29 May 2025

Doping, Carrier Concentration, and Mobility

Charge neutrality, donor and acceptor doping, Fermi-level displacement, mobility, and conductivity.

bsc semester-iv mj-7 semiconductor-devices doping mobility

Let $N_D^+$ and $N_A^-$ denote ionized donor and acceptor densities. The positive charge density is $q(p+N_D^+)$ and the magnitude of the negative charge density is $q(n+N_A^-)$. Electrical neutrality therefore requires

\[q(p+N_D^+)-q(n+N_A^-)=0, \qquad \boxed{p+N_D^+=n+N_A^-}.\]

When shallow dopants are fully ionized, $N_D^+=N_D$ and $N_A^-=N_A$. In non-degenerate thermal equilibrium the second relation is $np=n_i^2$, so neutrality and mass action determine both carrier concentrations, including compensation by dopants of the opposite type.

Compensated n-type material

For $N_D>N_A$, define the positive net donor density

\[N=N_D-N_A.\]

Neutrality gives $n-p=N$. Eliminating $p$ with $p=n_i^2/n$ produces

\[n-\frac{n_i^2}{n}=N \quad\Longrightarrow\quad n^2-Nn-n_i^2=0.\]

Only the positive quadratic root is physical:

\[\boxed{n=\frac{N+\sqrt{N^2+4n_i^2}}{2}}, \qquad \boxed{p=\frac{n_i^2}{n}}.\]

If $N\gg n_i$, expansion of the square root gives $n\simeq N+n_i^2/N\simeq N$ and $p\simeq n_i^2/N$. With

\[n=n_i\exp\!\left(\frac{E_F-E_i}{k_BT}\right),\]

the Fermi-level displacement is

\[\boxed{E_F-E_i=k_BT\ln\!\left(\frac{n}{n_i}\right)>0}.\]

Compensated p-type material

For $N_A>N_D$, define $P=N_A-N_D>0$. Now $p-n=P$, and the same elimination gives

\[\boxed{p=\frac{P+\sqrt{P^2+4n_i^2}}{2}}, \qquad \boxed{n=\frac{n_i^2}{p}}.\]

Since $p=n_i\exp[(E_i-E_F)/(k_BT)]$,

\[\boxed{E_F-E_i=-k_BT\ln\!\left(\frac{p}{n_i}\right)<0}.\]

The majority-carrier concentration is always positive; the sign appears in the displacement of $E_F$ from $E_i$. If heavy doping moves $E_F$ close to or inside a band, the non-degenerate exponential relations and $np=n_i^2$ approximation must be replaced by Fermi-Dirac integrals.

Temperature dependence of carrier concentration

Shallow-doped material has three characteristic temperature ranges:

In freeze-out, only a fraction of the dopants is ionized, so $N_D^+<N_D$ or $N_A^-<N_A$ and the majority-carrier density rises rapidly with temperature.
In the extrinsic range, ionization is nearly complete and the majority concentration is approximately the net dopant density: $n\simeq N_D-N_A$ or $p\simeq N_A-N_D$.
In the intrinsic range, thermal pair generation dominates and $n\simeq p\simeq n_i$.

For parabolic bands with approximately temperature-independent effective masses,

\[n_i=\sqrt{N_cN_v}\exp\!\left(-\frac{E_g}{2k_BT}\right), \qquad N_c,N_v\propto T^{3/2},\]

and hence

\[\boxed{n_i\propto T^{3/2}\exp\!\left(-\frac{E_g}{2k_BT}\right)}.\]

Even while the majority density is nearly constant in the extrinsic range, the minority density remains strongly temperature dependent through $p=n_i^2/n$ or $n=n_i^2/p$.

Mobility and scattering

In the relaxation-time model, a carrier of charge $q_c$ and effective mass $m^*$ obeys

\[m^*\frac{d\mathbf v}{dt}=q_c\mathbf E-\frac{m^*\mathbf v}{\tau}.\]

At steady state, $d\mathbf v/dt=0$, so

\[\mathbf v_d=\frac{q_c\tau}{m^*}\mathbf E, \qquad \boxed{\mu=\frac{\lvert q_c\rvert\tau}{m^*}}.\]

The mobility $\mu$ is defined as a positive magnitude. Thus $\mathbf v_n=-\mu_n\mathbf E$ for electrons and $\mathbf v_p=+\mu_p\mathbf E$ for holes. If independent lattice and ionized-impurity scattering processes have relaxation times $\tau_L$ and $\tau_I$, their collision rates add:

\[\frac1\tau=\frac1{\tau_L}+\frac1{\tau_I} \quad\Longrightarrow\quad \boxed{\frac1\mu=\frac1{\mu_L}+\frac1{\mu_I}}.\]

Lattice scattering normally strengthens as temperature rises, while ionized-impurity scattering strengthens as dopant density rises. Over limited temperature ranges, simple models give $\mu_L\propto T^{-3/2}$ and $\mu_I\propto T^{3/2}/N_{\rm ion}$; these powers are approximations, not universal laws. Conductivity remains

\[\boxed{\sigma=q(n\mu_n+p\mu_p)},\]

so doping can increase carrier density even while impurity scattering reduces mobility.

Solved Problems

1. Exact carrier balance in compensated n-type material

At $300\ \mathrm K$, let $N_D=8.00\times10^{21}\ \mathrm{m^{-3}}$, $N_A=2.00\times10^{21}\ \mathrm{m^{-3}}$, and $n_i=1.00\times10^{16}\ \mathrm{m^{-3}}$. Take $\mu_n=0.135\ \mathrm{m^2\,V^{-1}s^{-1}}$ and $\mu_p=0.0480\ \mathrm{m^2\,V^{-1}s^{-1}}$. Find $n$, $p$, $E_F-E_i$, $\sigma$, and $\rho$.

