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)
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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3af7c690-a47b-4d59-94b3-357a882b23e2%2Feda2db8a-e6be-42e3-b076-d5a8ed49b6b4%2F3ojewqf_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3af7c690-a47b-4d59-94b3-357a882b23e2%2Feda2db8a-e6be-42e3-b076-d5a8ed49b6b4%2Fuj744n6_processed.png&w=3840&q=75)
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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