Semiconductors

Diffusion is the net flow of carriers from a region of high concentration to a region of low concentration, caused by their random thermal motion. It needs no electric field; only a concentration gradient.

Diffusion process

Consider electrons with concentration $n(x)$ falling with $x$. Each carrier moves randomly, but more carriers cross a plane from the crowded side than from the sparse side, so there is a net flux towards lower concentration.

  n(x)
   |\
   | \      net electron flow -->
   |  \     electron current  <--
   |   `-._
   |       `--.___
   +--------------- x

By Fick's law, the particle flux is proportional to the gradient:

$$\Gamma_e = -D_e \frac{dn}{dx}, \qquad \Gamma_h = -D_h \frac{dp}{dx}$$

where $D_e$, $D_h$ are diffusion coefficients (cm$^2$/s). Since electrons carry charge $-e$ and holes $+e$, the diffusion current densities are

$$J_{D,e} = e D_e \frac{dn}{dx}, \qquad J_{D,h} = -e D_h \frac{dp}{dx}$$

With an electric field $E$ also present, the total currents are

$$\begin{aligned} J_e &= e n \mu_e E + e D_e \frac{dn}{dx} \\ J_h &= e p \mu_h E - e D_h \frac{dp}{dx} \end{aligned}$$

Einstein relation

Consider an n-type semiconductor with non-uniform doping, so $n$ varies with $x$, in equilibrium (no applied voltage).

1. Electrons diffuse from high to low $n$, leaving positive donor ions behind. This builds an internal field $E$ that drives a drift current opposing diffusion.
2. In equilibrium, the Fermi level $E_F$ is flat and the net current is zero:

$$J_e = e n \mu_e E + e D_e \frac{dn}{dx} = 0 \quad (1)$$

3. The bands bend, since $E_c - E_F$ changes with $x$. For a non-degenerate semiconductor:

$$n(x) = N_c \exp\left[-\frac{E_c(x) - E_F}{kT}\right]$$ $$\frac{dn}{dx} = -\frac{n}{kT}\frac{dE_c}{dx} \quad (2)$$

4. The electron potential energy is $E_c = \text{const} - eV(x)$, and $E = -dV/dx$, so

$$\frac{dE_c}{dx} = -e\frac{dV}{dx} = eE \quad (3)$$

5. From (2) and (3): $\dfrac{dn}{dx} = -\dfrac{neE}{kT}$. Put this in (1):

$$\begin{aligned} e n \mu_e E - e D_e \frac{neE}{kT} &= 0 \\ \mu_e &= \frac{e D_e}{kT} \\ \frac{D_e}{\mu_e} &= \frac{kT}{e} \end{aligned}$$

The same steps for holes give $D_h/\mu_h = kT/e$. So

$$\frac{D_e}{\mu_e} = \frac{D_h}{\mu_h} = \frac{kT}{e}$$

This is the Einstein relation. At 300 K, $kT/e = 0.0259$ V. For Si with $\mu_e = 1350$ cm$^2$/Vs, $D_e = 0.0259 \times 1350 = 35$ cm$^2$/s. It links diffusion (random motion) and drift (field-driven motion), since both arise from the same scattering processes.

A non-degenerate semiconductor has a carrier concentration much smaller than the effective density of states ($n \ll N_c$, $p \ll N_v$), so the Fermi level lies in the band gap at least a few $kT$ from the band edges. A degenerate semiconductor is so heavily doped (above about $10^{19}$ cm$^{-3}$ in Si) that the Fermi level enters the conduction band (n$^+$) or valence band (p$^+$), and it behaves more like a metal.

   Non-degenerate n        Degenerate n+
  ---------- Ec          ~~~~~~~~~~ EF
  - - - - -  Ed          ---------- Ec
  .......... EF          ////////// impurity band
                          merged with CB
  ---------- Ev          ---------- Ev

The Einstein relation links the diffusion coefficient and drift mobility of a carrier: $D_e/\mu_e = kT/e$. It is derived by noting that, in equilibrium, drift and diffusion currents in a non-uniformly doped sample cancel.

Setup

Take an n-type semiconductor whose donor concentration falls with $x$, in thermal equilibrium (no applied voltage).

