Donor - acceptor energy levels in the band gap: Exercises

Before we begin, one shared picture ties every problem together. Keep it open.

Figure — Donor - acceptor energy levels in the band gap

Level 1 — Recognition

L1-Q1

In the figure above, one dashed line sits $0.045 eV$ below $E_{C}$ and another sits $0.045 eV$ above $E_{V}$ . Which is the donor level and which is the acceptor level?

Recall Solution

A donor gives an electron up to the conduction band, so its level must be a short hop below $E_{C}$ → the line $0.045 eV$ below $E_{C}$ is the donor level $E_{D}$ . An acceptor catches an electron from the valence band, so its level sits a short hop above $E_{V}$ → the line $0.045 eV$ above $E_{V}$ is the acceptor level $E_{A}$ . The rule: donor near the ceiling, acceptor near the floor.

L1-Q2

Silicon is doped with phosphorus (group V). Circle the correct words: the dopant is a (donor / acceptor); after ionizing it becomes a fixed (positive / negative) ion; it releases an (electron / hole).

Recall Solution

Phosphorus has five valence electrons — one more than silicon's four. It donates that extra electron, so it is a donor. Losing a negative electron leaves it positive ( $N_{D}^{+}$ ). What it releases is a free electron.

Level 2 — Application

L2-Q1

A donor level in silicon lies at $E_{D}$ with $E_{C} - E_{D} = 45 meV$ . If $E_{C} = 1.12 eV$ and $E_{V} = 0 eV$ , find $E_{D}$ measured from the valence band.

Recall Solution

We measure energies upward from $E_{V} = 0$ , so $E_{C} = 1.12 eV$ . The donor sits $45 meV = 0.045 eV$ below $E_{C}$ : $E_{D} = E_{C} - 0.045 = 1.12 - 0.045 = 1.075 eV .$ So the donor level is at $1.075 eV$ above the valence band — just below the $1.12 eV$ ceiling, exactly the "shallow" picture.

L2-Q2

An acceptor in the same silicon has $E_{A} - E_{V} = 45 meV$ . Find $E_{A}$ (from $E_{V} = 0$ ), and find the gap $E_{C} - E_{A}$ between the acceptor level and the conduction band.

Recall Solution

Acceptor sits $45 meV$ above $E_{V}$ : $E_{A} = E_{V} + 0.045 = 0 + 0.045 = 0.045 eV .$ Distance up to the conduction band: $E_{C} - E_{A} = 1.12 - 0.045 = 1.075 eV .$ Notice this big $1.075 eV$ is not the acceptor ionization energy — the acceptor's easy job is grabbing an electron from the valence band just $0.045 eV$ below it, not from the far-away conduction band.

L2-Q3

Thermal energy at room temperature is often quoted as $k T \approx 26 meV$ (at $T = 300 K$ ). Compare it to a $45 meV$ ionization energy: is one nudge of thermal energy enough on average?

Recall Solution

Compute the ratio: $\frac{E _{C} - E _{D}}{k T} = \frac{45}{26} \approx 1.73.$ A single average thermal packet ( $26 meV$ ) is smaller than the $45 meV$ needed. But "average" hides a spread — many carriers carry more than average, and there are enormous numbers of attempts each second. So even though $45 > 26$ , at room temperature the level is still almost fully ionized because the shortfall is small (less than a factor of 2). If the level were $500 meV$ deep, that would be a different story.

Level 3 — Analysis

L3-Q1

Two donors are proposed for silicon: dopant X with $E_{C} - E_{D} = 45 meV$ , and dopant Y with $E_{C} - E_{D} = 260 meV$ . At room temperature ( $k T = 26 meV$ ), the fraction of un-ionized donors grows roughly like $e^{- (E_{C} - E_{D}) / k T}$ getting larger as the level gets deeper. Which dopant gives more free electrons at room temperature, and by roughly what factor does the "un-ionized tendency" differ?

Recall Solution

The exponential $e^{- Δ/ k T}$ measures how easy ionization is: bigger exponent → easier.

X: $Δ/ k T = 45/26 \approx 1.73$ , so $e^{- 1.73} \approx 0.177$ .
Y: $Δ/ k T = 260/26 = 10$ , so $e^{- 10} \approx 4.54 \times 1 0^{- 5}$ . The ratio of these Boltzmann factors: $\frac{e ^{- 1.73}}{e ^{- 10}} = e^{8.27} \approx 3.9 \times 1 0^{3} .$ Dopant X ionizes far more readily — its Boltzmann factor is about 3900× larger, meaning vastly more free electrons at room temperature. This is why engineers choose shallow dopants (P, As) and avoid deep-level impurities.

L3-Q2

The "hydrogen-like" model says the extra donor electron is bound with energy $E_{ion} = 13.6 eV \times \frac{m ^{*} / m _{0}}{ε _{r}^{2}},$ where $13.6 eV$ is the binding energy of a real hydrogen atom, $m^{*} / m_{0}$ is the electron's effective mass ratio inside the crystal, and $ε_{r}$ is silicon's dielectric constant (how much the crystal weakens electric forces). Using $m^{*} / m_{0} = 0.26$ and $ε_{r} = 11.7$ , estimate the donor ionization energy in meV.

