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EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 02

EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 02. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/. Law of Mass Action. n o p o = n i 2 n i 2 = N c N v exp(E g /kT) = 1E10/cm 3 N c = 2.8E19/cm3, N v = 1.04E19/cm3 and n i = 1E10/cm3.

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EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 02

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  1. EE 4345 - Semiconductor Electronics Design Project Spring 2002 - Lecture 02 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

  2. Law ofMass Action nopo = ni2 ni2 = NcNvexp(Eg/kT)= 1E10/cm3 Nc = 2.8E19/cm3, Nv = 1.04E19/cm3 and ni = 1E10/cm3

  3. Classes ofsemiconductors • Intrinsic: • no = po = ni, if Na and Nd << ni • ni =[NcNvexp(Eg/kT)]1/2 = 1E10/cm3 • not easy to get, since ni represents 1 part in 1E13 impurity level, Nsi~1E23/cm3 • n-type: • nno > pno, since Nd > Na • nno = Nd - Na = N, and pno=ni2/nno=ni2/|N|

  4. Classes ofsemiconductors • p-type: • ppo > nno, since Na > Nd • ppo = Na - Nd = -N, and npo= ni2/ppo=ni2/|N| • Compensated: no=po=ni, w/ Na- = Nd+ > 0 • Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants

  5. Equilibriumconcentrations • Charge neutrality requires q(po + Nd+) + (-q)(no + Na-) = 0 • Assuming complete ionization, so Nd+ = Nd and Na- = Na • Gives two equations to be solved simultaneously 1. Mass action, no po = ni2, and 2. Neutrality po + Nd = no + Na

  6. Carrier Mobility • In an electric field, Ex, the velocity (since ax = Fx/m* = qEx/m*) is vx = axt = (qEx/m*)t, and the displ x = (qEx/m*)t2/2 • If every tcoll, a collision occurs which “resets” the velocity to <vx(tcoll)> = 0, then <vx> = qExtcoll/m* = mEx

  7. Exp. mobility modelfor P, As and B in Si

  8. Drift Current • The drift current density (amp/cm2) is given by the point form of Ohm Law J = (nqmn+pqmp)(Exi+ Eyj+ Ezk), so J = (sn + sp)E =sE, where s = nqmn+pqmp defines the conductivity • The net current is

  9. Drift currentresistance • Given: a semiconductor resistor with length, l, and cross-section, A. What is the resistance? • As stated previously, the conductivity, s = nqmn + pqmp • So the resistivity, r = 1/s = 1/(nqmn + pqmp)

  10. Drift currentresistance (cont.) • Consequently, since R = rl/A R = (nqmn + pqmp)-1(l/A) • For n >> p, (an n-type extrinsic s/c) R = l/(nqmnA) • For p >> n, (a p-type extrinsic s/c) R = l/(pqmpA)

  11. Net silicon (ex-trinsic) resistivity • Since r = s-1 = (nqmn + pqmp)-1 • The net conductivity can be obtained by using the model equation for the mobilities as functions of doping concentrations. • The model function gives agreement with the measured s(Nimpur)

  12. Net silicon extrresistivity (cont.)

  13. Semiconductor Equilibrium Conditions • Law of Mass Action: nopo = ni2 • n-type: nno = Nd, and pno = ni2/Nd Rn = l/[(nnoqmnoA)] • p-type: ppo = Na, and npo = ni2/Na Rp = l/[ppoqmpoA]

  14. Examplecalculations • For Nd = 3.2E16/cm3, ni = 1.4E10/cm3 no = Nd = 3.2E16/cm3 po = ni2/Nd , (po is always ni2/no) = (1.4E10/cm3)2/3.2E16/cm3 = 6.125E3/cm3 (comp to ~1E23 Si) • For po = Na = 4E17/cm3, no = ni2/Na = (1.4E10/cm3)2/4E17/cm3 = 490/cm3

  15. Diffusion ofCarriers (cont.)

  16. Total currentdensity

  17. O O O O O O + + + - - - Induced E-fieldin the D.R. Ex N-contact p-contact p-type CNR n-type chg neutral reg Depletion region (DR) Exposed Donor ions Exposed Acceptor Ions W x -xpc -xp xn xnc 0

  18. Depletion approx.charge distribution r +Qn’=qNdxn +qNd [Coul/cm2] -xp x -xpc xn xnc Charge neutrality => Qp’ + Qn’ = 0, => Naxp = Ndxn -qNa Qp’=-qNaxp [Coul/cm2]

  19. Soln to Poisson’sEq in the D.R. Ex xn -xp x -xpc xnc -Emax

  20. Comments on theEx and Vbi • Vbi is not measurable externally since Ex is zero at both contacts • The effect of Ex does not extend beyond the depletion region • The lever rule [Naxp=Ndxn] was obtained assuming charge neutrality. It could also be obtained by requiring Ex(x=0-dx) = Ex(x=0+dx) = Emax

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