 Ec  \__
        \___
            \_______         (band bends)
 EF ---------------------   (flat in equilibrium)

 n high              n low
   diffusion of e- ---->
   drift of e-     <----  (due to built-in E)

• Electrons diffuse towards low $n$, leaving positive donor ions behind.
• This creates an internal electric field $E$, which drives electrons back by drift.
• In equilibrium the net electron current is zero:

$$J_e = e n \mu_e E + e D_e \frac{dn}{dx} = 0 \quad (1)$$

Electron concentration and band bending

For a non-degenerate semiconductor,

$$n(x) = N_c \exp\left[-\frac{E_c(x) - E_F}{kT}\right]$$

Since $E_F$ is constant,

$$\frac{dn}{dx} = -\frac{n}{kT}\frac{dE_c}{dx} \quad (2)$$

The conduction band edge follows the electron potential energy, $E_c(x) = \text{const} - eV(x)$. With $E = -dV/dx$:

$$\frac{dE_c}{dx} = eE \quad (3)$$

From (2) and (3):

$$\frac{dn}{dx} = -\frac{n e E}{kT} \quad (4)$$

Result

Substitute (4) in (1):

$$\begin{aligned} e n \mu_e E - e D_e \frac{n e E}{kT} &= 0 \\ \mu_e &= \frac{e D_e}{kT} \\ \frac{D_e}{\mu_e} &= \frac{kT}{e} \end{aligned}$$

This is the Einstein relation for electrons. A similar result holds for holes: $D_h/\mu_h = kT/e$.

At 300 K, $kT/e \approx 25.9$ mV. For Si, $\mu_e = 1350$ cm$^2$V$^{-1}$s$^{-1}$ gives $D_e \approx 35$ cm$^2$/s. Thus, once mobility is measured (e.g. from Hall and conductivity data), the diffusion coefficient follows directly.

In an intrinsic semiconductor, $n = p = n_i$ with

$$n_i = \sqrt{N_c N_v}\,\exp\left(-\frac{E_g}{2kT}\right), \quad N_c = 2\left(\frac{2\pi m_e^* kT}{h^2}\right)^{3/2}, \quad N_v = 2\left(\frac{2\pi m_h^* kT}{h^2}\right)^{3/2}$$

Data: $m_e^* = 1.08m_e$, $m_h^* = 0.60m_e$, $m_e = 9.109\times10^{-31}$ kg, $k = 1.381\times10^{-23}$ J/K, $h = 6.626\times10^{-34}$ J s, $T = 300$ K, $E_g = 1.10$ eV, $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$. Then $kT = 0.02586$ eV.

Effective densities of states

$$\begin{aligned} N_c &= 2\left[\frac{2\pi (1.08 \times 9.109\times10^{-31})(1.381\times10^{-23})(300)}{(6.626\times10^{-34})^2}\right]^{3/2} \\ &= 2.82\times10^{25}\ \text{m}^{-3} = 2.82\times10^{19}\ \text{cm}^{-3} \\ N_v &= N_c\left(\frac{0.60}{1.08}\right)^{3/2} = 1.17\times10^{19}\ \text{cm}^{-3} \end{aligned}$$

Intrinsic concentration

$$\begin{aligned} \frac{E_g}{2kT} &= \frac{1.10}{2 \times 0.02586} = 21.27 \\ \sqrt{N_c N_v} &= \sqrt{2.82\times10^{19} \times 1.17\times10^{19}} = 1.813\times10^{19}\ \text{cm}^{-3} \\ n_i &= 1.813\times10^{19} \times e^{-21.27} \\ &= 1.813\times10^{19} \times 5.80\times10^{-10} \\ &= 1.05\times10^{10}\ \text{cm}^{-3} \end{aligned}$$

Intrinsic resistivity

$$\begin{aligned} \sigma_i &= e n_i (\mu_e + \mu_h) \\ &= 1.602\times10^{-19} \times 1.052\times10^{10} \times (1350 + 450) \\ &= 3.03\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ \rho_i &= \frac{1}{\sigma_i} = 3.30\times10^{5}\ \Omega\,\text{cm} \end{aligned}$$

Answer: $n_i \approx 1.05\times10^{10}$ cm$^{-3}$; $\rho_i \approx 3.3\times10^{5}$ $\Omega$ cm ($3.3\times10^{3}$ $\Omega$ m).

Built-in potential

Consider an abrupt pn junction with acceptor concentration $N_a$ on the p-side and donor concentration $N_d$ on the n-side. Assumptions: abrupt junction, full ionisation, non-degenerate doping, depletion approximation (no free carriers in the depletion region), equilibrium.

