Semiconductor Fundamentals

Six worked problems on intrinsic and extrinsic semiconductors: the intrinsic carrier concentration and how steeply it climbs with temperature, the mass-action law in n-type, p-type and compensated silicon, conductivity and resistivity of doped material, choosing a donor density to hit a specified resistivity, the position of the Fermi level, and a silicon-versus-germanium comparison.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

A bar of pure (intrinsic) silicon is held at 300 K. Take \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \), \( \mu_n = 1350~\text{cm}^2/\text{V}\cdot\text{s} \), \( \mu_p = 480~\text{cm}^2/\text{V}\cdot\text{s} \) and a band gap \( E_g = 1.12~\text{eV} \) that may be treated as constant over the range of interest.

1. Find the intrinsic conductivity and resistivity at 300 K.

2. Estimate the intrinsic concentration at 400 K.

3. By how many kelvin must the temperature rise, near 300 K, to double \( n_i \)?

• \( n_i(300~\text{K}) = 1.5\times10^{10}~\text{cm}^{-3} \), \( E_g = 1.12~\text{eV} \)

• \( \mu_n = 1350 \), \( \mu_p = 480~\text{cm}^2/\text{V}\cdot\text{s} \), \( q = 1.602\times10^{-19}~\text{C} \)

• \( k = 8.617\times10^{-5}~\text{eV/K} \); mobilities are taken as fixed in part (b).

Part (a) — Intrinsic conductivity

In intrinsic material electrons and holes are created in pairs, so \( n = p = n_i \) and both carry current:

\( \sigma_i = 4.40\times10^{-6}~\text{S/cm} \), \( \rho_i \approx 227~\text{k}\Omega\cdot\text{cm} \)

Part (b) — \( n_i \) at 400 K

The equilibrium product \( n_i^2 = N_C N_V e^{-E_g/kT} \) carries a \( T^3 \) prefactor from the effective densities of states, so

\( n_i(400~\text{K}) \approx 5.19\times10^{12}~\text{cm}^{-3} \), about 346 times the 300 K value

Part (c) — Temperature rise that doubles \( n_i \)

Differentiate the logarithm of the same expression:

\( n_i \) doubles for roughly every \( 9.0~\text{K} \) rise near room temperature

Note how weak intrinsic conduction is: only a fraction \( 3.0\times10^{-13} \) of the silicon atoms are ionised at 300 K, which is why intrinsic silicon behaves almost as an insulator. The exponential term dominates the temperature behaviour completely: of the 346-fold increase between 300 K and 400 K, the \( T^{3/2} \) prefactor supplies a factor of only 1.54.

Three silicon samples are held at 300 K, where \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \). All dopants are fully ionised.

1. Sample A is doped with \( N_D = 1\times10^{16}~\text{cm}^{-3} \) phosphorus atoms. Find \( n \) and \( p \).

2. Sample B is doped with \( N_A = 5\times10^{17}~\text{cm}^{-3} \) boron atoms. Find \( p \) and \( n \).

3. Sample C is counter-doped: \( N_D = 2\times10^{16} \) and \( N_A = 1.2\times10^{16}~\text{cm}^{-3} \). Find \( n \) and \( p \), and state the sample type.

• \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \), so \( n_i^2 = 2.25\times10^{20}~\text{cm}^{-6} \)

• Complete ionisation, thermal equilibrium, no external excitation.

Relations used

Charge neutrality and the mass-action law hold together:

Whenever the net doping greatly exceeds \( n_i \), the majority concentration equals the net doping and the minority concentration follows from \( np = n_i^2 \).

Part (a) — n-type sample

\( n = 1\times10^{16}~\text{cm}^{-3} \), \( p = 2.25\times10^{4}~\text{cm}^{-3} \)

Part (b) — p-type sample

\( p = 5\times10^{17}~\text{cm}^{-3} \), \( n = 4.50\times10^{2}~\text{cm}^{-3} \)

Part (c) — Compensated sample

Donors and acceptors cancel first; only the excess dopes the crystal:

The square-root form is exact; since \( (N_D-N_A)/2 \gg n_i \) it collapses to \( n \approx N_D - N_A \) to well within one part in \( 10^{11} \).

\( n = 8.00\times10^{15}~\text{cm}^{-3} \), \( p = 2.81\times10^{4}~\text{cm}^{-3} \) — the sample is n-type

Doping does not add carriers to the product — it only redistributes them. Raising the majority concentration by a factor of 50 from sample A to sample B pushes the minority concentration down by the same factor of 50, which is exactly why heavy doping suppresses minority-carrier injection.

A silicon sample is doped with \( N_D = 5\times10^{16}~\text{cm}^{-3} \) donors. At this doping level the mobilities are \( \mu_n = 1200~\text{cm}^2/\text{V}\cdot\text{s} \) and \( \mu_p = 400~\text{cm}^2/\text{V}\cdot\text{s} \); \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \).

1. Find the conductivity and resistivity of the sample.

2. What fraction of the conductivity is carried by holes?

3. Compare the resistivity with that of intrinsic silicon, \( \rho_i = 2.27\times10^{5}~\Omega\cdot\text{cm} \).

• \( N_D = 5\times10^{16}~\text{cm}^{-3} \), fully ionised, so \( n \approx N_D \)

• \( \mu_n = 1200 \), \( \mu_p = 400~\text{cm}^2/\text{V}\cdot\text{s} \)

• \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \)

Step 1 — Both carrier concentrations

Step 2 — Conductivity

\( \sigma = 9.612~\text{S/cm} \), \( \rho = 0.104~\Omega\cdot\text{cm} \)

Step 3 — Hole contribution

Holes carry a fraction \( 3.0\times10^{-14} \) of the current — utterly negligible

Step 4 — Comparison with intrinsic silicon

Doping lowers the resistivity by a factor of about \( 2.19\times10^{6} \)

Only a fraction \( 1.0\times10^{-6} \) of the silicon atoms have been replaced by donors — roughly one atom in a million — yet the conductivity rises by more than six orders of magnitude. This controllability, not the absolute conductivity, is what makes silicon useful.

