Chapter 13 - 1 CHAPTER 13 Solutions for Exercises E13.1 The...

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Unformatted text preview: 1 CHAPTER 13 Solutions for Exercises E13.1 The emitter current is given by the Shockley equation: = 1 exp T BE ES E V v I i For operation with 1 exp have we , >> >> T BE ES E V v I i , and we can write T BE ES E V v I i exp Solving for BE v , we have mV 4 . 718 10 10 ln 26 ln 14 2 = = ES E T BE I i V v V 2816 . 4 5 7184 . = = = CE BE BC v v v 9804 . 51 50 1 = = + = mA 804 . 9 = = E C i i A 1 . 196 = = C B i i E13.2 = 1 0.9 9 0.99 99 0.999 999 E13.3 mA 5 . = = C E B i i i 95 . / = = E C i i 19 / = = B C i i E13.4 The base current is given by Equation 13.8: = = 1 026 . exp 10 961 . 1 1 exp ) 1 ( 16 BE T BE ES B v V v I i which can be plotted to obtain the input characteristic shown in Figure 13.6a. For the output characteristic, we have B C i i = provided that 2 V. 0.2 ely approximat CE v For V, 0.2 C CE i v falls rapidly to zero at . = CE v The output characteristics are shown in Figure 13.6b. E13.5 The load lines for V 0.8- and V 8 . in = v are shown: As shown on the output load line, we find V. . 1 and V, 5 V, 9 min max CE CEQ CE V V V 3 E13.6 The load lines for the new values are shown: As shown on the output load line, we have V. . 3 and V, 7 V, 8 . 9 min max CE CEQ CE V V V 4 E13.7 Refer to the characteristics shown in Figure 13.7 in the book. Select a point in the active region of the output characteristics. For example, we could choose the point defined by mA 5 . 2 and V 6 = = C CE i v at which we find A. 50 = B i Then we have . 50 / = = B C i i (For many transistors the...
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Chapter 13 - 1 CHAPTER 13 Solutions for Exercises E13.1 The...

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