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L4 January 271 Semiconductor Device Modeling and Characterization EE5342, Lecture 4-Spring 2005 Professor Ronald L. Carter

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Presentation on theme: "L4 January 271 Semiconductor Device Modeling and Characterization EE5342, Lecture 4-Spring 2005 Professor Ronald L. Carter"— Presentation transcript:

1 L4 January 271 Semiconductor Device Modeling and Characterization EE5342, Lecture 4-Spring 2005 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

2 L4 January 272 Web Pages *You should be aware of information at R. L. Carter’s web page –www.uta.edu/ronc/ EE 5342 web page and syllabus –www.uta.edu/ronc/5342/syllabus.htm University and College Ethics Policies –http://www.uta.edu/studentaffairs/judicialaffairs/ –www.uta.edu/ronc/5342/Ethics.htm Submit a signed copy to Dr. Carter

3 L4 January 273 First Assignment e-mail to listserv@listserv.uta.edu –In the body of the message include subscribe EE5342 This will subscribe you to the EE5342 list. Will receive all EE5342 messages If you have any questions, send to ronc@uta.edu, with EE5342 in subject line.

4 L4 January 274 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

5 L4 January 275 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

6 L4 January 276 For N d > N a >Let N = N d -N a, and (taking the + root) n o = (N)/2 + {[N/2] 2 +n i 2 } 1/2 For N d+ = N d >> n i >> N a we have >n o = N d, and >p o = n i 2 /N d Equilibrium conc n-type

7 L4 January 277 For N a > N d >Let N = N d -N a, and (taking the + root) p o = (-N)/2 + {[-N/2] 2 +n i 2 } 1/2 For N a- = N a >> n i >> N d we have >p o = N a, and >n o = n i 2 /N a Equilibrium conc p-type

8 L4 January 278 Position of the Fermi Level E fi is the Fermi level when n o = p o E f shown is a Fermi level for n o > p o E f < E fi when n o < p o E fi < (E c + E v )/2, which is the mid- band

9 L4 January 279 E F relative to E c and E v Inverting n o = N c exp[-(E c -E F )/kT] gives E c - E F = kT ln(N c /n o ) For n-type material: E c - E F =kTln(N c /N d )=kTln[(N c P o )/n i 2 ] Inverting p o = N v exp[-(E F -E v )/kT] givesE F - E v = kT ln(N v /p o ) For p-type material: E F - E v = kT ln(N v /N a )

10 L4 January 2710 E F relative to E fi Letting n i = n o gives  E f = E fi n i = N c exp[-(E c -E fi )/kT], so E c - E fi = kT ln(N c /n i ). Thus E F - E fi = kT ln(n o /n i ) and for n- typeE F - E fi = kT ln(N d /n i ) Likewise E fi - E F = kT ln(p o /n i ) and for p- type E fi - E F = kT ln(N a /n i )

11 L4 January 2711 Locating E fi in the bandgap Since E c - E fi = kT ln(N c /n i ), and E fi - E v = kT ln(N v /n i ) The sum of the two equations gives E fi = (E c + E v )/2 - (kT/2) ln(N c /N v ) Since N c = 2.8E19cm -3 > 1.04E19cm -3 = N v, the intrinsic Fermi level lies below the middle of the band gap

12 L4 January 2712 Sample calculations E fi = (E c + E v )/2 - (kT/2) ln(N c /N v ), so at 300K, kT = 25.86 meV and N c /N v = 2.8/1.04, E fi is 12.8 meV or 1.1% below mid-band For N d = 3E17cm -3, given that E c - E F = kT ln(N c /N d ), we have E c - E F = 25.86 meV ln(280/3), E c - E F = 0.117 eV =117meV ~3x(E c - E D ) what N d gives E c -E F =E c /3

13 L4 January 2713 Equilibrium electron conc. and energies

14 L4 January 2714 Equilibrium hole conc. and energies

15 L4 January 2715 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

16 L4 January 2716 Carrier mobility (cont.) The response function  is the mobility. The mean time between collisions,  coll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few. Hence  thermal = q  thermal /m*, etc.

17 L4 January 2717 Carrier mobility (cont.) If the rate of a single contribution to the scattering is 1/  i, then the total scattering rate, 1/  coll is

18 L4 January 2718 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

19 L4 January 2719 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 )

20 L4 January 2720 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)

21 L4 January 2721 Drift current resistance (cont.) Note: for an extrinsic semiconductor and multiple scattering mechanisms, since R = l/(nq  n A) or l/(pq  p A), and (  n or p total ) -1 =   i -1, then R total =  R i (series Rs) The individual scattering mechanisms are: Lattice, ionized impurity, etc.