The net donor density is $N=6.00\times10^{21}\ \mathrm{m^{-3}}$. Therefore

\[\begin{aligned} n&=\frac{N+\sqrt{N^2+4n_i^2}}2 =6.00000000002\times10^{21}\ \mathrm{m^{-3}},\\ p&=\frac{n_i^2}{n}=1.667\times10^{10}\ \mathrm{m^{-3}}. \end{aligned}\]

Using $k_BT=0.0258520\ \mathrm{eV}$,

\[E_F-E_i=k_BT\ln(n/n_i)=+0.3440\ \mathrm{eV}.\]

The positive sign places $E_F$ above $E_i$, as required for n-type material. Finally,

\[\sigma=q(n\mu_n+p\mu_p)=129.8\ \mathrm{S\,m^{-1}}, \qquad \rho=7.706\times10^{-3}\ \Omega\,\mathrm m.\]

Checks: $n-p=N_D-N_A$ and $np=n_i^2$ to the displayed precision, while $\sigma\rho=1$.

2. Combined lattice and impurity mobility

For electrons, suppose $\mu_L=0.180\ \mathrm{m^2\,V^{-1}s^{-1}}$ and $\mu_I=0.450\ \mathrm{m^2\,V^{-1}s^{-1}}$. If $n=3.00\times10^{21}\ \mathrm{m^{-3}}$ and holes are negligible, find the total mobility and resistivity.

Matthiessen’s rule gives

\[\mu=\left(\frac1{0.180}+\frac1{0.450}\right)^{-1} =0.1286\ \mathrm{m^2\,V^{-1}s^{-1}}.\]

Thus

\[\sigma=qn\mu=61.80\ \mathrm{S\,m^{-1}}, \qquad \rho=1.618\times10^{-2}\ \Omega\,\mathrm m.\]

Mobility and resistivity are positive magnitudes. The combined mobility is smaller than either limiting mobility, which checks the added-scattering interpretation.

Descriptive Questions

Derive the charge-neutrality equation for a semiconductor containing electrons, holes, ionized donors, and ionized acceptors.
Explain compensation and derive the exact majority- and minority-carrier concentrations for n-type material.
Describe the freeze-out, extrinsic, and intrinsic temperature ranges of a doped semiconductor.
Derive the relaxation-time expression for mobility and explain Matthiessen’s rule.

Numerical Problems

A compensated sample has measured equilibrium concentrations $n=5.00\times10^{20}\ \mathrm{m^{-3}}$ and $p=2.00\times10^{11}\ \mathrm{m^{-3}}$. Its total implanted impurity density is $N_D+N_A=1.40\times10^{21}\ \mathrm{m^{-3}}$, and all impurities are ionized. Infer $N_D$, $N_A$, and the acceptor compensation percentage $100N_A/N_D$.

Final answer: $N_D=9.500\times10^{20}\ \mathrm{m^{-3}}$, $N_A=4.500\times10^{20}\ \mathrm{m^{-3}}$, and $100N_A/N_D=47.37\%$.
A semiconductor has $E_g=1.12\ \mathrm{eV}$ and $n_i(300\ \mathrm K)=1.00\times10^{16}\ \mathrm{m^{-3}}$. Assuming temperature-independent effective masses, use $n_i\propto T^{3/2}\exp[-E_g/(2k_BT)]$ to estimate $n_i(400\ \mathrm K)$.

Final answer: $n_i(400)/n_i(300)=(400/300)^{3/2}\exp!\left[\frac{E_g}{2k_B}\left(\frac{1}{300}-\frac{1}{400}\right)\right]=346.2$, so $n_i(400\ \mathrm K)=3.462\times10^{18}\ \mathrm{m^{-3}}$.
An electron has $m^*=0.26m_0$ and $\tau=2.00\times10^{-13}\ \mathrm s$. Find its mobility magnitude.

Final answer: $\mu=0.1353\ \mathrm{m^2\,V^{-1}s^{-1}}$.
If lattice-limited mobility is $0.140\ \mathrm{m^2\,V^{-1}s^{-1}}$ at $300\ \mathrm K$ and varies as $T^{-3/2}$, find it at $400\ \mathrm K$.

Final answer: $\mu_L(400\ \mathrm K)=0.09093\ \mathrm{m^2\,V^{-1}s^{-1}}$.
A mobility measurement gives $\mu=0.120\ \mathrm{m^2\,V^{-1}s^{-1}}$ while the independently known lattice-limited mobility is $\mu_L=0.200\ \mathrm{m^2\,V^{-1}s^{-1}}$. Use Matthiessen’s rule to extract the impurity-limited mobility $\mu_I$.

Final answer: $\mu_I=[1/\mu-1/\mu_L]^{-1}=0.300\ \mathrm{m^2\,V^{-1}s^{-1}}$.
At $300\ \mathrm K$, the lattice- and impurity-limited mobilities are $\mu_L=0.400\ \mathrm{m^2\,V^{-1}s^{-1}}$ and $\mu_I=0.0500\ \mathrm{m^2\,V^{-1}s^{-1}}$. If $\mu_L\propto T^{-3/2}$ and $\mu_I\propto T^{3/2}$, find the temperature at which the two limits are equal and the total mobility there.

Final answer: Equating the limits gives $(T/300)^3=0.400/0.0500=8$, hence $T=600\ \mathrm K$. Both limiting mobilities are then $0.1414\ \mathrm{m^2\,V^{-1}s^{-1}}$, so Matthiessen’s rule gives $\mu=0.07071\ \mathrm{m^2\,V^{-1}s^{-1}}$.

All symbolic reductions and numerical values are reproducible in the accessible Maxima worksheet for doping, carrier concentration, and mobility.