This preview shows page 1. Sign up to view the full content.
Unformatted text preview: is neutral and has a small cross section for holes (typically the radius of the defect itself, or about 10−15 cm 2). 3a) Assume an N type silicon sample with an RG center concentration of 1012 cm 3. (Assume that the RG centers are located near the middle of the bandgap, as is typically the case.) Assume room temperature, so that υth ≈ 107 cm/s. Compute the minority hole lifetime assuming that the defects are donor like. Solution: Fermi level is well above the trap energy, so the traps are filled, which means neutral for donor like traps. This will give a small cross section, because there is no Coulombic attraction for holes. So (assuming that the thermal velocity is 107 cm/s): 1
1
τp =
= −15
= 10−4 s = 100 µs 7
12
c p N T 10 × 10 × 10 ECE 606 3 Spring 2013 Mark Lundstrom 2/24/2013 τ p = 100 µs 3b) Repeat 3a) assuming that the defects are acceptor like. (Assume that the RG centers are located near the middle of the bandgap, as is typically the case.) When acceptor like traps are filled, they are negatively charged, which creates a strong attraction for holes. The cross section will be large. 1
1
τp =
=
= 3.33 × 10−7 s = 0.33 µs −13
7
12
c p N T 3 × 10 × 10 × 10 τ p = 0.33 µs 4) The Fermi function gives the probability that a state in the conduction or valence band is occupied. One might think that the probability that a state in the forbidden gap (i.e. a trap or recombination center) would also be given by the Fermi function, but this is not quite right. Begin with eqn. (5.9a) in ASF, simplify it for equilibrium, and obtain the correct result. Solution: From eqn. (5.9a): rN = cn pT n...
View
Full
Document
 Fall '08
 Staff

Click to edit the document details