Explanation Given: The current gain β is 100 and early voltage V A is 100 V . The circuit for transistor is shown in Figure 1. The circuit parameters are shown below. V C C = 9 V R S = 10 k Ω R 1 = 27 k Ω R 2 = 15 k Ω R C = 2.2 k Ω R E = 1.2 k Ω R L = 2 k Ω Concept used: The expression for small signal voltage gain A v is written below. A v = − g m V π ( r o ∥ R C ∥ R L ) The expression for small signal current gain A i is written below. A i = − g m ( R T H ∥ r π ) ( r o ∥ R C r o ∥ R C + R L ) Calculation: The Thevenin resistance R T H is calculated as, R T H = R 1 ∥ R 2 = R 1 R 2 R 1 + R 2 = 27 ⋅ 15 27 + 15 ≈ 9.64 k Ω The Thevenin voltage is calculated below. V T H = V C C R 2 R 1 + R 2 = 9 ⋅ 15 15 + 27 = 3.214 V Apply dc analysis and KVL in base emitter loop. V T H − V B E ( on ) − I B Q R T H − I E Q R E = 0 V T H − V B E ( on ) − I B Q R T H − ( 1 + β ) I B Q R E = 0 I B Q ( R T H + ( 1 + β ) R E ) = V T H − V B E ( on ) I B Q = V T H − V B E ( on ) R T H + ( 1 + β ) R E Substitute 3.214 for V T H , 0.7 for V B E ( on ) , 9.64 for R T H , 100 for β and 1.2 for R E in above equation. I B Q = 3

Microelectronics: Circuit Analysis and Design

4th Edition

ISBN: 9780073380643

Author: Donald A. Neamen

Publisher: McGraw-Hill Companies, The

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Chapter 6, Problem 6.20P

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The input resistance Ri , small signal voltage gain Av and small signal current gain Ai .

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Microelectronics: Circuit Analysis and Design

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The input resistance R i , small signal voltage gain A v and small signal current gain A i .

Solution Summary: The circuit for transistor is shown in Figure 1. The expression for small signal voltage gain A_v is written below.

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The input resistance Ri , small signal voltage gain Av and small signal current gain Ai .

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Chapter 6 Solutions