Power Electronic Devices

1. Power Supply Systems Electrical Energy Conversion and Power Systems Universidad de Oviedo Power Electronic Devices Semester 1 Lecturer: Javier Sebastián

2. Research Group Power supply Systems (Sistemas Electrónicos de Alimentación) Javier Sebastián Dr. Electrical Engineer (Ingeniero Industrial)Full professor Room 3.1.21Edificio nº 3, Campus Universitario de Viesques 33204 Gijón (Asturias). SpainPhone (direct): 985 18 20 85 Phone (secretary): 985 18 20 87Fax: 985 18 21 38 E-mail: sebas@uniovi.es Web: http://www.unioviedo.es/sebas/

3. Outline • Review of the physical principles of operation of semiconductor devices. • Thermal management in power semiconductor devices. • Power diodes. • Power MOSFETs. • The IGBT. • High-power, low-frequency semiconductor devices (thyristors).

4. D G S Outline Review of physical principles of semiconductors Power electronics devices

5. Previous requirements • Basic electromagnetic theory. • Basic circuit theory. • The operation of basic electronics devices in circuits. The student must understand the behaviour of the following electronics devices in simple circuits: • Diodes. • Bipolar Junction Transistors, both PNP and NPN types. • Field Effect Transistor, especially enhancement-mode Metal-Oxide-Semiconductor Field Effect Transistors (MOSFETs), both in N-channel- and P-channel types.

6. Electrical Energy Conversion and Power Systems Universidad de Oviedo Lesson 1 - Review of the physical principles of operation of semiconductor devices Semester 1 - Power Electronics Devices

7. Outline • Review of the physical principles of operation of semiconductor devices: • Basic concepts about semiconductor materials: band diagrams, intrinsic and extrinsic semiconductors, mechanisms for electric current conduction and continuity equation and its use in simple steady-state and transient situations. • Basic concepts about PN junctions: Equilibrium conditions, forward- and reverse-biased operation and calculation of the current flow when biased. • Reverse-biased voltage limits of PN junctions. • PIN junctions. • Conductivity modulation. • Transient effects in PN junctions in switching-mode operation. • Metal-semiconductor junctions.

8. Energy level in a semiconductor as a function of inter-atomic spacing At 0 K, empty - - Energy of electrons - - - - - - - - Inter-atomic spacing At 0 K, filled Actual spacing

9. Energy of electrons - - - - Concept of band diagram Empty at 0 K 4 states/atom Conduction band Eg Band gap Valence band 4 electrons/atom Filled at 0 K

10. Band structure for insulators, semiconductors and metals at 0 K Conduction band Conduction band Conduction band Band gap Eg Band gap Eg Overlap Valence band Valence band Valence band Semiconductor Eg=0.5-2 eV Insulator Eg= 5-10 eV Metals No Eg

11. Si Si Si Si - - - - - - - - - - - - - - - - - + - - - - Band structure for semiconductors at room temperature. Concept of “hole” Conduction band Eg Valence band + Semiconductor Eg=0.5-2 eV Visualizationusingthebondingmodel • Someelectronsjumpfromthevalence band totheconduction band. They are chargecarriersbecausethey can movefromoneatomtoanother. • Theemptystate in thevalence band isreferredto as a “hole”. • Theholeshave positive charge. They are alsochargecarriers.

12. Si Si - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Concepts of generation and recombination Recombination Generation Si Si Eg Eg - - Si Si Si Si + + - - - - + + 12

13. ------- Si Si Si Si Si Si Si Si - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + - Why both holes and electrons are electric charge carriers? • In general, there will be electric current due to both electrons and holes • Example: when there is an electric field in the semiconductor lattice -

14. How many electrons and holes are there in 1 cm3? • The number of these electrons and holes strongly depend on both Eg and the room temperature. It is called intrinsic concentration and it is represented as “ni”. • The concentration of electrons in the conduction band (negative charge carriers) is represented as “n”. The concentration of holes in the valence band (positive charge carriers) is represented as “p”. • Obviously n = p = ni in this type of semiconductors (intrinsic semiconductors) • Some examples of the value of ni at room temperature: Taking into account the number of bonds of valence band electrons in 1cm3 of silicon, only one bond is broken for each amount of 1013 unbroken bonds (at room temperature)

