1. Semiconductor pn junctions

2. semiconductor pn junction context Figure 8.1-2 pn junction representations.

3. pn junction in forward bias:  J =  0 – V Forward bias:  J =  0 – V. Reduction of junction potential lowers E-field barrier.

4. pn junction forward bias: Thermal statistics Equilibrium: Forward bias: (n n = n n0 ) low-level injection. Only the minority-carrier levels are appreciably affected.

5. Low-level injection: Minority-carrier levels affected. Figure 8.5-2: The quasi-neutral regions (QNR)

6. Low-level injection: Injected carrier profiles

7. Injected carriers and diffusion Figure 7.7-1a. Concept of carrier injection with losses due to recombination

8. Carrier recombination: Recombination time constants Recombination of p-type carriers Recombination of n-type carriers

9. Carrier flux change (Fick’s laws) Figure 7.7-3 Carrier flow in/out for a one-dimensional slice Change in the total count N within the slice G = generation rate R = recombination rate

10. Diffusion and recombination (p-type example) Flux F recast as flow/area (Flux due to diffusion)

11. Steady-state flux balance of recombination Since recombination Then

12. Steady-state flux balance of recombination Solution: For which L p = Recombination length for p-type: Similarly L n = Recombination length for n-type:

13. SOLUTION: The mobility for n-type carriers in a material of ionized impurity density 5 × 10 16 #/cm 3, according to equation (7.3-7a) is: EXAMPLE: Determine the diffusion length for electrons injected into a p-type material doped with 5 × 10 16 #/cm 3 of Boron, assuming recombination time for the electrons t n = 200 ns. Assume T = 300K. = 905 cm 2 /Vs Then D n =  n V T = 905 ×.02585 = 23.4 cm 2 /s And = 21.6  m

14. Low-level injection

15. J = J n + J p

16. Low-level injection

17. EXAMPLE E8.5-1: An abrupt silicon pn junction is formed by an ion implant of N A = 10 17 #/cm 3 into an n-type substrate of impurity level N D = 10 15 #/cm 3. Determine: (a) Built-in potential  0, (b) reverse saturation current J S for recombination time constants  n =  p = 20ns (c) Current density level J for V = 0.6V. Assume default temperature (= 300K). (a) = 0.693V

18. (b) reverse saturation current J S for recombination time constants  n =  p = 20ns Both types of carriers exist on each side of the junction N A side: p p, n p N D side: n n, p n ∴ find (per heuristic formula)  n and  p on both sides of junction

19. The Shockley equation refers to the carriers that are injected into the other side. Hence the mobilites of interest are  n in the N A side and  p on the N D side, which are  n = 777cm 2 /Vs and  p = 458cm 2 /Vs, respectively. From the mobilities the diffusion coefficients are D n =  n V T = 777 ×.0259 = 20.1 cm 2 /s D p =  p V T = 458 ×.0259 = 11.84 cm 2 /s

20. From which the recombination lengths are = 6.34 × 10 -4 cm = 6.34  m = 4.86 × 10 -4 cm = 4.86  m

21. Then the reverse saturation current is = (1.6 × 10 -7 pC) = 888pA/cm 2 it is times like these that a spreadsheet would be a friend. = 36 × [(3.17 × 10 -13 ) + (2.43 × 10 -11 )] = 8.88 × 10 -10 A/cm 2