618 326

1. 618 326 Physics of electronic materials and devices I Lecture 7

2. Donors and Acceptors We can obtain the Fermi level dependence on temperature for three cases: • Very low temperature • Intermediate temperature • Very high temperature.

3. Donors and Acceptors • Very low temperature

4. Donors and Acceptors

5. Donors and Acceptors • Intermediate temperature

6. Donors and Acceptors

7. Donors and Acceptors • Very high temperature • In this case, all donors are ionized and electrons are excited from valence band to conduction band. • This is acting like an intrinsic semiconductor or EF = Ei. • It may be useful to express electron and hole densities in terms of intrinsic concentration ni and the intrinsic Fermi level Ei.

8. Donors and Acceptors • From , we have

9. Donors and Acceptors • Similarly to p-type, we have • This is valid for both intrinsic and extrinsic semiconductors under thermal equilibrium.

10. n-type semiconductor (a) Schematic band diagram. (b) Density of states. (c) Fermi distribution function (d) Carrier concentration. Note that np = ni2.

11. Donors and Acceptors • We have learned how to find new position of Fermi level for extrinsic semiconductors. • Now let us consider the new electron density in case of both donors ND and acceptors NA are present simultaneously. • The Fermi level will adjust itself to preserve overall charge neutrality as (1)

12. Donors and Acceptors • By solving (1) with , the equilibrium electron and hole concentrations in an n-type semiconductors yield

13. Donors and Acceptors • Similarly to p-type semiconductors, the electron and hole concentrations are expressed as

14. Donors and Acceptors • Generally, in case of all impurities are ionized, the net impurity concentration ND – NA is larger than the intrinsic carrier concentration ni; therefore, we may simply rewrite the above relationship as

15. Donors and Acceptors • The figure shows electron density in Si as a function of temperature for a donor concentration of ND= 1015 cm-3.

16. Donors and Acceptors • At low temperature, not all donor impurities could be ionized and this is called “Freeze-out region” since some electrons are frozen at the donor level.

17. Donors and Acceptors • As the temperature increased, all donor impurities are ionized and this remains the same for a wide range of temperature. • This region is called “Extrinsic region”.

18. Donors and Acceptors • Until the temperature is increased even higher and it reaches a point where electrons are excited from valence band. • This makes the intrinsic carrier concentration becomes comparable to the donor concentration. • At this region, the semiconductors act like an intrinsic one.

19. Degenerate semiconductor • If the semiconductors are heavily doped for both n- or p-type, EF will be higher than EC or below EV, respectively. • The semiconductor is referred to as degeneratesemiconductor. • This also results in the reduction of the bandgap.

20. Degenerate semiconductor • The bandgap reduction Eg for Si at room temperature is expressed by where the dopingN is in the unit of cm-3.

21. Example 1 • Si is doped with 1016 arsenic atoms/cm3. Find the carrier concentration and the Fermi level at room temperature (300K). Soln • At room temperature, complete ionization of impurity atoms is highly possible, then we have n = ND = 1016 cm-3.

22. Example 1 • Soln The Fermi level measured from the bottom of the conduction band is

23. Example 1 • Soln • The Fermi level measured from the intrinsic Fermi level is

24. Direct recombination • When a bond between neighboring atoms is broken, an electron-hole pair is generated. • The valence electron moves upward to the conduction band due to getting thermal energy. • This results in a hole being left in the valence band.

25. Direct recombination • This process is called carrier generation with the generation rate Gth (number of electron-hole pair generation per unit volume per time). • When an electron moves downward from the conduction band to the valence band to recombine with the hole, this reverse process is called recombination. • The recombination rate represents by Rth.

26. Direct recombination • Under thermal equilibrium, the generation rate Gth equals to the recombination rate Rth to preserve the condition of • The direct recombination rate R can be expressed as where  is the proportionality constant.

27. Direct recombination • Therefore, for an n-type semiconductor, we have where nn0 and pn0 represent electron and hole densities at thermal equilibrium.

28. Direct recombination • If the light is applied on the semiconductor, it produces electron-hole pairs at a rate GL, the carrier concentrations are above their equilibrium values. • The generation and recombination rates become where n and p are the excess carrier concentrations

29. Direct recombination • n = p to maintain the overall charge neutrality. • The net rate of change of hole concentration is expressed as

30. Direct recombination • In steady-state, dpn/dt = 0. From (7) we have where U is the net recombination rate. Substituting (3) and (5) into (8), this yields

31. Direct recombination • For low-level injection p, pn0 << nn0, (9) becomes • where p is called excess minority carrier lifetime .

32. Direct recombination • We may write pn in the function of t as

33. Example 2 • A Si sample with nn0 = 1014 cm-3 is illuminated with light and 1013electron-hole pairs/cm3 are created every microsecond. If n = p = 2s, find the change in the minority carrier concentration.

34. Example 2 • Soln Before illumination: After illumination:

35. Continuity Equation • We shall now consider the overall effect when drift, diffusion, and recombination occur at the same time in a semiconductor material. • Consider the infinitesimal slice with a thickness dx located at x shown in the figure.

36. Continuity Equation • The number of electrons in the slice may increase because of the net current flow and the net carrier generation in the slice. • Therefore, the overall rate of electron increase is the sum of four components: the number of electrons flowing into the slice at x, the number of electrons flowing out at x+dx, the rate of generated electrons, and the rate of recombination.

37. Continuity Equation • This can be expressed as • where A is the cross-section area and Adx is the volume of the slice.

38. Continuity Equation • By expanding the expression for the current at x + dx in Taylor series yields • Thus, we have the basic continuity equation for electrons and holes as

39. Continuity Equation • We can substitute the total current density for holes and electrons and (10) into (14).

40. Continuity Equation • For low-injection condition, we will have the continuity equation for minority carriers as

41. The Haynes-Shockley Experiment • This experiment can be used to measure the carrier mobility μ. • The voltage source establishes an electric field in the n-type semiconductor bar. Excess carriers are produced and effectively injected into the semiconductor bar at contact (1). Then contact (2) will collect a fraction of the excess carriers drifting through the semiconductor bar.

42. The Haynes-Shockley Experiment • After the pulse, the transport equation given by equation (15) can be rewritten as • If there is no applied electric field along the bar, the solution is given by

43. The Haynes-Shockley Experiment • N is the number of electrons or holes generated per unit area. If an electric field is applied along the sample, an equation (16) will becomes

44. The Haynes-Shockley Experiment

45. Example 3 • In Haynes-Shockley experiment on n-type Ge semiconductor, given the bar is 1 cm long, L =0.95 cm, V1 = 2 V, and time for pulse arrival = 0.25 ns. Find mobility μ.

46. Example 4 • In a Haynes-Shockley experiment, the maximum amplitudes of the minority carriers at t1 = 100 μs and t2 = 200 μs differ by a factor of 5. Calculate the minority carrier lifetime.

47. Example 4 • Soln