advanced34 min studyLesson 515 in Physics

Separate Schottky barrier height from band bending

Use ideal contact alignment as a starting model and identify pinning, depletion, and tunneling limits.

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What you’ll understand

  • Distinguish barrier energy, built-in voltage, and bulk Fermi-level separation.
  • Estimate a one-sided contact depletion width.
  • Explain why pinning and heavy doping can defeat naive work-function predictions.

An electrical contact is part of a device's physics, not merely an ideal wire attached after the calculation. When a metal touches a semiconductor, charge rearranges until equilibrium electrochemical potentials agree. The resulting interface may rectify, or it may pass current almost symmetrically. Predicting which occurs requires separating the energy barrier at the interface from the width of the bent-band region and from the transport process that carries charge across it.

Keep all contact energies on one reference

The metal work function WmW_m is the energy between its Fermi level and vacuum. Semiconductor electron affinity χ\chi is the energy between the conduction edge and vacuum. In the ideal Schottky–Mott picture, aligning vacuum levels before equilibration gives an electron barrier ΦBn=Wm−χ\Phi_{Bn}=W_m-\chi. These symbols denote energies, often tabulated in electronvolts. A corresponding hole barrier is the remainder of the semiconductor gap, provided the same band-edge picture applies and the interface has no additional gap-changing structure.

The electron barrier is not identical to the bulk-to-interface band bending. In neutral n-type bulk the conduction edge already lies above the Fermi level by Δb=Ec−EF\Delta_b=E_c-E_F. Only the difference between barrier energy and this bulk separation becomes the built-in electrostatic drop times q. The distinction matters even in an ideal contact: changing doping changes the bulk separation and depletion width without necessarily changing the ideal interface barrier set by the metal and semiconductor.

ΦBn=Wm−χ,ΦBp=Eg−ΦBn,qVbi=ΦBn−Δb,Δb=(Ec−EF)bulk\Phi_{Bn}=W_m-\chi,\qquad \Phi_{Bp}=E_g-\Phi_{Bn},\qquad qV_{\rm bi}=\Phi_{Bn}-\Delta_b,\qquad \Delta_b=(E_c-E_F)_{\rm bulk}
W_m, χ, Φ_Bn, Φ_Bp, E_g, and Δ_b are energies. If they are quoted in eV, their numerical energy differences correspond to volts after division by elementary charge. These are ideal alignment relations; the depleted n-type example assumes positive band bending.↔ Scroll wide equations horizontally.

Connect the interface to electrostatics and transport

A depleted n-type semiconductor contains positively charged ionized donors near the contact, balanced by negative charge on the metal. Most of the electrostatic drop lies in the semiconductor because the metal screens over a short distance. For uniform complete ionization, integrating Poisson's equation gives a one-sided depletion width. The relevant voltage is semiconductor band bending, not the full electron barrier energy divided by q. An accumulation contact has different mobile charge and cannot use the depletion formula unchanged.

In a thermionic-emission regime, carriers with sufficient energy cross the barrier. The reverse saturation scale is exponentially sensitive to barrier energy, with a temperature-squared prefactor in the elementary three-dimensional model. A forward current expression also assumes a specified voltage drop across the contact and neglects complicating interface transport. The Richardson constant depends on effective mass and material assumptions; treating it as a universal numerical constant for every semiconductor or interface is unjustified. Barrier lowering and tunneling can further alter the apparent temperature dependence.

W=2εVbiqND,Js=A∗T2e−ΦBn/(kBT),J=Js(eqV/(kBT)−1)W=\sqrt{\frac{2\varepsilon V_{\rm bi}}{qN_D}},\qquad J_s=A^*T^2 e^{-\Phi_{Bn}/(k_BT)},\qquad J=J_s\left(e^{qV/(k_BT)}-1\right)
The width formula is for an abrupt, uniformly doped depleted n-type contact at equilibrium. A* is a Richardson prefactor in A m⁻² K⁻², T is kelvin, and Φ_Bn and k_BT must be in matching energy units. The current law assumes thermionic emission and negligible other voltage drops, not arbitrary contacts.↔ Scroll wide equations horizontally.

Recognize why real contacts depart from ideal alignment

Interface states, chemical bonding, and dipoles can pin the barrier so that changing metal work function produces much less barrier change than Schottky–Mott alignment predicts. Pinning is a statement about the interface's electronic response, not a claim that the bulk semiconductor Fermi level cannot move with doping. The ideal formula remains a useful reference for identifying which physical effects must be added, but is not a guaranteed quantitative barrier prediction from two tabulated material numbers.

