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Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011 Professor Ronald L. Carter

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Presentation on theme: "Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011 Professor Ronald L. Carter"— Presentation transcript:

1 Semiconductor Device Modeling and Characterization – EE5342 Lecture 09– Spring 2011 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

2 ©rlc L09-14Feb20112 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.

3 ©rlc L09-14Feb20113 Second Assignment Submit a signed copy of the document that is posted at www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf

4 ©rlc L09-14Feb20114 Additional University Closure Means More Schedule Changes Plan to meet until noon some days in the next few weeks. This way we will make up for the lost time. The first extended class will be Monday, 2/14. The MT changed to Friday 2/18 The P1 test changed to Friday 3/11. The P2 test is still Wednesday 4/13 The Final is still Wednesday 5/11.

5 MT and P1 Assignment on Friday, 2/18/11 Quizzes and tests are open book –must have a legally obtained copy-no Xerox copies. –OR one handwritten page of notes. –Calculator allowed. A cover sheet will be published by Wednesday, 2/16/11. ©rlc L09-14Feb20115

6 6 Energy bands for p- and n-type s/c p-type EcEc EvEv E Fi E Fp q  p = kT ln(n i /N a ) EvEv EcEc E Fi E Fn q  n = kT ln(N d /n i ) n-type

7 ©rlc L09-14Feb20117 Making contact in a p-n junction Equate the E F in the p- and n-type materials far from the junction E o (the free level), E c, E fi and E v must be continuous N.B.: q  = 4.05 eV (Si), and q  = q   E c - E F EoEo EcEc EfEf E fi EvEv q  (electron affinity) qFqF qq (work function)

8 ©rlc L09-14Feb20118 Band diagram for p + -n jctn* at V a = 0 EcEc E fN E fi EvEv EcEc E fP E fi EvEv 0 xnxn x -x p -x pc x nc q  p < 0 q  n > 0 qV bi = q(  n -  p ) *N a > N d -> |  p | >  n p-type for x<0 n-type for x>0

9 ©rlc L09-14Feb20119 A total band bending of qV bi = q(  n -  p ) = kT ln(N d N a /n i 2 )is necessary to set E fP = E fN For -x p < x < 0, E fi - E fP < -q  p, = |q  p | so p < N a = p o, (depleted of maj. carr.) For 0 < x < x n, E fN - E fi < q  n, so n < N d = n o, (depleted of maj. carr.) -x p < x < x n is the Depletion Region Band diagram for p + -n at V a =0 (cont.)

10 ©rlc L09-14Feb201110 Depletion Approximation Assume p << p o = N a for -x p < x < 0, so  = q(N d -N a +p-n) = -qN a, -x p < x < 0, and p = p o = N a for -x pc < x < -x p, so  = q(N d -N a +p-n) = 0, -x pc < x < -x p Assume n << n o = N d for 0 < x < x n, so  = q(N d -N a +p-n) = qN d, 0 < x < x n, and n = n o = N d for x n < x < x nc, so  = q(N d -N a +p-n) = 0, x n < x < x nc

11 ©rlc L09-14Feb201111 Poisson’s Equation The electric field at (x,y,z) is related to the charge density  =q(N d -N a -p-n) by the Poisson Equation:

12 ©rlc L09-14Feb201112 Poisson’s Equation For n-type material, N = (N d - N a ) > 0, n o = N, and (N d -N a +p-n)=-  n +  p +n i 2 /N For p-type material, N = (N d - N a ) < 0, p o = -N, and (N d -N a +p-n) =  p-  n-n i 2 /N So neglecting n i 2 /N, [  =(N d -N a +p-n)]

13 ©rlc L09-14Feb201113 Quasi-Fermi Energy

14 ©rlc L09-14Feb201114 Quasi-Fermi Energy (cont.)

15 ©rlc L09-14Feb201115 Quasi-Fermi Energy (cont.)

