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chapter-10-solutions - Chapter 10 Pgob 10.1 I — Sketch...

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Unformatted text preview: Chapter 10 Pgob. 10.1 ' I — Sketch band diagrams for a junction wet/z a degenerate p sule and E;- - BC on the 11 side. Forward bias Reverse bias This is called a backward di direction (due to tunneling), forward bias. ode because it. conducts freely in the reverse but. the current remains small for low voltage 133 \J’ Prob. 10.2 Explain what determines the peak tunneling voltage V; of a tunnel diode. For the given tunnel diode, (a) calculate the minimum forward bias at which tunneling through E, occurs. (b) Calculate the maximum fomard bias for tunneling via E.. (c) Sketch the I-V curve ﬁyr this diode. Er 7/7/7/7/7/ ’ ‘ The sizes of Evp —— E; and EF— Em determine the voltage required to align the most ﬁlled states opposite empty states. (a) Tunneling through the level begins when En, - EFF = E, i.e., at a forward bias of 0.3 V. (b) Tunneling through the level ends when Em. -— Evp = E, i.e., when Er" - EFF = V = 0.3 + 0.1+ 0.1 = 0.5 V. (c) Band-to-band tunneling is maximum when En. - 15;, = 0.1 V and is essentially zero when En, - Er], = 0.2 V. I 134 134 Prob. 10g ‘ (a) Relate ap/Bt (where p is space charge density) to a and es neglecting recombination. (b) Show space charge p(t) decays exponentially with time constant 73. (c) Find the RC time constant of a eample. J=a£= ~<7VV _‘ 6!" 2_v_ P V.J———8t . V ‘ —— e (a) V.J=—~UV2V: thus ~§a§=GP/€ (b) p=poe"’l’d‘ where nze/a L EA (C) 12:33:, L 614. 6 RC=(U7)(T>-;=Td L 135 135 Prob. 10.4 Find the criterion for negative conductivity in terms of mobilities and clcc~ tron concentrations in the F and L bands of GaAs. . = 05 = q lI-‘I‘nl‘ + MM] 5 = q {Mnr + I‘L("0“n1‘)l 5 d] _ dnr dﬂr dVL d5 — q lﬂrnr + mm] + £15 (ﬁr - Pbl’d—g— + fir-EE- + TILE .‘ dno _ sxuce E5,— _ 0. This is negative when E (#r -pL) dnr/dS + 8 (up dpr/d£+nL dpL/di') < _ #rnr + Pill]. 1 If we assume ur r- A/E and [l]; = 8/8, = wA/f," and {if-Li’- : «8/83, and the condition is (If (A - B) (inf/d3 —- S“(n1~A + 11LB)< _— £41er + S‘lnLB 5(A ~ 13) drip/(15 nrA + 711,8 1 ~l<~—1 B is less than A, since [1,; < pr. Thus dnp/dE must be negative. That is, the conductivity is negative only while electrons are being transferred from the lower lying I ' valley into the upper 1; valley. 136 136 Prob. 1M ; Estimate d-c power dissipated per unit volume in a. 5 pm GaAs Gunn diode biased just below threshold. / (a) n01, c: 10”, no : 10"/(5x10“) = 2x10” cm~3 r, = L/v, .—. 5x10‘4/1a” _—. 5x10?” 3» v (b) P = N = (qnuvdAxem 213L— =1.6><10“9 x mm” x 2x107 x 3x103 2 2 x 107 W/cm3 For a. device with higher frequency, 11 must be smaller. Thus, L must be smaller and no correspondingly larger. As a result, the power dissipation P, which is proportional to nuL, is about the same encording to this simple analysis. However, the power (lensity, which is proportional to no, is larger for higher frequency devices. This is important, since P/AL is critical to heat dissipation requirements. £392. 10.6 (a) Calculaic the ratio of the densities of states in the F and L conduction bands in 041245. (b) Assuming a Boltzmann distribution, find the ratio of electron concentra— tions in these bands at 300 K. (c) What is the equivalent temperature of an electron in. the L minima .9 (a) Nc = 2(27rkTm;/h2)3/2 £5 _ mil“ 3” _ [Er/2 ~ 3 23/2 —— 23 5 N1, - mm ‘ 0.067 ‘ ‘ “ ' (b) :‘i : 23.5 raw-Wm” = 23-5 8“” = 2‘2 x 10—4 1‘ We notice that the upper (L) valley is essentially empty in equilibrium at 300 K, campared with the lower (P) valley. ((2) T = k“(0.0259 + 0.30) = 0‘326/(8.62x10‘5) = 3782 K 137 137 ...
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chapter-10-solutions - Chapter 10 Pgob 10.1 I — Sketch...

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