1 / 9

Tutorial #4 – selected problems

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.

lise
Download Presentation

Tutorial #4 – selected problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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:

  2. The intrinsic carrier concentration is given by Eg=1.43eV ni=1.87 x 106 cm-3

  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

  4. 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

  5. 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.

  6. 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

  7. 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.

  8. 3. Show that for acceptor doping concentration NA the equilibrium hole concentration p0 is Solution: =0 n0p0= ni2

  9. negative root not realistic

More Related