Many solids conduct electricity There are electrons that are not bound but are able to move through the crystal. Conducting solids fall into two main classes; metals and semiconductors. and increases by the addition of small amounts of impurities (semiconductors do opposite). In a metal, the conductivity  (1/resistivity ) normally gets smaller (or resistivity increases) at higher temperatures. Why? Why (discussed in book)? Electron-phonon scattering Smaller because only excited electrons contribute Why smaller?

Impurties have a big effect on semiconductors Why? Antimony doped germanium Impurties have a big effect on semiconductors Why? Another example, 1 boron in 100,000 silicon atoms increases the conductivity by 1000 over pure silicon. From page 565, Ashcroft Convert to per cubic angstrom 10^17 goes to 10^-7 (bottom ones are metallic, note X axis is 1/T) Impurities can add states within the band gap that makes excitation (and thus conduction) much more likely

Doping produces an extrinsic semiconductor (rather than intrinsic) Semiconductors can be easily doped Doping is the incorporation of [substitutional] impurities into a semiconductor in a controlled manner In a doped material, Ne is not equal to Nh Mott Transition: Impurities wavefunctions overlap, causing the material to become a metal Electron density at which this “metal insulator transition” occurs? aB ~ 50 Å & lattice constant, a ~ 2.5 Å. Need b ~ aB = 20a . i.e. one donor per 203 = 8000 semiconductor atoms Also happens for high density Excitons  forms plasma f(r) aB + + b

Donors Let’s use Silicon (Si) as an example Substitute one Si (Group IV) atom with a Group V atom (e.g. As or P) Si atoms have four valence electrons that participate in covalent bonding When a Group V atom replaces a Si atom, it will use four of its electrons to form the covalent bonding What happens with the remaining electron?

Donors The remaining electron will not be very tightly bound, and can be easily ionized at T > 0K Ionized electron is free to conduct In terms of the band structure, this electron is now in the conduction band Such Group V impurities are called Donors, since they “donate” electrons into the Conduction Band Semiconductors doped by donors are called n-type semiconductors (extra electrons with negative charge)

This crystal has been doped with a pentavalent impurity. The free electrons in n type silicon support the flow of current.

Hydrogenic Donors An electron added to an intrinsic semiconductor at T=0 would go into the lowest empty state i.e. at the bottom of the conduction band. When one adds a donor atom at T=0 the extra electron is bound to the positive charge on the donor atom (~ hydrogen) The electron bound to the positive ion is in a state just below the conduction band, where ED is the binding energy. Conduction Band Ec ED +ve ion -e Valence Band Ev

Donors: Energy Levels How much is ED? Such impurities create “shallow” levels - levels within the band gap, close to the conduction band The small ionization energy means a sizable fraction of donor atoms will be ionized at room temperature  How much is ED? Because most of the electrons are busy in bonds

Treat Similar to Hydrogen Employ the solution for hydrogen atom Substitute the effective mass for the electron mass The field of the charge representing the impurity must be reduced by the static dielectric constant of the semiconductor (dielectric constant for free space, κ = 1) me*  ~0.15 me and  ~15, ED ~10 meV. Relatively small when you compare to ~1-2 eV band gap for many semiconductors me*  ~0.15 me and  ~15, ED ~10 meV.

Acceptors Again use example: silicon (Si) -ve ion +e Acceptors Again use example: silicon (Si) Substitute one Group III atom (e.g. Al or In) Si atoms have 4 electrons for covalent bonding When a Group III atom replaces a Si atom, it cannot complete a tetravalent bond scheme A hole is formed. If the hole leaves the impurity, the core would be negatively charged, so the hole created is then attracted to the negative core, but can migrate We can say that this impurity atom accepted an electron, so called Acceptors (“donate holes”) Called p-type semiconductors since they contribute positive charge carriers

This crystal has been doped with a trivalent impurity. The holes in p type silicon contribute to the current. Note that the hole current direction is opposite to electron current so the electrical current is in the same direction

Acceptor: Energy Levels Such impurities create “shallow” levels - levels that are very close to the valence band Energy to “ionize” the atom is still small They are similar to “negative” hydrogen atoms Such impurities are called hydrogenic acceptors

Acceptor energy levels are bigger. Why? Examples Since holes are generally heavier than electrons, the acceptor levels are deeper (larger) than donor levels Why the range? The valence band has a complex structure and this formula is too simplistic to give accurate values for acceptor energy levels Acceptor energy levels are bigger. Why? Ge: 10 meV Si: 45 – 160 meV GaAs: 25 – 30 meV ZnSe: 80 – 114 meV GaN: 200 – 400 meV Acceptor and donor impurity levels are often called ionization energies or activation energies Why acceptor energies bigger?

