1. L04,... June 11,...1 Electronics I EE 2303/602 - Summer ‘01 Lectures 04,... Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

2. L04,... June 11,...2 Silicon Covalent Bond (2D Repr) Each Si atom has 4 nearest neighbors Si atom: 4 valence elec and 4+ ion core 8 bond sites / atom All bond sites filled Bonding electrons shared 50/50 _ = Bonding electron

3. L04,... June 11,...3 Si Energy Band Structure at 0 K Every valence site is occupied by an electron No electrons allowed in band gap No electrons with enough energy to populate the conduction band

4. L04,... June 11,...4 Si Bond Model Above Zero Kelvin Enough therm energy ~kT(k=8.62E-5eV/K) to break some bonds Free electron and broken bond separate One electron for every “hole” (absent electron of broken bond)

5. L04,... June 11,...5 Band Model for thermal carriers Thermal energy ~kT generates electron-hole pairs At 300K E g (Si) = 1.124 eV >> kT = 25.86 meV, Nc = 2.8E19/cm3 > Nv = 1.04E19/cm3 >> n i = 1E10/cm3

6. L04,... June 11,...6 Donor: cond. electr. due to phosphorous P atom: 5 valence elec and 5+ ion core 5th valence electr has no avail bond Each extra free el, -q, has one +q ion # P atoms = # free elect, so neutral H atom-like orbits

7. L04,... June 11,...7 Band Model for donor electrons Ionization energy of donor E i = E c -E d ~ 40 meV Since E c -E d ~ kT, all donors are ionized, so N D ~ n Electron “freeze- out” when kT is too small

8. L04,... June 11,...8 Acceptor: Hole due to boron B atom: 3 valence elec and 3+ ion core 4th bond site has no avail el (=> hole) Each hole adds -(- q) and has one -q ion #B atoms = #holes, so neutral H atom-like orbits

9. L04,... June 11,...9 Classes of semiconductors Intrinsic: n o = p o = n i, since N a &N d << n i =[N c N v exp(E g /kT)] 1/2,(not easy to get) n-type: n o > p o, since N d > N a p-type: n o < p o, since N d < N a Compensated: n o =p o =n i, w/ N a - = N d + > 0 Note: n-type and p-type are usually partially compensated since there are usually some opposite-type dopants

10. L04,... June 11,...10 Equilibrium concentrations Charge neutrality requires q(p o + N d + ) + (-q)(n o + N a - ) = 0 Assuming complete ionization, so N d + = N d and N a - = N a Gives two equations to be solved simultaneously 1. Mass action, n o p o = n i 2, and 2. Neutralityp o + N d = n o + N a

11. L04,... June 11,...11 Carrier Mobility In an electric field, E x, the velocity (since a x = F x /m* = qE x /m*) is v x = a x t = (qE x /m*)t, and the displ x = (qE x /m*)t 2 /2 If every  coll, a collision occurs which “resets” the velocity to = 0, then = qE x  coll /m* =  E x

12. L04,... June 11,...12 Exp. mobility model for P, As and B in Si

13. L04,... June 11,...13 Drift Current The drift current density (amp/cm 2 ) is given by the point form of Ohm Law J = (nq  n +pq  p )(E x i+ E y j+ E z k), so J = (  n +  p )E =  E, where  = nq  n +pq  p defines the conductivity The net current is

14. L04,... June 11,...14 Drift current resistance Given: a semiconductor resistor with length, l, and cross-section, A. What is the resistance? As stated previously, the conductivity,  = nq  n + pq  p So the resistivity,  = 1/  = 1/(nq  n + pq  p )

15. L04,... June 11,...15 Drift current resistance (cont.) Consequently, since R =  l/A R = (nq  n + pq  p ) -1 (l/A) For n >> p, (an n-type extrinsic s/c) R = l/(nq  n A) For p >> n, (a p-type extrinsic s/c) R = l/(pq  p A)

16. L04,... June 11,...16 Net silicon (ex- trinsic) resistivity Since  =  -1 = (nq  n + pq  p ) -1 The net conductivity can be obtained by using the model equation for the mobilities as functions of doping concentrations. The model function gives agreement with the measured  (N impur )

