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Unformatted text preview: (b) Employing Eq. (5.12). Vbi = 5[(E‘rEF)¢sidc+(EFEi)nsidc] ('YwG/zmr + 50/4) = #65614 + 2n) = $0.12) + 2(0.0259)
0.89 v 5.1 (a) Because NM > Mg and p = n; expl(EiEp)/k7' 5 NA far ﬁrm the junction. it follows
that (EiEp)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‘EFblsidc(Ei—EF)p2side]
= 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) pplot
is caused by a depletion of holes on the higherdoped NMside of the junction, leading to a
net charge associated with the ionized acceptors. The plus charge on the NAZside 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 nside
of the junction in excess of NM. [The hole execs on the NAzside 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) 7.31x105 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 ...xSxpandx2xn
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 nregion 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 $131..
0 ...x>xn Substituting into Poisson's equation then yields 48 «INA/K590 ...—xp s x S 0
a: qND/ZKsco ...OSxSxo
l qND/Kse‘o "JO 5 x S xn Separating variables and integrating from the depletion region edges where 8 = 0 to
arbitrary points in the pregion and x“ > 10 nregion yield the same relationships and results
as in the standard stepjunction analysis. To obtain the 0 S x 5 x0 solution. we can either
integrate from x = 0 where 8 is known from the p~region solution to an arbitrary point in
theO Sx Sxoregion. or we can start the integration atx =x.. where Sis known from the
x“ >xo uregion solution and integrate backward into the lighterdoped nregion. Taking
the former approach we can write 00:) X d8 = (IND dx M580 0 2(0)
Of
No ~4NA 4ND
8m=£t0)+" 1:: x + x ...0$x$
ZKseo Ksm p ZKSEI) 1° If the integration is perforated from x = x. backward into the lighter doped nregion. 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))(xnxo/2x/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
iregion. the electric
ﬁeld is constant in that
region. One expects a stepjunction type
solution outside the
iregion.) (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 pin diode just like a stepjunction 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 pregions yield the same relationships and results as in the
step junction analysis. In the Iregion, 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 pregions again ields the same relationships and
results as in the step junction analysis. Introducing é—xﬂ) I 8i and V(~xJ2) 5 Vi. we
note that in the iregion 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)[(xpIi/2)(Xp+xﬂ+2x)l "art/2 5 I S Kin
\ VbaVA  («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) = Nounx512)
and (qNA/zxsm)t(xrx¢2)up+3x/2n  VbiVA —(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_Vv =0
(xp X./2) +~A+No(xp IJZ) q NAWMND“ m— A) Finally. solving the quadratic equation yields In
+[(NNDxi )2 +—§312K —LN (VaV0] 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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 Fall '10
 OzdalBoyraz
 Quadratic equation, Electric charge, Pn junction, depletion region, Vbi

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