ISEGORIA / MATH ENCYCLOPEDIA
Semiconductor physics: bands, doping, junctions
Silicon conducts badly on its own and superbly with one impurity atom in a million, and putting two kinds of doping side by side makes a one-way valve for current. How temperature and doping set the electron and hole densities through the Fermi level, how a p–n junction bends the bands and builds a field with no battery attached, and the Shockley diode equation that follows, including the same device run backwards as a solar cell.
Before you begin: Solid-state physics and statistical mechanics
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Carriers, doping and the Fermi level
In a semiconductor the electrons that conduct live in the conduction band, above a gap of 1.12 eV in silicon, and the missing electrons in the valence band below behave as positive holes. Both densities are set by one number, the Fermi level E_F, through the Fermi–Dirac function: n = N꜀ exp(−(E꜀ − E_F)/kT) and p = Nᵥ exp(−(E_F − Eᵥ)/kT), so their product np = nᵢ² does not depend on doping at all. Phosphorus adds donors that give up an electron 45 meV below the band; the Fermi level then moves wherever charge neutrality demands. On the log scale at the left, the occupation curves become straight lines, and where they cross the band edges they read off n and p. Press Play to warm the crystal from 20 K and watch it pass from freeze-out to the extrinsic plateau to intrinsic behaviour.
Worked example. For silicon at 300 K, \(E_g=1.1245\) eV, \(N_c=2.86\times10^{19}\) and \(N_v=3.10\times10^{19}\,\mathrm{cm^{-3}}\), so \(n_i=1.07\times10^{10}\,\mathrm{cm^{-3}}\). Dope with \(10^{16}\) phosphorus atoms per cm³: 99.6% are ionised, \(n=1.0\times10^{16}\), the Fermi level sits \(kT\ln(N_c/n)=0.206\) eV below the conduction band, and the holes are reduced to \(p=n_i^2/n=1.1\times10^4\,\mathrm{cm^{-3}}\), a trillion times fewer. At 50 K only 5% of the donors are ionised. By 700 K, \(n_i=2.7\times10^{16}\) has overtaken the doping and the crystal behaves as if undoped.
Watch out. The densities use the Boltzmann approximation, valid while the Fermi level stays a few \(kT\) inside the gap; above about \(10^{18}\,\mathrm{cm^{-3}}\) silicon becomes degenerate, the donor levels merge into an impurity band and the gap itself narrows, none of which is modelled. The band gap follows Varshni’s fit \(E_g(T)=1.17-4.73\times10^{-4}T^2/(T+636)\) eV, and \(N_c\), \(N_v\) scale as \(T^{3/2}\); the donor and acceptor levels are taken as 45 meV for phosphorus and boron. The Fermi level is a property of the whole crystal, not an energy that any particular electron has.
Why is the product np independent of the doping, when doping changes n by a factor of a million?
Because multiplying the two Boltzmann factors cancels the Fermi level: \(np=N_cN_v\,e^{-(E_c-E_F)/kT}e^{-(E_F-E_v)/kT}=N_cN_v\,e^{-E_g/kT}\). Physically it is a law of mass action for the reaction electron + hole \(\rightleftharpoons\) nothing: generation across the gap happens at a rate set only by temperature, recombination at a rate proportional to \(np\), and in equilibrium the two balance. Adding donors raises \(n\), which speeds up recombination until \(p\) has fallen in exact proportion.
2. The p–n junction
Put p-type and n-type silicon in contact and electrons diffuse from the n side, where they are plentiful, into the p side, and holes the other way. They leave behind the fixed charges of their ionised dopants, a depletion region that is positive on the n side and negative on the p side, and its field pushes back until diffusion and drift balance. In equilibrium the Fermi level is flat, so the bands must bend by the built-in potential V_bi. A bias V lowers (forward) or raises (reverse) the barrier by qV, splits the quasi-Fermi levels by the same amount, and narrows or widens the depletion region. Drag the p side up and down to apply a bias.
