06 Electrical properties Flashcards
- Give two equation forms of Ohm’s law.
U = R*I = V
U=V=[Volts = J/C], [Ohms = V/A], [Ampere = C/s]
J = σ*E
J…current density = current/surface area = I/A [A/m²]
σ…conductivity
electr. field intensity E = V/l
l. ..sepearation distance
- Given the electrical resistance, as well as length and cross-sectional area of a specimen, compute its resistivity and conductivity.
- Compute the electric field intensity given the voltage drop across a specified distance l.
E = V/l
- Make the distinction between electronic and ionic conduction.
an electric current results from the motion of free electrons, which are accelerated in response to an applied electric field
in ionic materials, there may also be a net motion of ions, which also makes a contribution to the conduction process
- Describe the formation of electron energy bands as a large number of atoms, initially widely separated and isolated from one another, are gradually brought together, and allowed to bond to one another such that a crystalline solid is formed.
solid consist of a large number of atoms, say N, initially separated and then brought together to form the atomic arrangement found in crystalline materials.
As the atoms come within close proximity of one another, e- are acted upon, or perturbed (gestört), by the e- and nuclei of adjacent atoms. This influence is such that each distinct atomic state may split into a series of closely spaced e- states to form what is termed an electron energy band.
- Describe the four possible electron band structures for solid materials.
for metals, two band structure types are possible, empty e- states are adjacent to filled ones
band strcutures for semidonductors and insulators are similar - both have a forbidden energy band gap that, at 0K, lies between a filled valence band and an empty conduction band. The gap is rel. wide (>2eV or >6eV acc. to Prof. Holl.) for insulators and relativel narrow (<=6eV) for semiconductors
- Briefly describe the electron excitation events that produce free electrons/holes in metals, semiconductors (intrinsic and extrinsic), and insulators.
an electron becomes free by being excited (e.g. temperature) from a filled state to an available empty state at a higher energy (than the Fermi energy Ef). These are the e- that participate in the conduction process and are called free e-. Holes have a energy lower than Ef and are found in semicond. and insulators
rel. small energies are required for e- excitations in metals giving rise to a large numbers of free e-
greater energies are required in semiconductors and insulators, which accounts for their lower free e- concentration and thus smaller conductivity values
- Calculate the mobility of an electron, given its drift velocity and the magnitude of the electric field.
the scattering phenomenon is manifested as a resistenceto the passage of an electric current. Several parameters are used to describe this s.a. the drivt velocity and the mobility of an e-
the drift velocity vd represents the average electron veloctiy in the direction of the force imposed by the applied field. It´s directionaly proportional to the E-field.
the constant of proportionality ue is called electron mobility and is an indication of the frequency of scattering events [m²/Vs]
- Calculate the electrical conductivity of a metal, given the number of free electrons per unit volume, the electron mobility, and the electrical charge on an electron.
σ = n * |e| * ue
n…number of free or conducting e- per unit volume
|e|…abs. magnitude of the electrical charge on an e-, 1.6E-19 C
- (a) Cite three sources of electron scattering centers for metals.
(b) Write Matthiessen’s rule in equation form.
a) Because crystalline defects serve as scattering centers for conduction e- in metals, increasing their number raises the resistivity. The concentration of these imperfections depends on temperature, composition, and the degree of cold work of a metal specimen. In fact the total resistivity of a metal has the following resistivity contributions:
- thermal vibrations ρt
- impurities ρi
- plastic deformation ρd
b) see Fig.
- (a) Briefly explain the nature and source of the safety problem that exists at the connection point between aluminum and copper wires.
(b) Describe the best procedure for making these connections safe.
a) aluminium has a higher thermal expansion than Cu
thermal cycling of the wires and connectors let the material expand and contract, different in magnitud for Al and Cu
After some time this causes the connectors to loosen, increasing resistivity -> increased heating by the e- current
Al oxidizes more readily than Cu -> risk of fire
b) replace Al with Cu (most expensive)
crimp connector repair unit at each Al-Cu connection, a special designed metal sleeve
- Distinguish between intrinsic and extrinsic semiconducting materials.
In semiconductors, additionally to the free e-, holes (missing e- in the valence band) may also participate in the conduction process
intrinsic behavior: the e- properties are inherent (angeboren, innewohnend) in the pure material, and e- and h-concentrations are equal
extrinsic behavior: electrical behavior is dictated by impurities, extra. semicond. may be either n- or p-type depending on whether e- or holes, respectively, are the predominant charge carrier
- Cite two examples for each of the Groups IVA, IIIA-VA, and IIB-VIA semiconducting materials.
IVA: Si, Ge
IIIA-VA: GaAs, InSb
IIB-VIA: CdS, ZnTe
The wider the EN difference between the elements the wider the energy band gap.
- Describe the formation of a hole in terms of electron excitations in semiconductors.
hole = missing e- in the valence band when e- is excited
(see question 7. )
- For n-type extrinsic semiconduction:
(a) Describe the excitation of a donor electron in terms of both electron bonding and energy band models.
(b) Compute the electrical conductivity given the electron mobility, the number of free electrons per unit volume, and the electronic charge.
a) donor impurities like P, As and Pb (VA group: 5 valence e-) have one e- in excess if added as a substitutional impurity in Si, the other 4 e- are covalently bonded to the Si, the fifth one is loosely bonded (<0.01 eV), thus it is easily remove and can become a free e-
the donor state (electron) is located just below the bottom of the conduction band
b) because number of e- far exceeds number of holes (n>>p) the equation reduces to:
σ = n * |e| * µe