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3. The answer is already given. I want you to explain them in plain English. BJT Biasing

3. The answer is already given. I want you to explain them in plain English. BJT Biasing

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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3. The answer is already given. I want you to explain them in plain English.

BJT Biasing

15. For the voltage-divider bias configuration of Fig. 125, determine: IBo b. Icơ VCEg d. Vc- e. VE- f. Vg. а. с. е. 16 V 3.9 kQ 62 kN o Vc VB VCE, B= 80 IBQ o VE 9.1 kN 0.68 k2 FIG. 125
Transcribed Image Text:15. For the voltage-divider bias configuration of Fig. 125, determine: IBo b. Icơ VCEg d. Vc- e. VE- f. Vg. а. с. е. 16 V 3.9 kQ 62 kN o Vc VB VCE, B= 80 IBQ o VE 9.1 kN 0.68 k2 FIG. 125
BRĘ 2 10R2 (80)(0.68 kQ) > 10(9.1 kQ) 54.4 kΩ 91 Κ (No!) 15. (a) Use exact approach: Rh = R1 || R2 = 62 k2 || 9.1 k2 = 7.94 kQ R,Vcc _ (9.1 k2)(16 V) Eh = = 2.05 V R, + R, 9.1 k2+ 62 k2 ETh - VBE Ry + (B+1)Rg 7.94 k2 + (81)(0.68 k2) I Bo 2.05 V – 0.7 V = 21.42 μΑ (b) Ic, = BIB, = (80)(21.42 µA) = 1.71 mA (c) VCE, = Vcc - Ic, (Rc + RE) = 16 V – (1.71 mA)(3.9 k2 + 0.68 kQ) = 8.17 V (d) Ve = Vcc - lRC = 16 V - (1.71 mA)(3.9 k2) = 9.33 V (e) VE = IRE = 1¢RE = (1.71 mA)(0.68 kQ) = 1.16 V (f) VB = VE + VBE = 1.16 V + 0.7 V = 1.86 V
Transcribed Image Text:BRĘ 2 10R2 (80)(0.68 kQ) > 10(9.1 kQ) 54.4 kΩ 91 Κ (No!) 15. (a) Use exact approach: Rh = R1 || R2 = 62 k2 || 9.1 k2 = 7.94 kQ R,Vcc _ (9.1 k2)(16 V) Eh = = 2.05 V R, + R, 9.1 k2+ 62 k2 ETh - VBE Ry + (B+1)Rg 7.94 k2 + (81)(0.68 k2) I Bo 2.05 V – 0.7 V = 21.42 μΑ (b) Ic, = BIB, = (80)(21.42 µA) = 1.71 mA (c) VCE, = Vcc - Ic, (Rc + RE) = 16 V – (1.71 mA)(3.9 k2 + 0.68 kQ) = 8.17 V (d) Ve = Vcc - lRC = 16 V - (1.71 mA)(3.9 k2) = 9.33 V (e) VE = IRE = 1¢RE = (1.71 mA)(0.68 kQ) = 1.16 V (f) VB = VE + VBE = 1.16 V + 0.7 V = 1.86 V
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