p = {470.5 [1+(N i 2.23E17) ]}+44.9, in cm 2 /V-sec. 14. For electrons in silicon doped primarily with phosphorous, assume

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1 Final - EE 5340/001 (print last name) (print first name) SOLUTION Wednesday, December 15, 2010, 11:00 AM, 108 Nedderman Hall 2 hr 30 min allowed (last four digits of your student #) ( if new) Instructions: 1. Do your own work. PLEASE DO NOT REMOVE THE STAPLE ON THIS EXAM. 2. You may use either a legal copy of the text OR reference text. You may NOT pass a book or note sheet to another student. You may NOT use class notes. Do not use previously solved problems. 3. Calculator allowed. You may NOT share a calculator with another student. 4. Use values given on this cover sheet. If a value is not given, explicitly state definitions and assumptions that you use. In all cases, the notation, men = m 10 n. 5. Where possible, calculate parameters rather than read them from a graph. 6. Do all work in the spaces provided on this exam paper. If you write on the back of a sheet, make the notation "PTO" in your solution in order to assure that material written on the back of the page is evaluated for a grade. AN EXTRA BLANK SHEET IS ATTACHED AT THE BACK OF THE EXAM. 7. Show all calculations, making numerical substitutions and giving numerical results where possible. 8. Write answers in space given. 9. Unless stated otherwise, T = 300K, V t = mv (to agree with M&K k and q values) 10. Unless otherwise stated, the material is silicon (300K) with n i = 1.45E10 cm -3 N c = 2.8E19 cm -3 q Si = 4.05 ev E g,si = ev. N v = 1.04E19 cm For the work function of poly silicon, use n+ = si = 4.05 V p+ = Si + E g,si /q = V. 12. For minority carrier (either electrons or holes) lifetime in silicon, see the figures on page 3. In the case of moderate compensation, the horizontal axis can be assumed to be the total impurity concentration, N i. 13. For holes in silicon doped primarily with boron, assume p = {470.5 [1+(N i 2.23E17) ]}+44.9, in cm 2 /V-sec. 14. For electrons in silicon doped primarily with phosphorous, assume n = {1414 [1+(N i 9.2E16) ]}+68.5, in cm 2 /V-sec. 15. For electrons in silicon doped primarily with arsenic, assume n = {1417 [1+(N i 9.68E16) ]}+52.2, in cm 2 /V-sec. (In 13 through 15, N i = the total impurity concentration in n- or p-type material, compensated or not. and the graphs of these relationships are on page 2) 16. Metal gate work functions should be assumed to be M,Al = 4.1 V for aluminum, M,Pt = 5.3 V for platinum, M,Au = 4.75 V for gold 17. The electron affinity of SiO 2 is SiO2 = 0.95 V. 18. Planck constant h = 6.625E-34 J-s = 4.135E-15 ev-s, (1 ev = 1.602E-19 Joule). 19. free electron mass m o = 9.11E-28 g. 20. Boltzmann constant, k = 1.38E-23 J/K 21. Electron charge, q = 1.602E-19 Coulomb 22. Permittivity of free space, o = 8.854E-14 Fd/cm 23. Relative permittivity of silicon, r = Relative permittivity of silicon dioxide, rox = The breakdown voltage of an abrupt (step) junction (asymmetrical or one-sided) diode with doping on the lightly doped side of N B is V B = 60(Eg/1.1) 3/2 (10 16 /N B ) 3/4 V. The critical field for breakdown is modeled as E crit = (120V qn B /( r 0 )) 1/2 (Eg/1.1) 3/4 (10 16 /N B ) 3/8 26. Each part is worth [x] points, as given in the problem. 27. Turn your cell phone off. Page 1

2 28. Mobility vs. impurity concentration for Si. (From Muller and Kamins.) 29. Critical electric field vs. doping concentration for Si. (From Muller and Kamins.) Page 2

3 Regarding comment 12: Minority carrier lifetime relationships given in Law, Solley and Burk, Self- Consistent Model of Minority-Carrier Lifetime, Diffusion Length, and Mobility, IEEE ELECTRON DEVICE LETTERS, VOL. 12, NO. 8, AUGUST Page 3

4 1. A p-n junction has N d = 2E15 cm -3 on the n-type side (assume the charge neutral region width is 700 µm), and N a = 3E18 cm -3 on the p-side (assume the charge neutral region width is 0.3 µm). a. What is the built-in voltage, V bi, of this device at T = 300 K? Vt = mv Nd = 2.00E+15 cm^-3 Na = 3.00E+18 cm^-3 Vbi = Vt*ln(Na*Nd/ni^2) = 801 mv a. [7] The built-in voltage V bi = Volts. b. What is the depletion region width at V a = -1.0 Volts and T = 300 K? Neff = Na*Nd/(Na+Nd) = 2.00E+15 cm^-3 ersi = 11.7 e0 = 8.85E-14 Fd/cm W = sqrt(2*ersi*e0*(vbi-va)/(q*neff)) W = microns b. [7] The depletion region width = μm. c. What is the peak electric field in the depletion region at V a = -1.0 Volts and T = 300 K? Em = 2*(Vbi-Va)/W = 3.34E+04 V/cm c. [6] The peak electric field, E m = V/cm. Page 4

