ENEE 313, Spr. ’09 Quiz VI - Apr 1, 2009Name:1. Consider a piece of uniformly-doped p-type material, with NA= 1 × 10171/cm3. In this material,ni= 10101/cm3, µn=1000 cm2/V.sec and µp=500 cm2/V.sec.(a) (2 pts.) Calculate n0and p0.p-type material: p0= NA= 1 × 10171/cm3; n0= n2i/p0= 1031/cm3.(b) (3 pts.) Calculate electron and hole conductivities. σn= qµnnnet; similar for holes.σn= qµnnnet= 1.6 × 10−19× 103× 103= 1.6 × 10−131/(Ω.cm)σp= qµppnet= 1.6 × 10−19× 5 × 102× 1017= 8 1/(Ω.cm)(c) (5 pts.) The material is illuminated until an excess carrier concentration δn = δp = 10151/cm3is created uniformly throughout the sample. Calculate the new electron and hole conductivities,and compare with the previous case. Which type of carrier conductivity rose more, that ofminorities or that of majorities?Now, nnet= n0+ δn = 103+ 1015≈ 10151/cm3and pnet= p0+ δp = 1017+ 1015= 1.01 × 10171/cm3. Then the new conductivities areσn= qµnnnet= 1.6 × 10−19× 103× 1015= 0.16 1/(Ω.cm)σp= qµppnet= 1.6 × 10−19× 5 × 102× 1.01 × 1017= 8.08 1/(Ω.cm)The hole conductivity rises 1% while the electron conductivity goes to 1015times its previousvalue. The minority conductivity, therefore, rose much more.(Turn over for the next question).12. (5 pts.) Use the material in the last case (item c) in Question 1. Assume that this is a direct-bandgap material and that the light is turned off at t = 0. Remember that excess minority carrierconcentrations recombine to return the material to its equilibrium carrier concentrations, followingan exponential decay:δn(t) = δn(0)e(−t/τn)where τnis the minority carrier lifetime. Assume that τn=100 ns here.Sketch the electron conductivity σn(t) vs. time t for this material between t = 0 and t = 500ns.Mark the conductivity, at least roughly, at t = 0 and t = 100ns.We calculated that at t = 0, σn(0) = 0.16 1/(Ω.cm). From that point on, as the excess electronconcentration δn decays exponentially in time, the conductivity will also follow this decay toeventually end up back at the 1.6 × 10−131/(Ω.cm) level set by
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