Design problem. An n-type silicon resistor layer must have a resistivity of \( 0.050~\Omega\cdot\text{cm} \) at 300 K. The electron mobility itself falls as the doping rises, following the measured values below (interpolate logarithmically in \( N_D \)):

1. Determine the donor concentration required.

2. Express the result as one donor atom per how many silicon atoms (\( 5\times10^{22}~\text{cm}^{-3} \)).

• Target \( \rho = 0.050~\Omega\cdot\text{cm} \Rightarrow \sigma = 20.0~\text{S/cm} \)

• Minority holes are negligible, so \( \sigma \approx q N_D \mu_n \)

• \( \mu_n \) is itself a function of \( N_D \) — the design equation is implicit.

Step 1 — Set up the design equation

Because \( \mu_n \) depends on the answer, solve by iteration: assume a mobility, compute \( N_D \), read back the mobility at that \( N_D \), and repeat.

Step 2 — Iterate

The loop settles on

Step 3 — Check

\( N_D \approx 1.9\times10^{17}~\text{cm}^{-3} \) of phosphorus

Step 4 — Doping fraction

About one donor atom for every \( 2.6\times10^{5} \) silicon atoms

Had the mobility been held at its lightly doped value of 1350, the answer would have come out as \( 9.25\times10^{16}~\text{cm}^{-3} \) — low by about 52 per cent. Below roughly \( 10^{17}~\text{cm}^{-3} \) resistivity falls almost inversely with doping, but beyond that ionised-impurity scattering eats the gain, and pushing resistivity much lower costs disproportionate doping.

Silicon at 300 K has \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \), \( E_g = 1.12~\text{eV} \) and \( kT/q = 25.85~\text{mV} \). Assume the intrinsic level \( E_i \) sits at midgap.

1. Locate \( E_F \) relative to \( E_i \) in a sample doped \( N_D = 1\times10^{16}~\text{cm}^{-3} \), and relative to \( E_C \).

2. Locate \( E_F \) in a sample doped \( N_A = 5\times10^{17}~\text{cm}^{-3} \).

3. The n-type sample of part (a) is heated to 400 K, where \( n_i = 5.19\times10^{12}~\text{cm}^{-3} \). Where is \( E_F \) now?

• \( n = n_i e^{(E_F-E_i)/kT} \) and \( p = n_i e^{(E_i-E_F)/kT} \) (non-degenerate statistics)

• \( kT = 25.85~\text{meV} \) at 300 K, \( kT = 34.47~\text{meV} \) at 400 K

• Dopants fully ionised; \( E_i \) at midgap, i.e. \( E_C - E_i = 0.560~\text{eV} \)

Part (a) — n-type sample

\( E_F \) lies \( 0.347~\text{eV} \) above \( E_i \), i.e. \( 0.213~\text{eV} \) below \( E_C \)

Part (b) — p-type sample

\( E_F \) lies \( 0.448~\text{eV} \) below \( E_i \), i.e. \( 0.112~\text{eV} \) above \( E_V \)

Part (c) — Same n-type sample at 400 K

The doping is unchanged, so \( n \) is still \( 1\times10^{16}~\text{cm}^{-3} \), but \( n_i \) has risen by more than two orders of magnitude:

\( E_F \) has fallen back to only \( 0.261~\text{eV} \) above \( E_i \)

Each decade of doping moves \( E_F \) by \( kT\ln 10 = 59.5~\text{meV} \), so the Fermi level is a logarithmically slow indicator of doping. Heating drives \( E_F \) toward midgap because the thermally generated carriers begin to rival the doping — carried far enough, that is intrinsic breakdown of device action.

Compare intrinsic silicon and intrinsic germanium at 300 K.

1. Find the intrinsic conductivity and resistivity of each.

2. By what factor is germanium the better conductor, and how much of that factor comes from \( n_i \) rather than from mobility?

3. What donor density would give silicon the same conductivity as intrinsic germanium?

• Intrinsic material: \( n = p = n_i \), so \( \sigma_i = q n_i(\mu_n+\mu_p) \)

• \( q = 1.602\times10^{-19}~\text{C} \), \( T = 300~\text{K} \)

Part (a) — Conductivities

\( \rho_{\text{Si}} = 2.27\times10^{5}~\Omega\cdot\text{cm} \), \( \rho_{\text{Ge}} = 43.0~\Omega\cdot\text{cm} \)

Part (b) — Where the factor comes from

Germanium conducts about \( 5.28\times10^{3} \) times better; the carrier-density ratio \( 1.67\times10^{3} \) supplies almost all of it, mobility only \( 3.17 \)

The carrier ratio traces directly to the band gaps: \( \exp[(1.12-0.72)/(2\times0.02585)] \approx 2.3\times10^{3} \), which is the right order.

Part (c) — Doping silicon to match

In n-type silicon the holes are negligible, so \( \sigma \approx q N_D \mu_n \):

\( N_D \approx 1.07\times10^{14}~\text{cm}^{-3} \) — one donor per \( 4.7\times10^{8} \) silicon atoms

Germanium's narrow gap is why it conducts better when pure and also why it was abandoned: \( n_i \) grows so fast with temperature that leakage swamps the device above about \( 75~^\circ\text{C} \), whereas silicon works to \( 150~^\circ\text{C} \) and beyond.