Homework 4 , (1) 1+( NA +N D , (2)

Homework 4. Problem. Find the resistivity ρ (in ohm-cm) for piece of Si doped with both cceptors (N A = 9 cm 3 ) nd donors (N D = 6 cm 3 ). Since the electron nd hole mobilities depend on the concentrtion of the dopnts, use the following empriricl expressions to evlute them: µ n = 232 + 8 +( NA +N D 8 6 ).9, () µ p = 48 + 447 where the mobilities re expressed in cm 2 /V s nd N A nd N D in cm 3. Solution. Subtituting for the vlues of N A nd N D in the given equtions, nd +( NA +N D.3 6 ).76, (2) µ n = 247.6 cm 2 /Vs, µ p =5.84 cm 2 /Vs. Since σ = en D µ n + en A µ p nd ρ =/σ, wehve ρ =.22 2 Ωcm. 2. Problem. Consider smple of p-type Si doped with N A = 8 cm 3 nd N D =. Over length of µm the electron concentrtion drops from 6 cm 3 to 3 cm 3. Using Eq. () bove, clculte the current density due to diffusion lone. ECE69 Spring 2

Solution. The electron current due to diffusion lone is: J diff = ed n dn dx. Since from Einstein reltion ed n = k b Tµ n, ssuming dn/dx 6 cm 3 / 4 cm nd using µ n from the eqution given in the first problem, we hve µ n = 342.8 cm 2 /Vs nd J diff = 4.65 A/cm 2. 3. Problem. A p-n Si junction is formed between n-type region with N D = 2 8 cm 3 nd p-type region with N A = 5 6 cm 3. Find: () the width of the depletion region on the n side, x n (in µm), (b) the width of the depletion region on the p side, x p (in µm), (c) the built-in potentil V bi (in ev), (d) the frction of V bi which drops over the n-side of the junction, (e) the frction of V bi which drops over the p-side of the junction. Finlly, using the results just obtined, plot (to scle s ccurtely s you cn) the bnd digrm of the junction t equilibrium. Solution. () From ( ) ( 2ɛkB T NA N D W = e 2 ln + ) /2, N A N D we get W =.54 µm nd (b) Similrly, n 2 i x n = WN A N A + N D =.375 µm. x p = WN D N A + N D =.5 µm. ECE69 Spring 2 2

(c) The built-in potentil is given by: V bi = k BT e ln ( ) NA N D n 2 i =.892 V. (d) The potentil drop on the n side of the junction is: V bi,n = V bix n W =.2 V. (d) On the other end, the potentil drop on the p side of the junction is: V bi,p = V bix p W =.87 V. 4. Problem. ASi p-n junction hs sturted reverse current I s = 4 At 3 K. Determine the forwrd bis required to get current of () 5 µand (b) 2.5 ma. Solution. () From the expression for the diode current ( ) ev I = I s exp k B T nd from the fct tht I s = 4 Awe hve V =.5784 V when I =5µA. (b) Similrly, for I =2.5mA we hve V =.6797 V. 5. Problem. The sturted reverse current of GAs p-n junction is 5 2 A. Clculte the current under forwrd bis of ().8 V nd (b).2 V. Solution. () As in the previous problem, we hve I =.29 µafor V =.8Vnd(b) I =.66AforV =.2 V. ECE69 Spring 2 3

6. Problem. Derive Eq. (49) of the Lecture Notes, Prt 2. Solution. We require: J(x = l n ) = J = J { + dx α p exp Now let s multiply both sides of this eqution by: l J exp n, } exp (3) so tht Eq. (3) becomes: exp = + dx α p exp, (4) or = exp dx α p exp. (5) Now let s dd nd substrct α n in the integrnd of the lst term: = exp + dx (α n α p ) exp dx α n exp. (6) ECE69 Spring 2 4

The second term on the right-hnd side cn be integrted immeditely, since in generl: b ) ( dx f(x) exp dx f(x ) = exp = exp ( ( ) dx f(x b ) b ) dx f(x ). So, Eq. (6) becomes: = exp dx α n exp + exp = dx α n exp (7) so tht l n x = dx α n exp, (8) which is Eq. (49) of the Lecture Notes, Prt 2. Note tht the result bove is completely generl. We hve not mde use of the ssumption α n = α p,norof ny rbitrry ssumption of the type x exp dx(α n α p ) = exp (α n α p )(x + l p ), which is only vlid if α n nd α p do not depend on x, nd, finlly, it is vlid lso fr from brekdown, so it does not require ny condition of the sort ln l dxα =. p 7. Problem. (). Plot the Zener tunneling current J Zener given by Eq. (425) of the Lecture Notes s function of pplied bis V using m =.32 m el =.32 9. 3 kg nd E G =.ev (the gp of Si). Assume, ECE69 Spring 2 5

