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This document discusses conductivity and mobility in semiconductors. It explains that an applied electric field causes charge carriers like electrons to drift in a direction, with their drift velocity depending on mobility. Mobility decreases with increasing temperature or impurity concentration. The document also covers how conductivity is calculated from charge carrier concentration and mobility, and how the Hall effect can be used to determine these parameters by applying a magnetic field.

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Electrical Science-2 (15B11EC211) Unit-4 Introduction to Semiconductor Lecture-4 Dr. Akansha Bansal OLED Technology Solar Cells
Conductivity and Mobility β€’ The charge carriers in a solid are in constant motion, even at thermal equilibrium. β€’ At room temperature, for example, the thermal motion of an individual electron may be visualized as random scattering from lattice vibrations, impurities, other electrons, and defects. Fig. 1(a) Random thermal motion of an electron in a solid [1]
Conductivity and Mobility If an electric field Ex is applied in the x-direction, each electron experiences a net force -qEx from the field. This force averaged over all the electrons, however, is a net motion of the group in x- direction. As a result of this electrostatic force , the electrons would be accelerated and the velocity increases indefinitely with time, however due to the inelastic collision with ions electrons loses energy and a steady state condition is reached where a finite value of drift speed vd is attained. This drift velocity is in the direction opposite to that of the electric field. 𝑣π‘₯ = βˆ’πœ‡π‘›πΈπ‘₯ Ex Field Fig. 1b Well directed drift velocity with an applied electric field.
Conductivity and Mobility β€’ The quantity πœ‡π‘›, called the electron mobility, describes the ease with which electrons drift in the material. β€’ Mobility is a very important quantity in characterizing semiconductor materials and in device development. β€’ The mobility can be expressed as the average particle drift velocity per unit electric field. We have, μ𝑛 = βˆ’ 𝑣π‘₯ 𝐸π‘₯ β€’ The units of mobility are cm2/V-s
Effect of Temperature on Mobility The two basic types of scattering mechanisms that influence electron and hole mobility are: Lattice scattering Impurity scattering Fig.2 Temperature dependence of mobility with both lattice and impurity scattering
Effect of Temperature on Mobility β€’ Lattice scattering: A carrier moving through the crystal is scattered by a vibration of the lattice, resulting from the temperature. The frequency of such scattering events increases as the temperature increases, since the thermal agitation of the lattice becomes greater. Therefore, we should expect the mobility to decrease as the sample is heated. β€’ Impurity scattering: Scattering from crystal defects such as ionized impurities becomes the dominant mechanism at low temperatures. Since the atoms of the cooler lattice are less agitated and a slowly moving carrier is likely to be scattered more strongly by an interaction with a charged ion than is a carrier with greater momentum, impurity scattering events cause a decrease in mobility with decreasingtemperature. The mobilities due to two or more scattering mechanisms add inversely: 1 ΞΌ = 1 ΞΌ1 + 1 ΞΌ2 … … … …
Effect of Doping on Mobility As the concentration of impurities increases, the effects of impurity scattering are felt at higher temperatures. For example, the electron mobility πœ‡π‘› of intrinsic silicon at 300 K is 1350 cm2/(V-s).With a donor doping concentration of 1017 cm-3, however, πœ‡π‘› is 700 cm2/(V-s). Thus, the presence of the 1017 ionized donors/cm3 introduces a significant amount of impurity scattering. Fig.3 Variation of mobility with total doping impurity concentration (N0 + Nd) for Si at 300 K.
High Field Effects Upper limit is reached for the carrier drift velocity in a high field and this limit occurs near the mean thermal velocity (107 cm/s) and represents the point at which added energy imparted by the field is transferred to the lattice rather than increasing the carrier velocity. The result of this scattering limited velocity is a fairly-constant current at high field. Fig.4 Saturation of electron drift velocity at high electric fields for Si
Example Example 1: How long does it take an average electron to drift 1 ΞΌm in pure Si at an electric field of 100V/cm? Repeat for 105V/cm. Assume ΞΌn = 1350 cm2/V-s Solution: Low Field: 𝑣𝑑 = μ𝑛𝐸 = 1350 x 100=1.35 x 105 cm/s t = L/ 𝑣𝑑 = 10-4/1.35 x 105 = 0.74ns High Field: Scattering limited velocity 𝑣𝑠 = 107 cm/s t = L/ 𝑣𝑠 = 10-4/107 = 10ps
Current Density β€’ If n electrons are contained in a length L of conductor, and if it takes an electron a time t seconds to travel distance of L meter in the conductor. A 𝐼 = π‘π‘ž 𝑑 = π‘π‘žπ‘£ 𝐿 As L/t is the average, or drift speed, v m/s of the electrons. Current Density denoted by symbol J, is the current per unit area of the conducting medium. β€’ Assuming a uniform current distribution 𝐽 = 𝐼 𝐴 Where, J is Amperes/Square meter A- Cross-sectional area of the conductor in meters L
Conductivity 𝐽 = π‘π‘žπ‘£ 𝐿𝐴 n = 𝑁 𝐿𝐴 n=no. of electrons per cubic meter 𝐽 = π‘›π‘žπ‘£ 𝐽 = π‘›π‘žπ‘£ = π‘›π‘žΞΌπΈ = σ𝐸 (1) Where, Οƒ = π‘›π‘žΞΌ is the conductivity of the metal in (ohm-meter)-1] Eq. 1 is recognized as Ohm’s law, namely, the conduction current is proportional to applied voltage.