16 ©rlc L09-14Feb201116 Induced E-field in the D.R. The sheet dipole of charge, due to Q p ’ and Q n ’ induces an electric field which must satisfy the conditions Charge neutrality and Gauss’ Law* require thatE x = 0 for -x pc < x < -x p and E x = 0 for -x n < x < x nc h 0h 0

17 ©rlc L09-14Feb201117 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

18 ©rlc L09-14Feb201118 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 ]

19 ©rlc L09-14Feb201119 1-dim soln. of Gauss’ law x xnxn x nc -x pc -x p ExEx -E max

20 ©rlc L09-14Feb201120 Depletion Approxi- mation (Summary) For the step junction defined by doping N a (p-type) for x 0, the depletion width W = {2  (V bi -V a )/qN eff } 1/2, where V bi = V t ln{N a N d /n i 2 }, and N eff =N a N d /(N a +N d ). Since N a x p =N d x n, x n = W/(1 + N d /N a ), and x p = W/(1 + N a /N d ).

21 ©rlc L09-14Feb201121 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 -

22 ©rlc L09-14Feb201122 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

23 ©rlc L09-14Feb201123 Junction C (cont.) The C-V relationship simplifies to

24 ©rlc L09-14Feb201124 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

25 ©rlc L09-14Feb201125 Arbitrary doping profile If the net donor conc, N = N(x), then at x n, the extra charge put into the DR when V a ->V a +  V a is  Q’=-qN(x n )  x n The increase in field,  E x =-(qN/  )  x n, by Gauss’ Law (at x n, but also const). So  V a =-(x n +x p )  E x = (W/  )  Q’ Further, since N(x n )  x n = N(x p )  x p gives, the dC/dx n as...

26 ©rlc L09-14Feb201126 Arbitrary doping profile (cont.)

27 ©rlc L09-14Feb201127 Arbitrary doping profile (cont.)

28 ©rlc L09-14Feb201128 Arbitrary doping profile (cont.)

29 ©rlc L09-14Feb201129 Arbitrary doping profile (cont.)

30 ©rlc L09-14Feb201130 Debye length The DA assumes n changes from N d to 0 discontinuously at x n, likewise, p changes from N a to 0 discontinuously at -x p. In the region of x n, the 1-dim Poisson equation is dE x /dx = q(N d - n), and since E x = -d  /dx, the potential is the solution to -d 2  /dx 2 = q(N d - n)/  n x xnxn NdNd 0

31 ©rlc L09-14Feb201131 Debye length (cont) Since the level E Fi is a reference for equil, we set  = V t ln(n/n i ) In the region of x n, n = n i exp(  /V t ), so d 2  /dx 2 = -q(N d - n i e  /Vt ), let  =  o +  ’, where  o = V t ln(N d /n i ) soN d - n i e  /Vt = N d [1 - e  /Vt-  o/Vt ], for  -  o =  ’ <<  o, the DE becomes d 2  ’/dx 2 = (q 2 N d /  kT)  ’,  ’ <<  o

32 ©rlc L09-14Feb201132 Debye length (cont) So  ’ =  ’(x n ) exp[+(x-x n )/L D ]+con. and n = N d e  ’/Vt, x ~ x n, where L D is the “Debye length”

33 ©rlc L09-14Feb201133 Debye length (cont) L D estimates the transition length of a step-junction DR (concentrations N a and N d with N eff = N a N d /(N a +N d )). Thus, For V a =0, & 1E13 < N a,N d < 1E19 cm -3 13% DA is OK

34 ©rlc L09-14Feb201134 Example An assymetrical p+ n junction has a lightly doped concentration of 1E16 and with p+ = 1E18. What is W(V=0)? Vbi=0.816 V, Neff=9.9E15, W=0.33  m What is C’ j ? = 31.9 nFd/cm2 What is L D ? = 0.04  m

35 ©rlc L09-14Feb201135 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. 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. 3 Physics of Semiconductor Devices, Shur, Prentice- Hall, 1990.


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