SHW5 - (b Employing Eq(5.12 Vbi =...

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Unformatted text preview: (b) Employing Eq. (5.12). Vbi = 5-[(E‘r-EF)¢sidc+(EF-Ei)n-sidc] ('YwG/zmr + 50/4) = #65614 + 2n) = $0.12) + 2(0.0259) 0.89 v 5.1 (a) Because NM > Mg and p = n; expl(Ei-Ep)/k7'| 5 NA far firm the junction. it follows that (Ei-Ep)x<o > (Er—Es)” The energy band diagram must therefore be of the fonn (b) Given that the dopings are nondcgcnctate, the same development leading to Eq. (5.10) can be employed. except nan) E Rig/Nu n(-Ip) E rug/NM which when substituted into Eq. (5.8) yields vbi = H: {VAL} q A2 Alternatively, one can write Vbi = filial—“F Ei(+°'°)] = #kEi‘EFbl-sidc-(Ei—EF)p2-side] = qlllen(NM/ni)—len(NA2/ui)] = quIan/NAZ) Note that, as must be the case. Vhi —> 0 if NM = N“. (c) V X 8 X p X (It should be emphasized that the above are tough sketches. The exact functional dependencies mnnot be deduced employing a graphical approach.) (d) . It is true that the minus charge shown on the x < 0 portion of the part (c) p-plot is caused by a depletion of holes on the higher-doped NM-side of the junction, leading to a net charge associated with the ionized acceptors. The plus charge on the NAZ-side of the junction. however. cannot be attributed to ionized donors. There is no donor doping! The only other source of a positive charge is holes—There is a hole concentration on the n-side of the junction in excess of NM. [The hole execs on the NAz-side of the junction can actually be inferred from the energy band diagram in part (a).] Since p > NAz near the junction, we cannot invoke the depletion approximation. SA (a) Vbi = 521 NAM; = (00259)] w = (L614 v 4 n3 (102“) (b) 3.655Xl0'5 cm " q NA(NA+ND) 7.31x10-5 cm x“ = [Km—"gyulm q ND(NA+ND) W = xn+xp= 1.10Xl0‘4cm I9 15 -s (c) 8(0) = _q~_0xn . _W = _|,|2x 104 wan [(5:1) (I l.8)(8.85X10‘”) :9 I5 -5 2 (d) m) = qNA ,3 a W = 0.205 v ZKsa) (2m 1.3)(s.8sxlo"4) (c) p x 8 5.8 Essentially, the solution is just a superposition of the step junction solution forx < 0 and the linearly graded solution forx > 0. p x X n -qN A E -Xp Xn 1 Linear / k Quadratic dependence like Fig. S.l3(c) 5.2 (a) The carrier concentrations are assumed to be negligible compared to the net doping concentration in a region —xp 51: 5x" straddling the metallurgical junction. 1he charge density outside the depletion region is taken to be identically zero. (b) We must have p = q(NDWA) for —xp 5 x S x“ and the total (+) charge must equal the total (-) charge. p I In (G) Since based on the depletion approximation (INC. (l —e‘“‘) 4pran p z 0 ...xS-xpandx2xn substitution into Poisson's equation gives flo—(t — em) -x,, 51: En. g , “Sm dx 0 ...xS—xpand121n Separating variables and integrating from the —xp depletion region edge when: 8 = 0 to an mbiuzuy puiuu in the depletion legit)", one obtains No - - . qNo . -ccr' ‘ qNo -az w: gm=q_ (1_eax)dx=__x+e_ __,+e__+,_.e_n qu) Ksm a 4', K530 a: p a ) .,P or _ 4N0 8(1) ‘ [(I+XP)+i-(e m—ea‘P)] -—xpran 5.19 (a) Given that xn > m for all applied biases of interest. then surely xn > x0 under equilibrium (VA 2 0) conditions. Now. the derivation of the Eq. (5.8) V5; relationship is valid for an arbitrary nondcgenerate doping profile. Moreover. n(—xp) = naleA. and with x“ > 10, nun) = ND far the given junction profile. just like in a standard step junction. Assuming the p- andxn >xo n-region dopings to be nondegenerate. the standard Eq. (5.10) result therefore applies. . = kl NAND Vb. 4 ll‘( n I (lhis problem points out that only the dopings at the depletion region edges are relevant in determining V51.) (b) Since p = «ND—NA) for —xp S x S xn and zero elsewhere under the depletion approximation. we conclude (c) Invoking the depletion approximation gives I 0 ...x < —x,, —qNA ...