No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]
Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jordan Professor of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, Urbana, Illinois, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India 4.1 Conductors and Semiconductors 4.1-3 Material Media can be classified as (1) Conductors and Semiconductors electric property (2) Dielectrics (3) Magnetic materials – magnetic property Conductors and Semiconductors Conductors are based upon the property of conduction, the phenomenon of drift of free electrons in the material with an average drift velocity proportional to the applied electric field 4.1-4 electron cloud free electrons + bound elecrons nucleus In semiconductors, conduction occurs not only by electrons but also by holes – vacancies created by detachment of electrons due to breaking of covalent bonds with other atoms The conduction current density is given by J c E Ohm’s Law at a point 4.1-5 conductivity (S/m) e Ne e h N h e e N e e conductors semiconductor s Mobility Nh,e = Density of holes (h) or electrons (e) 4.1-6 Ohm’s Law V El V Jc = E = l A I Jc A V l l V I A V IR Ohm’s Law l R= A A l E, Jc I V 4.1-7 D4.1 I 0.1 Jc 10 A m A 10 –4 (a) For cu, 5.8 10 S m Jc 10 E 17.24 V m 5.8 10 (b) h e Ne e 1700 3600 10 2.5 1013 106 1.602 10 19 2.1229 S m Jc 103 E 471.1 V m 2.1227 4.1-8 (c) l FromR 1 A l 10 Sm –6 RA 10 Jc 10 E 3.14 mV m 10 4.1-9 Conductor in a static electric field E E 4.1-10 – – – – – S = –0E0 E0az z=d S0 z=d z=0 z=0 + + + + + S = E – – – – – S = –0E0 z=d + + + + + S = 0E0 z=0 – – – – – S = –0E0 + + + + + S = 0E0 E =0 + + + + + – – – – – E=– –S0 S0 – a z E0 a z 0 0 S0 E0 S0 0 a z 4.1-11 P4.3 (a) S0 S1 S2 Ei = S1 z S2 S1 S2 Ei – az a z 0 2 2 S1 S2 S0 4.1-12 (b) E i1 = S11 S12 Ei2 = S21 S22 S11 S12 S1 S21 S22 S2 Write two more equations and solve for the four unknowns