Question bank — N-type doping with donor atoms (phosphorus, arsenic)

False. Every freed mobile electron ($-$) is matched by a fixed ionised donor ($N_D^{+}$), so the crystal stays electrically neutral. "N" names the sign of the movers, not the whole block.

False. The donor ion is locked into a lattice site — it is immobile. Only the freed electron moves; the $N_D^{+}$ core stays put and just sits there being positive.

False. Holes actually drop. Mass-action ($np=n_i^2$) is fixed by temperature, so if $n$ rises to $\approx N_D$, then $p=n_i^2/N_D$ must fall.

False. It moves ==up, toward $E_C$==. More electrons near the conduction band shift the "average filling level" upward — that upward move is the fingerprint of N-type. See Fermi level and its shift with doping.

False. Phosphorus (group V, a donor) makes free electrons; boron (group III, an acceptor) makes free holes. They are opposites — see P-type doping with acceptor atoms (boron).

False. The donor binding energy (real value $\approx 45$ meV for phosphorus in Si) is close to thermal energy $k_{B}T\approx 26$ meV, so nearly all donors are ionised. That's the regime we always assume.

False. The four bonding electrons sit deep in covalent bonds ($\sim$1 eV scale); the fifth is bound by only $\sim$45 meV — over an order of magnitude weaker. That looseness is the whole doping mechanism.

False. Atom count is irrelevant; free-carrier count matters. Only $n_i\approx 1.5\times 10^{10}$/cm³ carriers exist (1 in a trillion atoms), so pure Si conducts poorly. See Intrinsic vs Extrinsic Semiconductors.

There is no hole to fill. The donor puts its electron straight into the conduction band ($E_C$) as a free carrier; donors don't create holes, they reduce them ($p=n_i^2/N_D$).

At high doping, mobility ${\mu }_n$ drops (more ionised-impurity scattering), so $\sigma$ rises sub-linearly, not proportionally — see the curve in the figure above. Detail in Conductivity and carrier mobility.

Inside the crystal you must use the effective mass $m^{\ast }\approx 0.26m_0$ and the dielectric constant ${\epsilon }_r\approx 11.7$. The ${\epsilon }_r^2\approx 137$ in the denominator is exactly what crushes 13.6 eV down to a few tens of meV.

Holes never reach zero; mass-action forces $p=n_i^2/N_D>0$. They are just the tiny minority carriers, outnumbered by electrons by $\sim 10^{11}$.

You can drop $p$ in the conductivity sum $\sigma =q(n{\mu }_n+p{\mu }_p)$ because $p{\mu }_p\ll n{\mu }_n$, but holes still matter physically — they dominate reverse leakage and are essential once you build a PN junction.

It sits just below $E_C$ (only a few tens of meV down, vs the full $\sim$1.12 eV gap). Being so close to $E_C$ is why a tiny thermal nudge frees its electron.

Silicon needs exactly 4 electrons per site for its bonds. A group-V atom bonds with 4 and has one left over — that spare, weakly-held electron becomes the free carrier. Group IV would have none spare.

The donor loses one electron's worth of screening, leaving a net $+1$ core. The spare electron orbits that $+1$ charge just like hydrogen's one electron orbits its proton — that's why the Bohr formula applies.

The electron isn't in vacuum — it moves through silicon, which screens the Coulomb pull (${\epsilon }_r$) and alters how the electron responds to force (effective mass $m^{\ast }$). Ignoring both would over-predict the binding energy by $\sim$500×.

Because typically $N_D\gg n_i$: the donor electrons vastly outnumber the thermal ones, so the $+p$ term in $n=p+N_D$ is negligible. See Mass-action law and carrier concentrations.

It comes from the balance of generation and recombination, which depends only on temperature (and the material), not on impurity count. Doping shifts how $n$ and $p$ split, but their product is pinned.

Conductivity $\sigma$ depends on free-carrier count, which jumps from $n_i$ to $N_D$. Net charge depends on the balance of $+$ and $-$, which stays zero because each electron is matched by a fixed donor ion.

The approximation $n\approx N_D$ breaks down. You must solve $n=p+N_D$ with $np=n_i^2$ exactly, and the thermal carriers are no longer negligible — the material behaves near-intrinsic.

Donors "freeze out" — thermal energy can no longer ionise them, so electrons fall back onto the donor levels ($E_D$) and $n$ drops well below $N_D$. The near-full ionisation we assume requires room temperature.

Thermal generation floods the crystal with intrinsic pairs; once $n_i\gtrsim N_D$ the doping is swamped and $n\approx p\approx n_i$. The material loses its N-type character and behaves intrinsic again — the extrinsic-to-intrinsic transition, the high-temperature mirror of freeze-out.

The Fermi level $E_F$ pushes into or above $E_C$, donor levels merge into a band, and simple non-degenerate formulas fail. Conductivity keeps rising but ${\mu }_n$ falls hard due to impurity scattering.

The net doping wins by compensation: effectively $N_D-N_A$. If donors exceed acceptors it's N-type; if acceptors exceed, it's P-type. This overlap is exactly what forms a PN junction.

Yes, it degenerates gracefully: $n\approx N_D$ collapses to the intrinsic case where $n=p=n_i$. The extrinsic picture smoothly reduces to intrinsic silicon.