15. - - - - - - - - - - - - - - - - - - - - - - - - Sb Al Concept of extrinsic semiconductors: doping semiconductor materials • Can we have different concentration of electrons and holes? • The answer is yes. We need to introduce “special” impurities into the crystal: • Donors: atoms from column V of the Periodic Table. We obtain an extra electron for each atom of donor. • Acceptors: atoms from column III of the Periodic Table. We obtain an extra hole for each atom of acceptor. - - Si Si - Si Si - + 1 Donor 1 5 2 2 - Si Si - - + Acceptor 4 - 3 - 3

16. - - - - - - - - - - - - - - - - - - - - - - - - 5 Al- Sb+ N-type and P-type semiconductors - - Si - Si Si - Si • N-type semiconductor: • Majority carriers are electrons. • Minority carriers are holes. • Positively-charged atoms of donor (positive ions). • P-type semiconductor: • Majority carriers are holes. • Minority carriers are electrons. • Negatively-charged atoms of acceptor (negative ions). + 1 Donor 1 2 2 Si - Si - Acceptor 4 - - 3 3

17. Al- Al- Al- Al- Al- Thermalgeneration Al- Al- Al- Al- Al- - - - - - - - - - - - - - + + + + + + + + + + + + + Thermalgeneration Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Charges in N-type and P-type semiconductors P-typesilicon Acceptorions (negativeions) hole electron Donorions(positive ions) N-typesilicon

18. Al- Al- Al- Al- Al- Al- - - - - - - - - + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Charge carries in N-type and P-type semiconductors • Concentration of majority carriers: pP • Concentration of minority carriers: nP • Mass action law: pP·nP =ni2 • (only valid at equilibrium) P-type • Concentration of majority carriers: nN • Concentration of minority carriers: pN • Mass action law: nN·pN =ni2 • (only valid at equilibrium) Very important equations!!! N-type

19. Al- Al- Al- Al- Al- Al- - - - - - - - - + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Static charges in N-type and P-type semiconductors • Concentration of acceptors: NA • (only negative static charges in a P-type semiconductor) P-type • Concentration of donors: ND • (only positive static charges in a N-type semiconductor) N-type

20. Al- Al- Al- Al- Al- Al- - - - - - - - - + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Neutrality in N-type and P-type semiconductors • Silicon, aluminium and antimony were neutral before being used Þ The extrinsic semiconductor must be neutral, too. • Positive charges in volume V: pP·V • Negative charges in volume V: nP·V + NA·V • Neutrality: pP = nP + NA P-type • Negative charges in volume V: nN·V • Positive charges in volume V: pN·V + ND·V • Neutrality: nN = pN + ND Very important equations!!! N-type

21. Al- Al- Al- Al- Al- Al- - - - - - - - - + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Calculating the concentration of electrons and holes (I) • Neutrality: pP = nP + NA • Mass action law: pP·nP =ni2 • 2 known (NA and ni) and 2 unkown (pP and nP) variables Þ can be solved P-type • Neutrality: nN = pN + ND • Mass action law: nN·pN =ni2 • 2 known (ND and ni) and 2 unkown (nN and pN) variables Þ can be solved N-type

22. Al- Al- Al- Al- Al- Al- - - - - - - - - + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Calculating the concentration of electrons and holes (II) • Frequent case: quite heavy doped semiconductors • NA >> ni • Neutrality: pP» NA • Mass action law: nP»ni2/NA P-type • ND >> ni • Neutrality: nN» ND • Mass action law: pN»ni2/ND Very useful equations!!! N-type

23.    jn E jp - - - - - - - - + + + + + + + + Mechanisms to conduct electric current: Drift (I) • Semiconductors can conduct electric current due to the presence of an electric field E - - - - - + + + + +   • jp_Drift = q·p·p·Eis the current density of holes due to drift. • jn_Drift = q·n·n·E is the current density of electrons due to drift.  

24. Mechanisms to conduct electric current: Drift (II) q = magnitude of the electronic charge (1.6·10-19 coulombs). p = hole mobility. n = electron mobility. p = hole concentration. n = electron concentration. E = electric field.   • jp_Drift = q·p·p·E • jn_Drift = q·n·n·E  

25.  jn_Diff - - - - - - - - - - - - - - - - - - 1 2 n2 < n1 n1 Mechanisms to conduct electric current: Diffusion (I) Electrons have migrated due to “diffusion” (you can see the same phenomenon in gases)

26.  jn_Diff - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 2 n2 < n1 n1   n Mechanisms to conduct electric current: Diffusion (II) • If we maintain a different concentration of electrons, we also maintain the motion of electrons in the lattice

27.  jn_Diff - - - - - - - - - - - - - - - - - - - - - - 1 2 n2 < n1 n1   n Mechanisms to conduct electric current: Diffusion (III) The current density is proportional to the electron concentration gradient:   jn_Diff = q·Dn· n  Dn = electron diffusion coefficient.

