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# MSE-630 Week 3 - PowerPoint PPT Presentation

MSE-630 Week 3. Conductivity, Energy Bands and Charge Carriers in Semiconductors. Objectives:. To understand conduction, valence energy bands and how bandgaps are formed To understand the effects of doping in semiconductors

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### MSE-630 Week 3

Conductivity, Energy Bands and Charge Carriers in Semiconductors

• To understand conduction, valence energy bands and how bandgaps are formed

• To understand the effects of doping in semiconductors

• To use Fermi-Dirac statistics to calculate conductivity and carrier concentrations

• To understand carrier mobility and how it is influenced by scattering

• To introduce the idea of “effective mass”

• To see how we can use Hall effect to determine carrier concentration and mobility

Charge carriers follow a random path unless an external field is applied. Then, they acquire a drift velocity that is dependent upon their mobility,mnand the strength of the field,x

Vd = -mnx

The average drift velocity, vav is dependent

Upon the mean time between collisions, 2t

Current density, J, is the rate at which charges, cross any plane perpendicular to the flow direction.

J = -nqvd = nqmnx = sx

n is the number of charges, and

q is the charge (1.6 x 10-19 C)

The total current density depends upon the total charge carriers, which can be ions, electrons, or holesJ = q(nmn + pmp) x

OHM’s Law: V = IR

Resistance, R(W) is an extrinsic quantity. Resistivity, r(Wm), is the corresponding intrinsic property.

r = R*A/l

Conductivity, s, is the reciprocal of resistivity: s(Wm)-1 = 1/r

When we add carriers by doping, the number of additional carrers, Nd, far exceeds those in an intrinsic semiconductor, and we can treat conductivity ass = qNdmd

In general, np=ni2, where niis the intrinsic concentration of carriers at a given temperature (ni = 1.5 x 1010 cm-3 in Si) In a doped semiconductor, charges balance, thus Na+n = Nd+p. For an n-type semiconductor, n ~ Nd, and p ~ ni2/Nd

As the distance between atoms decreases, the energy of each orbital must split, since according to Quantum Mechanics we cannot have two orbitals with the same energy.

The splitting results in “bands” of electrons. The energy difference between the conduction and valence bands is the “gap energy” We must supply this much energy to elevate an electron from the valence band to the conduction band. If Eg is < 2eV, the material is a semiconductor.

In the ground state, at 0K, all the electrons have energy less than Ef, the Fermi energy. If we add energy we can boost an electron into the conduction band. If we add dopants, we can enhance the number of positive or negative carriers, vastly increasing conductivity

Adding a column 5 atom to Si (e.g., As) adds a 5 less than Ef, the Fermi energy. If we add energy we can boost an electron into the conduction band. If we add dopants, we can enhance the number of positive or negative carriers, vastly increasing conductivityth, loosely bound electron. The ionization energy is very low, and it can be boosted to the conduction band easily. This is an electron donor

If we add a material that is short one electron (e.g., B, valence 3) we will create an electron “sink” in the material. This results in “holes” that migrate in the valence band. Holes carry positive charge.

In an intrinsic semiconductor, elevating an electron to the conduction band leaves behind a hole in the valence band, or creates an “electron-hole pair”, or EHP

Fermi-Dirac statistics less than Ef, the Fermi energy. If we add energy we can boost an electron into the conduction band. If we add dopants, we can enhance the number of positive or negative carriers, vastly increasing conductivity

• Conductivity in semiconductors is described by Fermi-Dirac statistics. In an intrinsic semiconductor, the conductivity is:

For an extrinsic semiconductor, corresponding probability of an electron is

Where DE = Ec-Ed for an n-type, and DE = Ea-Ev for a p-type

At low temperatures, the conduction is provided by the dopants, and the slope is defined by DE/k. When the available electrons/holes is depleted in the “exhaustion” range, the slope is constant. When the temperature becomes high enough, energies exceed Eg, and the semiconductor moves into the intrinsic conduction range, with a slope of Eg/2k.

Effective Mass dopants, and the slope is defined by

In general, the curve of Energy vs. k is non-linear, with E increasing as k increases.

E = ½ mv2 = ½ p2/m = h2/4pm k2

We can see that energy varies inversely with mass. Differentiating E wrt k twice, and solving for mass gives:

Effective mass is significant because it affects charge carrier mobility, and must be considered when calculating carrier concentrations or momentum

Effective mass and other semiconductor properties may be found in Appendix III

In reality, band structures are highly dependent upon crystal orientation. This image shows us that the lowest band gap in Si occurs along the [100] directions, whil for GaAs, it occurs in the [111]. This is why crystals are grown with specific orientations.

The diagram showing the constant energy surface (3.10 (b)), shows us that the effective mass varies with direction. We can calculate average effective mass from:

Hall Effect crystal orientation. This image shows us that the lowest band gap in Si occurs along the [100] directions, whil for GaAs, it occurs in the [111]. This is why crystals are grown with specific orientations.

When a magnet field is appplied perpendicular to the direction in which a charged particle (electron or hole) is moving, the particle will be deflected as shown

To counter the flow of particles in the x-direction, we apply a field xy=vxBzso that the net force is zero. The applied field is called the Hall effect, and the resulting voltage, VH=xyd

The force on the particle will be F = q(x+vXB)

In the x-direction, the force will be Fy = q(xy+vxXBz)

Drift velocity for an electron in the x-direction is: crystal orientation. This image shows us that the lowest band gap in Si occurs along the [100] directions, whil for GaAs, it occurs in the [111]. This is why crystals are grown with specific orientations.

<vy>=-Jy/qn

where J is the current density, n is the number of carriers and q is the charge

Defining the Hall coefficient, RH=1/qn, then

• xy = vxBz = - Jx/qn Bz = RHJxBz

• and

Measuring the resistance gives the resistivity:

r (Wm)= Rwt/L = (Vcd/Ix)/(L/wt)

Since conductivity, s = 1/r = qmnn, the mobility is:

mn = s/qn = (1/r)/q(1/qRH) = RH/r