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Unformatted text preview: Chapter 8 Prob. §.l
For the p‘in phocodiode of Fig. 8—”? ((1) Explain why this photodetector does not have gain . An EHP created within W by
absorption of a photon is
collected as the c” is swept to n and the h"L to p. Since
only one. EHP is collected per *— "'— photon, there is no gain. 5 ((7) Explain how making the device more sensitive in low light levels degrades
its speed If W is made wider to receive more photons, the transit time to collect the
BB? is longer (degraded response speed). If W is kept the same and the area A is increased to receive more pho» tons, the capacitance is increased and again the response speed is degraded~ ('6) If this photodiode is to be used to detect light with /\ = 0.6 am. what
material would you use 3? Vth substrate would you grow this material on? EM = 1.24/06 2 2.07eV
From Fig. 113 , In.5Ga.5P grown on GaAs has E9 z 2 eV
AlAsjsSbﬂ grown on III? has E, a: 1.95 eV Both of these have Ey slightly smaller than the photon energy. 120 120 grab};
A Si solar cell has .4 = 4 cm", I”. = 31? 11A and W’ = 1 pm. 0' g“, r: IO‘SEIIP/cmss within L, 2 L,. = ‘2 pm of the junction, ﬁnd I,c and Wm.
From (81), 1..c = 10,, = quW(L,+L,‘+W) = 1.6x 1019 x4x10‘3(5x10“‘) = 0.32 mA
From Eq. (33), V“ = %1n(1 + 1041".) = 0.02591n(1+ w)
= 0.24V Prob. 8.3
Plot I,V for a 51' solar cell with [m = 5 nA and I“ = 200mA. \ From Eq.(82), kT 1+1}c __ [+0.2
 q ln(1+ I”! )—0.0259]n(1+5x10_9)
I (mA) v V (volt)
~200 0
~190 0.376
—180 0.39
——1.60 0.41 '10“)
—120 0.43 200
~80 0.44
—40 0.448
0 0.453
100
V0101!)
0.2 0.4
121 121 Prob. 8.4
Ii’epmzt Prof}. 8.3 wiih a 1 Q scrim reszstance. For a given current; 1 and terminal voltage Vm the voltage I
across the diode is reduced by IR +——.
W
; l
11, mm I r .L R "
v» V M V;
30 w u j g 1M(€q(§l.—IR)/kT __‘ ‘u I“: : 5 x 10—9(c(Vo—;}/0.0259 _' __ 0.2 i ~KmA) 0 ‘ 0.453
*0.04 ‘ 0.403 200
Wong; 0.36
—G.1‘2 0.31
«new 0.25
4.1:; 0.21 :00 ’ a ’ V(volt) Prob. 8.5 0.2 0.4 How can several semiconductors be used in a solar cell .9 In a GaAs cell ( .1525 ~ 1.4 8V), 3 top layer of AlGaAs ( Es ~ ‘2 (N) can
be grown latticewmatched to keep generated carriers from the surface. re—
ducing surface recombination One might also use a second cell below with
E9 «1; 1.4 eV (e.g., Si} to absorb the light passed through the GaAs band gap, Prob. 8.6
Find the photocurrent A! in terms of 7,; and r, for a sample dominated by p". A0 2:: q;.r.,,An = qpngoprn 122 122 AI = WAR = VAAa/L = Vqungoan/L The transit time is AI = q/1.Lga,{,r,I IT, Prob. 8.7
What ternary alloys produce A = 680 nm ?
1.24
Eg .— — LSQCV From Fig. 36(c),
Alea‘qu : x = 0.32 From Fig. 811.