17. L04,... June 11,...17 Net silicon extr resistivity (cont.)

18. L04,... June 11,...18 Semiconductor Equilibrium Conditions Law of Mass Action: n o p o = n i 2 n-type: n no = N d, and p no = n i 2 /N d R n = l/ [(n no q  no A)] p-type: p po = N a, and n po = n i 2 /N a R p = l /[p po q  po A]

19. L04,... June 11,...19 Example calculations For N d = 3.2E16/cm 3, n i = 1.4E10/cm 3 n o = N d = 3.2E16/cm 3 p o = n i 2 /N d, (p o is always n i 2 /n o) = (1.4E10/cm 3 ) 2 /3.2E16/cm 3 = 6.125E3/cm 3 (comp to ~1E23 Si) For p o = N a = 4E17/cm 3, n o = n i 2 /N a = (1.4E10/cm 3 ) 2 /4E17/cm 3 = 490/cm 3

20. L04,... June 11,...20 Diffusion of Carriers (cont.)

21. L04,... June 11,...21 Total current density

22. L04,... June 11,...22 Induced E-field in the D.R. xnxn x -x p -x pc x nc O - O - O - O + O + O + Depletion region (DR) p-type CNR ExEx Exposed Donor ions Exposed Acceptor Ions n-type chg neutral reg p-contact N-contact W 0

23. L04,... June 11,...23 Depletion approx. charge distribution xnxn x -x p -x pc x nc  +qN d -qN a +Q n ’=qN d x n Q p ’=-qN a x p Charge neutrality => Q p ’ + Q n ’ = 0, => N a x p = N d x n [Coul/cm 2 ]

24. L04,... June 11,...24 Soln to Poisson’s Eq in the D.R. xnxn x -x p -x pc x nc ExEx -E max

25. L04,... June 11,...25 Comments on the E x and V bi V bi is not measurable externally since E x is zero at both contacts The effect of E x does not extend beyond the depletion region The lever rule [N a x p =N d x n ] was obtained assuming charge neutrality. It could also be obtained by requiring E x (x=0  x  E x (x=0  x)  E max

26. L04,... June 11,...26 Effect of V > 0 Define an external voltage source, V a, with the +term at the p-type contact and the -term at the n-type contact For V a > 0, the V a induced field tends to oppose E x due to DR For V a < 0, the V a induced field tends to add to E x due to DR Will consider V a < 0 now

27. L04,... June 11,...27 Effect of V > 0

28. L04,... June 11,...28 One-sided p+n or n+p jctns If p + n, then N a >> N d, and N a N d /(N a + N d ) = N eff --> N d, and W --> x n, DR is all on lightly d. side If n + p, then N d >> N a, and N a N d /(N a + N d ) = N eff --> N a, and W --> x p, DR is all on lightly d. side The net effect is that N eff --> N -, (- = lightly doped side) and W --> x -

29. L04,... June 11,...29 Junction Capacitance The junction has +Q’ n =qN d x n (exposed donors), and (exposed acceptors) Q’ p =-qN a x p = -Q’ n, forming a parallel sheet charge capacitor.

30. L04,... June 11,...30 Junction C (cont.) This Q ~ (V bi -V a ) 1/2 is clearly non- linear, and Q is not zero at V a = 0. Redefining the capacitance,

31. L04,... June 11,...31 Junction C (cont.) xnxn x -x p -x pc x nc  +qN d -qN a +Q’ n =qN d x n Q’ p =-qN a x p Charge neutrality => Q’ p + Q’ n = 0, => N a x p = N d x n  Q’ n =qN d  x n  Q’ p =-qN a  x p  Q j =  Q’ n A

32. L04,... June 11,...32 Depletion Capacitance

33. L04,... June 11,...33 Junction C (cont.) The C-V relationship simplifies to

34. L04,... June 11,...34 Junction C (cont.) If one plots [C j ] -2 vs. V a Slope = -[(C j0 ) 2 V bi ] -1 vertical axis intercept = [C j0 ] -2 horizontal axis intercept = V bi C j -2 V bi VaVa C j0 -2

35. L04,... June 11,...35 Practical Junctions Junctions are formed by diffusion or implantation into a uniform concentration wafer. The profile can be approximated by a step or linear function in the region of the junction. If a step, then previous models OK. If not, 1/2 --> M, 1/3 < M < 1/2.

