Prévia do material em texto
Semiconductor Physics and Devices: Basic Principles, 4 th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ B BB nC x neD J 0 4 1319 10 1059.220106.1 or 828.0nCJ A/cm 2 Assuming a long collector t BC C nOC pC V V L peD J exp where 4 16 2102 1025.2 10 105.1 C i cO N n p cm 3 and 710215 COCC DL 310732.1 cm Then 3 419 10732.1 1025.215106.1 pCJ 0259.0 6.0 exp or 359.0pCJ A/cm 2 The collector current is AJJI pCnCC 310359.0828.0 or 19.1CI mA The emitter current is 310828.0 AJI nCE or 828.0EI mA _______________________________________ 12.27 (a) BB T Lxcosh 1 and T T 1 BB Lx T 0.01 0.10 1.0 10.0 0.99995 0.995 0.648 0.0000908 19,999 199 1.84 0 (b) For BE DD , BE LL , BE xx , we have EBBOEO NNnp 1 1 1 1 and 1 EB NN 0.01 0.10 1.0 10.0 0.990 0.909 0.50 0.0909 99 9.99 1.0 0.10 (c) For 10.0BB Lx , the value of is unreasonably large, which means that the base transport factor is not the limiting factor. For 0.1BB Lx , the value of is very small, which means that the base transport factor will probably be the limiting factor. If 01.0EB NN , the emitter injection efficiency is probably not the limiting factor. If, however, 01.0EB NN , then the current gain is small and the emitter injection efficiency is probably the limiting factor. _______________________________________ 12.28 We have BBB BOB sO LxL neD J tanh Now 3 17 2102 1025.2 10 105.1 B i BO N n n cm 3 and 71025 BOBB DL 4108.15 cm Then 8.157.0tanh108.15 1025.225106.1 4 319 sOJ or 1010287.1 sOJ A/cm 2 Now