618 326

1. 618 326 Physics of electronic materials and devices I Lecture 10

2. Junction Breakdown • When a huge reverse voltage is applied to a p-n junction, the junction breaks down and conducts a very large current. • Although, the breakdown process is not naturally destructive, the maximum current must be limited by an external circuit to avoid excessive junction heating. • There are two mechanisms dealing with the breakdown: tunneling effect and avalanche multiplication.

3. Tunneling effect • If a very high electric field is applied to a p-n junction in the reverse direction, a valence electron can make a transition from the valence band to the conduction band by penetrating through the energy bandgap called tunneling. • The typical field for Si and GaAs is about 106 V/cm or higher.

4. Tunneling effect • Tunneling

5. Tunneling effect • To achieve such a high field, the doping concentration for both p- and n-regions must be very high such as more than 5 x 1017 cm-3. • The breakdown voltage for Si and GaAs junctions about 4Eg/qis the result of the tunneling effect.

6. Junction Breakdown • With the breakdown voltage is more than 6Eg/q, the breakdown mechanism is the result of avalanche multiplication. • As the voltage is in between 4Eg/q and 6Eg/q, the breakdown is due to a mix of both tunneling effect and avalanche multiplication

7. Avalanche multiplication • Let consider a p+-n one-sided abrupt junction with a doping concentration of ND1017 cm-3 or less is under reverse bias. • As an electron in the depletion region gains kinetic energy from a high electric field, the electron gains enough energy and acceleration in order to break the lattice bonds creating an electron-hole pair, when it collides with an atom.

8. Avalanche multiplication • A new electron also receives a high kinetic energy from the electric field to create another electron-hole pair. This continue the process creating other electron-hole pairs and is called avalanche multiplication.

9. Avalanche multiplication • Assume that In0 is incident current to the depletion region at x = 0. • If the avalanche multiplication occurs, the electron current In will increase with distance through the depletion region to reach a value of Mn In0 at W, where Mn is the multiplication factor. (1)

10. Avalanche multiplication

11. Avalanche multiplication • The breakdown voltage VBfor one-sided abrupt junctions can be found by • The breakdown voltage for linearly graded junctions is expressed as

12. Avalanche multiplication • The critical field Ec can be calculated for Si and GaAs by using the plot in the figure.

13. Example 1 • Calculate the breakdown voltage for a Si one-sided p+-n abrupt junction withND = 5 x 1016cm-3. • SolnFrom the figure, at the given NB, Ec is about 5.7 x 105 V/cm.

14. Breakdown voltage

15. Punch-through • Assume the depletion layer reaches the n-n+ interface prior to breakdown. • By increase the reverse bias further, the device will break down. • This is called the punch-through.

16. Punch-through • The critical field Ec is the same as the previous case, but the breakdown voltage VBfor this punch-through diode is (4) • Punch-through occurs when the doping concentration NB is considerably low as in a p+--n+ or p+--n+ diode, where  stands for a lightly doped p-type and  stands for a lightly doped n-type.

17. Breakdown voltage Breakdown voltage for p+-π-n+ and p+-v-n+junctions. W is the thickness of the lightly doped region.

18. Example 2 • For a GaAs p+-n one-sided abrupt junction with ND = 8 x 1014 cm-3, calculate the depletion width at breakdown. If the n-type region of this structure is reduced to 20 μm, calculate the breakdown voltage if r for GaAs is 12.4.

19. Example 2 • SolnFrom the figure, we can find VB is about 500 V (VB >> Vbi)

20. Example 2 • Soln

21. Heterojunction • A heterojunction is defined as a junction formed by two semiconductors with different energy bandgaps Eg, different dielectric permittivities s, different work function qs, and different electron affinities qχ.

22. Heterojunction

23. Heterojunction • The difference energy between two conduction band edges and between two valence band edges are represented by EC and EV, respectively, as • where Eg is the difference energy bandgap of two semiconductors.

