Donor - acceptor energy levels in the band gap: Visual walkthrough

Our target result (we will build every piece of it):

E_{C} - E_{D} = hydrogen binding 13.6 eV \times \frac{m ^{*} / m _{0}}{ε _{r}^{2}}

Don't worry about any symbol here yet — by Step 5 you will have met each one on its own picture.

Step 1 — Picture the leftover charge

WHAT. A phosphorus atom (group V, five outer electrons) sits in a silicon crystal where a silicon atom (four outer electrons) used to be. Four of phosphorus's electrons lock into the four covalent bonds around it, exactly like silicon did. That leaves one extra electron and, on the phosphorus core, one extra unit of positive charge that the four bonds don't cancel.

WHY. Before we can talk about energy levels, we need to know what object we are finding the energy of. The object is: one negative electron loosely tied to one leftover positive charge.

PICTURE. Look at the figure. Black = the ordinary silicon lattice and its bonds. Red = the one extra electron, orbiting the phosphorus core which carries a net $+ 1$ .

Figure — Donor - acceptor energy levels in the band gap

Step 2 — Recall the hydrogen atom we are copying

WHAT. A real hydrogen atom is one electron bound to one proton. Physics already knows its binding energy — the energy you must supply to rip the electron completely free:

E_{H} = value 13.6 energy unit eV

WHY. We reach for the hydrogen result because our object in Step 1 is structurally a hydrogen atom — one $+ q$ , one $- q$ . Rather than solve the whole problem again, we borrow the known answer and then correct it for the fact that our "hydrogen atom" lives inside silicon, not in empty space. That correction is Steps 3–4.

PICTURE. Left: real hydrogen in vacuum — a small tight orbit, electron held hard (13.6 eV to remove). Right: a preview of our "hydrogen in silicon" — a huge loose orbit, electron held weakly. Same shape, very different scale. The red orbit is the electron's path.

Recall The eV unit we will use

How many meV is $13.6 eV$ ? Answer: $13.6 \times 1000 = 13600 meV$ (see the eV definition box at the top).

Step 3 — Correction 1: silicon screens the pull (the $ε_{r}$ factor)

WHAT. In vacuum the $+ q$ core pulls the electron with full strength. Inside silicon, the crystal's own electrons shift slightly and partly cancel that pull — like the attraction is happening through a soft, polarizable jelly. We measure this weakening with silicon's relative permittivity $ε_{r}$ . For silicon $ε_{r} \approx 11.7$ : the effective pull is about $11.7$ times weaker.

WHY this tool. We need one honest number for "how much does the surroundings soften the attraction?" That number is exactly what $ε_{r}$ is defined to be — the factor by which a material weakens the electric force between charges compared to vacuum. So it is the right (and only) tool for this correction.

PICTURE. Same two charges, but now the black silicon atoms around them are drawn reacting: their electron clouds lean toward the core, throwing up a counter-pull. The red field lines from core to electron are drawn thin/faded to show the pull is muted.

How it enters the energy — from the Bohr formula. The known hydrogen binding energy comes from the Bohr model:

E = \frac{m q ^{4}}{8 ε ^{2} h ^{2}}

$m$ — the electron mass, $q$ — its charge, $h$ — Planck's constant (a fixed constant of nature).
$ε$ — the permittivity (how easily the medium lets electric field through); it sits squared in the denominator.

Because $ε$ appears as $ε^{2}$ in the bottom, replacing vacuum by silicon ( $ε \to ε_{r} ε_{0}$ ) divides the energy by $ε_{r}^{2}$ :

E \propto \frac{1}{ε _{r}^{2}} \Rightarrow divide 13.6 eV by ε_{r}^{2} \approx 137

$ε_{r}$ — how many times weaker the pull is inside silicon.
The square — read straight off the $ε^{2}$ in the Bohr denominator above.

Step 4 — Correction 2: the electron feels light (the $m^{*}$ factor)

WHAT. An electron travelling through the periodic lattice does not respond to forces like a bare electron would. It behaves as if it has a different mass, the effective mass $m^{*}$ , usually smaller than the true mass $m_{0}$ . For silicon roughly $m^{*} / m_{0} \approx 0.26$ .

WHY this tool. Look again at the Bohr formula from Step 3: the mass $m$ sits on top, so $E \propto m$ — a heavier electron sits in a tighter, deeper orbit. Our electron is not a bare vacuum electron; it is a lattice electron, so we replace $m$ by $m^{*}$ ( $m_{0} \to m^{*}$ ). Effective mass is precisely the bookkeeping tool that lets us keep using the simple hydrogen formula inside a crystal.

PICTURE. Two identical orbits side by side: a heavy electron pulled into a tight orbit (deep well), a red light electron drifting in a wide orbit (shallow well). Lighter ⇒ looser ⇒ easier to free.

How it enters the energy. Since $E \propto m$ in the Bohr result, swapping $m \to m^{*}$ multiplies by the ratio $m^{*} / m_{0}$ :

E \propto m^{*} \Rightarrow multiply by \frac{m ^{*}}{m _{0}} \approx 0.26

$m^{*}$ — the mass the electron acts like inside the lattice.
$m_{0}$ — the true free-electron mass, used as the yardstick.
Ratio $< 1$ ⇒ binding energy shrinks further.

