2.2.1Doping & PN Junctions

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

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WHY do we dope silicon at all?

WHY dope? To control conductivity by design. We deliberately add impurity atoms so we get exactly as many free carriers as we want. Doping is what makes transistors and diodes possible.


WHAT is a donor atom?

HOW it produces a free electron (from first principles):

  1. The donor sits on a silicon site. It needs only 4 electrons to bond, so the 5th is unbonded.
  2. This 5th electron is weakly attracted to the +1 core left behind (the donor loses one electron worth of screening → net +1 charge). It orbits like a mini hydrogen atom inside the crystal.
  3. Using the hydrogen-model binding energy we can show it takes almost no energy to free it (derivation below), so at room temperature it is already free.
  4. Result: one extra mobile electron per donor, and one fixed positive donor ion ND+N_D^+ left behind (it does NOT move).

Deriving the donor binding energy (Feynman-style, from scratch)

WHY derive it? So you believe the electron is nearly free — not just memorise it.

Model the 5th electron + donor core as a hydrogen atom, but living inside silicon. Two things change vs. vacuum:

  • The Coulomb attraction is screened by silicon's dielectric constant εr11.7\varepsilon_r \approx 11.7, so replace ε0εrε0\varepsilon_0 \to \varepsilon_r\varepsilon_0.
  • The electron moves through the lattice with an effective mass m0.26m0m^* \approx 0.26\,m_0, not the free mass.

The Bohr ground-state energy in vacuum is EH=m0e48ε02h2=13.6 eV.E_H = \frac{m_0 e^4}{8\varepsilon_0^2 h^2} = 13.6\ \text{eV}.

Applying the two substitutions, every ε0εrε0\varepsilon_0 \to \varepsilon_r\varepsilon_0 (appears squared) and m0mm_0 \to m^*:

ED=13.6 eV×m/m0εr2\boxed{E_D = 13.6\ \text{eV}\times\frac{m^*/m_0}{\varepsilon_r^{2}}}

Interpret it: thermal energy at room temperature is kBT0.026 eVk_BT \approx 0.026\ \text{eV}. Since EDkBTE_D \approx k_BT, virtually every donor is ionised at room temperature → each donor gives one free electron. This is the regime we always assume.

(Real measured values: P in Si ≈ 45 meV, As ≈ 54 meV — same order; our hydrogen model is beautifully close for a one-line estimate.)


Band-diagram / energy picture

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

Carrier concentration — the numbers you actually use

HOW we get the concentrations (from mass-action law):

Detailed balance of generation & recombination gives the temperature-only mass-action law: np=ni2.n\,p = n_i^{2}. Charge neutrality with fully ionised donors (ND+NDN_D^+ \approx N_D, and NA=0N_A=0): n=p+ND.n = p + N_D.

Since typically NDniN_D \gg n_i, we get nNDn \approx N_D, then p=ni2/np = n_i^2/n:


Common mistakes (steel-manned)


Active recall

Recall Test yourself (hide answers)
  • Why does a group-V atom donate an electron? → It only needs 4 for bonds; the 5th is spare and weakly bound.
  • What stays behind when the electron leaves? → A fixed, immobile positive donor ion ND+N_D^+.
  • Which way does EFE_F move? → Up, toward ECE_C.
  • Majority carrier in N-type? → electrons.
  • Estimate pp if ND=1017N_D=10^{17}, ni=1.5×1010n_i=1.5\times10^{10}. → ni2/ND=2.25×103 cm3n_i^2/N_D = 2.25\times10^3\ \text{cm}^{-3}.
Recall Feynman: explain to a 12-year-old

Imagine a big dance where every silicon person is holding hands with 4 neighbours — nobody has a free hand, so nobody can move around: no "flow". Now sneak in a few phosphorus people who have 5 hands. They hold 4 hands like everyone else, but one hand is left waving free. That free hand wanders off around the room and can carry messages (electricity). We didn't make the room charged — for every wandering hand, the phosphorus person keeps a "+" sticker where the hand used to be. Lots of wandering hands = electricity flows easily. That's N-type silicon.


