133 from eqs 3131 and 3132 i1 2 07 13 a 100

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Unformatted text preview: For the BJT circuit in Fig. 3.43, β = 150 and VBE = 0.7 V. Find vo . Solution: 1 kΩ We can solve this problem in two ways. One way is by direct analysis of the circuit in Fig. 3.43. Another way is by replacing the transistor with its equivalent circuit. METHOD 1 We can solve the problem as we solved the problem in the previous example. We apply KVL to the input and output loops as shown in Fig. 3.44(a). For loop 1, 2 = 100 × 103 I1 + 200 × 103 I2 + 100 kΩ + 2V − Figure 3.43 ⇒ 200 kΩ − (3.13.1) For loop 2, VBE = 0.7 = 200 × 103 I2 vo For Example 3.13. I2 = 3.5 µA (3.13.2) For loop 3, | v v −vo − 1000IC + 16 = 0 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents + 16 V − 106 PART 1 DC Circuits or vo = 16 − 1000IC (3.13.3) From Eqs. (3.13.1) and (3.13.2), I1 = 2 − 0.7 = 13 µA, 100 × 103 IB = I1 − I2 = 9.5 µA IC = βIB = 150 × 9.5 µA = 1.425 mA Substituting for IC in Eq. (3.13.3), vo = 16 − 1.425 = 14.575 V IC 1 kΩ + I1 IB 100 kΩ Vo I2 + 2V − Loop 1 200 kΩ Loop 3 − + 16 V − Loop 2 (a) 100 kΩ I1 IB I2 + 2V −...
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This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.

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