MidTerm - EE 5342 (print last name) (print first name) Solution 11:00 AM, Wednesday February 10, 2010, 108 Nedderman Hall 80 minutes allowed (last four digits of your student #) (e-mail if new) Instructions: 1. Do your own work. 2. You may use a legal copy of the Massobrio and Antognetti text, you may write notes in your text, and 1 page of additional notes. You may NOT pass a book or note sheet to another student. 3. Calculator allowed. You may NOT share a calculator with another student. 4. Where values or equations are given on this cover sheet, use them in lieu of any other source. If a value is not given, explicitly state definitions and assumptions that you use. 5. Where necessary, calculate parameters if they cannot be read 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 = 25.843 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 qχ Si = 4.05 ev E g,si = 1.124 ev. N v = 1.04E19 cm -3 11. For the work function of poly silicon, use φ n+ = χ si = 4.05 V φ p+ = χ Si + E g,si /q = 5.174 V. 12. For minority carrier (either electrons or holes) lifetime in silicon, use the relationship τ min = (4.5E-5 sec)/(1 + N i /1E17 + (N i /5E17) 2 ), where N i = the total impurity concentration in cm -3 13. For holes in silicon doped primarily with boron, assume μ p = {470.5 [1+(N i 2.23E17) 0.719 ]}+44.9, in cm 2 /V-sec. 14. For electrons in silicon doped primarily with phosphorous, assume μ n = {1414 [1+(N i 9.2E16) 0.711 ]}+68.5, in cm 2 /V-sec. 15. For electrons in silicon doped primarily with arsenic, assume μ n = {1417 [1+(N i 9.68E16) 0.680 ]}+52.2, in cm 2 /V-sec. (In 12 through 16, N i = the total impurity concentration in n- or p-type material, compensated or not). 16. Metal gate work functions should be assumed to be φ M,Al = 4.1 V for aluminum, φ M,Pt = 5.3 V for platinum, φ M,Au = 4.75 V for gold 17. The electron affinity of SiO 2 is χ 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, ε o = 8.854E-14 Fd/cm 23. Relative permittivity of silicon, ε r = 11.7 24. Relative permittivity of silicon dioxide, ε rox = 3.9 25. 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 /(ε r ε 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
Be sure to note the special conditions for statements 12 through 15 after statement 15 on page 1. 1. A region of silicon has 7E15/cm 3 phosphorous and 1E17/cm 3 boron. a. Calculate the majority carrier concentration and type. Na > Nd, so po = -(Nd-Na) = -(7E16/cm3-1E17/cm3) po = 9.3E+16 /cm3 Answer 1a[7] Majority carrier concentration = 9.3E+16 /cm 3, type =. b. Calculate the minority carrier concentration and type. no = ni^2/po no = 2.3E+03 /cm3 Answer 1b[7] Minority carrier concentration = 2.3E+03 /cm 3, type =. c. What is the majority carrier mobility. The net impurity concentration is Na+Nd = 1.07E+17 /cm3 So from the chart, MUp = 310 cm2/v-sec Answer 1c[7] Majority carrier mobility = 310 cm 2 /V-sec. d. What is the minority carrier diffusion coefficient. The net impurity concentration is Na+Nd = 1.07E17/cm3, so from the chart Dn = 18 cm2/sec Answer 1d[7] Minority carrier diffusion coefficient = 18 cm 2 /sec. e. This material is made into a diode with the doping on the other side being 4E15/cm 3. What is the contact potential of this diode? Vbi = Vt*ln(Na*Nd/ni^2) Vbi = 0.729 V Answer 1e[8] Contact potential = 0.729 V. f. What is the capacitance per unit area for this diode when the applied voltage is zero volts? From the chart, for N- = 4E15/cm3, and Vbi-Va = 700 mv C'jo = 22.5 nf/cm2 Answer 1f[8] Capacitance per unit area = 21 n Fd/cm. Page 2
2. A region of silicon has n = 2E17/cm 3 and a majority carrier mobility of 550 cm 2 /V-sec, and a minority carrier mobility of 250 cm 2 /V-sec. Assume that the total impurity concentration is 2.8E17/cm 3. a. What is the minority carrier lifetime in this region? From the chart, for Ni = 2.8E17/cm3 TAUp = 9E-6 sec Answer 2a[8] The minority carrier lifetime is 9E-6 sec. b. What is the minority carrier diffusion length in this region? Lp = (Dp*TAUp) = (250cm2/V-sec*.025843V*9e-6sec) Lp = 7.63E-03 cm Answer 2b[8] The minority carrier diffusion length is 7.63E-03 cm. c. What is the saturation current density for minority current injection into this region from an P+ anode region into the cathode region described above (assume a short diode with charge neutral region width 2 μm)? Is = q*ni^2*dmin/(nd*xcnr) = q*ni^2*mumin*vt/(nd*xcnr) Is = 5.44E-12 A/cm2 Answer 2c[8] The saturation current density is 5.44E-12 A/cm2. d. If the charge neutral region of the cathode region in part c is changed to 500 μm long, what does the saturation current density become? Xcnr will be replaced by Lp*tanh(Xcnr/Lp) = 7.63E-03 cm, so this is a long diode since Lp = 76.3 microns << Xcnr = 500 microns. Is = q*ni^2*dmin/(nd*lmin*tanh(xcnr/lmin) Is = 1.43E-13A/cm2 Answer 2d[8] The saturation current density is 1.43E-13 A/cm2. Page 3
3 A diode has a saturation current of 1.2E-14 A. a Evaluate the diode current at 820 mv. id = IS*(exp(vD/Vt)-1) id = 0.723 A Answer 3a[8] The diode current is 723 ma. b Evaluate the diode conductance at 820 mv. gd = id/(n*vt) gd = 28.0 S Answer 3b[8] The diode conductance is 28E+3 ma/v. c The diode is a n+p diode with p = 3E15/cm 3 in the ( short 2 μm CNR and uncompensated) anode region. What is the diffusion capacitance of this diode at 820 mv. Dn,min = 33.6 cm2/sec from the chart TAUtransit = Wcnr^2/(2*Dmin) TAUtransit = 5.95E-10 sec Cdiff = gd*tau = 1.67E-08 Fd Answer 3c[8] The diffusion capacitance is 1.67E-08 Fd. Page 4
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Note: Be sure to recognize that the scale on the left side is φ i - V a = V bi - V a Page 6
Figures on pp. 5-7 from Device Electronics for Integrated Circuits, 3/E by Richard S. Muller and Theodore I. Kamins, Copyright 2003 John Wiley & Sons. Inc. Page 7
Figures on this page from Physics of Semiconductor Devices, M. Shur, Prentice-Hall, 1990. Page 8