Worked example. With \(N_A=10^{17}\) and \(N_D=10^{16}\,\mathrm{cm^{-3}}\) at 300 K, \(V_{bi}=0.02585\ln(10^{33}/1.14\times10^{20})=0.770\) V. With \(\varepsilon=11.7\varepsilon_0\) the depletion width is \(W=0.331\) µm, of which \(x_n=0.301\) µm lies in the lightly doped n side and only \(0.030\) µm in the p side, and the peak field is \(47\) kV/cm. A reverse bias of 5 V widens it to \(0.91\) µm and raises the field to \(127\) kV/cm; the capacitance \(\varepsilon/W\) falls from \(31\) to \(11\) nF/cm².
Watch out. This is the depletion approximation: the region is taken as completely empty of mobile carriers and the neutral regions as perfectly neutral, with sharp edges between. It is excellent in reverse bias and fails in strong forward bias, where injected carriers fill the junction. The quasi-Fermi levels are drawn flat across the depletion region, the usual low-injection picture. The breakdown field of roughly 300 kV/cm used for the warning in the readout is indicative: the true avalanche threshold depends on the doping and on the width of the high-field region.
Why does the depletion region reach mainly into the lightly doped side?
Because the whole junction must stay neutral: the negative charge \(qN_Ax_p\) uncovered on the p side has to equal the positive charge \(qN_Dx_n\) on the n side, so \(x_n/x_p=N_A/N_D\). The lightly doped side has fewer dopant ions per unit length, so it has to be emptied over a longer stretch to supply the same charge. In a one-sided junction such as \(p^+n\) almost all of the width, the voltage drop and the capacitance are set by the light side, which is why device designers specify that doping most carefully.
3. The diode, and the solar cell
Forward bias lowers the barrier by qV, and the number of carriers with enough energy to cross it rises by the Boltzmann factor exp(qV/kT); reverse bias leaves only the small, voltage-independent flow of minority carriers that wander to the junction and are swept across. That is the Shockley equation. The saturation current Iₛ is proportional to nᵢ², so it grows steeply with temperature, and the forward voltage at a fixed current falls by about 2 mV per kelvin. Shine light on the junction and every absorbed photon adds a carrier pair that the field separates: the whole curve shifts down by the photocurrent, and in the fourth quadrant the diode is a power source. Drag the operating point and the illumination.
Worked example. With \(I_s=10^{-12}\) A at 300 K, an ideal diode (\(n=1\)) needs \(V=0.0259\ln(10^9)=0.536\) V to carry 1 mA, and each further decade of current costs \(59.5\) mV. Heating it by 10 K multiplies \(I_s\) by about 5, and at 1 mA the voltage falls by \(2.5\) mV/K. Under light giving \(I_{ph}=40\) mA the open-circuit voltage is \(0.0259\ln(1+4\times10^{10})=0.631\) V, the best load draws \(38.2\) mA at \(0.551\) V, and \(P_{\max}=21.0\) mW gives a fill factor of \(0.834\), as Green’s formula \(FF\approx(v_{oc}-\ln(v_{oc}+0.72))/(v_{oc}+1)\) with \(v_{oc}=qV_{oc}/kT\) predicts.
Watch out. Real diodes add a series resistance that bends the forward curve over at high current, recombination in the depletion region that makes the ideality factor \(n\) approach 2 at low current, and reverse breakdown, none of which is drawn. The temperature dependence assumes only \(I_s\propto T^3e^{-E_g(T)/kT}\); real devices often show \(I_s\) roughly doubling every 10 K because of other contributions. A silicon solar cell also loses much of the sunlight to thermalisation and to photons with less energy than the gap, which is why the photocurrent here is a free parameter rather than computed.
Why does a solar cell’s open-circuit voltage fall when it gets hot, even though the light is unchanged?
At open circuit the forward diode current exactly cancels the photocurrent, \(I_s(e^{qV_{oc}/kT}-1)=I_{ph}\). The photocurrent barely depends on temperature, but \(I_s\propto n_i^2\) grows by orders of magnitude, so a smaller forward bias is enough to push the same cancelling current: \(V_{oc}=(kT/q)\ln(I_{ph}/I_s)\) falls by roughly 2 mV per kelvin per cell. Panels on a hot roof lose several per cent of their output this way, which is why their datasheets quote a temperature coefficient of power.