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The transistor in the circuit shown in Fig. P4.64 is biased to operate in the active mode. Assuming that B is very large, find the collector bias current Ic. Replace the transistor with the small-signal equivalent circuit model of Fig. 4.27(b) (remem- ber to replace the de power supply with a short circuit). Analyze the resulting amplifier equiva- lent circuit to show that

The transistor in the circuit shown in Fig. P4.64 is biased to operate in the active mode. Assuming that B is very large, find the collector bias current Ic. Replace the transistor with the small-signal equivalent circuit model of Fig. 4.27(b) (remem- ber to replace the de power supply with a short circuit). Analyze the resulting amplifier equiva- lent circuit to show that

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Please help with this homework problem. The topic is Bipolar Junction Transistors (BJTs). I'm having a difficult time understanding how to replace the transistor with the small-signal equivalent. 

*4.64 The transistor in the circuit shown in Fig. P4.64 is biased to operate in the active mode. Assuming that B is very large, find the collector bias current Ic. Replace the transistor with the small-signal equivalent circuit model of Fig. 4.27(b) (remen- ber to replace the de power supply with a short circuit). Analyze the resulting amplifier equiva- lent circuit to show that RE %3D Vi R: + r. - «Rc Vi RE + +15 V Rc = 4.3 kn 100 kN 8. Vi 100 kn RE = 6.8 kn Fig. P4.64
Transcribed Image Text:*4.64 The transistor in the circuit shown in Fig. P4.64 is biased to operate in the active mode. Assuming that B is very large, find the collector bias current Ic. Replace the transistor with the small-signal equivalent circuit model of Fig. 4.27(b) (remen- ber to replace the de power supply with a short circuit). Analyze the resulting amplifier equiva- lent circuit to show that RE %3D Vi R: + r. - «Rc Vi RE + +15 V Rc = 4.3 kn 100 kN 8. Vi 100 kn RE = 6.8 kn Fig. P4.64
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