When the two sides are joined, holes diffuse from p to n and electrons from n to p. They leave behind uncovered negative acceptor ions on the p-side and positive donor ions on the n-side. These form a space charge (depletion) region whose field opposes further diffusion. In equilibrium the Fermi level is constant through the junction and the bands bend by $eV_0$.

        p-side  |  depletion  |  n-side
                | -Wp  0  +Wn |
  Ec ___________
                \_____________
  EF ............................. (flat)
  Ev ___________   eV0
                \_____________

In equilibrium, carrier concentrations at the two ends of the depletion region follow the Boltzmann factor:

$$\frac{p_{p0}}{p_{n0}} = \exp\left(\frac{eV_0}{kT}\right)$$

With $p_{p0} = N_a$ and $p_{n0} = n_i^2/N_d$:

$$\begin{aligned} \frac{N_a N_d}{n_i^2} &= \exp\left(\frac{eV_0}{kT}\right) \\ V_0 &= \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right) \end{aligned}$$

(The same result follows from $eV_0 = E_{Fn} - E_{Fp}$ before contact, using $E_{Fn} - E_{Fi} = kT\ln(N_d/n_i)$ and $E_{Fi} - E_{Fp} = kT\ln(N_a/n_i)$.)

Depletion width

Charge density: $\rho = -eN_a$ for $-W_p < x < 0$ and $\rho = +eN_d$ for $0 < x < W_n$.

  rho
  +eNd     +-----+
           |     |
 ---+------+-----+--- x
 -Wp|   0        Wn
    +------+
  -eNa

1. Charge neutrality:

$$N_a W_p = N_d W_n$$

2. Poisson's equation: $\dfrac{dE}{dx} = \dfrac{\rho}{\varepsilon}$, with $\varepsilon = \varepsilon_0\varepsilon_r$. Integrating from $-W_p$ (where $E = 0$) gives a field that is maximum at the junction:

$$E_0 = -\frac{eN_d W_n}{\varepsilon} = -\frac{eN_a W_p}{\varepsilon}$$

3. Potential: $V_0$ is the area under the triangular field plot:

$$V_0 = -\frac{1}{2}E_0 W = \frac{eN_d W_n W}{2\varepsilon}, \quad W = W_n + W_p$$

4. From neutrality, $W_n = \dfrac{N_a W}{N_a + N_d}$. Substituting:

$$\begin{aligned} V_0 &= \frac{e N_a N_d W^2}{2\varepsilon (N_a + N_d)} \\ W &= \sqrt{\frac{2\varepsilon (N_a + N_d) V_0}{e N_a N_d}} \end{aligned}$$

   E(x)
  ---+-----------+---- x
  -Wp \         / Wn
       \       /
        \     /
         \   /
          \ /
          -E0

The depletion region extends mostly into the lightly doped side ($W_n/W_p = N_a/N_d$). With reverse bias $V_r$, replace $V_0$ by $V_0 + V_r$.

Resistance of a cube with side $L = 1$ cm: $R = \rho L/A = \rho \times \dfrac{1\ \text{cm}}{1\ \text{cm}^2}$, so $R$ in $\Omega$ equals $\rho$ in $\Omega$ cm.

Data: $n_i = 1.45\times10^{10}$ cm$^{-3}$, $N_{Si} = 5\times10^{22}$ cm$^{-3}$, $e = 1.602\times10^{-19}$ C. Assumed room-temperature mobilities for Si: $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$ (at such low doping, impurity scattering is negligible).

(a) Pure silicon

$$\begin{aligned} \sigma &= e n_i(\mu_e + \mu_h) \\ &= 1.602\times10^{-19} \times 1.45\times10^{10} \times 1800 \\ &= 4.18\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 2.39\times10^{5}\ \Omega \end{aligned}$$

(b) Doped with 1 As per $10^9$ Si atoms

Arsenic is a donor:

$$\begin{aligned} N_d &= \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3} \gg n_i \\ n &\approx N_d = 5\times10^{13}\ \text{cm}^{-3} \\ p &= \frac{n_i^2}{n} = \frac{(1.45\times10^{10})^2}{5\times10^{13}} = 4.2\times10^{6}\ \text{cm}^{-3} \ (\text{negligible}) \\ \sigma &= e n\mu_e = 1.602\times10^{-19} \times 5\times10^{13} \times 1350 \\ &= 1.081\times10^{-2}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 92.5\ \Omega \end{aligned}$$

(c) Also doped with 1 B per billion Si atoms

Boron is an acceptor: $N_a = 5\times10^{13}$ cm$^{-3}$ $= N_d$.

