Density of valence band holes

Pierrot’s Semiconductor Fundamentals gives a derivation of the number of conduction band electrons in a semiconductor. Here is the analogous math for the density of holes in the valence band. Let \(f(E)\) be the Fermi function, which gives the probability that a state of a specific energy \(E\) will be occupied by an electron when the Fermi level is \(E_F\):

\[ f(E)=\frac1{1+e^{(E-E_F)/kT}}. \]

Let \(g_v(E)\) be the density of states function, so that \(g_v(E)\,dE\) gives the number of states per unit volume in the range \((E,E+dE)\):

\[ g_v(E)=\frac{\sqrt{2}(m_p^{*})^{3/2}}{\pi^2\hbar^3}\sqrt{E_v-E}, \]

where \(m_p^{*}\) is the effective mass of the hole and \(E_v\) is the energy at the top of the valence band. This means that the number of states occupied by holes per unit of volume will be given by

\[ p=\int_{-\infty}^{E_v} g_v(E)\bigl(1-f(E)\bigr)\,dE. \]

In a non-degenerate semiconductor, the Fermi level \(E_F\) is many multiples of \(kT\) above the energies \(E\) we consider; for example, in silicon at 300K the quantity \((E-E_F)/kT\approx-22\). This lets us simplify our expression for \(f(E)\):

\[ f(E)=\frac1{1+e^{(E-E_F)/kT}}\approx1-e^{(E-E_F)/kT}. \]

So our integral becomes

\[ p=\frac{\sqrt{2}(m_p^{*})^{3/2}}{\pi^2\hbar^3} \int_{-\infty}^{E_v} \sqrt{E_v-E}\,e^{(E-E_F)/kT}\,dE. \]

Because the integral involves a power times an exponential, integrated to infinity, we attempt to make the integrand fit the definition of the gamma function,

\[ \Gamma(p+1)=\int_0^\infty\eta\,^p\,e^{-\eta}\,d\eta. \]

By multiplying the integrand by a constant, we can add something to \((E-E_F)/kT\) and multiply something into \(\sqrt{E_v-E}\). With that in mind, we choose \(\eta=(E_v-E)/kT\): we can freely multiply a \(1/kT\) into the square root and can freely shift \((E-E_F)/kT\) to \(-(E_v-E)/kT\). Also, when \(E=E_v\), we have \(\eta=0\). The substitution gives

\[ \begin{align*} p&=\frac{\sqrt{2}(m_p^{*})^{3/2}}{\pi^2\hbar^3} \sqrt{kT}\int_{-\infty}^{E_v}\sqrt{\frac{E_v-E}{kT}}\, \exp\left(-\frac{E_v-E}{kT}\right)\exp\left(\frac{E_v-E_f}{kT}\right)dE\\ &=\frac{\sqrt{2}(m_p^{*})^{3/2}}{\pi^2\hbar^3} (kT)^{3/2}\left[\int_0^\infty\sqrt{\eta}\,e^{-\eta}\,d\eta\right] \exp\left(\frac{E_v-E_F}{kT}\right)\\ &=\frac{\sqrt{2}(m_p^{*}kT)^{3/2}}{2\pi^{3/2}\hbar^3} \exp\left(\frac{E_v-E_F}{kT}\right). \end{align*} \]

The most important result from the derivation is the \(\exp((E_v-E_F)/kT)\) factor, because it lets us relate the ratio between different hole densities \(p_1/p_2\) to their Fermi levels \(E_{F2}-E_{F1}\). For example, if we know the hole density \(p_i\) of an intrinsic semiconductor, with intrinsic Fermi level \(E_i\), we can easily calculate the hole density for any other non-degenerate Fermi level:

\[ p=p_i\exp\left(\frac{E_i-E_F}{kT}\right). \]

Also, the relationship between \(p\) and \(E_F\) can be used both ways: if we control the hole density by doping, we can calculate the corresponding Fermi level in the semiconductor.

Etymologies

The mineral fluorite or fluorspar, with chemical formula CaF2, was used as a flux for iron smelting; its name, a Latinization of the German Flussspat, came from the same verb as the word flux: fluere, flūxus. It was this mineral that gave the root to the element name fluorine, and also to the term fluorescence.

However, the fluorescence of fluorite is not due to the fluorine but to mineral impurities, for example, blue from bivalent europium, yellow–green from bivalent ytterbium, and red from bivalent samarium or thulium (see Przibram, Nature, 1937). Interestingly, europium, ytterbium, and samarium are the only lanthanides with stable +2 aqueous oxidation states, although in 2013 it was shown possible to synthesize all lanthanides in the +2 state by MacDonald et. al.

Carmine, crimson, and kermes all derive from an Arabic word qirmiz, which refers to insects of the genus we now call Kermes. The related noun cochineal comes from the Greek κόκκος, referring to the kermes oak Quercus coccifera on which the insect feeds. Indeed, the word vermilion comes from Latin vermis, “worm,” specifically, Kermes vermilio.