36. L04,... June 11,...36 Law of the junction (injection of minority carr.)

37. L04,... June 11,...37 Carrier Injection and diff. xnxn -x pc 0 ln(carrier conc) ln N a ln N d ln n i ln n i 2 /N d ln n i 2 /N a x nc -x p x ~V a /V t

38. L04,... June 11,...38 Ideal diode equation I = I s [exp(V a /nV t )-1], I s = I sn + I sp

39. L04,... June 11,...39 Diffnt’l, one-sided diode conductance VaVa IDID Static (steady- state) diode I-V characteristic VQVQ IQIQ

40. L04,... June 11,...40 Diffnt’l, one-sided diode cond. (cont.)

41. L04,... June 11,...41 Charge distr in a (1- sided) short diode Assume N d << N a The sinh (see L12) excess minority carrier distribution becomes linear for W n << L p  p n (x n )=p n0 expd( V a /V t ) Total chg = Q’ p = Q’ p = q  p n (x n )W n /2 xnxn x x nc  p n (x n ) W n = x nc - x n Q’ p pnpn

42. L04,... June 11,...42 Charge distr in a 1- sided short diode Assume Quasi- static charge distributions Q’ p = Q’ p = q  p n (x n )W n /2 d  p n (x n ) = (W/2)* {  p n (x n,V a +  V) -  p n (x n,V a )} xnxn x x nc  p n (x n,V a ) Q’ p pnpn  p n (x n,V a +  V)  Q’ p

43. L04,... June 11,...43 Cap. of a (1-sided) short diode (cont.)

44. L04,... June 11,...44 Diode equivalent circuit (small sig) IDID VDVD VQVQ IQIQ  is the practical “ideality factor”

45. L04,... June 11,...45 Small-signal eq circuit C diff C depl r diff C diff and C depl are both charged by V a = V Q VaVa

46. L04,... June 11,...46 Diode Switching Consider the charging and discharging of a Pn diode –(N a > N d ) –W d << Lp –For t < 0, apply the Thevenin pair V F and R F, so that in steady state I F = (V F - V a )/R F, V F >> V a, so current source –For t > 0, apply V R and R R I R = (V R + V a )/R R, V R >> V a, so current source

47. L04,... June 11,...47 Diode switching (cont.) + + VFVF VRVR D R RFRF Sw R: t > 0 F: t < 0 V F,V R >> V a

48. L04,... June 11,...48 Diode charge for t < 0 xnxn x nc x pnpn p no

49. L04,... June 11,...49 Diode charge for t >>> 0 (long times) xnxn x nc x pnpn p no

50. L04,... June 11,...50 Equation summary

51. L04,... June 11,...51 Snapshot for t barely > 0 xnxn x nc x pnpn p no Total charge removed, Q dis =I R t

52. L04,... June 11,...52 I(t) for diode switching IDID t IFIF -I R tsts t s +t rr - 0.1 I R

53. L04,... June 11,...53 Reverse bias junction breakdown Avalanche breakdown –Electric field accelerates electrons to sufficient energy to initiate multiplication of impact ionization of valence bonding electrons –field dependence shown on next slide Heavily doped narrow junction will allow tunneling - see Neamen*, p. 274 –Zener breakdown

54. L04,... June 11,...54 E crit for reverse breakdown (M&K**) Taken from p. 198, M&K**

55. L04,... June 11,...55 Reverse bias junction breakdown Assume -V a = V R >> V bi, so V bi -V a -->V R Since E max = 2(V bi -V a )/W, when E max = E crit BV =  (E crit ) 2 /(2qN - )

56. L04,... June 11,...56 BV for reverse breakdown (M&K**) Taken from Figure 4.13, p. 198, M&K** Breakdown voltage of a one-sided, plan, silicon step junction showing the effect of junction curvature. 4,5

57. L04,... June 11,...57 References * Semiconductor Physics and Devices, 2nd ed., by Neamen, Irwin, Boston, 1997. **Device Electronics for Integrated Circuits, 2nd ed., by Muller and Kamins, John Wiley, New York, 1986.