Answered step by step
Verified Expert Solution
Question
1 Approved Answer
5. Consider a Si non bipolar junction transistor at T = 300 K with the following parameters. Doping densities in the emitter and base: NDE
5. Consider a Si non bipolar junction transistor at T = 300 K with the following parameters. Doping densities in the emitter and base: NDE is unknown, NAB = 1 x 1016 cm-3 Diffusion coefficients: DpE = 12.3 cm2/s, DnB = 34.6 em2/s Carrier Lifetimes: TDE = 5 x 107 s, InB = co (i.e. there is no recombination in the base) Quasi-neutral base width: XB = 1 um Further assume that there is no recombination in the depletion region and that the emitter region is infinitely long. The transistor is biased into an active mode with VBE = 0.5 V and VBC is large. If the common-emitter current gain, B, is measured to be 500, what must the hole current density in the emitter, JpE, be?Physical Constants and Material Properties Table B.2 | Conversion factors Prefixes 1 A (angstrom) = 10- cm = 10-10 m 10-15 1 pum (micrometer) = 10 *cm femto- 10 12 1 mil = 10- in. = 25.4 um pico- =P 10-9 2.54 cm = 1 in. nano- =n 10-6 1 ev = 1.6 X 10-" J micro- 10- 1 J = 10 erg milli- = m 10+3 kilo- =K mega- = M 10+9 giga- = G 10+12 tera = T Table B.3 | Physical constants Avogadro's number NA = 6.02 X 10+23 atoms per gram molecular weight Boltzmann's constant k = 1.38 x 10 23 J/K = 8.62 X 10 eV/K Electronic charge e = 1.60 X 10-19 C (magnitude) Free electron rest mass mo = 9.11 X 10- kg Permeability of free space Mo = 47 X 10 7 H/m Permittivity of free space 6) = 8.85 X 10 " F/cm = 8.85 X 10-12 F/m Planck's constant h = 6.625 X 10 * J-s = 4.135 x 10 15 eV-s 2 TT h = h = 1.054 x 10 # J-s Proton rest mass M = 1.67 X 10-2 kg Speed of light in vacuum c = 2.998 X 10" cm/s Thermal voltage (7 = 300 K) V, = KI = 0.0259 V KT = 0.0259 evTable B.A | Silicon, gallium arsenide, and germanium properties (7 - 300 K) Property SI Atoms (em ) GRAS Ge Atomic weight 5.0 X 1037 4.42 x 10" 4,42 x 10" Crystal structure 28.09 144.63 72.60 Density (glem' ) Diamond Zineblende Diamond Lattice constant (A) 2.33 5.32 5.33 Melting point ("C) 5.43 5.65 5.65 1415 Dielectric constant 1238 937 11.7 Bandgap energy (eV) 13,1 160 1.12 1.42 0.66 Electron affinity. X (V) 4.01 4.07 4.13 Effective density of states in 2.8 X 10" 4.7 X 10" 1.04 x 10 conduction band, N. (cm ') Effective density of states in 1.04 X 10 7.0 X 10' 6.0 X 10" valence band, N, (cm ') Intrinsic carrier concentration (em ) 1.5 X 10 1.8 X 10 2.4 X 10" Mobility (cm /V-5) Electron, #. 1350 8500 3900 Hole, Pp 480 400 1900 Effective mass (") Electrons m; = 0.98 0.067 1.64 m; = 0.19 0.082 Holes If = 0.16 0.082 0.044 my= 0.49 0.45 0.28 Density of states effective mass Electrons ma 1.08 0.067 0.55 Holes . 0.56 0.48 0.37 Conductivity effective mass Electrons m. 0.26 0.067 0.12 Holes 0.37 0.34 0.21
Step by Step Solution
There are 3 Steps involved in it
Step: 1
Get Instant Access to Expert-Tailored Solutions
See step-by-step solutions with expert insights and AI powered tools for academic success
Step: 2
Step: 3
Ace Your Homework with AI
Get the answers you need in no time with our AI-driven, step-by-step assistance
Get Started