From the given circuit, find: the upper cutoff frequency of the system. a) 8.6 MHz, b) 10.77 MHz, c) 690.15 kHz, d) 801 kHz the cutoff frequency due to CE a) 222 Hz, b) 327 Hz, c) 489 Hz, d) 577 Hz Input Impedance Zi a) 2 kohms, b) 1.32 kohms, c) 3.45 kohms, d) 2.22 kohms
From the given circuit, find: the upper cutoff frequency of the system. a) 8.6 MHz, b) 10.77 MHz, c) 690.15 kHz, d) 801 kHz the cutoff frequency due to CE a) 222 Hz, b) 327 Hz, c) 489 Hz, d) 577 Hz Input Impedance Zi a) 2 kohms, b) 1.32 kohms, c) 3.45 kohms, d) 2.22 kohms
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Question
From the given circuit, find:
- the upper cutoff frequency of the system. a) 8.6 MHz, b) 10.77 MHz, c) 690.15 kHz, d) 801 kHz
- the cutoff frequency due to CE a) 222 Hz, b) 327 Hz, c) 489 Hz, d) 577 Hz
- Input Impedance Zi a) 2 kohms, b) 1.32 kohms, c) 3.45 kohms, d) 2.22 kohms
![Amplifier's Frequency Response
Vcc
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
RC
CE = 20 uF
Cbc = 4 pF
Cbe = 36 pF
Cce-1 pF
Cwi = 6 pF
Cwo -8 pF
Beta = 100
Сс
Cce
Rg
Cw
RL
V; R2 Cw*
Cbe
CE
Rg
Determine the upper cutoff frequency, f2, of the
system.
а.
8.6 MHz
b.
10.77 MHz
с.
690.15 kHz
d.
801 kHz
+](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58b8da05-0ead-4003-ab5f-e9764c613df9%2F4d2f8953-b728-4164-a423-b78671cba56c%2Fcs1a5f_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 = 2.2 kohm
Cs = 10 uF
Cc = 1 uF
RC
CE = 20 uF
Cbc = 4 pF
Cbe = 36 pF
Cce-1 pF
Cwi = 6 pF
Cwo -8 pF
Beta = 100
Сс
Cce
Rg
Cw
RL
V; R2 Cw*
Cbe
CE
Rg
Determine the upper cutoff frequency, f2, of the
system.
а.
8.6 MHz
b.
10.77 MHz
с.
690.15 kHz
d.
801 kHz
+
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