What group are donor atoms from and how many valence electrons?
Group V, 5 valence electrons (e.g. P, As, Sb).
Why does a donor produce a free electron?
Only 4 of its 5 valence electrons form bonds; the 5th is weakly bound and thermally freed at room temperature.
What is left behind after a donor is ionised?
A fixed, immobile positive donor ion ND+N_D^+ (crystal stays neutral overall).
Formula for donor ionisation energy via hydrogen model?
ED=13.6 eV(m/m0)/εr226E_D = 13.6\text{ eV}\cdot (m^*/m_0)/\varepsilon_r^2 \approx 26 meV for Si.
Why is nearly every donor ionised at room temperature?
ED26E_D\approx26 meV kBT\approx k_BT at 300 K, so thermal energy suffices to free the electron.
State the mass-action law.
np=ni2np = n_i^2 (holds at fixed temperature).
Charge-neutrality equation for N-type?
n=p+NDn = p + N_D (with NA=0N_A=0, donors fully ionised).
Approximate n and p in N-type when NDniN_D\gg n_i?
nNDn\approx N_D, pni2/NDp\approx n_i^2/N_D.
Majority and minority carriers in N-type?
Majority = electrons; minority = holes.
Is N-type silicon electrically charged?
No — neutral overall; mobile electrons balanced by fixed positive donor ions.
Which way does the Fermi level shift with donor doping?
Upward, toward the conduction band ECE_C.
Compute p for ND=1016N_D=10^{16}, ni=1.5×1010n_i=1.5\times10^{10} cm⁻³.
p=ni2/ND=2.25×104p=n_i^2/N_D=2.25\times10^{4} cm⁻³.

Connections

Concept Map

only 1 in trillion free carriers

motivates

adds

examples

4 electrons bond

5th electron leftover

modelled as

screened by dielectric and effective mass

freed at room temp

leaves behind

balanced by

net result

carries negative charge

Intrinsic silicon group IV

Poor conductor

Doping by design

Donor atom group V

Phosphorus and Arsenic

Normal covalent bonds

Weakly bound electron

Hydrogen atom in crystal

Tiny binding energy

Free mobile electron

Fixed positive ion N_D plus

N-type crystal neutral overall

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, pure silicon mein har atom ke paas 4 valence electrons hote hain aur wo apne 4 padosiyon ke saath covalent bonds bana leta hai — koi bhi electron free nahi bachta, isliye pure silicon current achhe se conduct nahi karta. Ab jab hum ismein phosphorus ya arsenic (Group V, yaani 5 valence electrons wale) atom daalte hain, to unke 4 electron to bond bana lete hain, par 5th electron extra bach jaata hai. Ye extra electron bahut kamzori se bandha hota hai.

Kitna kamzor? Humne hydrogen-atom model se nikala ki iski binding energy sirf ~26 meV hai (formula: 13.6eV×(m/m0)/εr213.6\,\text{eV}\times (m^*/m_0)/\varepsilon_r^2). Aur room temperature par thermal energy kBTk_BT bhi ~26 meV hoti hai — matlab lagbhag har donor ka electron free ho jaata hai. Jab electron chhod ke jaata hai, peeche ek fixed positive donor ion (ND+N_D^+) reh jaata hai, isliye crystal overall neutral hi rehta hai. "N-type" ka matlab hai negative charge carry karne wale electrons majority mein hain — iska matlab crystal charged nahi hai (ye galti sab karte hain!).

Numbers ki baat karein to, jab NDniN_D \gg n_i, tab electrons nNDn \approx N_D ho jaate hain, aur holes p=ni2/NDp = n_i^2/N_D (mass-action law np=ni2np=n_i^2 se). Isliye electrons majority carrier aur holes minority carrier kehlaate hain. Iska faayda? Conductivity lakhon guna badh jaati hai, controlled tareeke se — aur yahi cheez transistors aur diodes banane ka base hai. Band diagram mein bas yaad rakho: donor level EDE_D, conduction band ECE_C ke bilkul neeche baithta hai, aur Fermi level EFE_F upar ki taraf khisak jaata hai — yahi N-type ki pehchaan hai.

Go deeper — visual, from zero

Test yourself — Doping & PN Junctions

Connections