13 the transistor circuit in fig 345 has 80 and vbe 07

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Unformatted text preview: 200 kΩ 1 kΩ C B + 0.7 V 150 IB Vo − + 16 V − E (b) Figure 3.44 Solution of the problem in Example 3.13: (a) method 1, (b) method 2. M E T H O D 2 We can modify the circuit in Fig. 3.43 by replacing the transistor by its equivalent model in Fig. 3.40(b). The result is the circuit shown in Fig. 3.44(b). Notice that the locations of the base (B), emitter (E), and collector (C) remain the same in both the original circuit in Fig. 3.43 and its equivalent circuit in Fig. 3.44(b). From the output loop, vo = 16 − 1000(150IB ) But IB = I1 − I2 = 2 − 0.7 0.7 − = (13 − 3.5) µA = 9.5 µA 3 100 × 10 200 × 103 and so | v v vo = 16 − 1000(150 × 9.5 × 10−6 ) = 14.575 V | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis 107 PRACTICE PROBLEM 3.13 The transistor circuit in Fig. 3.45 has β = 80 and VBE = 0.7 V. Find vo and io . 20 kΩ Answer: −3 V, −150 µA. 30 kΩ + 1V − Figure 3.45 + io + 20 kΩ vo VBE − − For Practice Prob. 3.13. 3.10 SUMMARY 1. Nodal analysis is the...
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