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Tutorial #4 – selected problems. Problem 1 (Problem 3.7 Streetman) Given m n * and m p * from Table 3-1 p 65, calculate the effective densities of states N c and N v for GaAs at 300K (assume m n * and m p * do not vary with temperature). Calculate the intrinsic carrier concentration.

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## Tutorial #4 – selected problems

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**Tutorial #4 – selected problems**Problem 1 (Problem 3.7 Streetman) Given mn* and mp* from Table 3-1 p 65, calculate the effective densities of states Nc and Nv for GaAs at 300K (assume mn* and mp* do not vary with temperature). Calculate the intrinsic carrier concentration. Answer: mn*=0.067m mp*=0.48m T=300K The effective densities of states are given by:**The intrinsic carrier concentration is given by**Eg=1.43eV ni=1.87 x 106 cm-3**Problem 2**For silicon doped to 1014 donors/cm3: Calculate the temperature where the total electron carrier density in this semiconductor is equal to twice the donor concentration. (This gives a good estimate of where the sample changes from “extrinsic” to “intrinsic” behavior.) Answer: If the material is to remain electrostatically neutral, the sum of the positive charges (holes and ionized donor atoms) must balance the sum of the negative**charges (electrons and ionized acceptor atoms):**p0 + Nd+ = n0 + Na- Nd+ = 1014 cm-3 Na- = 0 cm-3 n0 = 2 Nd+ cm-3 p0 = n0 – Nd+ = 1014 (2-1) = 1014 cm-3 ni2 = n0p0**where h=6.63 x 10-34 J·s**k=1.38 x 10-23 J/K = 8.62 x 10-5 eV/K mn*= 1.1·(9.11 x 10-31 kg) mp*=0.56·(9.11 x 10-31 kg) Eg = 1.11eV So, We can solve for T in various ways. One way is to let a calculator solve the equation.**Another way is to solve for T using Excel or**Quattro Pro. TTpower3/2 exp(-6438.5/T) (Tpower3/2) * exp(-6438.5/T) 300 5196.2 4.779E-10 2.483E-06 400 8000 1.022E-07 0.0008177 500 11180 2.556E-06 0.0285789 510 11517 3.29E-06 0.0378966 513.9 11650 3.621E-06 0.0421863 514 11653 3.63E-06 0.0423016 515 11687 3.719E-06 0.0434697**You can start with e.g. T = 300 K. Check last**column to see how close you are to 0.0422. Repeat for 400 K and 500 K. Increase to 510 K, and then 515 K. Decrease to 514 K. Fine tune by trying 513.9 K. This value gives a product close to 0.0422. Therefore, T=513.9K.**3. Show that for acceptor doping concentration NA**the equilibrium hole concentration p0 is Solution: =0 n0p0= ni2

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