The acceptors capture all the electrons given by the donors (full compensation). The net doping $N_d - N_a = 0$, so the crystal behaves as intrinsic: $n = p = n_i$.

$$R \approx 2.39\times10^{5}\ \Omega$$

Answer: Pure Si: $R = 2.39\times10^{5}$ $\Omega$. With 1 As per $10^9$: $R = 92.5$ $\Omega$. With 1 As and 1 B per $10^9$: compensated, $R$ returns to about $2.39\times10^{5}$ $\Omega$.

Donor dopants in n-type silicon

An n-type semiconductor is made by adding a small amount of a pentavalent (group V) impurity such as P, As or Sb to Si or Ge.

1. The donor atom replaces a Si atom in the lattice. Four of its five valence electrons form covalent bonds with the four neighbouring Si atoms.
2. The fifth electron has no bond to join. It stays only weakly bound to the positive donor ion, like the electron in a hydrogen atom but in a medium of high permittivity ($\varepsilon_r = 11.9$) and with a small effective mass.
3. Its binding (ionisation) energy is therefore tiny, about 0.045 eV for P in Si (hydrogen model: $E_b = 13.6\,\frac{m_e^*/m_e}{\varepsilon_r^2}$ eV).
4. In the band diagram this electron occupies a donor level $E_d$ just below $E_c$ (about 0.05 eV).
5. At room temperature $kT = 0.026$ eV is enough to ionise almost every donor, so each donor gives one electron to the conduction band and becomes a fixed positive ion $D^+$. Hence $n \approx N_d$.
6. By the mass action law $np = n_i^2$, the hole concentration falls far below $n_i$, so electrons are the majority carriers.

            Ec ----------------------
                 e-  e-  e-   (free)
            Ed  - + - + - + -  ~0.05 eV
                (ionised donors)

            EF  .......... (near Ec)

            Ev ----------------------

     Si -- Si -- Si
     |     |     |
     Si -- P+ -- Si    e- (5th electron,
     |     |     |         loosely bound)
     Si -- Si -- Si

Proof of $\sigma = ne\mu_e$

Consider a bar of n-type material of cross-section $A$, with electron concentration $n$, in an electric field $E$.

1. The field accelerates electrons, but collisions with the lattice limit them to an average drift velocity $v_d$ opposite to $E$. The drift mobility is defined as

$$\mu_e = \frac{v_d}{E} \;\Rightarrow\; v_d = \mu_e E$$

(From the Drude model, $v_d = \frac{e\tau}{m_e^*}E$, so $\mu_e = \frac{e\tau}{m_e^*}$, where $\tau$ is the mean free time.)

2. In time $\Delta t$, electrons travel $v_d\Delta t$. All electrons in the volume $A v_d \Delta t$ cross a given section. Charge crossing:

$$\Delta Q = e\,n\,A\,v_d\,\Delta t$$

3. Current density:

$$J = \frac{\Delta Q}{A\,\Delta t} = n e v_d = n e \mu_e E$$

4. Ohm's law in point form is $J = \sigma E$. Comparing:

$$\sigma = n e \mu_e$$

If holes are also present, their contribution adds: $\sigma = e(n\mu_e + p\mu_h)$. In n-type material $n \gg p$, so $\sigma \approx ne\mu_e \approx N_d e\mu_e$.

Example: Si with $N_d = 10^{16}$ cm$^{-3}$, $\mu_e = 1350$ cm$^2$/Vs gives $\sigma = 10^{16}\times1.602\times10^{-19}\times1350 = 2.16$ S/cm.

Formulas (abrupt junction, depletion approximation):

$$V_0 = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}, \quad W = \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}, \quad W_n = \frac{N_a W}{N_a+N_d}, \quad W_p = \frac{N_d W}{N_a+N_d}$$

Constants: $kT/e = 0.02586$ V at 300 K, $e = 1.602\times10^{-19}$ C, $\varepsilon = \varepsilon_r\varepsilon_0 = 11.9 \times 8.854\times10^{-14} = 1.054\times10^{-12}$ F/cm (silicon assumed).