Carrier Concentrations in Extrinsic Semiconductors The carrier densities in extrinsic semiconductors can be very high Depends on doping levels and ionization energy of the dopants Often both types of impurities are present If the total concentration of donors (ND) is larger than the total concentration of acceptors (NA) have an n-type semiconductor In the opposite case have a p-type semiconductor How can we measure whether our material is more n or p type?

Bringing Two Semiconductors Together All solid-state electronic and opto-electronic devices are based on doped semiconductors. In many devices the doping and hence the carrier concentrations vary. In the following section we will consider the p-n junction which is an important part of many semiconductor devices and which illustrated a number of key effects Bringing Two Semiconductors Together

What happens? Electrons Holes p-type semiconductor n-type semiconductor Electrons EC E E C C m m m E E E E V V V Holes Consider bringing into contact p-type and n-type semiconductors. n-type semiconductor: Chemical potential, m below bottom of conduction band p-type semiconductor: Chemical potential, m above top of valence band. Electrons diffuse from n-type into p-type filling empty valence states. Do they fill all of them?

Does this continue until they cancel out? Note: n and p sides are reversed in this diagram compared to others.. c02f19 Does this continue until they cancel out? Each e- that departs from the n side leaves behind a positive ion. Electrons enter the P side and create neg. ion. Originally lots of free carriers of opposite types in each material. Then joined. The immediate vincinity of the junction is depleted of free carriers. Redo this slide when I have time (so maybe never) Note: Absence of electrons and holes close to interface -- This is called the depletion region.

Without applied voltage, an equilibrium is reached in which a potential difference is formed across the junction. This potential difference is called built-in potential. P-type is connected with positive terminal. The positive potential applied to the p-type material repels the holes, while the negative potential applied to the n-type material repels the electrons, pushing carriers toward the junction and reducing the width of the depletion zone. With increasing voltage, the depletion zone eventually becomes thin enough that the zone's electric field cannot counteract charge carrier motion across the p–n junction. Such a junction is often operated in forward or reverse bias. Let us consider the effect of this added voltage.

Applications of p-n junctions Excellent diodes, which can be used for rectification (converter of AC to DC). Light emitting diodes (LEDs) and lasers: In forward bias one has an enhanced recombination current. For direct band gap semiconductors, light is emitted. Solar cells: If photons with hn>Eg are absorbed in the depletion region, get enhanced generation current. Photon energy can be converted to electrical power. Also temperature sensors

The MOS-FET Cgate = K0A / d In the MOS device, the gate electrode, gate oxide, and silicon substrate from a capacitor. High capacitance is required to produce high transistor current. Cgate = K0A / d K = dielectric constant, 0 = permittivity of vacuum A = area of capacitor, d = dielectric oxide thickness A to B The most common type of transistor is the metal–oxide–semiconductor field-effect transistor (MOSFET, MOS-FET, or MOS FET) is a device used to amplify or switch electronic signals. Since the 1960’s, complementary metal oxide semiconductor (CMOS) technology, including transistors, has relied on silicon dioxide for its gate dielectric (shown in green in the diagram).

Making Computers Smaller Cgate = K0A / d K = dielectric constant, A = area of capacitor, d = oxide thickness Area Capacitance Area Speed For today’s technology, the area of the gate oxide is being decreased in order to increase speed. The shorter electrons have to travel, the faster they can get there. However, decreasing the area also decreases the capacitance and the only was to counteract this, without changing the material, is to decrease the dielectric thickness. Yet, silicon dioxide can only be reduced so much without problems. Typical MOSFET channel lengths were once several micrometres; but now, as the gate oxide thickness goes below 2 nm, there is high gate leakage current due to the tunneling of electrons through the oxide. Though this may not be as great of a problem for machines with continual power supply, such as desktop computers, current leakage will highly effect machines, such as laptops or cell phones, which run off batteries. Capacitance Thickness

Replacement Oxides High dielectric constant Low leakage current Works well with current Si technology Many materials have been tried but none are as cheap and easy to manipulate as existing SiO2. Insulators that have a larger dielectric constant than silicon dioxide (referred to as high-k dielectrics), (e.g. hafnium oxides) are being used to reduce the gate leakage from the 45 nanometer technology onwards. Increasing the dielectric constant of the gate oxide allows a thicker layer while maintaining a high capacitance (capacitance is proportional to dielectric constant and inversely proportional to dielectric thickness). All else equal, a higher dielectric thickness reduces the quantum tunneling current through the dielectric between the gate and the channel.

How can we study potential replacement materials? Silicon (p doped)