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29. For the voltage feedback network of Fig. 130, a. Ic. b. Vc- VẸ- d. VCE-

29. For the voltage feedback network of Fig. 130, a. Ic. b. Vc- VẸ- d. VCE-

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

Vcc - VBE 30 V – 0.7 V = 12.47 µA 29. (a) Ig = Rg + B(Rc + RE) 550 k2 +180(8.2 k2 + 1.8 k2) Ic = Bls = (180)(12.47 µA) = 2.24 mA (b) Vc = Vcc- IRc = 30 V – (2.24 mA)(8.2 k2) = 30 V – 18.37 V = 11.63 V (c) VE = IgRE = I¢R£ = (2.24 mA)(1.8 k2) = 4.03 V %3D %3D (d) VCE = Vcc - IɖRc+ R£) = 30 V – (2.24 mA)(8.2 kQ + 1.8 kQ) = 7.6 V
Transcribed Image Text:Vcc - VBE 30 V – 0.7 V = 12.47 µA 29. (a) Ig = Rg + B(Rc + RE) 550 k2 +180(8.2 k2 + 1.8 k2) Ic = Bls = (180)(12.47 µA) = 2.24 mA (b) Vc = Vcc- IRc = 30 V – (2.24 mA)(8.2 k2) = 30 V – 18.37 V = 11.63 V (c) VE = IgRE = I¢R£ = (2.24 mA)(1.8 k2) = 4.03 V %3D %3D (d) VCE = Vcc - IɖRc+ R£) = 30 V – (2.24 mA)(8.2 kQ + 1.8 kQ) = 7.6 V
29. For the voltage feedback network of Fig. 130, determine: a. Ic. b. Vс c. VẸ. d. VCE- o 30 V 8.2 k2 330 k2 220 k2 Vc 10 μF o Vo 5 µF Ic 10 μF Vio VCE = 180 VE 1.8 k2 5 μF FIG. 130
Transcribed Image Text:29. For the voltage feedback network of Fig. 130, determine: a. Ic. b. Vс c. VẸ. d. VCE- o 30 V 8.2 k2 330 k2 220 k2 Vc 10 μF o Vo 5 µF Ic 10 μF Vio VCE = 180 VE 1.8 k2 5 μF FIG. 130
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