28.  jp_Diff + + + + + + + + + + + + + + + + + + + + + + 1 2 p2 < p1 p1  p  Mechanisms to conduct electric current: Diffusion (IV) The current density is proportional to the hole concentration gradient:   jp_Diff = -q·Dp· p  Dp = hole diffusion coefficient.

29. Mechanisms to conduct electric current: Diffusion (V) q = magnitude of the electronic charge (1.6·10-19 coulombs). Dp = hole diffusion coefficient. Dn = electron diffusion coefficient. Ñp = hole concentration gradient. Ñn = electron concentration gradient.       jp_Diff= -q·Dp· p jn_Diff= q·Dn· n  

30. Summary of conduction mechanisms   • Drift currents depend on the carrier concentration and on the electric field. • Diffusion currents do not depend on the carrier concentration, but on the carrier concentration gradient.  • jp_Drift = q·p·p·E • jn_Drift = q·n·n·E      jp_Diff = -q·Dp· p jn_Diff = q·Dn· n  

31. A B Continuity equations (I) There are some relationships between spatial and time variations of carrier concentrations because electrons and holes cannot mysteriously appear and disappear at a given point, but must be transported to or created at the given point via some type of ongoing action. • The concentration of holes can be time-changing due to: • Different current density of holes across “A” and “B”. • Excess of carriers over the equilibrium (mass action law). • Generation of electron-hole pairs by radiation (light) .

32. - - - Light - A A A B B B + + + + + + Continuity equations (II) jn_B • Different current density of holes across “A” and “B”. jp_B jp_A • Excess of carriers over the equilibrium (mass action law). • Generation of electron-hole pairs by radiation (light) .

33.   ·jp/q p/t = GL- [p(t)-p]/p  -   ·jn/q  + n/t = GL- [n(t)-n]/n Continuity equations (III) Taking into account the three effects, we obtain the continuity equation for holes: Variation due to the excess of carriers over the equilibrium Total time variation of holes Variation due to the generation of electron-hole pairs by light Variation due to the different current density of holes across “A” and “B” GL: rate of generation of electron-hole pairs by light. tp: hole minority-carrier lifetime. p: hole concentration in steady-state. Similarly, we can obtain the continuity equation for electrons:

34.   ·jp/q p/t = GL- [p(t)-p]/p  - N + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Time evolution of an “excess” of minority carries (I) • We generate an “excess” of electron-hole pairs by injecting light into a piece of N-type silicon and we reach the steady-state. Þ p0= GL·p + p 0 0 p p0 N • Now the light injected disappears. We want to compute the time evolution of the hole concentration afterwards. N p0 p(t) p

35. Tangent line p0   Samearea ·jp/q p/t = GL- [p(t)-p]/p  - p(t) p p t p Time evolution of an “excess” of minority carries (II) • We can also compute the time evolution of the hole concentration from the continuity equation: 0 0 After integrating Þp(t) = p+(p-p)·e-tp • Physical interpretation: There is an appreciable increase of holes during 3-5 times tp.

36. N   ·jp/q p/t = GL- [p(t, x)-p]/p  - xN x + + + + + + + + + + + + + + + + + + Spatial evolution of an “excess” of minority carries (I) • We constantly inject an “excess” of holes into a surface of a piece of N-type silicon and we reach the steady-state. No electric field exists and the hole current is due to diffusion. p0 0 0 p 0 Þ 0 = - [p(x)-p]/p + Dp·2[p(x)-p]/x2 After integrating Þp(x) = p + C1·e-x/Lp + C2·ex/Lp where : Lp=(Dp· p)1/2 is the minority hole diffusion length

37. p0 p(x) p x xN Spatial evolution of an “excess” of minority carries (II) p(x) = p + C1·e-x/Lp + C2·ex/Lp xN: length of the N-type crystal Lp: hole diffusion length • Cases: • a) Lp << xN (wide crystal): Þ p(x) = p+(p-p)·e-xLp(decay exponentially). • b) Lp >> xN (narrow crystal): Þ p(x) = p+(p-p)·(xN-x)/xN(decay linearly). • c) Other cases Þhyperbolic sine evolution. Lp >> xN (narrow) p0 Lp << xN (wide) p(x) p Tangent line x xN Lp

38. P-typesilicon N-typesilicon Al- Al- Al- Al- Al- Al- Al- Al- - - - - - - - - - - Barrier to avoid carrier diffusion + + + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Concept of PN junction (I) What happens if we remove the barrier?

