1. EE 5340 Semiconductor Device Theory Lecture 10 – Fall 2010 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc

2. L10 22Sep102 Test 1 – W 29Sep10 11 AM Room 108 Nedderman Hall Covering Lectures 1 through 10 Open book - 1 legal text or ref., only. You may write notes in your book. Calculator allowed A cover sheet will be included with full instructions. For examples see http://www.uta.edu/ronc/5340/tests/.

3. L10 22Sep103 Ideal n-type Schottky depletion width (V a =0) xnxn x qN d  Q’ d = qN d x n x  ExEx -E m xnxn (Sheet of negative charge on metal)= -Q’ d

4. L10 22Sep104 Debye length n x xnxn NdNd 0

5. L10 22Sep105 Effect of V  0

6. L10 22Sep106 Schottky diode capacitance xnxn x qN d -Q-  Q Q’ d = qN d x n x  ExEx -E m xnxn  Q’

7. L10 22Sep107 Schottky Capacitance (continued) 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 =  i C j -2 ii VaVa C j0 -2

8. Diagrams for ideal metal-semiconductor Schottky diodes. Fig. 3.21 in Ref 4. L10 22Sep108

9. 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 9

10. L10 22Sep10 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) 10

11. L10 22Sep10 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 11

12. L10 22Sep10 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.) 12

13. L10 22Sep10 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 13

14. L10 22Sep10 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 Due to Charge neutrality Q p ’ + Q n ’ = 0, => N a x p = N d x n [Coul/cm 2 ] 14

15. L10 22Sep10 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 ≈0≈0 15

16. L10 22Sep10 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 16

17. L10 22Sep10 Induced E-field in the D.R. (cont.) Poisson’s Equation  E =  / , has the one-dimensional form, dE x /dx =  / , which must be satisfied for  = -qN a, -x p < x < 0, and  = +qN d, 0 < x < x n, with E x = 0 for the remaining range 17

18. L10 22Sep10 Soln to Poisson’s Eq in the D.R. xnxn x -x p -x pc x nc ExEx -E max 18

19. L10 22Sep10 Soln to Poisson’s Eq in the D.R. (cont.) 19

20. L10 22Sep10 Soln to Poisson’s Eq in the D.R. (cont.) 20

21. L10 22Sep10 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 21

22. L10 22Sep10 Sample calculations V t  25.86 mV at 300K  =  r  o = 11.7*8.85E-14 Fd/cm = 1.035E-12 Fd/cm If N a  5E17/cm 3, and N d  2E15 /cm 3, then for n i  1.4E10/cm 3, then what is V bi = 757 mV 22

23. L10 22Sep10 Sample calculations What are N eff, W ? N eff, = 1.97E15/cm 3 W = 0.707 micron What is x n ? = 0.704 micron What is E max ? 2.14E4 V/cm 23

24. L10 22Sep1024 References 1 Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. See Semiconductor Device Fundamentals, by Pierret, Addison- Wesley, 1996, for another treatment of the  model. 2 Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981. 3 Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, 1997. 4 Device Electronics for Integrated Circuits, 3/E by Richard S. Muller and Theodore I. Kamins. © 2003 John Wiley & Sons. Inc., New York.