Vcc = 12 V R2 Rc 300 k2 22 k2 R3 R1 RỊ vo 1 k2 Rc R3, 1 k2 100 k2 Rg = R||R2 22 k2 R1. 104 k2 100 kQ RE C3→0 Signal 160 k2 13 k. source (b) (a) R1 R; Rg Rc Ra Ube RB Ube RL Vo T&mUbe 8m Ube RL (d) Figure 13.18 (a) Common-emitter amplifier circuit employing a bipolar transistor; (b) ac equivalent circuit for the common- emitter amplifier in part (a); the common-emitter connection should now be evident; (c) ac equivalent circuit with the bipolar transistor replaced by its small-signal model; (d) final equivalent circuit for ac analysis of the common-emitter amplifier.
Vcc = 12 V R2 Rc 300 k2 22 k2 R3 R1 RỊ vo 1 k2 Rc R3, 1 k2 100 k2 Rg = R||R2 22 k2 R1. 104 k2 100 kQ RE C3→0 Signal 160 k2 13 k. source (b) (a) R1 R; Rg Rc Ra Ube RB Ube RL Vo T&mUbe 8m Ube RL (d) Figure 13.18 (a) Common-emitter amplifier circuit employing a bipolar transistor; (b) ac equivalent circuit for the common- emitter amplifier in part (a); the common-emitter connection should now be evident; (c) ac equivalent circuit with the bipolar transistor replaced by its small-signal model; (d) final equivalent circuit for ac analysis of the common-emitter amplifier.
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
Verify the bias point values used as shown by directly calculating the Q-point.
![Vcc = 12 V
R2
Rc
300 k2
22 k2
R3
R1
RỊ
vo
1 k2
Rc
R3,
1 k2
100 k2
Rg = R||R2
22 k2
R1.
104 k2
100 kQ
RE
C3→0
Signal
160 k2
13 k.
source
(b)
(a)
R1
R;
Rg
Rc
Ra
Ube
RB
Ube
RL Vo
T&mUbe
8m Ube
RL
(d)
Figure 13.18 (a) Common-emitter amplifier circuit employing a bipolar transistor; (b) ac equivalent circuit for the common-
emitter amplifier in part (a); the common-emitter connection should now be evident; (c) ac equivalent circuit with the bipolar
transistor replaced by its small-signal model; (d) final equivalent circuit for ac analysis of the common-emitter amplifier.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2a071594-6e5e-497f-84d6-0a8b102048c4%2F447aa059-a415-4c4e-97b7-e7d41762bc32%2Fuxe2cvw.png&w=3840&q=75)
Transcribed Image Text:Vcc = 12 V
R2
Rc
300 k2
22 k2
R3
R1
RỊ
vo
1 k2
Rc
R3,
1 k2
100 k2
Rg = R||R2
22 k2
R1.
104 k2
100 kQ
RE
C3→0
Signal
160 k2
13 k.
source
(b)
(a)
R1
R;
Rg
Rc
Ra
Ube
RB
Ube
RL Vo
T&mUbe
8m Ube
RL
(d)
Figure 13.18 (a) Common-emitter amplifier circuit employing a bipolar transistor; (b) ac equivalent circuit for the common-
emitter amplifier in part (a); the common-emitter connection should now be evident; (c) ac equivalent circuit with the bipolar
transistor replaced by its small-signal model; (d) final equivalent circuit for ac analysis of the common-emitter amplifier.
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