential curve?

The field rises linearly (slope $\rho/\kappa\varepsilon_0 > 0$0). The potential is a parabola. Where the field is negative the potential is increasing, because $d\psi/dx = -\mathcal{E}$−E.

5. Reading a band diagram

The idea. A band diagram plots allowed electron energies against position. Electrons roll down the conduction band, holes float up the valence band, and a flat Fermi level means equilibrium.

The lines you will see:

• $E_C$EC​, the conduction-band edge: the lowest energy of a free electron.
• $E_V$EV​, the valence-band edge: the highest energy of a free hole. $E_C - E_V = E_g = 1.1\,\mathrm{eV}$ in silicon.
• $E_i$Ei​, the intrinsic level: near mid-gap; where $E_F$ would sit in undoped silicon.
• $E_F$EF​, the Fermi level: the energy where a state is occupied with probability one half. Above $E_i$ means n-type, below means p-type.

Rules for reading one:

1. All three band lines move together. $E_C$EC​, $E_i$ and $E_V$ shift by $-q\psi(x)$−qψ(x), so their shape is the potential turned upside down.
2. Slope means field. A tilted band means an electric field ($\mathcal{E} = \frac{1}{q}\frac{dE_C}{dx}$E=q1​dxdEC​​). Curved bands mean the field is changing, and that means charge (Section 4).
3. A flat $E_F$ through the whole device means thermal equilibrium: no net current. When a bias is applied, $E_F$ splits into separate levels, offset by $q$ times the applied voltage. The lecturer calls these quasi-Fermi levels.
4. Distance between $E_F$ and $E_i$ gives the carrier concentration (Section 3). Where bands bend so that $E_i$ crosses below $E_F$EF​, a p-type region locally looks n-type. That is inversion, the key to the MOS transistor (Chapter 3).

Check yourself In a band diagram, E_C slopes downward from left to right. Which way do electrons drift, and which way does the field point?

Electrons roll downhill in energy, to the right. Since $\mathcal{E} = \frac1q\frac{dE_C}{dx} < 0$0, the field points to the left. Electrons drift against the field, which is consistent.

6. Capacitance is a slope, dQ/dV

The idea. Capacitance tells you how much charge moves when the voltage changes a little. It is a derivative, not a ratio.

For a parallel-plate capacitor $Q = CV$CV, so slope and ratio agree, and per unit area

$$C = \frac{\kappa\varepsilon_0}{d} \quad (\mathrm{F/cm^2}).$$

In semiconductor devices the charge is usually a nonlinear function of voltage. For example, a junction’s depletion charge grows like $\sqrt{V}$V​. The capacitance that matters for circuits is then the small-signal value at the operating point:

$$C = \left.\frac{dQ}{dV}\right|_{V_0}.$$

The ratio $Q/V$ gives a different, wrong number. Chapter 3 shows that a reverse-biased junction behaves exactly like a parallel-plate capacitor whose plate spacing is the depletion width $W$W. Since $W$ changes with bias, so does $C$C.

Check yourself A device's charge is Q = A√V. Is its capacitance at V larger or smaller than Q/V?

$dQ/dV = A/(2\sqrt V)$A/(2V​), which is half of $Q/V = A/\sqrt V$A/V​. Using the ratio would overestimate the capacitance by a factor of 2.

7. Volume versus sheet quantities

The idea. A concentration counts per cubic centimetre. A charge stored in a thin layer counts per square centimetre. You go from one to the other by multiplying by a thickness.

Kind	Example	Unit
volume	doping $N_A$NA​, electron concentration $n$	$\mathrm{cm^{-3}}$
sheet (per area)	depletion charge $Q = qN_AW$qNA​W, inversion charge, $C_{ox}$	$\mathrm{C/cm^2}$C/cm2, $\mathrm{F/cm^2}$

Worked example. A p region doped $N_A = 10^{17}\,\mathrm{cm^{-3}}$ is depleted to a depth of $0.1\,\mu\mathrm{m}$0.1μm. What charge per area does it hold?

1. $Q = qN_AW = 1.602\times10^{-19} \times 10^{17} \times 10^{-5}\,\mathrm{cm}$10−5cm.
2. $Q = 1.60\times10^{-7}\,\mathrm{C/cm^2}$10−7C/cm2, from $10^{12}$ acceptor ions per $\mathrm{cm^2}$cm2.

Common mistake. Setting a sheet charge equal to a volume concentration. “Surface electron concentration equals $N_A$NA​” (both in cm⁻³, Chapter 3) is a different statement from “inversion charge per area”, which is in C/cm².

Formulas to remember

• Units: cm-based; $1\,\mu\mathrm{m}=10^{-4}\,\mathrm{cm}$10−4cm, $1\,\mathrm{nm}=10^{-7}\,\mathrm{cm}$10−7cm; $\varepsilon_0 = 8.85\times10^{-14}\,\mathrm{F/cm}$10−14F/cm; $\kappa_s = 11.7$11.7, $\kappa_{ox} = 3.9$3.9.
• $kT = 26\,\mathrm{meV}$26meV, $kT/q = 26\,\mathrm{mV}$ at 300 K; $n_i = 10^{10}\,\mathrm{cm^{-3}}$1010cm−3; $E_g = 1.1\,\mathrm{eV}$1.1eV.
• $E = -q\psi$−qψ: higher potential means lower electron energy.
• $E_F - E_i = kT\ln(n/n_i)$kTln(n/ni​); about 60 meV per decade.
• $d\mathcal{E}/dx = \rho/\kappa\varepsilon_0$ρ/κε0​, $\mathcal{E} = -d\psi/dx$−dψ/dx; constant charge → linear field → quadratic potential.
• $C = dQ/dV$dQ/dV; parallel plate $C = \kappa\varepsilon_0/d$ per area.
• Sheet charge $Q = qNW$ (C/cm²).