39. P-side N-side Al- Al- Al- Al- Al- Al- Al- Al- - - - - - - - - - - + + + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Concept of PN junction (II) Holes begin to diffuse from the P-side to the N-side. Similarly, electrons diffuse from the N-side to the P-side Are all the carriers to be diffused?

40. Al- Al- Al- Al- Al- Al- Al- Al- - - - - - - - - - - + + + + + + + + + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Concept of PN junction (III) Are all the carriers to be diffused? P-side N-side Non-neutral N-type region, but positively charged Non-neutral P-type region, but negatively charged Is this situation “the final situation”? The answer is no

41. P-side N-side Al- Al- Al- Al- Al- Al- Al- Al- - - - - - - - - - - - +  + + + + + + + + + + E Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Concept of PN junction (IV) An electric field appears just in the boundary between both regions (we call this boundary metallurgical junction)

42. Al- Al- Al- Al- Al- Al- Al- Al- - - -  - + + + + E Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Concept of PN junction (V) • Now, we do zoom over the metallurgical junction P-side N-type Due to diffusion (® ¬) Due to drift (electric field) (¬®) The electric field limits the carrier diffusion

43. Al- Al- Al- Al- Al- Al- Al- Al- - - - - - - - + Neutral N-type region (electronsare balanced by positive ions ) Neutral P-type region (holes are balanced by negative ions )  + + + + + + E Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Concept of PN junction (VI) • Steady-state situation near the metallurgical junction Depletion region, or space charge region, or transition region Unbalanced charge exists because carriers barely exist

44. Metallurgical junction N-side (neutral) P-side (neutral) + - Concept of PN junction (VII) • Summary and terminology  E Many electrons, but neutral Many holes, but neutral V0 Depletion, or transition, or space charge region (non neutral) There is space charge and, therefore, there are electric field E and voltage V0. However, there are almost no charge carriers 

45. - + - + - + - - - + N-side Due to drift - + Due to diffusion - + + + Due to drift - + Due to diffusion P-side - + jp_Diff jp_Drift jn_Drift jn_Diff Computing the built-in voltage V0 (I) Net current passing through any section must be zero. As neither holes nor electrons are being accumulated in any parts of the crystal, net current due to holes is zero and net current due to electrons is zero. Currents must cancel each other Currents must cancel each other

46. pP(hole concentration in P-side ) (hole concentration in N-side )pN - + + - + - + - + - N-side V0 Due to diffusion + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Zona P Due to drift jp_Drift jp_Diff Computing the built-in voltage V0 (II) jp_Drift= - jp_Diff

47. Computing the built-in voltage V0 (III) Equations: jp_Drift= - jp_Diff jp_Drift= q·p·p·E jp_Diff= -q·Dp·dp/dx E = -dV/dx Therefore: dV = -(Dp/mp)·dp/p After integrating : V0= VN-side – VP-side = -(Dp/p)·ln(pN/pP) = (Dp/p)·ln(pP/pN) Repeating the same process with electrons, we obtain:  E V0= (Dn/n)·ln(nN/nP) k = Boltzmann constant. VT = 26 mV at 300 K. It could be demonstrated: Dp/p = Dn/n = kT/q = VT(Einstein relation)

48. - + + - + - + - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Computing the built-in voltage V0 (IV) Summary (I) nN pN pP nP V0 Zona P N-side: many electrons P-side: many holes Almost no holes or electrons Almost no holes Almost no electrons V0 = VT·ln(pP/pN) and also V0 = VT·ln(nN/nP)

49. P-side N-side + - ND, nN, pN NA, pP, nP V0 If ND >> ni(current case) nN = NDpN = ni2/ND If NA >> ni (current case) pP = NA nP = ni2/NA Computing the built-in voltage V0 (V) Summary (II) V0= VT·ln(pP/pN) and also V0= VT·ln(nN/nP) V0 = VT·ln(NA·ND/ni2), VT = 26 mV at 300 K

50. N-side P-side Al- Al- Al- Al- Al- Al- Al- Al- - - - WP0 WN0 NA ND W0 + + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Depletion width in P-side and in N-side Charge neutrality implies: NA·WP0 = ND·WN0 The heavier doped a side, the narrower the depletion region in that side