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# 5 - .e.L Solutions to Chapter 5 Problems Fundamentals of...

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Unformatted text preview: ....-,_......._.e._..._.-.L.;...-.., Solutions to Chapter 5 Problems Fundamentals of III-V Devices, by W. Liu Z l>(a) The depletion potential is total potential drop (”from V05 and V93) in the . 'e'mi conductor such that the active layer of thickness a is fully depleted. VP is the drain bias at which the active layer at the drain side is fully depleted. (b) s = 1 implies that. ¢hi - V63 = 4600; or. rpm - VT: (2)00. Therefore, VT: 4355,,- - \$00. '(C) V}: is the drain voltage such that the channel charge at the drain end pinches off. It is exactly equal to 1105531 given in Eq. (5-20). Hence. Vp : (000 - ebb,- + V55: 2.76 — 0.861 + 0 :1899 V. VT: q)!”- — (2)00 = 0.861 — 2.76 = — 1.899 V. (d) VP = (300 - (lib; + V03: 2.76 — 0.86] — 0.5 = 1.399 V. VT: ((0,; - @500 = 0.861 - 2.76 = - 1.899 V. 9a) 7 n = N4 = 2 x 1017 cm'3. Because NC for GaAs is.4.7 X 1017 cm‘3, n/NC, the y - is value of Fig. 1-21, is equal to 0.426. We read off the corresponding x axis value from the figure and determine that the Fermi level in the semiconductor is ~ 0.70.258 = 0.018 eV below the conduction band. With (15,, = 0.018'eV, and (:53 given as 0.9 eV. \$1,; calculated from'Eq. (5-2) is 0.9 - 0.018 = 0.882 eV. ago is calculated from Eq. (5-8) as 1.6e-l'9 >< 2e17 X (15006-8):2 /(2 x 1.159e—12): 3.1 V. lips-5m calculated from Eq- (5-20) is therefore V93,“ 2 3.1 - 0.882 + 0 = 2.18 V- Since V95 < V9333“, d of Eq. (5—33). is equal to (0.882 ~ 0 + 0.5)/3.1 = 0.446. s from Eq. (5-33) is equal to (0.882 — 0)/3.l = 0.285. Imam according to Eq. (5-41) is 166-151 X 15006-8 X 20.0e-4 X 2e17 x 3000 X 3.1 I le- 4 = 0.89 A. The drain current given by Eq. (5-40) is therefore 0.89 x ( 0.446 - 0.285 - 2/3-(0.446)3I2 + 2/3‘(0.285)3/2 ) = 0.0117 A. (b) V3353: calculated in part (a) is 2.218 V. Since VDS > VDS‘gat‘ d of Eq. (5-33) is equal to (0.882 - 0 + 2.218)/3.1 =1. 5 from Eq. (5-33) is still equal to (0.882 - 0)]3.1 = 0.285. [max found in part (a) is 0.89 A. The drain current given by Eq. (5—40) is therefore 0.89 x ( 1 - 0.285 - 2/3'(I)3"2 + 2/3-(0.285)3V2 ) = 0.088 A. 3. (a) According. to Eqs. (5-43) and (5-33), b(x) is: 0(1):!!(1- f¢s,dc(x) )- ' 4500 - Taking the derivatives on both sides of'the above equation, we obtain an identity in the differential form: . :__.a.___l__ db 2(1-b/Q) \$00 dame- (b) It is convenient to deﬁne a normalizing current. I0: 10 = qWHnNd \$00. The equation relating dais to the is obtained by combining Eqs. (5-29) and (5—31): ID = I _ \$3411: (x) d’¢.s,dc anl-LnNa' \$00 dx - In terms of the normalized current, the above equation is expressed as: 41 4- IS 1.91 .M' _ —— dxa+1 111+ ———.-———— dxa-I-I uI K uzI 5“ )5-0I x 96'1 ~53 I _ WI ‘21 )6-0I x III J3 ) ‘SI ‘(36—9) 'bEI no paseq ‘uoavnbs Jauw atIL (p) "P'I PUB g-I seldunexg uI punoI e'q mas I‘ng Io uonquualap sq; 'oo- = x mfg - 93 SI I‘qu 91914:“ b . b . - -IN ———= L”? 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5 - .e.L Solutions to Chapter 5 Problems Fundamentals of...

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