Clade, a monophyletic taxon, that is, comprising a common ancestor and all its descendants, from Greek κλάδος, shoot or branch. Spume, froth, foam (used of the sea), from Latin spūma. Weft, a weaving term, coordinate with warp; from Old English wefan, to weave. Ombré, of clothing, a graduation from light to dark; past tense of French ombrer from Latin umbra, shadow. Svelte, slender, through French from Italian svelto < svellere < Latin vellere, to stretch. Carrel, small space for studying, reading; from Old French carole in a sense indicating a ring shape; perhaps from Latin chorus or corolla. Reindeer, from Old Norse hreinn reindeer, dýr animal; compare Old English hrān, dēor with the same meanings. Vitriol, metal sulfates, from Latin vitriolum < vitrum glass, referring to crystals or crystalline appearance; blue vitriol is a copper sulfate, while green vitriol is an iron sulfate.

Field outside a solenoid

Farley and Price, Am. J. Phys, 2001, show that the field just outside a solenoid of finite length \(L\) has relatively constant magnitude of order \(L^{-2}\). I understand the argument in section II but not in section III, which requires a better grasp on mutual inductance than I have.

In section II, they consider the closed surface consisting of a disk of radius \(R\) through the midpoint of the solenoid (perpendicular to its axis), and a hemisphere bounded by the disk which encloses one end of the solenoid. We take \(R\gg L\), so that to the hemisphere the solenoid appears to be a magnetic dipole. The field of that dipole at a distance \(R\) decreases as \(R^{-3}\), while the area of the hemisphere increases as \(R^2\), so as \(R\to\infty\), the flux through the hemisphere goes to zero.

The total flux (hemisphere and disk) must be zero because \(\nabla\cdot\mathbf B=0\); so the flux through the disk must also go to zero as \(R\to\infty\). The disk cuts through the solenoid, so there is certainly a flux of magnitude \(\mu_0nIA\) due to the interior magnetic field; that means that the flux from the exterior field must amount to the same magnitude and the opposite sign. How do we account for that?

Farther than some distance \(r_\mathrm{min}\gg L\), the field must again look like a dipole; the contribution to the disk flux from that far field is

\[ -\frac{\mu_0nIAL}{4\pi}\int_{r_\mathrm{min}}^\infty r^{-3}\,2\pi r\,dr =-\mu_0nIA\frac L{2r_\mathrm{min}}. \]

But \(L/2r_\mathrm{min}\ll1\), so the far field can’t account for the flux. This shows that the close field, at distances \(r<L\), provides a flux of magnitude \(\mu_0nIA\) across an area \(\propto L^2\); so the field itself must be proportional to \(L^{-2}\).

Callaghan and Maslen, in NASA Technical Note D-465, derive the full magnetic field due to an ideal finite solenoid. We can show directly from their formula that the field outside is \(\propto L^{-2}\).

Taking equation (8), we choose \(z=0\) to look at the center of the solenoid, and set \(a=1\) to nondimensionalize the integral (in other words, \(r\) and \(L\) are measured in units of \(a\)). This gives

\[ B_z=\mu_0nI\frac1{2\pi}\int_0^\pi \left(\frac{1-r\cos\theta}{r^2-2r\cos\theta+1}\right) \left(\frac{L}{\sqrt{L^2/4+r^2-2r\cos\theta+1}}\right) d\theta. \]

We can use the binomial series to expand the second factor when \(L\gg r\):

\[ \begin{align*} L\left[\frac{L^2}4+r^2-2r\cos\theta+1\right]^{-1/2} &=2\left[1+\frac4{L^2}\left(r^2-2r\cos\theta+1\right)\right]^{-1/2}\\ &\approx2\left(1-\frac2{L^2}(r^2-2r\cos\theta+1)\right). \end{align*} \]

The integral becomes

\[ B_z=\mu_0nI\frac1\pi\int_0^\pi\left[ \frac{1-r\cos\theta}{r^2-2r\cos\theta+1} -\frac2{L^2}+\frac2{L^2}\,r\cos\theta \right]d\theta. \]

The first term evaluates to \(\pi\) inside the solenoid and vanishes outside. The third term always vanishes under integration. Adding the necessary factors of \(a\) back in, we get

\[ \begin{align*} B_{z,\mathrm{int}}&=\mu_0nI(1-2a^2/L^2),\\ B_{z,\mathrm{ext}}&=\mu_0nI(-2a^2/L^2). \end{align*} \]

Pronouncing the suffix -ine

These words were gathered from my computer’s /usr/share/dict/words file; the pronunciations are from the OED, mostly following the American variants.

Some words are almost always pronounced with /-iːn/: amine, astatine, bromine, chlorine, fluorine; cuisine, dopamine, figurine, gasoline, histamine, libertine, machine, marine, mezzanine, nectarine, praline, pristine, quarantine, ravine, routine, saline, saltine, sardine, tambourine, tangerine, trampoline, vaccine, wolverine.

These ones almost always take /-ɪn/: destine, determine, examine, illumine, imagine; adrenaline, discipline, doctrine, engine, ermine, famine, feminine, genuine, heroine, jasmine, masculine, medicine, peregrine, saccharine, sanguine.