GaA51_,P, : x = 0.32
From Fig. 1‘13,
InxGaqu : X a 0.4. 123 123 Prob,8.8
((1) Why mm! a mlar cell be operated in the 4m quadrant of the junction I—V
characteristic? Power is consumed in 1" (+1, +V} and 3"1 (—1, V) quadrants. Power is generated in 4m(1',
+ V) quadrant. p 71
1/ (b) What is the advantage ofa quaternary alloy in ﬁtbricating LEDS forﬁbcr optics? Bandgap (Eg) and therefore waveiength 0») can be chosen whiie lattice constant (21) is
adjusted for epitaxial growth on convenient substrates, {0) W71}; is a revenmbt'ased GaAs pn junction not a good photoa’etecmrfor light nfk «1
1pm? ‘7 .. hci .. 42Li§£9fi§§i9ﬁ
1”“ x 10'4 Since ES = 1.43 «N for GaAs, the photon is not absorbed. 124 124 Prob. 8.9
For a uniformly illuminated pan diode with gap in steady state, find (a) 5p(x,,), (b) [pow
and 1pm; 525p §&_§’2e
2 — 2
dx” 2:}, 0,, 2
’30pr DP 5p(x,,) = 36"" ’ ’v + L2
atxn :0, 5p(0) : Apn.ThusB= Apn ~351M Dp\ (a) 5P(X,, ) = [pn(eqlr’1k7' _1)_. gapz‘; [DP k—IJLF
dBp n 1 2 i ‘
dxn " “"Z;[Apn “ gopl‘p’ Dple + gopLﬁ, lop 1pm :0): .41 1,54qu” ~1)qALpgop This corresponds to Eq.(8~2) for 22,, << p,1 except that the component due to generation on
the p side is not included. Prob. 8.10 An illuminated Si solar cell has a shortcircuit current of 100 mA and an opencircuit
voltage of 0.8V. and a ﬁll factor of 0. 7‘ What is the maximum power delivered to a load
by this cell? PM a (m 1me (().s)(100)(0.7) = 56 mW Prgh. 8.11
For a solar cell with I“ under illumination and 13;, in the dark. ﬁnd an expression for the voltage at maximum power and solve graphicallvfor 1,}, = 1.5 m1 and I}c = l 00mA to get
Vmp and the maximum power. Eq.(8—2) can be written as 125 3..., I = lth(qu,kT ‘1)” 13c
IV = 1,,,(e""’*"" 4W — 1ch dUV) r q m
7;— = 1,,,(e4”" —1)+;;1,,,Ve4””‘r — I” = o (a) (l+—I}%Vmp)qu"/kr =1+1£a It):
For [so >> I”, and I’m], >> kT/q, (b) ln(qup/kT) + qup IkT = mum/1d,) —3
ﬁzm=ae7xlo7
1,}, 1.5X10‘ [nigg = 18
1:1.
1m: =18  x, where x = qup/kT
(c) The solution is x = 15.3 Vm’, = 15.3 x0.0259 = 0.396V l =10‘9e‘53 ~10" = —96mA
(d) P = ——IV = 37.9mW 18
189:
A x
15.3 18
W For the solar celi of Prob. 8.11, plot the I—V curve and draw the maximum power
rectangle. V = 0.0259ln[l +3331.) 1.5x10‘9 126 126 I (mA) V (mV) 0 467
25 459
50 449
~75 431
90 407
95 389
98 365
400 0
100 mp.
2‘
E
’7 50
0 0.2 0.4
V(volts)
Prob. 8.13
ao WW1 ._ i
5 § S ‘
u. 3 3
u. ’ l
24 g i
; E
.08 1 3
j 3
o " ,WWWWWE
0.0 1.0 2.0 3.0 4.0 5,0
Sctics stisumce (Q) 127 127 Prob. 8.14
From Fig. 113, what epitaxial IayerAs‘ubstrate combination would you choose for an LED with 3. = 1.55 pm? Repeatfar 1.3 pm Assume the lattice constant varies iinearly with composition between the binary limits.
For 7» = 1.55 pm, use GaAso,SSbo_5 on an InP substrate.
For A = 1.3 gm, use InGaAsP on InP. w
How does degeneracy prevent absorption of the emission wavelength .9 Since absorption requires promotion of an electron from a. ﬁlled state in the
valence band to an empty state in the conduction band, Fig. 819 shows
that photons with iw > (Fn— F,) are absorbed. On the other hand, emitted
photons have [w < (Fn  [7,). This is true only in the inversion region, and
absorption becomes important away from the neighborhood of the junction. 128 128 Prob. 8.16 Show 812 := 1321 at high temperature, Bzzmﬂm) = A21n2+321n2P(V12} B12: A21 +321] {ﬁrmww
9(1’12) As T » co, e‘(E’“E‘)/H > 1 Thus as Mun) —* 00, 312 = 321 P292. 8.17
Use Planck ’3 radiation law to ﬁnd A21 [8”,
A21 n2 712 [A21 ]
—_— __.... _ .—.. = —— — 197‘
[10/12) B” m +7;l P012} Bu + MM) ex“ hVn/ )
A21 axing? m = 9(V12)(exp(hm/kT)—1}= 63 Prob. 8.18
Estimate minimum 11 = p for population inversion in GaAs. Fn—Fp = E, = 1.43 eV
For n n p, FnEg : Eg—F, 2' 0.715
n = p = m awn—Eula = 105 Buns/0.0259 2 1018 cm‘3 ’I 29 129 . " My”. _ ...
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 Spring '08
 Banjeree

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