24. Heterojunction

25. Heterojunction • Generally, heterojunction has to be formed between semiconductors with closely matched lattice constants. • For example, the AlxGa1-xAs material is the most important material for heterojunction. • When x = 0, the bandgap of GaAs is 1.42 eV with a lattice constant of 5.6533 Å at 300 K. • When x = 1, the bandgap of AlAs is 2.17 eV with a lattice constant of 5.6605 Å.

26. Heterojunction • We clearly see that the lattice constant is almost constant as x increased. The total built-in potential Vbi can be expressed by (7) • where N1 and N2 are the doping concentrations in semiconductor 1 and 2, respectively.

27. Heterojunction • The depletion widths x1and x2 can be found by (8)

28. Example 3 • Consider an ideal abrupt heterojunction with a built-in potential of 1.6 V. The impurity concentrations in semiconductor 1 and 2 are 1016 donors/cm3 and 3 x 1019 acceptors/cm3, and the dielectric constants are 12 and 13, respectively. Find the electrostatic potential and depletion width in each material at thermal equilibrium.

29. Example 3 • Soln • Note: Most of the built-in potential is in the semiconductor with a lower doping concentration and also its depletion width is much wider.

30. Metal-Semiconductor Junctions • The MS junction is more likely known as the Schottky-barrier diode. • Let’s consider metal band and semiconductor band diagram before the contact.

31. Metal-Semiconductor Junctions • When the metal and semiconductor are joined, electrons from the semiconductor cross over to the metal until the Fermi level is aligned (Thermal equilibrium condition). • This leaves ionized donors as fixed positive charges that produce an internal electric field as the case of one-sided p-n junction.

32. Metal-Semiconductor Junctions • At equilibrium, equal number of electrons across the interface in opposite directions. • Hence, no net transport of charge, electron current Ie equals to zero. • The built-in voltage Vbi = m - s. • The barrier for electrons to flow from the metal to semiconductor is given by qb = q(m - χs) or it is called the barrier heightof MS contact.

33. Metal-Semiconductor Junctions Barrier height Built-in voltage When a voltage is applied, the barrier height remains fixed but the built-in voltage changes as increasing when reverse biased and decreasing when forward biased.

34. Metal-Semiconductor Junctions • Reverse bias

35. Metal-Semiconductor Junctions • Few electrons move across the interface from metal to semiconductor due to a barrier, but it is harder for electrons in the semiconductor to move to the metal. • Hence, net electron transport is caused by electrons moving from metal to semiconductor. • Electron current flows from right to left which is a small value.

36. Metal-Semiconductor Junctions • Forward bias

37. Metal-Semiconductor Junctions • Few electrons move across the interface from metal to semiconductor, but many electrons move across the interface from semiconductor to metal due to the reduced barrier. • Therefore, net transport of charge flows from semiconductor to metal and electron current flows from left to right.

38. Schottky-diode • Under forward bias, the electrons emitted to the metal have greater energy than that of the metal electrons by about e(m - χs). • These electrons are called hot-carrier since their equivalent temperature is higher than that of electrons in the metal. • Therefore, sometimes, Schottky-diode is called “hot-carrier diode”.

39. Schottky-diode • This leads to the thermionic emission with thermionic current density under forward bias as

40. Schottky-diode • This behavior is referred to a rectification and can be described by an ideal diode equation of (10) • where V positive for forward bias and negative for reverse bias.

41. Schottky-diode • The space-charge region width of Schottky diode is identical to that of a one-sided p-n junction. • Therefore, under reverse bias, they can contain the charges in their depletion region and this is called Schottky diode capacitance.

42. Schottky-diode v.s. pn junction diode

43. Example 4 • A Schottky junction is formed between Au and n-type semiconductor of ND = 1016 cm-3. Area of junction = 10-3 cm2 and me*= 0.92 m0. Work function of gold is 4.77 eV and eχs = 4.05 eV. Find current at VF = 0.3 volts.

44. Example 4 • Soln

45. Example 5 • Si-Schottky diode of 100 μm diameter has (1/C2) v.s. VR slope of 3 x 1019F-2V-1. Given r = 11.9 for Si. Find NB for this semiconductor.

46. Example 5 Soln