Vcc = 20V R1= 40 kohm R2- 10 kohm Rc -4 kohm RE-2 kohm Rs-1 kohm RL = 22 kohm Cs= 10 uF Cc=1 uF CE= 20 uF Cbc = 4 pF Cbe - 36 pF Cce-1 pF Cwi-6 pF Cwo =8 pF C; Beta = 100
Vcc = 20V R1= 40 kohm R2- 10 kohm Rc -4 kohm RE-2 kohm Rs-1 kohm RL = 22 kohm Cs= 10 uF Cc=1 uF CE= 20 uF Cbc = 4 pF Cbe - 36 pF Cce-1 pF Cwi-6 pF Cwo =8 pF C; Beta = 100
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
determine the higher and lower cut off frequency
![Amplifier's Frequency Response
Vcc
Vcc = 20V
R1 = 40 kohm
R2 = 10 kohm
Rc -4 kohm
RE- 2 kohm
Rs-1 kohm
RL = 22 kohm
Rc
Cs = 10 uF
Cc=1 uF
CE= 20 uF
R1
Cc
Cbc = 4 pF
Cbe = 36 pF
Cce-1 pF
Cwi= 6 pF
Cwo =8 pF
Beta = 100
Cce
CW
R1
R2 Cw;
Cbe
+
CE
RE
+](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99c75b86-2ede-407a-bd08-39f46bc80978%2F846d897c-1322-4c9e-b537-c5917d53339b%2Fi47ai1n_processed.png&w=3840&q=75)
Transcribed Image Text:Amplifier's Frequency Response
Vcc
Vcc = 20V
R1 = 40 kohm
R2 = 10 kohm
Rc -4 kohm
RE- 2 kohm
Rs-1 kohm
RL = 22 kohm
Rc
Cs = 10 uF
Cc=1 uF
CE= 20 uF
R1
Cc
Cbc = 4 pF
Cbe = 36 pF
Cce-1 pF
Cwi= 6 pF
Cwo =8 pF
Beta = 100
Cce
CW
R1
R2 Cw;
Cbe
+
CE
RE
+
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