Given: $N_a = 10^{18}$ cm$^{-3}$, $N_d = 10^{16}$ cm$^{-3}$, $n_i = 1.45\times10^{10}$ cm$^{-3}$, $T = 300$ K.

Built-in potential

$$\begin{aligned} \frac{N_aN_d}{n_i^2} &= \frac{10^{18}\times10^{16}}{(1.45\times10^{10})^2} = 4.76\times10^{13} \\ V_0 &= 0.02586 \times \ln(4.76\times10^{13}) = 0.02586 \times 31.49 \\ &= 0.814\ \text{V} \end{aligned}$$

Total depletion width

$$\begin{aligned} W &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times (10^{18}+10^{16}) \times 0.814}{1.602\times10^{-19} \times 10^{18} \times 10^{16}}} \\ &= 3.29\times10^{-5}\ \text{cm} = 0.329\ \mu\text{m} \end{aligned}$$

Width on each side

$$\begin{aligned} W_n &= \frac{N_a}{N_a+N_d}W = \frac{10^{18}}{1.01\times10^{18}} \times 0.329 = 0.326\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W = \frac{10^{16}}{1.01\times10^{18}} \times 0.329 = 3.26\times10^{-3}\ \mu\text{m} = 3.26\ \text{nm} \end{aligned}$$

Check: $N_aW_p = 10^{18}\times3.26\times10^{-7} = 3.26\times10^{11}$ cm$^{-2}$ $= N_dW_n$. Since this is a p$^+$n junction, almost all the depletion region lies in the lightly doped n-side. (The peak field is $E_0 = eN_dW_n/\varepsilon = 4.95\times10^4$ V/cm.)

Answer: $V_0 = 0.814$ V; $W_n = 0.326$ $\mu$m; $W_p = 3.26$ nm; $W = 0.329$ $\mu$m.

Formulas (abrupt junction, depletion approximation):

$$V_0 = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2}, \quad W = \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}}, \quad W_n = \frac{N_a W}{N_a+N_d}, \quad W_p = \frac{N_d W}{N_a+N_d}$$

Constants: $kT/e = 0.02586$ V at 300 K, $e = 1.602\times10^{-19}$ C, $\varepsilon = \varepsilon_r\varepsilon_0 = 11.9 \times 8.854\times10^{-14} = 1.054\times10^{-12}$ F/cm (silicon assumed).

Given: $N_d = 10^{17}$ cm$^{-3}$ (n-side), $N_a = 10^{16}$ cm$^{-3}$ (p-side). $n_i$ and $T$ are not given; assume silicon at 300 K with $n_i = 1.45\times10^{10}$ cm$^{-3}$.

(i) Built-in potential

$$\begin{aligned} \frac{N_aN_d}{n_i^2} &= \frac{10^{16}\times10^{17}}{(1.45\times10^{10})^2} = 4.76\times10^{12} \\ V_0 &= 0.02586 \times \ln(4.76\times10^{12}) = 0.02586 \times 29.19 \\ &= 0.755\ \text{V} \end{aligned}$$

(ii) Depletion widths

$$\begin{aligned} W_0 &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times (10^{16}+10^{17}) \times 0.755}{1.602\times10^{-19} \times 10^{16} \times 10^{17}}} \\ &= 3.30\times10^{-5}\ \text{cm} = 0.330\ \mu\text{m} \\ W_n &= \frac{N_a}{N_a+N_d}W_0 = \frac{10^{16}}{1.1\times10^{17}} \times 0.3305 = 0.0300\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W_0 = \frac{10^{17}}{1.1\times10^{17}} \times 0.3305 = 0.300\ \mu\text{m} \end{aligned}$$

The depletion region lies mostly in the lightly doped p-side.

(iii) Field at the metallurgical junction

$$\begin{aligned} |E_0| &= \frac{eN_dW_n}{\varepsilon} = \frac{1.602\times10^{-19} \times 10^{17} \times 3.005\times10^{-6}}{1.054\times10^{-12}} \\ &= 4.57\times10^{4}\ \text{V/cm} = 4.57\ \text{MV/m} \end{aligned}$$

It points from the n-side to the p-side ($E_0 = -4.57\times10^4$ V/cm if $x$ is measured from p to n). Check: $\frac{1}{2}|E_0|W_0 = 0.5 \times 4.57\times10^4 \times 3.305\times10^{-5} = 0.755$ V $= V_0$.