The following almost always end with /-aɪn/: combine, confine, decline, define; affine, alpine, asinine, bovine, canine, divine, equine, feline, iodine, leonine, lupine, opine, porcine, saturnine, serpentine, sibylline, supine, turpentine, valentine, vulpine.

Some take /-iːn/ or /-ɪn/: adenine, benedictine, margarine. Others take /-iːn/ or /-aɪn/: byzantine, clementine, strychnine. Still others take /-ɪn/ or /-aɪn/: alkaline, aquiline, carmine, crystalline, elephantine, turbine, ursine.

The pronunciation is not determined by etymology: for example, marīnus, peregrīnus, canīnus; also dētermināre, dēclīnāre.

Rodrigues’ formula in terms of the cross product matrix

Suppose we want to rotate a vector \(\mathbf v\) an angle \(\theta\) about an axis \(\mathbf k\) (a unit vector). Let us decompose \(\mathbf v\) into components parallel and perpendicular to \(\mathbf k\): we write \(\mathbf v=\mathbf v_\parallel+\mathbf v_\perp\). The parallel component \(\mathbf v_\parallel\) should be left unchanged by the rotation, but what about the perpendicular component?

A factor of \(\cos\theta\) of it will remain in the same direction, while a factor of \(\sin\theta\) of it will be rotated into a third direction: the one which is perpendicular to both \(\mathbf k\) and \(\mathbf v_\perp\), namely \(\mathbf k\times\mathbf v_\perp\). So, calling our rotation \(R\), we have:

\[ R\mathbf v=\mathbf v_\parallel+(\cos\theta)\mathbf v_\perp +(\sin\theta)(\mathbf k\times\mathbf v_\perp). \]

We can represent \(\mathbf v_\parallel\) and \(\mathbf v_\perp\) in terms of cross products of \(\mathbf k\) and \(\mathbf v\); this will let us write our result in matrix form. First, notice that

\[ \mathbf k\times\mathbf v_\parallel=0 \]

so

\[ \mathbf k\times\mathbf v=\mathbf k\times(\mathbf v_\parallel+\mathbf v_\perp)=\mathbf k\times\mathbf v_\perp. \]

This takes care of the last term. Then, notice that \(\mathbf k\times(\mathbf k\times\mathbf v_\perp)\) has the effect of twice rotating \(\mathbf v_\perp\) by 90° about the axis of rotation, so we have

\[ \mathbf k\times(\mathbf k\times\mathbf v_\perp)=-\mathbf v_\perp \implies\mathbf v_\perp=-\mathbf k\times(\mathbf k\times\mathbf v) \]

for the second term. Finally, in the first term, \(\mathbf v_\parallel=\mathbf v-\mathbf v_\perp=\mathbf v+\mathbf k\times(\mathbf k\times\mathbf v)\). We can now introduce the notation \(K\mathbf v=\mathbf k\times\mathbf v\), defining the skew-symmetric matrix \(K\) corresponding to \(\mathbf k\); this lets us write that

\[ R\mathbf v=\mathbf v+K^2\mathbf v-(\cos\theta)K^2\mathbf v+(\sin\theta)K\mathbf v, \]

so, collecting terms, we have

\[ R=I+(\sin\theta)K+(1-\cos\theta)K^2. \]

Writing the cross product as a symbolic determinant

The triple product \(\det(\mathbf a,\mathbf b,\mathbf c)\) gives the oriented volume spanned by the vectors \(\mathbf a\), \(\mathbf b\), and \(\mathbf c\). Let us try to separate the dependence on \(\mathbf a\) from that on \(\mathbf b\) and \(\mathbf c\). A cofactor expansion along the coefficients of \(\mathbf a\) lets us rewrite the determinant as a dot product:

\[ \begin{align*} \begin{vmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{vmatrix} &=a_1\begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix} -a_2\begin{vmatrix}b_1&c_1\\b_3&c_3\end{vmatrix} +a_3\begin{vmatrix}b_1&c_1\\b_2&c_2\end{vmatrix}\\ &=(a_1,a_2,a_3)\cdot\left( \begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix}, -\begin{vmatrix}b_1&c_1\\b_3&c_3\end{vmatrix}, \begin{vmatrix}b_1&c_1\\b_2&c_2\end{vmatrix} \right)\\ &=\mathbf a\cdot(\mathbf b\times\mathbf c) \end{align*} \]

where we can use the last line as the definition of the cross product. It is a collection of the cofactors of the determinant \(\det(\mathbf a,\mathbf b,\mathbf c)\), expanded along the vector \(\mathbf a\), into a product which only depends on \(\mathbf b\) and \(\mathbf c\). From this perspective, the “symbolic determinant”

\[ \begin{vmatrix} \mathbf i&b_1&c_1\\ \mathbf j&b_2&c_2\\ \mathbf k&b_3&c_3 \end{vmatrix} \]

is just a way to collect the cofactors of a matrix into a separate vector.