22 L4 January 2722 Exp. mobility model function for Si 1 ParameterAsPB  min 52.268.544.9  max 14171414470.5 N ref 9.68e169.20e162.23e17  0.6800.7110.719

23 L4 January 2723 Exp. mobility model for P, As and B in Si

24 L4 January 2724 Carrier mobility functions (cont.) The parameter  max models 1/  lattice the thermal collision rate The parameters  min, N ref and  model 1/  impur the impurity collision rate The function is approximately of the ideal theoretical form: 1/  total = 1/  thermal + 1/  impurity

25 L4 January 2725 Carrier mobility functions (ex.) Let N d = 1.78E17/cm3 of phosphorous, so  min = 68.5,  max = 1414, N ref = 9.20e16 and  = 0.711. Thus  n = 586 cm2/V-s Let N a = 5.62E17/cm3 of boron, so  min = 44.9,  max = 470.5, N ref = 9.68e16 and  = 0.680. Thus  n = 189 cm2/V-s

26 L4 January 2726 Lattice mobility The  lattice is the lattice scattering mobility due to thermal vibrations Simple theory gives  lattice ~ T -3/2 Experimentally  n,lattice ~ T -n where n = 2.42 for electrons and 2.2 for holes Consequently, the model equation is  lattice (T) =  lattice (300)(T/300) -n

27 L4 January 2727 Ionized impurity mobility function The  impur is the scattering mobility due to ionized impurities Simple theory gives  impur ~ T 3/2 /N impur Consequently, the model equation is  impur (T) =  impur (300)(T/300) 3/2

28 L4 January 2728 Mobility Summary The concept of mobility introduced as a response function to the electric field in establishing a drift current Resistivity and conductivity defined Model equation def for  (N d,N a,T) Resistivity models developed for extrinsic and compensated materials

29 L4 January 2729 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 )

30 L4 January 2730 Net silicon extr resistivity (cont.)

31 L4 January 2731 Net silicon extr resistivity (cont.) Since  = (nq  n + pq  p ) -1, and  n >  p, (  = q  /m*) we have  p >  n Note that since 1.6(high conc.) <  p /  n < 3(low conc.), so 1.6(high conc.) <  n /  p < 3(low conc.)

32 L4 January 2732 Net silicon (com- pensated) res. For an n-type (n >> p) compensated semiconductor,  = (nq  n ) -1 But now n = N = N d - N a, and the mobility must be considered to be determined by the total ionized impurity scattering N d + N a = N I Consequently, a good estimate is  = (nq  n ) -1 = [Nq  n (N I )] -1

33 L4 January 2733 Equipartition theorem The thermodynamic energy per degree of freedom is kT/2 Consequently,

34 L4 January 2734 Carrier velocity saturation 1 The mobility relationship v =  E is limited to “low” fields v < v th = (3kT/m*) 1/2 defines “low” v =  o E[1+(E/E c )  ] -1/ ,  o = v 1 /E c for Si parameter electrons holes v 1 (cm/s) 1.53E9 T -0.87 1.62E8 T -0.52 E c (V/cm) 1.01 T 1.55 1.24 T 1.68  2.57E-2 T 0.66 0.46 T 0.17

35 L4 January 2735 v drift [cm/s] vs. E [V/cm] (Sze 2, fig. 29a)

36 L4 January 2736 Carrier velocity saturation (cont.) At 300K, for electrons,  o = v 1 /E c = 1.53E9(300) -0.87 /1.01(300) 1.55 = 1504 cm 2 /V-s, the low-field mobility The maximum velocity (300K) is v sat =  o E c = v 1 = 1.53E9 (300) -0.87 = 1.07E7 cm/s

37 L4 January 2737 Diffusion of carriers In a gradient of electrons or holes,  p and  n are not zero Diffusion current,  J =  J p +  J n (note D p and D n are diffusion coefficients)

38 L4 January 2738 Diffusion of carriers (cont.) Note (  p) x has the magnitude of dp/dx and points in the direction of increasing p (uphill) The diffusion current points in the direction of decreasing p or n (downhill) and hence the - sign in the definition of  J p and the + sign in the definition of  J n

39 L4 January 2739 Diffusion of Carriers (cont.)

40 L4 January 2740 Current density components

41 L4 January 2741 Total current density

42 L4 January 2742 Doping gradient induced E-field If N = N d -N a = N(x), then so is E f -E fi Define  = (E f -E fi )/q = (kT/q)ln(n o /n i ) For equilibrium, E fi = constant, but for dN/dx not equal to zero, E x = -d  /dx =- [d(E f -E fi )/dx](kT/q) = -(kT/q) d[ln(n o /n i )]/dx = -(kT/q) (1/n o )[dn o /dx] = -(kT/q) (1/N)[dN/dx], N > 0

43 L4 January 2743 Induced E-field (continued) Let V t = kT/q, then since n o p o = n i 2 gives n o /n i = n i /p o E x = - V t d[ln(n o /n i )]/dx = - V t d[ln(n i /p o )]/dx = - V t d[ln(n i /|N|)]/dx, N = -N a < 0 E x = - V t (-1/p o )dp o /dx = V t (1/p o )dp o /dx = V t (1/N a )dN a /dx

44 L4 January 2744 The Einstein relationship For E x = - V t (1/n o )dn o /dx, and J n,x = nq  n E x + qD n (dn/dx) = 0 This requires that nq  n [V t (1/n)dn/dx] = qD n (dn/dx) Which is satisfied if

45 L4 January 2745 References *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989. **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago. M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.

46 L4 January 2746 References 1 Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. 2 Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.


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