Answer: $V_0 = 0.755$ V; $W_0 = 0.330$ $\mu$m, $W_n = 0.030$ $\mu$m, $W_p = 0.300$ $\mu$m; $|E_0| = 4.57\times10^4$ V/cm.

The Fermi level $E_F$ is the energy at which the probability of occupation by an electron is exactly one half, from the Fermi-Dirac function

$$f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)}$$

In a semiconductor it usually lies in the band gap, where there are no states, but it still fixes how many electrons and holes are present.

Importance

1. Sets carrier concentrations. For a non-degenerate semiconductor:

$$n = N_c\exp\left[-\frac{E_c - E_F}{kT}\right], \qquad p = N_v\exp\left[-\frac{E_F - E_v}{kT}\right]$$

The closer $E_F$ is to $E_c$, the more electrons; the closer to $E_v$, the more holes.

2. Shows the type of semiconductor.

 Intrinsic       n-type          p-type
 --- Ec          --- Ec          --- Ec
                 ... EF
 ... EF = EFi    (near Ec)
 (mid-gap)                       ... EF
 --- Ev          --- Ev          --- Ev  (near Ev)

• Intrinsic: $E_{Fi}$ near mid-gap, $E_{Fi} = \frac{E_c+E_v}{2} + \frac{3}{4}kT\ln\frac{m_h^*}{m_e^*}$.
• n-type: $E_F - E_{Fi} = kT\ln(N_d/n_i)$, above mid-gap.
• p-type: $E_{Fi} - E_F = kT\ln(N_a/n_i)$, below mid-gap.

3. Measures doping. The shift of $E_F$ from $E_{Fi}$ tells the doping level; heavy doping pushes $E_F$ into a band (degenerate semiconductor).

4. Constant in equilibrium. In any system in thermal equilibrium, $E_F$ is the same everywhere. This rule gives band bending, the built-in potential of a pn junction ($eV_0 = E_{Fn} - E_{Fp}$ before contact) and metal-semiconductor contact potentials.

5. Applied voltage. An applied voltage $V$ separates the Fermi levels on the two sides by $eV$. $dE_F/dx = 0$ means no net current; a gradient in $E_F$ drives current.

6. Temperature behaviour. As $T$ rises, $E_F$ of an extrinsic semiconductor moves towards mid-gap, showing the change to intrinsic behaviour.

7. Work function. $E_F$ fixes the work function $\Phi = E_{vac} - E_F$, which decides contact behaviour (ohmic or Schottky).

When both donors and acceptors are present, they compensate each other and the larger one decides the type.

Given: $N_d = 10^{16}$ cm$^{-3}$ (P), $N_a = 10^{17}$ cm$^{-3}$ (B). Assume Si at 300 K, $n_i = 1.45\times10^{10}$ cm$^{-3}$, full ionisation.

Since $N_a > N_d$, the sample becomes p-type. The donors' electrons fill $10^{16}$ acceptors; the rest of the acceptors create holes.

$$\begin{aligned} p &= N_a - N_d = 10^{17} - 10^{16} = 9\times10^{16}\ \text{cm}^{-3} \\ n &= \frac{n_i^2}{p} = \frac{(1.45\times10^{10})^2}{9\times10^{16}} = 2.34\times10^{3}\ \text{cm}^{-3} \end{aligned}$$

Answer: $p = 9\times10^{16}$ cm$^{-3}$, $n \approx 2.3\times10^{3}$ cm$^{-3}$ (the sample is now p-type).

The carrier concentration in a semiconductor depends strongly on temperature because carriers are created thermally, either by ionising impurities or by breaking covalent bonds across the band gap.

Intrinsic semiconductor

Electron-hole pairs are created when thermal energy lifts electrons across $E_g$:

$$n_i = \sqrt{N_cN_v}\,\exp\left(-\frac{E_g}{2kT}\right), \quad N_c, N_v \propto T^{3/2}$$

so $n_i \propto T^{3/2}e^{-E_g/2kT}$. The exponential dominates: for Si, $n_i$ roughly doubles for every 8 to 10 K rise near room temperature.

Extrinsic (n-type) semiconductor

A plot of $\ln n$ against $1/T$ shows three regions:

1. Low temperature (ionisation or freeze-out region): only some donors are ionised.

$$n = \left(\frac{1}{2}N_cN_d\right)^{1/2}\exp\left(-\frac{\Delta E}{2kT}\right), \quad \Delta E = E_c - E_d$$

2. Medium temperature (extrinsic or saturation region): all donors are ionised, $n \approx N_d$, nearly constant. This range (about 150 K to 500 K for Si with $N_d = 10^{15}$ cm$^{-3}$) is where devices work.

