Explanation Given: The collector resistance R C is 2 k Ω , the transconductance g m is 50 mA / V and the current gain β is 100 . The output signal v o is 1.2 sin ω t V . Concept used: The expression for instantaneous base emitter voltage v b e is written below. v b e = i b r π …… (1) The expression or instantaneous base current i b ( t ) is written below. i b = i c β …… (2) Calculation: The quiescent collector current is calculated as, g m = I C Q V T I C Q = g m V T = 50 ⋅ 26 = 1.3 mA Apply KVL in the collector loop as follows: v o + i c R C = 0 i c = − v o R C = − 1.2 sin ω t 2 i c ( t ) = − 0.6 sin ω t mA Substitute − 0.6 sin ω t for i c and 100 for β in equation (2)

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 6.11P

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The instantaneous base emitter voltage vbe(t) and the instantaneous base current ib(t) .

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Chapter 6, Problem 6.10P

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Chapter 6 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 6 - The circuit parameters for the circuit in Figure...Ch. 6 - For the circuit in Figure 6.3, assume transistor...Ch. 6 - For the circuit in Figure 6.14(a), let =90 ,...Ch. 6 - Using the circuit and transistor parameters given...Ch. 6 - Consider the circuit in Figure 6.18. The circuit...Ch. 6 - Repeat Example 6.4 if the quiescent collector...Ch. 6 - For the circuit in Figure 6.31, let RE=0.6k ,...Ch. 6 - Prob. 6.6EP Ch. 6 - The parameters of the circuit shown in Figure 6.28...Ch. 6 - For the circuit shown in Figure 6.31, let =100 ,...

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The instantaneous base emitter voltage v b e ( t ) and the instantaneous base current i b ( t ) .

Solution Summary: The author describes the instantaneous base emitter voltage v_be(t), the transconductance, and the current gain.

The instantaneous base emitter voltage vbe(t) and the instantaneous base current ib(t) .

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Chapter 6 Solutions