Heavy doping makes the depleted region narrow and can permit substantial tunneling through a barrier whose height remains nonzero. Such a contact may have a small, approximately linear resistance over the operating range and be called ohmic. Low barrier height is therefore not the only route to a useful contact. Conversely, a nominally favorable metal can still have large contact resistance because of contamination, an interfacial layer, or current crowding. Contact resistance, rectification, and ideal barrier height describe related but distinct properties.

A barrier is larger than its band bending

Use ideal alignment with Wm=4.8 eVW_m=4.8\,\mathrm{eV}, χ=4.0 eV\chi=4.0\,\mathrm{eV}, Eg=1.2 eVE_g=1.2\,\mathrm{eV}, and bulk Δb=0.20 eV\Delta_b=0.20\,\mathrm{eV}. Let ND=1022 m−3N_D=10^{22}\,\mathrm{m^{-3}}, ε=10−10 F/m\varepsilon=10^{-10}\,\mathrm{F/m}, and q=1.6×10−19 Cq=1.6\times10^{-19}\,\mathrm C. Find both carrier barriers, the built-in drop, and the depletion width.

  1. The electron barrier is ΦBn=4.8−4.0=0.80 eV\Phi_{Bn}=4.8-4.0=0.80\,\mathrm{eV}. The hole barrier is ΦBp=1.2−0.8=0.40 eV\Phi_{Bp}=1.2-0.8=0.40\,\mathrm{eV}. These are interface energy thresholds under the ideal model.

  2. Subtract the bulk edge-to-Fermi separation: qVbi=0.80−0.20=0.60 eVqV_{\rm bi}=0.80-0.20=0.60\,\mathrm{eV}, giving Vbi=0.60 VV_{\rm bi}=0.60\,\mathrm V. Using 0.80 V would incorrectly include the bulk separation in the electrostatic drop.

  3. The depletion width square is 2(10−10)(0.60)/[(1.6×10−19)(1022)]=7.5×10−14 m22(10^{-10})(0.60)/[(1.6\times10^{-19})(10^{22})]=7.5\times10^{-14}\,\mathrm{m^2}. Hence W=2.74×10−7 m=0.274 μmW=2.74\times10^{-7}\,\mathrm m=0.274\,\mu\mathrm m.

The result

The ideal electron and hole barriers are 0.80 and 0.40 eV, while the semiconductor band-bending drop is only 0.60 V. The depleted semiconductor width is approximately 0.274 μm under the stated ionization and electrostatic assumptions.

Explore the model

Interactive lab

Interactive lab · function

Separate potential value from its gradient

Reset and set both controls to coefficient a = 1 and position x = 1; predict value 1 and tangent slope 2. Reset and set both controls to a = 0.5 and x = 1; predict value 0.5 and slope 1. This normalized parabola illustrates a change in a potential segment and its derivative. It does not compute work functions, interface pinning, depletion width, tunneling, or thermionic current.

Separate potential value from its gradientThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn interface barrier energy and an extended electrostatic potential profile require different information.

Check your understanding

Why can heavy doping make a contact approximately ohmic even if its interface barrier height stays nonzero?

Try it yourself

Keep the example's bulk separation, doping, and permittivity fixed, but suppose the electron barrier falls from 0.80 to 0.75 eV0.75\,\mathrm{eV}. At kBT=0.025 eVk_BT=0.025\,\mathrm{eV}, find the change in built-in drop, depletion width, and thermionic saturation current, assuming an unchanged Richardson prefactor and valid thermionic transport.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Subtract the same bulk separation, scale width with the square root of band bending, and compare the two barrier exponentials.

Compare with the solution

The new built-in drop is 0.75−0.20=0.55 V0.75-0.20=0.55\,\mathrm V. Width becomes 0.2740.55/0.60=0.262 μm0.274\sqrt{0.55/0.60}=0.262\,\mu\mathrm m. The saturation-current ratio is e(0.80−0.75)/0.025=e2=7.39e^{(0.80-0.75)/0.025}=e^2=7.39. A modest barrier-energy change thus produces a large thermionic-current change while changing depletion width only slightly. The comparison holds the prefactor and bulk statistics fixed; a real metal substitution might also alter interface states and invalidate those controlled assumptions.

Take it with you

  • Barrier height and semiconductor band bending differ by the bulk edge-to-Fermi separation.
  • Schottky–Mott alignment is an ideal interface model, not a universal prediction.
  • A narrow nonzero barrier can transmit by tunneling and yield an ohmic contact.

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