3. High temperature (intrinsic region): thermal generation across $E_g$ exceeds $N_d$; $n \approx n_i$ and the material behaves as intrinsic. The changeover temperature $T_i$ is where $n_i \approx N_d$.

 ln n
   |        ionisation   extrinsic    intrinsic
   |          region      region       region
   |                                    /
   |                                   / slope
   |          ___________________     /  -Eg/2k
   |        /                    \__/
   |       / slope -ΔE/2k   n = Nd
   |      /
   +------|------------------|----------- 1/T
         1/Ts               1/Ti
   (high T at left, low T at right)

Fermi level

As $T$ rises, $E_F$ moves from near $E_d$ (low $T$) towards mid-gap (high $T$), because the extrinsic carriers become a smaller fraction of the total.

The upper limit of device operation is set by $T_i$; wide band gap materials (GaAs, SiC) have higher $T_i$ and can work at higher temperatures.

In a doped semiconductor, conductivity $\sigma = e(n\mu_e + p\mu_h)$. Both factors, the carrier concentration and the drift mobility, change with temperature.

Carrier concentration vs temperature

For n-type doping $N_d$:

1. Ionisation region (low T): $n = \left(\tfrac{1}{2}N_cN_d\right)^{1/2}\exp(-\Delta E/2kT)$; $n$ rises as more donors ionise.
2. Extrinsic region: all donors ionised, $n \approx N_d$ (constant).
3. Intrinsic region (high T): $n \approx n_i \propto T^{3/2}\exp(-E_g/2kT)$; rises steeply.

 ln n
   |        ionisation   extrinsic    intrinsic
   |          region      region       region
   |                                    /
   |                                   / slope
   |          ___________________     /  -Eg/2k
   |        /                    \__/
   |       / slope -ΔE/2k   n = Nd
   |      /
   +------|------------------|----------- 1/T
         1/Ts               1/Ti
   (high T at left, low T at right)

Drift mobility vs temperature

Mobility $\mu = e\tau/m^*$ depends on the mean time between scattering events. Two scattering processes dominate:

1. Lattice (phonon) scattering: atomic vibrations grow with $T$, so $\tau$ falls:

$$\mu_L \propto T^{-3/2}$$

2. Ionised impurity scattering: a slow carrier is deflected more by charged dopant ions; faster (hotter) carriers are deflected less:

$$\mu_I \propto \frac{T^{3/2}}{N_I}$$

where $N_I$ is the ionised impurity concentration.

3. Matthiessen's rule combines them:

$$\frac{1}{\mu} = \frac{1}{\mu_L} + \frac{1}{\mu_I}$$

At low $T$, impurity scattering controls $\mu$ (rises with $T$); at high $T$, lattice scattering controls it (falls with $T$). So $\mu$ has a peak, and heavier doping lowers the curve and shifts the peak to higher $T$.

  log μ
    |         .-''-.          lightly doped
    |       /       `-.
    |     /    .--.    `-.
    |   /    /     `-.    `-. ~T^-3/2
    |  /   /  heavily  `-.
    | /  /    doped       `-.
    |/ / ~T^3/2
    +--------------------------- log T

Result

Conductivity follows mainly $n$: rises at low $T$, roughly flat (falling slightly with $\mu$) in the extrinsic range, then rises sharply in the intrinsic range. So both carrier concentration and mobility are strongly temperature dependent, and device behaviour must account for both.

Minority carrier suppression is the reduction of the minority carrier concentration far below $n_i$ when a semiconductor is doped. Adding donors increases electrons and at the same time decreases holes (and vice versa for acceptors).

Reason

In thermal equilibrium, the mass action law holds for a non-degenerate semiconductor:

$$np = n_i^2$$

$n_i^2$ depends only on the material and temperature, not on doping. So if doping raises the majority concentration, the minority concentration must fall in proportion. Physically, the large number of electrons increases the recombination rate with holes, so fewer holes survive.