Recall Solution

Why this formula? A real hydrogen atom binds its electron with $13.6 eV$ . Inside silicon, two things make the donor's grip far weaker:

The crystal's dielectric constant $ε_{r} = 11.7$ screens the $+ 1$ charge, and screening scales the energy by $1/ ε_{r}^{2}$ .
The electron behaves as if lighter, effective mass ratio $0.26$ , which scales energy by that ratio.

Plug in: $E_{ion} = 13.6 \times \frac{0.26}{11. 7 ^{2}} = 13.6 \times \frac{0.26}{136.89} .$ $= 13.6 \times 0.0018994 = 0.02583 eV \approx 25.8 meV .$ So the model predicts a shallow level of about $26 meV$ — the right order of magnitude for real donors (tens of meV). The theory explains shallowness without any fitting.

Level 4 — Synthesis

L4-Q1

Silicon is doped with both phosphorus (donors, $N_{D} = 1 0^{16} cm^{- 3}$ ) and boron (acceptors, $N_{A} = 4 \times 1 0^{15} cm^{- 3}$ ). Assuming full ionization, is the material n-type or p-type, and what is the net free-carrier density from the dopants?

Recall Solution

Donors add electrons; acceptors add holes. When both are present they cancel (an electron from a donor drops into an acceptor). What matters is the difference. $N_{D} - N_{A} = 1 0^{16} - 4 \times 1 0^{15} = 1.0 \times 1 0^{16} - 0.4 \times 1 0^{16} = 6 \times 1 0^{15} cm^{- 3} .$ Since $N_{D} > N_{A}$ , donors win → the material is n-type, with net donor density $6 \times 1 0^{15} cm^{- 3}$ of un-cancelled electron-suppliers. This is called compensation: the boron compensates part of the phosphorus.

L4-Q2

Draw (describe) the band diagram for the L4-Q1 sample: which level(s) sit where, and which are ionized? Reference the shared figure and state the numeric positions using $E_{V} = 0$ , $E_{C} = 1.12 eV$ , both ionization energies $= 45 meV$ .

Recall Solution

The donor level $E_{D} = 1.075 eV$ sits just below $E_{C}$ (dashed line near the ceiling in the figure).
The acceptor level $E_{A} = 0.045 eV$ sits just above $E_{V}$ (dashed line near the floor).
All acceptors are filled by electrons dropping from donors (that is the compensation), so every acceptor is ionized to $N_{A}^{-}$ .
Of the donors, $N_{A} = 4 \times 1 0^{15}$ gave their electrons to acceptors; the remaining $6 \times 1 0^{15}$ donors give electrons to the conduction band. All donors end up ionized $N_{D}^{+}$ .
Net result: conduction band holds $\approx 6 \times 1 0^{15} cm^{- 3}$ free electrons. n-type, exactly as computed.

Level 5 — Mastery

L5-Q1

A germanium sample uses a donor with ionization energy $E_{C} - E_{D} = 10 meV$ . At what temperature $T$ does the thermal energy $k T$ equal this ionization energy? Use Boltzmann's constant $k = 8.617 \times 1 0^{- 5} eV/K$ . What does reaching this temperature physically signify?

Recall Solution

Set $k T = 10 meV = 0.010 eV$ and solve for $T$ : $T = \frac{0.010}{8.617 \times 1 0 ^{- 5}} = 116.1 K .$ So around $T \approx 116 K$ (well below room temperature, $300 K$ ), one unit of thermal energy already matches the donor's binding. Above this, essentially every donor is ionized — the sample is in its saturation / extrinsic regime with a flat carrier count. Because germanium donors are so shallow ( $10 meV$ ), even a cold sample is fully ionized, which is why Ge devices work down to low temperatures.

L5-Q2 (open synthesis)

You are handed a mystery silicon sample. Two impurities are present. When you cool it toward absolute zero, the free-electron count drops smoothly toward zero; but even at $77 K$ there are already many free electrons and no free holes. Argue whether the dominant dopant is a shallow donor or an acceptor, and estimate an upper bound on its ionization energy given that it is "well ionized" at $77 K$ .

Recall Solution

Type: free electrons and no holes → majority carriers are electrons → the sample is n-type, dominant dopant is a donor. Depth bound: "well ionized at $77 K$ " means the thermal energy at $77 K$ is at least comparable to the ionization energy. Compute $k T$ at $77 K$ : $k T = 8.617 \times 1 0^{- 5} \times 77 = 6.635 \times 1 0^{- 3} eV \approx 6.6 meV .$ For the level to be substantially ionized at that temperature, its ionization energy should be no more than a small multiple of $k T$ — an order-of-magnitude upper bound is roughly $E_{C} - E_{D} ≲ a few \times 6.6 meV \sim 20 - 30 meV .$ So the mystery donor is shallow, tens of meV at most — consistent with ordinary P or As in silicon. A deep donor ( $> 100 meV$ ) would freeze out and give very few electrons at $77 K$ , contradicting the observation.

Recall Quick self-check ladder

Donor near which edge? ::: The conduction band edge $E_{C}$ (a small hop below it). Acceptor ionization energy formula? ::: $E_{A} - E_{V}$ . Why divide, not subtract, Boltzmann factors? ::: Ratios measure "how many times easier"; $e^{a} / e^{b} = e^{a - b}$ . Net dopant density with both present? ::: $∣ N_{D} - N_{A} ∣$ , and its sign gives the type. What does $k T = E_{ion}$ mark? ::: The temperature scale for full ionization / saturation.

🇮🇳 Yeh exercises Hinglish mein →