NATIONAL UNIVERSITY of Computer & Emerging Sciences Chiniot-Faisalabad Campus Physics for Engineers Lecture 21 – 22 Department of Electrical Engineering Outline • Effect of Temperature on Semiconductor’s Resistivity • Diffusion and Drift Current Density • Electron and Hole Mobility • Width of Depletion Region • Width of n and p sides of the depletion region • Charge stored in the depletion region NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 2 Recall from PN Junction Diodes (1/2) p = n = ni p = n = ni2 (Both electrons and holes) • For N-type semiconductors nn ND pn ni2/ND • For P type semiconductors pp NA np ni2/NA • And barrier potential 𝑁𝐴 𝑁𝐷 𝑉0 = 𝑉𝑇 ln 𝑛𝑖 2 NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 3 Recall from PN Junction Diodes (2/2) • At equilibrium • No net current • Diffusion current = drift current (ID = IS) • During reverse bias • Barrier potential increases • No diffusion current • Only a small amount of drift current (IS) exists • During forward bias • Barrier potential decreases • Very large diffusion current (ID) • Very small drift current (IS) in the opposite direction I = ID – IS NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 4 Diffusion Current Density 𝑑𝑛(𝑥) 𝑑𝑝(𝑥) 𝐽𝑛 = 𝑞𝐷𝑛 𝐽𝑝 = −𝑞𝐷𝑝 𝑑𝑥 𝑑𝑥 • Dp and Dn are the diffusivity of holes and electrons respectively • p(x) and n(x) are concentrations of holes and electrons at point x • dp(x)/dx and dn(x)/dx are concentration gradients are point x for holes and electrons respectively NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 5 Example 1 Consider a bar of silicon in which a hole concentration profile described by is established a) b) c) 𝑝(𝑥) = 𝑝0 𝑒 − 𝑥 𝐿𝑝 Find the hole-current density at x = 0. Let p0 = 1016 /cm3, Lp = 1m and Dp = 12 cm2/s If the cross-sectional area of the bar is 100m2, find the current Ip. Find the hole-current density if the x is changed to 2 in part (a). Answers a) b) 192 A/cm2 192 A NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 6 Drift Current Density Jdrift = q(pp + nn)E • Where • • • • E = Electric Field q = charge on an electron p and n are concentrations of electrons and holes p and n are hole mobility and electron mobility respectively • It represents the degree of ease by which holes/electrons move through the silicon crystal in response to the electrical field E NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 7 Resistivity and Conductivity Jdrift = q(pp + nn)E We also write Jdrift = E Where is the conductivity So = q(pp + nn) The resistivity is defined as 1/ NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 8 Effect of Temperature on Conductivity • For a silicon diode, the conductivity is given as: = q(pp + nn) • As mobility is constant, the conductivity entirely depends on the concentration of holes and electrons. • With the increase in temperature the carrier concentration increases, hence the conductivity will increase. • The conductivity will also increase if we increase the doping concentration in p or n-type semiconductors • Hence, we can say that: 1 𝜌 𝑇 NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 9 Example 2 Find the resistivity at T = 300K of a) Intrinsic silicon b) P-type silicon with NA = 1016 /cm3 Use ni = 1.5 x 1010/cm3 and assume that for intrinsic silicon n = 1350 cm2/Vs and p = 480 cm2/Vs. And for doped silicon n = 1110 cm2/Vs and p = 400 cm2/Vs. Answer a) 2.28 x 105 cm b) 1.5625 cm NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 10 Example 3 Find the resistivity at T = 400K of an a) Intrinsic silicon b) P-type silicon with NA = 1016 /cm3 Assume that for intrinsic silicon n = 1350 cm2/Vs and p = 480 cm2/Vs. And for doped silicon n = 1110 cm2/Vs and p = 400 cm2/Vs. Answer 𝑛𝑖 = 𝐵𝑇 a) 427.44 cm b) 1.5605 cm NS1002 - Physics for Engineers 𝐸𝑔 − 3 2 2𝑘𝑇 𝑒 B = 7.3 x 1015 cm-3 K-3/2 Eg = 1.09 eV at 400K k = 8.62 x 10-5 eV/K Course Instructor: Muhammad Sajid Iqbal 11 Relationship Between D and • A simple but powerful relationship ties the diffusion constant with the mobility 𝐷𝑛 𝐷𝑝 = = 𝑉𝑇 𝜇𝑛 𝜇𝑝 • VT is the thermal voltage and VT = kT/q • The value of VT at 300K is 25.9 mV • The above relation is called Einstein Relationship NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 12 Width of and Charge Stored in the Depletion Region • The width of depletion region is given as • s is the permittivity of silicon = 1.04 x 10-12 F/cm • The width on n and p sides are given as • The charge stored on either side is given as NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 13 Example 4 • Consider a pn junction in equilibrium at room temperature (T = 300 K) for which the doping concentrations are NA = 1018/cm3 and ND = 1016/cm3 and the cross-sectional area A = 10-4 cm2. Calculate pp, np, nn, pn, V0, W, xn, xp and QJ. NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 14 The Current–Voltage Relationship of the Junction (1/3) • In forward bias • The current in the forward bias is essentially the diffusion current I = A(Jp + Jn) = Ip + In 𝑉 𝐷 𝑝 𝐼𝑝 = 𝐴𝑞𝑛𝑖 2 𝑒 𝑉𝑇 − 1 𝐿𝑝 𝑁𝐷 𝑉 𝐷 𝑛 𝐼𝑛 = 𝐴𝑞𝑛𝑖 2 𝑒 𝑉𝑇 − 1 𝐿𝑛 𝑁𝐴 • Lp and Ln are diffusion lengths of holes and electrons respectively • The smaller the value of diffusion length the faster the recombination NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 15 The Current–Voltage Relationship of the Junction (2/3) • In reverse bias 𝐼 = 𝐴𝑞𝑛𝑖 2 𝑉 𝑒 𝑉𝑇 𝐷𝑝 𝐷𝑛 + 𝐿𝑝 𝑁𝐷 𝐿𝑛 𝑁𝐴 −1 • for a negative V (reverse bias) the exponential term becomes essentially zero and the current across the junction becomes negative and constant and we know that this current must be IS 𝐼 = 𝐼𝑆 𝑉 𝑒 𝑉𝑇 −1 𝐼𝑆 = 𝐴𝑞𝑛𝑖 2 𝐷𝑝 𝐷𝑛 + 𝐿𝑝 𝑁𝐷 𝐿𝑛 𝑁𝐴 • In the reverse direction the current saturates at –IS so we call it as Saturation current NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 16 The Current–Voltage Relationship of the Junction (3/3) NS1002 - Physics for Engineers Course Instructor: Muhammad Sajid Iqbal 17