Example

Si at 300 K, $n_i = 1.45\times10^{10}$ cm$^{-3}$, doped with $N_d = 10^{16}$ cm$^{-3}$:

$$\begin{aligned} n &\approx N_d = 10^{16}\ \text{cm}^{-3} \\ p &= \frac{n_i^2}{n} = \frac{(1.45\times10^{10})^2}{10^{16}} = 2.1\times10^{4}\ \text{cm}^{-3} \end{aligned}$$

The hole concentration falls from $1.45\times10^{10}$ to $2.1\times10^4$ cm$^{-3}$, nearly a million times smaller.

Consequences

• Conductivity is due almost entirely to majority carriers: $\sigma \approx eN_d\mu_e$.
• Minority carriers, though few, control pn junction reverse saturation current and transistor action.

A pn junction is the boundary formed inside a single semiconductor crystal when one region is doped p-type (acceptors, $N_a$) and the adjacent region n-type (donors, $N_d$). It is the basic structure of diodes, BJTs, solar cells and LEDs.

   p-type           |           n-type
 + + + + +   - - |+ +   - - - - -
 + + + + +   - - |+ +   - - - - -
 holes (+)    depletion    electrons (-)
              region W
            <----E----

When formed, holes diffuse from p to n and electrons from n to p and recombine near the junction. This leaves a depletion region of fixed negative acceptor ions (p-side) and positive donor ions (n-side) with a built-in field from n to p, which stops further net diffusion.

Built-in potential

Consider an abrupt pn junction with acceptor concentration $N_a$ on the p-side and donor concentration $N_d$ on the n-side. Assumptions: abrupt junction, full ionisation, non-degenerate doping, depletion approximation (no free carriers in the depletion region), equilibrium.

When the two sides are joined, holes diffuse from p to n and electrons from n to p. They leave behind uncovered negative acceptor ions on the p-side and positive donor ions on the n-side. These form a space charge (depletion) region whose field opposes further diffusion. In equilibrium the Fermi level is constant through the junction and the bands bend by $eV_0$.

        p-side  |  depletion  |  n-side
                | -Wp  0  +Wn |
  Ec ___________
                \_____________
  EF ............................. (flat)
  Ev ___________   eV0
                \_____________

In equilibrium, carrier concentrations at the two ends of the depletion region follow the Boltzmann factor:

$$\frac{p_{p0}}{p_{n0}} = \exp\left(\frac{eV_0}{kT}\right)$$

With $p_{p0} = N_a$ and $p_{n0} = n_i^2/N_d$:

$$\begin{aligned} \frac{N_a N_d}{n_i^2} &= \exp\left(\frac{eV_0}{kT}\right) \\ V_0 &= \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right) \end{aligned}$$

(The same result follows from $eV_0 = E_{Fn} - E_{Fp}$ before contact, using $E_{Fn} - E_{Fi} = kT\ln(N_d/n_i)$ and $E_{Fi} - E_{Fp} = kT\ln(N_a/n_i)$.)

Depletion width

Charge density: $\rho = -eN_a$ for $-W_p < x < 0$ and $\rho = +eN_d$ for $0 < x < W_n$.

  rho
  +eNd     +-----+
           |     |
 ---+------+-----+--- x
 -Wp|   0        Wn
    +------+
  -eNa

1. Charge neutrality:

$$N_a W_p = N_d W_n$$

2. Poisson's equation: $\dfrac{dE}{dx} = \dfrac{\rho}{\varepsilon}$, with $\varepsilon = \varepsilon_0\varepsilon_r$. Integrating from $-W_p$ (where $E = 0$) gives a field that is maximum at the junction:

$$E_0 = -\frac{eN_d W_n}{\varepsilon} = -\frac{eN_a W_p}{\varepsilon}$$

3. Potential: $V_0$ is the area under the triangular field plot:

$$V_0 = -\frac{1}{2}E_0 W = \frac{eN_d W_n W}{2\varepsilon}, \quad W = W_n + W_p$$

4. From neutrality, $W_n = \dfrac{N_a W}{N_a + N_d}$. Substituting:

$$\begin{aligned} V_0 &= \frac{e N_a N_d W^2}{2\varepsilon (N_a + N_d)} \\ W &= \sqrt{\frac{2\varepsilon (N_a + N_d) V_0}{e N_a N_d}} \end{aligned}$$

   E(x)
  ---+-----------+---- x
  -Wp \         / Wn
       \       /
        \     /
         \   /
          \ /
          -E0

The depletion region extends mostly into the lightly doped side ($W_n/W_p = N_a/N_d$). With reverse bias $V_r$, replace $V_0$ by $V_0 + V_r$.