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chapter-8-solutions - Chapter 8 Prob .l For the p‘i-n...

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Unformatted text preview: Chapter 8 Prob. §.l For the p‘i-n 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. 1-13 , 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 (8-1), 1..c = 10,, = quW(L,+L,‘+W) = 1.6x 10-19 x4x10‘3(5x10“‘) = 0.32 mA From Eq. (3-3), 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.(8-2), 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. 3-6(c), Alea‘qu : x = 0.32 From Fig. 8-11. 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 revenm-bt'ased GaAs p-n 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 short-circuit current of 100 mA and an open-circuit 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‘5-3 ~10" = -—96mA (d) P = ——IV = 37.9mW 18 18-9: 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. 1-13, 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. 8-19 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, Fn-Eg : 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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