# Practice problem 312 for the transistor circuit in

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Unformatted text preview: ed by its equivalent model can we apply nodal analysis. EXAMPLE 3.12 Find IB , IC , and vo in the transistor circuit of Fig. 3.41. Assume that the transistor operates in the active mode and that β = 50. Solution: For the input loop, KVL gives −4 + IB (20 × 103 ) + VBE = 0 IC 100 Ω + 20 kΩ IB + + 4V − | v v Figure 3.41 | e-Text Main Menu Input loop VBE − vo − Output loop + 6V − For Example 3.12. | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis 105 Since VBE = 0.7 V in the active mode, IB = 4 − 0.7 = 165 µA 20 × 103 But IC = βIB = 50 × 165 µA = 8.25 mA For the output loop, KVL gives −vo − 100IC + 6 = 0 or vo = 6 − 100IC = 6 − 0.825 = 5.175 V Note that vo = VCE in this case. PRACTICE PROBLEM 3.12 For the transistor circuit in Fig. 3.42, let β = 100 and VBE = 0.7 V. Determine vo and VCE . 500 Ω + Answer: 2.876 V, 1.984 V. + 12 V − 10 kΩ + VCE VBE − + 5V − 200 Ω − + vo − Figure 3.42 For Practice Prob. 3.12. EXAMPLE 3.13...
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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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