This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**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. —> . ...

View
Full Document

- Fall '10
- OzdalBoyraz
- Quadratic equation, Electric charge, P-n junction, depletion region, Vbi