semisolpr13

semisolpr13 - Semiconductor Physics and Devices: Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 13 Solutions Manual Problem Solutions 199 Chapter 13 Problem Solutions 13.1 Sketch 13.2 Sketch 13.3 p-channel JFET & Silicon (a) V ea N xx x x PO a = = −− () 2 19 4 2 16 14 2 16 10 0 5 10 3 10 2117 88510 .. bg b g or VV PO = 579 . Now V x bi = L N M M O Q P P 00259 510 310 15 10 18 16 10 2 .l n . b gb g or bi = 0884 . so V PP O b i =− = 579 0884 or P = 4.91 (b) aha VV V eN bi DS GS a −=− ∈−+ L N M O Q P 2 12 af / (i) For V GS DS == 10 , Then ah x −= 05 10 4 +− L N M O Q P 0884 1 3 10 14 19 16 . . / xV DS a f b g or x 4 4.31 10 1884 10 DS b g . / or m 0215 . µ (ii) For V V GS DS . 5 m 00653 . (iii) For V V GS DS 15 m 0045 . which implies no undepleted region. 13.4 p-channel JFET & GaAs (a) V aN x PO a = = 2 2 20510 2131 88510 2 4 2 16 14 . b g or PO = 518 . Now V x bi = L N M M O Q P P 18 10 18 16 6 2 n . b gb g or bi = 135 so O b i = 518 135 or P = 383 . (b) eN bi DS GS a L N M O Q P 2 / (i) For V GS DS Then x 4 L N M O Q P 135 1 14 19 16 . . / DS b gb g or x 4 4.83 10 2.35 10 DS b g / which yields m 0163 . (ii) For V V GS DS . 5 m 0016 . (iii) For V V GS DS m 0096 . which implies no undepleted region.
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 13 Solutions Manual Problem Solutions 200 13.5 (a) V ea N PO d = 2 2 = −− () 16 10 0 5 10 8 10 2117 88510 19 4 2 16 14 .. xx x x bg b g or VV PO = 155 . (b) aha eN bi GS d −=− ∈− L N M O Q P 2 12 af / so 0 2 10 0 5 10 44 = L N M O Q P 8 10 14 19 16 . / xV V bi GS a f b g or 9 10 1618 10 10 10 V V bi GS =− . a f which yields V bi GS −= 556 . Now V x bi = L N M M O Q P P 00259 310 810 15 10 18 16 10 2 .l n . b gb g or bi = 0896 Then V GS 0897 556 GS 4.66 13.6 For GaAs: (a) V ea N PO d = 2 2 = 0 5 10 8 10 2131 88510 19 4 2 16 14 x x b g or PO = 138 . (b) eN bi GS d L N M O Q P 2 / 0 2 10 0 5 10 = L N M O Q P 14 19 16 . / V bi GS a f b g which can be written as 9 10 1811 10 10 10 V V bi GS . a f or V bi GS 4.97 Now V x bi = L N M M O Q P P 18 10 18 16 6 2 n . b gb g or bi = 136 Then GS bi = 4.97 4.97 or GS 361 . 13.7 (a) V ea N PO a = 2 2 = 0 3 10 3 10 19 4 2 16 14 x x b g or PO = 1863 . V x bi = L N M M O Q P P 510 18 16 6 2 n . or bi = 1352 Then V PP O b i = 1863 1352 or P = 0511 . (b) (i) eN bi GS a ∈+ L N M O Q P 2 / or ah x 03 10 4 . ( ) L N M O Q P 3 10 14 19 16 . . / x b gb g
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 13 Solutions Manual Problem Solutions 201 which yields ah x c m −= 4.45 10 6 (ii) x 03 10 4 . bg + () ( ) L N M O Q P 2131 88510 1351 1 16 10 3 10 14 19 16 12 .. . . / x xx b g which yields m 37 10 6 . which implies no undepleted region. 13.8 (a) n-channel JFET & Silicon V ea N x x PO d = = −− 2 19 4 2 16 14 2 0 35 10 4 10 2117 88510 b g b g or VV PO = 379 . Now V x bi = L N M M O Q P P 00259 510 410 15 10 18 16 10 2 .l n . b gb g or bi = 0892 . so that VVV Pb iP O =− = 0892 379 or P 2.90 (b) aha VV V eN bi DS GS d −=− ∈+− L N M O Q P 2 af / We have x 035 10 4 +− L N M M O Q P P 2 117 885 10 16 10 4 10 14 19 16 . . / ch c h xV V DS GS or x 4 −+ 324 10 10 / V DS GS b g (i) For V GS DS == 01 ,, m 0102 .
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This note was uploaded on 05/07/2008 for the course ECE E-302 taught by Professor Nabet during the Spring '08 term at Drexel.

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semisolpr13 - Semiconductor Physics and Devices: Basic...

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