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Semiconductor Physics and Devices: Basic Principles, 4 th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 19 14 106.1 5574.01085.87.112 2/1 151414 15 10510 1 10 105 or 41066.2 nx cm (c) For 30nx m, we have 19 14 4 106.1 1085.87.112 1030 Rbi VV 2/1 151414 15 10510 1 10 105 which becomes Rbi VV 76 10269.1109 We find 4.70RV V _______________________________________ 7.29 An pn junction with 1410aN cm 3 , (a) A one-sided junction and assume biR VV . Then 2/1 2 a Rs p eN V x or 1419 14 24 10106.1 1085.87.112 1050 RV which yields 193RV V (b) d a pn a d n p N N xx N N x x so 16 14 4 10 10 1050nx 41050.0 cm 50.0 m (c) 4max 105.50 15.19322 W VR or 4 max 1065.7 V/cm _______________________________________ 7.30 (a) 210 1517 105.1 102102 ln0259.0biV 7305.0 V (b) 2/1 2 Rbi ds VV Ne ACAC Rbi VV2 106.1 10 19 5 2/11514 1021085.87.11 Rbi VV C 1310287.1 (i) For 1RV V, 1410783.9 C F (ii) For 3RV V, 1410663.6 C F (iii) For 5RV V, 1410376.5 C F _______________________________________ 7.31 (a) 26 16 108.1 108 ln0259.0 d bi N V 20.1 0259.0 20.1 exp108.1108 2616 dN 151036.5 dN cm 3 (b) 2/1 2 daRbi das NNVV NNe ACAC 0.120.12 106.1 1010.1 19 12 A 2/1 1516 151614 1036.5108 1036.51081085.81.13 51056.7 A cm 2 (c) Rbi VV2 106.1 1056.71080.0 19 512 2/1 1516 151614 1036.5108 1036.51081085.81.13 Rbi VV 8 8 101585.2 100582.1 RRbi VVV 20.1161.4 96.2RV V _______________________________________