5 d. What is the minority carrier diffusion length on the n-side of the junction? Lm = sqrt(dm*taum), the minority carriers are holes, the impurity concentration is Nd = 2e15/cm^3 From the graphs given Dp(2e15) = 12.2 cm^2/sec TAUp(2e15) = 9.00E+03 n-sec = 9.00E-06 sec Lp = 105 microns d. [7] The minority carrier diffusion length on the n-side of the junction, L min = cm. e. What is the current density at V a = 775 mv and T = 300 K? State assumptions clearly. Assuming the hole current on the n-side is the major contribution to the current, Js,p = q*ni^2*dp/(nd*lp), since Lp < the charge neutral width. J = Js,p*(exp(Va/Vt)-1) Va = V J = 2.07E-02 A/cm^2 e. [7] The current density at V a = 775 mv and T = 300 K, J s = Fd/cm 2. f. What is the diffusion conductance per unit area, g d, at V a = 775 mv? g'd = J/Vt = 8.02E-04 A/(V-cm^2) f. [6] The conductance per unit area, g d = A/(V-cm 2 ). Page 5

6 2. An ideal prototype BJT structure has is intentionally doped with N de = 2E18 cm -3, N ab = 5E16 cm -3, N dc = 1E15 cm -3. Assume the charge neutral base width is 0.6E-4 cm. The emitter area and collector areas are 200E-8 cm 2. Assume phosphorous parameters when electron minority carrier values are needed. a. Estimate the emitter efficiency for this BJT. You may assume that the emitter charge neutral region width is 0.35E-4 cm. Assuming "short" parameters for both regions: Db = Dn(NaB) = 23.3 cm^2/sec, NaB = 5.00E+16 cm^-3, & TAUn = 6E3 nsec, Ln = 118 microns De = Dp(NdE) = 3.2 cm^2/sec, NdE = 2.00E+18 cm^-3, & TAUp = 330 nsec, Lp = 10 microns GAMMA ~ 1 - nb*de*xb/(ne*db*xe), Ln >> CNRbase and Lp >> CNRemitter GAMMA = a. [7] =. b. Calculate the base transport factor for this device. TAUb will be the minority electron lifetime for 5E16 doped material TAUB = 2.00E+04 n-sec = 2.00E-05 sec Lb = sqrt(db*taub) = 2.16E-02 cm ALPHA = 1 - (xb/lb)^2/2 = b. [7] T =. c. Calculate the IS parameter for this device. IS = q*ni^2*a*db/(nb*xb) xb = 0.6 micron = cm A = 2.00E-06 cm^2 IS = 5.23E-16 A c. [6] IS = A. Page 6

7 3. A nmos capacitor structure is to be made on N a = 8E14 cm -3 silicon material. The gate is n+ poly silicon and the gate oxide is 260 nm thick. a. What is the ms value for this structure? EgSi = ev CHIsi = 4.05 ev Nv = 1.04E+19 cm^-3 PHIgate = PHIm = CHIsi = 4.05 ev PHIsi = CHIsi + EgSi - k*t*ln(nv/ni) = ev PHIms = PHIm - PHIs = ev Answer a. [7] ms =. b. Calculate the oxide capacitance per unit area. C' = erox*e0/xox xox = 2.60E-05 cm C' = 1.33E-08 Fd/cm^2 Answer b. [7] C' Ox =. c. There is an interface charge density of q*1e11cm -2 at the Oxide/silicon interface. Calculate the flat band voltage for this structure. VFB = PHIms - Q'ss/C'ox VFB = V Answer c. [6] V FB =. Page 7

8 d. Calculate the capacitance per unit area at the deep accumulation condition. C'(deep accumulation) = C'ox C'(deep accumulation) = 1.33E-08 Fd/cm^2 Answer d. [6] C' accum =. e. Calculate the capacitance per unit area at the onset of strong inversion (threshold). Na = 8.00E+14 cm^-3 PHIp = Vt*ln(ni/Nd) = -282 mv xdmax = sqrt(2*e0*ersi*2* PHIp /(q*na)) = micron At OSI, the C' = (1/C'ox + xdmax/(e0*ersi))^-1 = 5.97E-09 Fd/cm^2 Answer e. [7] C' OSI =. f. Calculate the threshold voltage for this structure. VT = VFB + 2* PHIp - (-q*na*xdmax/c'ox) = V Answer f. [7] V T =. Page 8