Exercises — Intrinsic vs extrinsic semiconductors

Phosphorus is group V: it forms 4 bonds and has one electron left over, which is nearly free. That electron is donated to the conduction band without creating a matching hole. → The material is n-type. The majority carriers are electrons. What it looks like: see the figure below — the red dot is the spare, weakly-bound electron floating just below the conduction band.

By definition, breaking one bond makes exactly one electron and one hole together, so they are always created in pairs. Therefore $n=p=n_i.$

• Group III = one electron short → acceptor → p-type.
• Group V = one electron extra → donor → n-type. Mnemonic from the parent note: "DoNor gives Negative, Acceptor Adds A hole."

Step 1 (majority). Since $N_D=10^{16}\gg n_i=1.5\times 10^{10}$, thermal pairs are negligible next to the donated electrons: $n\approx N_D=1\times 10^{16}{\text{cm}}^{-3}.$ Step 2 (minority). The Law of Mass Action $np=n_i^2$ holds at equilibrium for any doping, so solve for $p$: $p=\frac{n_i^2}{n}=\frac{(1.5\times 10^{10}{)}^2}{1\times 10^{16}}=2.25\times 10^4{\text{cm}}^{-3}.$ Electrons outnumber holes by ~$10^{11}$ — thoroughly n-type.

Step 1. $N_A\gg n_i$, so each acceptor supplies one hole: $p\approx N_A=2\times 10^{17}{\text{cm}}^{-3}.$ Step 2. Mass action again: $n=\frac{n_i^2}{p}=\frac{(1.5\times 10^{10}{)}^2}{2\times 10^{17}}=1.125\times 10^3{\text{cm}}^{-3}.$

Why this formula: for pure material $n=p$, and combining $np=N_c{N}_v{e}^{-E_g/kT}$ with $n=p$ gives $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/2kT}$. Step 1 — prefactor: $\sqrt{N_c{N}_v}=\sqrt{2.8\times 10^{19}\times 1.04\times 10^{19}}=1.706\times 10^{19}{\text{cm}}^{-3}$. Step 2 — exponent: $\frac{-E_g}{2kT}=\frac{-1.12}{2\times 0.02585}=-21.66$, so $e^{-21.66}=3.92\times 10^{-10}$. Step 3 — multiply: $n_i\approx 1.706\times 10^{19}\times 3.92\times 10^{-10}\approx 6.7\times 10^9{\text{cm}}^{-3}.$ This is the right order of magnitude ($\sim 10^{10}$); the small difference from the quoted $1.5\times 10^{10}$ comes from temperature-dependent $N_c,N_v$ and a slightly smaller effective $E_g$.

Step 1 — sanity check. Here $N_A=5\times 10^9<n_i$. The shortcut $p\approx N_A$ requires $N_A\gg n_i$, which fails. We must solve properly. Step 2 — the two exact equations.

• Charge neutrality: positive charges = negative charges. Mobile holes + ionised... acceptors are negative when filled, so neutrality reads $p=n+N_A$ (holes balance electrons plus the fixed negative acceptor ions).
• Mass action: $np=n_i^2$. Step 3 — one quadratic. Substitute $n=n_i^2/p$: $p=\frac{n_i^2}{p}+N_A\Rightarrow p^2-N_{A}p-n_i^2=0.$ $p=\frac{N_A+\sqrt{N_A^2+4n_i^2}}{2}.$ Step 4 — numbers. $N_A^2=2.5\times 10^{19}$, $4n_i^2=4(2.25\times 10^{20})=9\times 10^{20}$, sum $=9.25\times 10^{20}$, root $=3.041\times 10^{10}$. $p=\frac{5\times 10^9+3.041\times 10^{10}}{2}=1.771\times 10^{10}{\text{cm}}^{-3}.$ $n=\frac{n_i^2}{p}=\frac{2.25\times 10^{20}}{1.771\times 10^{10}}=1.271\times 10^{10}{\text{cm}}^{-3}.$ Interpretation: $p$ and $n$ are almost equal — the sample is only weakly p-type ($p/n\approx 1.4$), not the ~$10^{11}$ ratio the naive shortcut would suggest.

Why compare to $N_D$: in the extrinsic region carriers are pinned at $N_D$; once thermal pairs $n_i$ catch up, they swamp the doping and control is lost. Step — evaluate at $T=600$ K. $kT=8.617\times 10^{-5}\times 600=0.05170$ eV. $\frac{-E_g}{2kT}=\frac{-1.12}{2\times 0.05170}=-10.83,e^{-10.83}=1.97\times 10^{-5}.$ $T^{3/2}=600^{1.5}=1.469\times 10^4.$ $n_i=3.1\times 10^{16}\times 1.469\times 10^4\times 1.97\times 10^{-5}\approx 8.97\times 10^{15}{\text{cm}}^{-3}.$ At 600 K, $n_i\approx 9\times 10^{15}\gg N_D=10^{15}$ — the device has already gone intrinsic by 600 K. So the true crossover $n_i=N_D$ sits below 600 K; the sample cannot be trusted much past a few hundred °C. This is exactly why silicon chips have a max operating temperature.