Step 5 — Assemble and get the number

WHAT. Stack both corrections onto the hydrogen value. This gives the donor ionization energy — the energy to lift the extra electron from its level $E_{D}$ up to the conduction-band edge $E_{C}$ :

E_{C} - E_{D} = 13.6 eV \times \frac{m ^{*} / m _{0}}{ε _{r}^{2}}

Read left to right, each symbol earned:

$E_{C} - E_{D}$ — the gap between the conduction edge and the donor level = ionization energy.
$13.6 eV$ — the vacuum hydrogen binding (Step 2).
$m^{*} / m_{0}$ — light-electron correction, makes it smaller (Step 4).
$ε_{r}^{2}$ — screening correction in the denominator, makes it much smaller (Step 3).

WHY. Multiplying and dividing the independent corrections is legal because each affects the hydrogen energy by its own clean factor ( $\propto m^{*}$ , $\propto 1/ ε_{r}^{2}$ ).

Plug in silicon numbers ( $m^{*} / m_{0} = 0.26$ , $ε_{r} = 11.7$ ):

E_{C} - E_{D} = 13.6 \times \frac{0.26}{11. 7 ^{2}} eV = 13.6 \times \frac{0.26}{136.9} eV \approx 0.0258 eV \approx 26 meV

PICTURE. The band-gap diagram: valence band and conduction band as black bars, the gap between them, and the red donor level drawn a tiny $\sim 26$ meV sliver below $E_{C}$ — exactly the "shallow" claim of the parent note, now with a derived number.

Step 6 — The mirror case: the acceptor level

WHAT. First name the two edges we will use. The valence-band edge $E_{V}$ is the top of the filled band (highest energy an electron can have while still bound in bonds). The acceptor level $E_{A}$ is the new state a group-III dopant (like boron) creates. Swap phosphorus (one extra electron) for boron (group III, one missing electron): the leftover core is now $- 1$ and it loosely binds a hole (a $+ q$ mobile vacancy). Same hydrogen math with a hole's effective mass gives $E_{A}$ a small distance above $E_{V}$ .

WHY show it. The parent note gives donor and acceptor symmetrically; skipping the acceptor would leave a case uncovered. The derivation is a mirror image, so we only flip the sign of every charge and reflect the diagram top-to-bottom.

PICTURE. The gap again, mirrored: this time the red acceptor level sits a small sliver above $E_{V}$ , ready to catch an electron from the valence band and leave a hole behind.

E_{A} - E_{V} = 13.6 eV \times \frac{m _{h}^{*} / m _{0}}{ε _{r}^{2}}

$E_{A} - E_{V}$ — acceptor ionization energy (how far the level sits above the valence edge).
$m_{h}^{*}$ — effective mass of the hole (replaces the electron's $m^{*}$ ).
Everything else identical ⇒ acceptor levels are also shallow, tens of meV.

Step 7 — Edge cases: what if the corrections weren't small?

WHAT. Our result is shallow because $ε_{r}$ is large and $m^{*}$ is small. Test the extremes:

Case	$ε_{r}$	$m^{*} / m_{0}$	Result	Meaning
Silicon (normal)	11.7	0.26	~26 meV	shallow — ionizes at room temp
No screening ( $ε_{r} \to 1$ )	1	0.26	~3.5 eV	deep — basically a real hydrogen atom, never ionizes
Heavy electron ( $m^{*} \to m_{0}$ )	11.7	1	~99 meV	deeper, harder to ionize
Both extreme	1	1	13.6 eV	exactly hydrogen — the formula's ceiling

WHY. Checking limits proves the formula behaves sensibly and shows which physical fact (strong screening + light electrons in real semiconductors) makes doping work. Remove either and dopants would freeze out.

PICTURE. Two gaps side by side: left a red shallow level hugging $E_{C}$ (real silicon); right a red deep level sitting near mid-gap (the $ε_{r} \to 1$ fantasy). The reader sees exactly why real semiconductors are the lucky case.

The one-picture summary

Hydrogen (13.6 eV) → soften the pull by $ε_{r}^{2}$ (÷137) → lighten the electron by $m^{*} / m_{0}$ (×0.26) → land ~26 meV below $E_{C}$ : a shallow, easily ionized donor level. Flip every sign for the acceptor.

Recall Feynman retelling — say it back in plain words

Put a phosphorus atom where a silicon atom was. Four of its electrons do the normal bonding job; one electron and one bit of positive charge are left over, attracting each other — that's just a tiny hydrogen atom hiding in the crystal. A real hydrogen atom holds its electron with 13.6 eV. But two things weaken our hidden hydrogen: (1) the silicon around it partly cancels the pull — that's the permittivity $ε_{r}$ , and since the Bohr energy has $ε^{2}$ in its denominator we divide by $ε_{r}^{2}$ , about 137; (2) the electron sailing through the lattice acts lighter than usual — the Bohr energy has mass on top, so with $m^{*} / m_{0} \approx 0.26$ we multiply by that. Combine: $13.6 \times 0.26/137 \approx 0.026$ eV, or 26 meV. That tiny number is how far below the conduction band the donor level sits — so tiny that room-temperature warmth (also ~26 meV) knocks the electron free. Boron does the exact mirror: a missing electron, a loosely bound hole, a shallow level just above the valence band. Real measured levels (P ~45, As ~54 meV) sit a bit deeper because of central-cell and valley effects we left out, but the hydrogen picture explains the order of magnitude. Kill the screening or make the electron heavy, and the level plunges deep — the dopant would stay frozen. Silicon is lucky on both counts, which is why doping works.