EE 5340 Semiconductor Device Theory Lecture 10 – Fall 2010 Professor Ronald L. Carter

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

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

L10 22Sep104 Debye length n x xnxn NdNd 0

L10 22Sep105 Effect of V  0

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

L10 22Sep10 Sample calculations V t  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

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

L10 22Sep1024 References 1 Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 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, Semiconductor Physics & Devices, 2nd ed., by Neamen, Irwin, Chicago, Device Electronics for Integrated Circuits, 3/E by Richard S. Muller and Theodore I. Kamins. © 2003 John Wiley & Sons. Inc., New York.