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VBB 5 V = RB = 56 ks RB = 56 k www VBB 5VVBE = 0.7 V. Voc -0.7V{ (a) 8-12 (b) or displav. Rc = 1 k ẞdc = 100 Voc = 15 V Rc = 1 k Ic = ẞdc 1B Vcc = 15 V

VBB 5 V = RB = 56 ks RB = 56 k www VBB 5VVBE = 0.7 V. Voc -0.7V{ (a) 8-12 (b) or displav. Rc = 1 k ẞdc = 100 Voc = 15 V Rc = 1 k Ic = ẞdc 1B Vcc = 15 V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Question

1. A BJT has IE (Emitter Current) = 5 mA and IB (Base Current) = 100 µA. What are IC

    (Collector Current), alpha (α),and  beta (β).

 

 

2. What is Beta (β), if Alpha (α) is 0.92?

 

 

3. Compute the base current (IB), collector current (IC), and the collector-emitter voltage

     (VCE).

VBB 5 V = RB = 56 ks RB = 56 k www VBB 5VVBE = 0.7 V. Voc -0.7V{ (a) 8-12 (b) or displav. Rc = 1 k ẞdc = 100 Voc = 15 V Rc = 1 k Ic = ẞdc 1B Vcc = 15 V
Transcribed Image Text:VBB 5 V = RB = 56 ks RB = 56 k www VBB 5VVBE = 0.7 V. Voc -0.7V{ (a) 8-12 (b) or displav. Rc = 1 k ẞdc = 100 Voc = 15 V Rc = 1 k Ic = ẞdc 1B Vcc = 15 V
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