Amplifier's Frequency Response Vcc=20V R1 = 40 kohm R2 = 10 kohm Rc = 4 kohm RE=2 kohm Rs = 1 kohm RL = 2.2 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 Beta = 100 + Rs V₂ C₂ + V₁ R₁ C. d. Vcc R₂ Cw₁ Calculate Rin(base). a. 1.576 ΚΩ b. 1.679 ΚΩ 1.812 ΚΩ 1.987 ΚΩ Cbc Rc Che RE Cc Cce CE Cwo ●●● RL
Amplifier's Frequency Response Vcc=20V R1 = 40 kohm R2 = 10 kohm Rc = 4 kohm RE=2 kohm Rs = 1 kohm RL = 2.2 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 Beta = 100 + Rs V₂ C₂ + V₁ R₁ C. d. Vcc R₂ Cw₁ Calculate Rin(base). a. 1.576 ΚΩ b. 1.679 ΚΩ 1.812 ΚΩ 1.987 ΚΩ Cbc Rc Che RE Cc Cce CE Cwo ●●● RL
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
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AMPLIFIER'S FREQUENCY RESPONSE 14
![Amplifier's Frequency Response
Vcc=20V
R1 = 40 kohm
R2 = 10 kohm
Rc = 4 kohm
RE=2 kohm
Rs = 1 kohm
RL = 2.2 kohm
Cs= 10 uF
Cc = 1 uF
CE=20 uF
Cbc = 4 pF
Cbe = 36 pF
Cce = 1 pF
Cwi=6pF
Cwo = 8 pF Cs
Beta = 100
+
Rs
V₂
+
V₁
R₁
Vcc
R₂ CW₁
Calculate Rin(base).
a. 1.576 ΚΩ
b.
1.679 ΚΩ
C.
1.812 ΚΩ
d. 1.987 ΚΩ
Rc
Che
RE
Сс
Cwo
4
RL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F29c8cf49-5423-4f4c-a559-4f2bc9841eaf%2F98df2dfb-f7e0-4570-9c58-cea0e175738d%2F1x5wxuc_processed.png&w=3840&q=75)
Transcribed Image Text:Amplifier's Frequency Response
Vcc=20V
R1 = 40 kohm
R2 = 10 kohm
Rc = 4 kohm
RE=2 kohm
Rs = 1 kohm
RL = 2.2 kohm
Cs= 10 uF
Cc = 1 uF
CE=20 uF
Cbc = 4 pF
Cbe = 36 pF
Cce = 1 pF
Cwi=6pF
Cwo = 8 pF Cs
Beta = 100
+
Rs
V₂
+
V₁
R₁
Vcc
R₂ CW₁
Calculate Rin(base).
a. 1.576 ΚΩ
b.
1.679 ΚΩ
C.
1.812 ΚΩ
d. 1.987 ΚΩ
Rc
Che
RE
Сс
Cwo
4
RL
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