for the field F the pproximte vlue F F mx, where the mximum field in the junction, F mx, is given by Eq. (364). Assume N A = 7 cm 3 nd N D = 9 cm 3 to estimte the width l p nd l n of the depletion regions. (b). Repet the clcultion, but now using E G =.64 ev (the gp of Ge) nd m =.22 m el. (c). Repet the clcultions in () nd (b), but now ssuming higher doping, N A = 9 cm 3 nd N D = 2 cm 3. Wht hppens? Why? (d). Considering tht the current VLSI technolgy requires incresingly higher doping, cn you foresee ny problems with the possible use of Ge (insted of Si) when the width of the depletion regions shrinks? Solution. Higher doping enhnces tunneling. Ge exhibits much lrger Zener tunneling current thn Si becuse of its smller bnd-gp. This drwbck will become more severe s the doping concentrtions increse. The figure below shows the result: ECE69 Spring 2 6

2 Ge J Zener (A/cm 2 ) 2 Si high doping Ge low doping Si 4 2 2 4 6 8 V pplied (V) Note tht the current is nonzero only under reverse bis such tht V < (E G V bi ) or under forwrd bis V >E G + V bi. For bis of smller mgnitude, V, tunneling cnnot occurr becuse there re no finl sttes vilble. The plot show only the current in reverse bis, since this mtters in devices. (Under forwrd bis the diode current is very lrge nywy nd tunneling is not concern.) NOTE: The most importnt dependence of the Zener tunneling current on the pplied reverse bis V is vi the depletion width l p ppering in the expression for the field F mx = en A l p /ɛ s. Mny of you hve forgotten this crucil dependence. 8. Problem. The current trends in VLSI technology lso demnd reduced pplied bises (i.e., smller V ). Without doing ny clcultion, do you think tht the strictest brekdown limittions will be due to impct ioniztion or to Zener tunneling? ECE69 Spring 2 7

Solution. Zener tunneling will dominte, since the energy threshold for impct-ioniztion is E G, while for Zener tunneling is E G ev bi,wherev bi is the built-in potentil of the drin/body junction in MOSFET. 9. Problem. Using Eqns. (45) nd (45), plot the totl chrge t the GAs-side of the GAs-Al x G x As heterojunction s function of the interfce potentil ψ i = eψ i /(k B T ). Indicte clerly the ccumultion, depletion, inversion, nd strong inversion regions. Refer to the discussion on pge 35 of the Notes for help. Solution. The equtions you were told to use ccount only for the chrge of the mjority crriers (electrons) nd donors. They do not ccount for minority crriers (holes), so tht it is impossible to obtin the chrge in inversion (wek or strong), unless the hole chrge-density is dded. The chrge density of the minority crriers is included in the equtions for the chrge t the interfce of n MOS cpcitor. So, one could use similr eqution, remembering tht in tht cse we considered p-type substrte, while here we re deling with n n-type substrte, so the plot should be flipped (s in the mirror imge ψ i ψ i ).. Problem. Derive Eq. (457) from Eq. (458). Solution. The potentil energy is The mximum of the potentil energy is given by e2 V im (z) = 6πɛ s z ef zz. (9) = dv im(z) dz The vlue of z t which this occurs is thus: = e 2 6πɛ s z 2 ef z. () z = ( ) e /2. () 6πɛ s F z ECE69 Spring 2 8

Inserting this into Eq. (9) we get: e2 V im (z ) = ef z z = e 6πɛ s z ( efz 4πɛ s ) /2, (2) so tht the brier-lowering (in V) will be Φ B =ef z /(4πɛ s ) /2.. Problem. Using the derivtion of Eq. (48) s guide, derive Eq. (482). Solution. The expression given in clss, Eq. (475) of the Notes, Prt 2, for the tunneling current from the metl to the semiconductor is: j MS = em M 2π 2 h 3 deef M (E) f S (E) exp { zt 2 } dz κ(z). (3) Approximting the potentil brrier with φ(z) φ B F zz, we hve for the tunneling distnce z t = (eφ B E)/(eF z) nd: zt dz κ z (z) = (2m S )/2 h zt dz (eφ B ef z E) /2. (4) Using the new integrtion vrible y = eφ B ef z E we get: so tht zt j MS = em M 2π 2 h 3 which is Eq. (482). dz κ z (z) = (2m S )/2 e hf z eφ B E dy y /2 = 2(2m S )/2 3e hf z (eφ B E)3/2, (5) { } deef M (E) f S (E) exp 4(2m S )/2 (eφ B 3e hf E)3/2 z, (6) ECE69 Spring 2 9