Example Example 2: A Si bar 0.1 cm long and 100 ΞΌm2 in cross sectional area is doped with 1017 cm-3 antimony. Find the current at 300K with 10V applied. Assume ΞΌn = 700cm2/V-s Solution: Οƒ = π‘›π‘žΞΌn = 1.6 x 1019 x 700 x 1017 = 11.2(Ξ©.cm)-1 ρ = 1/Οƒ =0.0893(Ξ©.cm) 𝑅 = ρ𝐿 𝐴 𝐼 = V 𝑅 = 10 0.0893 = 0.16𝐴
Conductivity of semiconductor Semiconductor contains two types of charge carriers β–« Negative(free electron) with mobility ΞΌn β–« Positive(hole) with mobility ΞΌp These particles move in opposite direction in an electric field E, but since they are of opposite sign, the current of each is in same direction. Hence, Current density J=(nΞΌn + pΞΌp)qE =ΟƒE n = magnitude of free electron concentration p = magnitude of hole concentration Οƒ = Conductivity Hence, Οƒ = (nΞΌn + pΞΌp)q
Hall Effect β€’ If a magnetic field is applied perpendicular to the direction in which holes drift in a p-type bar, the path of the holes tends to be deflected. Fig.1 The Hall effect [1]
Hall Effect Using vector notation, the total force on a single hole due to the electric and magnetic fields is F = q(E + v Γ— B) In the y-direction the force is Fy = q(Ey-vxBz) β€’ The important result of above equation is that unless an electric field Ey is established along the width of the bar, each hole will experience a net force (and therefore an acceleration) in the y- direction due to the qvxBz product. β€’ Therefore, to maintain a steady state flow of holes down the length of the bar, the electric field Ey must just balance the product qvxBz β—¦
Hall Effect Once the electric field Ey becomes as large as qvxBz, no net lateral force is experienced by the holes as they drift along the bar. The establishment of the electric field Ey is known as the Hall effect, and the resulting voltage VAB = Eyw is called the Hall voltage. 𝐸𝑦 = 𝐽π‘₯ π‘žπ‘0 𝐡𝑧 = 𝑅𝐻𝐽π‘₯𝐡𝑧 𝑅𝐻 = 1 π‘žπ‘0 𝑅𝐻 is called the Hall Coefficient
Hall Effect A measurement of the Hall voltage for a known current and magnetic field yields a value for the hole concentration p0 𝑝0 = 1 π‘žπ‘…π» = 𝐽π‘₯𝐡𝑧 π‘žπΈπ‘¦ = (𝐼π‘₯/𝑀𝑑)𝐡𝑧 π‘ž(𝑉𝐴𝐡/𝑀) = 𝐼π‘₯𝐡𝑧 π‘žπ‘‘π‘‰π΄π΅ If a measurement of resistance R is made, the sample resistivity ρ can be calculated ρ (Ξ©-cm) = 𝑅𝑀𝑑 𝐿 = 𝑉𝐢𝐷/𝐼π‘₯ 𝐿/𝑀𝑑 Mobility is simply the ratio of the Hall coefficient and the resistivity μ𝑝 = Οƒ π‘žπ‘0 = 1/ρ π‘ž(1/π‘žπ‘…π») = 𝑅𝐻 ρ
Example Example: Consider a semiconductor bar with w = 0.1mm, t =10 ΞΌm, and L = 5mm. For B=10-4 Wb/cm2 and a current of 1mA, we have VAB = - 2mV, VCD = 100mV. Find the type, concentration, and mobility of the majority carrier. Solution From the sign of VAB, we can see that the majority carriers are electrons: 𝑛0 = 𝐼π‘₯𝐡𝑧 π‘žπ‘‘(βˆ’π‘‰π΄π΅) = 10βˆ’310βˆ’4 1.6 π‘₯ 10βˆ’19π‘₯10βˆ’3π‘₯2π‘₯10βˆ’3=31.25 x 1017 cm-3 ρ = 𝑅𝑀𝑑 𝐿 = 𝑉𝐢𝐷/𝐼π‘₯ 𝐿/𝑀𝑑 = 0.1 10βˆ’3 0.5 0.01 π‘₯ 10βˆ’3 =0.002 Ξ©.cm μ𝑝 = 1 Οπ‘žπ‘0 = 1 (0.002)(1.6 π‘₯ 10βˆ’19)(3.125 π‘₯ 1017) = 10,000 π‘π‘š2 𝑉. 𝑠 βˆ’ 1