Step 1 — the idea of compensation. Donors give electrons; acceptors soak them up. Only the excess dopant survives as net majority carriers. Step 2 — net type. $N_D-N_A=8\times 10^{16}-5\times 10^{16}=3\times 10^{16}>0$, so donors win → n-type. Step 3 — majority. Charge neutrality with both dopants gives $n\approx N_D-N_A$: $n\approx 3\times 10^{16}{\text{cm}}^{-3}.$ Step 4 — minority via mass action. $p=\frac{n_i^2}{n}=\frac{(1.5\times 10^{10}{)}^2}{3\times 10^{16}}=7.5\times 10^3{\text{cm}}^{-3}.$ Interpretation: even though 5×10¹⁶ acceptors are present, they only cancel an equal number of donors — the sample behaves like lightly-doped n-type.

Why this formula: drift conductivity is (carrier density)×(charge)×(mobility): $\sigma =ne{\mu }_n$ for the majority electrons. Step 1 — conductivity. $\sigma =ne{\mu }_n=10^{16}\times 1.602\times 10^{-19}\times 1350=2.163(\Omega \cdot \text{cm}{)}^{-1}.$ Step 2 — resistivity. $\rho =\frac{1}{\sigma }=\frac{1}{2.163}=0.462\Omega \cdot \text{cm}.$ Compare: intrinsic Si has $\rho \approx 2\times 10^5\Omega \cdot \text{cm}$. Doping by 1 part in ~$5\times 10^6$ dropped the resistivity by nearly a million-fold — the whole point of doping.

Step 1 — invert the drift formula. $\sigma =1/\rho =pe{\mu }_p$, so $p=\frac{1}{\rho e{\mu }_p}=\frac{1}{0.10\times 1.602\times 10^{-19}\times 480}.$ $p=\frac{1}{7.690\times 10^{-18}}=1.300\times 10^{17}{\text{cm}}^{-3}.$ Step 2 — dopant. Since $p\gg n_i$, $N_A\approx p=1.30\times 10^{17}{\text{cm}}^{-3}$. Step 3 — minority carriers. $n=\frac{n_i^2}{p}=\frac{2.25\times 10^{20}}{1.30\times 10^{17}}=1.73\times 10^3{\text{cm}}^{-3}.$ We started from a bulk macroscopic measurement (a multimeter reading) and deduced a single atomic quantity: how many acceptor atoms were sprinkled in. That is the power of the model.

Step 1 — exact equations. Neutrality for n-type: $n=N_D+p$; mass action: $np=n_i^2$. Step 2 — quadratic in $n$. Sub $p=n_i^2/n$: $n^2-N_{D}n-n_i^2=0\Rightarrow n=\frac{N_D+\sqrt{N_D^2+4n_i^2}}{2}.$ Step 3 — numbers. $N_D^2=4\times 10^{20}$, $4n_i^2=9\times 10^{20}$, sum $=1.3\times 10^{21}$, root $=3.606\times 10^{10}$. $n=\frac{2\times 10^{10}+3.606\times 10^{10}}{2}=2.803\times 10^{10}{\text{cm}}^{-3}.$ $p=\frac{n_i^2}{n}=\frac{2.25\times 10^{20}}{2.803\times 10^{10}}=8.027\times 10^9{\text{cm}}^{-3}.$ Step 4 — ratio & error. $n/p=2.803\times 10^{10}/8.027\times 10^9\approx 3.5$: only mildly n-type. The naive $n\approx N_D=2\times 10^{10}$ underestimates the true $n$ by ~40%, because intrinsic pairs still contribute meaningfully. Shortcut fails when doping is not $\gg n_i$.

Doping $\gg n_i$, want majority? ::: $n\approx N_D$ (or $p\approx N_A$). Doping $\gg n_i$, want minority? ::: $p=n_i^2/N_D$ (or $n=n_i^2/N_A$). Doping $\sim n_i$? ::: Solve the neutrality quadratic $n^2-N_{D}n-n_i^2=0$. Both dopants present? ::: Net majority $\approx |N_D-N_A|$, then mass action for minority. Given resistivity, want doping? ::: Invert $\sigma =ne\mu \Rightarrow n=1/(\rho e\mu )$.