Determine VCE and Ic in the stiff voltage-divider biased transistor circuit of Figur if Bpc = 100. FIGURE 5-10 Vcc +10 V 10 k Re 1.0 kN R2 5.6 k RE 560 N tion The base voltage is R2 5.6 kN 10 V = 3.59 V VB = R+R Vcc = 15.6 kN So, VE = VB - VBE = 3.59 V – 0.7 V = 2.89 V and VE 2.89 V = 5.16 mA %3D %3D RE 560 2 Therefore, Ic = IE = 5.16 mA

Determine VCE and Ic in the stiff voltage-divider biased transistor circuit of Figur if Bpc = 100. FIGURE 5-10 Vcc +10 V 10 k Re 1.0 kN R2 5.6 k RE 560 N tion The base voltage is R2 5.6 kN 10 V = 3.59 V VB = R+R Vcc = 15.6 kN So, VE = VB - VBE = 3.59 V – 0.7 V = 2.89 V and VE 2.89 V = 5.16 mA %3D %3D RE 560 2 Therefore, Ic = IE = 5.16 mA

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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