—xp SXSO ' - qND/Z "051510 ‘ qND ...xo $131.. 0 ...x>xn Substituting into Poisson's equation then yields 48 «INA/K590 ...—xp s x S 0 a: qND/ZKsco ...OSxSxo l qND/Kse‘o "JO 5 x S xn Separating variables and integrating from the depletion region edges where 8 = 0 to arbitrary points in the p-region and x“ > 10 n-region yield the same relationships and results as in the standard step-junction analysis. To obtain the 0 S x 5 x0 solution. we can either integrate from x = 0 where 8 is known from the p~region solution to an arbitrary point in theO Sx Sxoregion. or we can start the integration atx =x.. where Sis known from the x“ >xo u-region solution and integrate backward into the lighter-doped n-region. Taking the former approach we can write 00:) X d8 = (IND dx M580 0 2(0) Of No ~4NA 4ND 8m=£t0)+" 1:: x + x ...0$x$ ZKseo Ksm p ZKSEI) 1° If the integration is perforated from x = x. backward into the lighter doped n-region. one obtains the equivalent result Thus the final result is -(qNA/Kseo)(Xp+x) ~xp s x s 0 ~(0/Ksco)(NAxp—NDx/2) 0 S I 5 x0 0f —(qNDIKsa))(xn-xo/2-x/2) 0 S x S xo -{qNo/Ks£o)(xn~xl Io S I S In (a) N A assumed greater (+ and — anus must be the same size.) (Since p = 0 in the i-region. the electric field is constant in that region. One expects a step-junction type solution outside the i-region.) (b) The derivation of the Eq. (5.8) Vb; relationship is valid for an arbitrary doping profile. Moreover, nun) x: No and n(—-xp) = nileA for the p-i-n diode just like a step-junction pn diode. Assuming the p— and n- region dopings to be nondegenerate. the standard Eq.($.10) result therefore applies and 4 "i2 (This problem points out that only the dopings at the depletion region edges are relevant in determining Vbi.) (c) Invoking the depletion approximation, we can write ...x<-xp -xp 5 x S -x5I2 -xi/ZSxSxi/2 anti/2‘3:qu ...I>xn Substitution into Poisson's equation yields d8 «INA/Kym ...—x', S x S —xilZ —= 0 ...—xi/2 S): S x;/2 d1 qND/Kse'o S x S In Separating variables and integrating from the depletion region edges where 8 = 0 to arbitrary points in the n- and p-regions yield the same relationships and results as in the step junction analysis. In the I-region, E = constant = £(«xg/2). Thus we conclude —(qNAIKsa))(xp+x) ...—x,, s x s ~xfl 8(1) ‘ —(qNAle£o)(xp Xi/2) xfl fix 5 Avg/2 ~(qND/Kseo)(xn—x) ...x-,/2 S x S xn Setting 80) = —dV/dx. separating variables. and integrating from the depletion region edges to arbitrary points in the n- and p-regions again ields the same relationships and results as in the step junction analysis. Introducing é—xfl) I 8i and V(~x-J2) 5 Vi. we note that in the i-region dV/dx = —£i and V0!) 1 I W n - dx‘ Vt all V(x) = Vi-&(x+x-/2) 01' Thus (qNA/ZKsm)(xp+x)2 ...-x', S x S —xi/2 V0!) = (qNA/ZKsm)[(xp-Ii/2)(Xp+xfl+2x)l "art/2 5 I S Kin \ Vba-VA - («IND/ZKseo)(zr..—x)2 "xi/2 s x s x“ To determine xn and II, we require 8(1) and V(x) to be continuous at x = 13/2, or NAtert/Z) = Noun-x512) and (qNA/zxsm)t(xrx¢2)up+3x/2n - Vbi-VA —(qNDIZKsm)(Xn~15/2)2 Solving for xrxilz from the first equation ditectly above. substituting into the second equation, and rearranging, gives a. 2 33,21 _. _KS£1_NL_V-v =0 (xp X./2) +~A+No(xp IJZ) q NAWMND“ m— A) Finally. solving the quadratic equation yields In +[(NNDxi )2 +—§312K -—LN (Va-V0] A+ND q NA(NA+ND) where the (+) root has been chosen because 1: 11/2 must be greater than zero. Note that the result here reduces to Eq. (5.34) ifx. —> . ...
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This note was uploaded on 12/13/2010 for the course ELECTRICAL EECS 170A taught by Professor Ozdalboyraz during the Fall '10 term at UC Irvine.

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SHW5 - (b Employing Eq(5.12 Vbi =...

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